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1l6hz0rdm | maths | 3d-geometry | plane-in-space | <p>A vector $$\vec{a}$$ is parallel to the line of intersection of the plane determined by the vectors $$\hat{i}, \hat{i}+\hat{j}$$ and the plane determined by the vectors $$\hat{i}-\hat{j}, \hat{i}+\hat{k}$$. The obtuse angle between $$\vec{a}$$ and the vector $$\vec{b}=\hat{i}-2 \hat{j}+2 \hat{k}$$ is :</p> | [{"identifier": "A", "content": "$$\\frac{3 \\pi}{4}$$"}, {"identifier": "B", "content": "$$\\frac{2 \\pi}{3}$$"}, {"identifier": "C", "content": "$$\\frac{4 \\pi}{5}$$"}, {"identifier": "D", "content": "$$\\frac{5 \\pi}{6}$$"}] | ["A"] | null | <p>If $${\overrightarrow n _1}$$ is a vector normal to the plane determined by $$\widehat i$$ and $$\widehat i + \widehat j$$ then</p>
<p>$${\overrightarrow n _1} = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
1 & 0 & 0 \cr
1 & 1 & 0 \cr
} } \right| = \widehat k$$</p>
<p>If $${\overrightarrow n _2}$$ is a vector normal to the plane determined by $$\widehat i - \widehat j$$ and $$\widehat i + \widehat k$$ then</p>
<p>$${\overline n _2} = \left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
1 & { - 1} & 0 \cr
1 & 0 & 1 \cr
} } \right| = - \widehat i - \widehat j + \widehat k$$</p>
<p>Vector $$\overrightarrow a $$ is parallel to $${\overrightarrow n _1} \times {\overrightarrow n _2}$$</p>
<p>i.e. $$\overrightarrow a $$ is parallel to $$\left| {\matrix{
{\widehat i} & {\widehat j} & {\widehat k} \cr
0 & 0 & 1 \cr
{ - 1} & { - 1} & 1 \cr
} } \right| = \widehat i - \widehat j$$</p>
<p>Given $$\overrightarrow b = \widehat i - 2\widehat j + 2\widehat k$$</p>
<p>Cosine of acute angle between</p>
<p>$$\overrightarrow a $$ and $$\overrightarrow b = \left| {{{\overrightarrow a \,.\,\overrightarrow b } \over {|\overrightarrow a |\,.\,|\overrightarrow b |}}} \right| = {1 \over {\sqrt 2 }}$$</p>
<p>Obtuse angle between $$\overrightarrow a $$ and $$\overrightarrow b = {{3\pi } \over 4}$$</p> | mcq | jee-main-2022-online-26th-july-evening-shift |
1l6jd01uv | maths | 3d-geometry | plane-in-space | <p>If the plane $$P$$ passes through the intersection of two mutually perpendicular planes $$2 x+k y-5 z=1$$ and $$3 k x-k y+z=5, k<3$$ and intercepts a unit length on positive $$x$$-axis, then the intercept made by the plane $$P$$ on the $$y$$-axis is :</p> | [{"identifier": "A", "content": "$$\\frac{1}{11}$$"}, {"identifier": "B", "content": "$$\\frac{5}{11}$$"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "7"}] | ["D"] | null | <p>$${P_1}:2x + ky - 5z = 1$$</p>
<p>$${P_2}:3kx - ky + z = 5$$</p>
<p>$$\because$$ $${P_1}\, \bot \,{P_2} \Rightarrow 6k - {k^2} + 5 = 0$$</p>
<p>$$ \Rightarrow k = 1,5$$</p>
<p>$$\because$$ $$k < 3$$</p>
<p>$$\therefore$$ $$k = 1$$</p>
<p>$${P_1}:2x + y - 5z = 1$$</p>
<p>$${P_2}:3x - y + z = 5$$</p>
<p>$$P:(2x + y - 5z - 1) + \lambda (3x - y + z - 5) = 0$$</p>
<p>Positive x-axis intercept = 1</p>
<p>$$ \Rightarrow {{1 + 5\lambda } \over {2 + 3\lambda }} = 1$$</p>
<p>$$ \Rightarrow \lambda = {1 \over 2}$$</p>
<p>$$\therefore$$ $$P:7x + y - 4z = 7$$</p>
<p>y intercept = 7.</p> | mcq | jee-main-2022-online-27th-july-morning-shift |
1l6m62lhe | maths | 3d-geometry | plane-in-space | <p>The foot of the perpendicular from a point on the circle $$x^{2}+y^{2}=1, z=0$$ to the plane $$2 x+3 y+z=6$$ lies on which one of the following curves?</p> | [{"identifier": "A", "content": "$$(6 x+5 y-12)^{2}+4(3 x+7 y-8)^{2}=1, z=6-2 x-3 y$$"}, {"identifier": "B", "content": "$$(5 x+6 y-12)^{2}+4(3 x+5 y-9)^{2}=1, z=6-2 x-3 y$$"}, {"identifier": "C", "content": "$$(6 x+5 y-14)^{2}+9(3 x+5 y-7)^{2}=1, z=6-2 x-3 y$$"}, {"identifier": "D", "content": "$$(5 x+6 y-14)^{2}+9(3 x+7 y-8)^{2}=1, z=6-2 x-3 y$$"}] | ["B"] | null | <p>Any point on $${x^2} + {y^2} = 1$$, $$z = 0$$ is $$p(\cos \theta ,\,\sin \theta ,\,0)$$</p>
<p>If foot of perpendicular of p on the plane $$2x + 3y + z = 6$$ is $$(h,k,l)$$ then</p>
<p>$${{h - \cos \theta } \over 2} = {{k - \sin \theta } \over 3} = {{l - 0} \over 1}$$</p>
<p>$$ = - \left( {{{2\cos \theta + 3\sin \theta + 0 - 6} \over {{2^2} + {3^2} + {1^2}}}} \right) = r$$ (let)</p>
<p>$$h = 2r + \cos \theta ,\,k = 3r + \sin \theta ,\,l = r$$</p>
<p>Hence, $$h - 2l = \cos \theta $$ and $$k - 3l = \sin \theta $$</p>
<p>Hence, $${(h - 2l)^2} + {(k - 3l)^2} = 1$$</p>
<p>When $$l = 6 - 2h - 3k$$</p>
<p>Hence required locus is</p>
<p>$${(x - 2(6 - 2x - 3y))^2} + {(y - 3(6 - 2x - 3y))^2} = 1$$</p>
<p>$$ \Rightarrow {(5x + 6y - 12)^2} + 4{(3x + 5y - 9)^2} = 1,\,z = 6 - 2x - 3y$$</p> | mcq | jee-main-2022-online-28th-july-morning-shift |
1l6nnoffv | maths | 3d-geometry | plane-in-space | <p>A plane P is parallel to two lines whose direction ratios are $$-2,1,-3$$ and $$-1,2,-2$$ and it contains the point $$(2,2,-2)$$. Let P intersect the co-ordinate axes at the points $$\mathrm{A}, \mathrm{B}, \mathrm{C}$$ making the intercepts $$\alpha, \beta, \gamma$$. If $$\mathrm{V}$$ is the volume of the tetrahedron $$\mathrm{OABC}$$, where $$\mathrm{O}$$ is the origin, and $$\mathrm{p}=\alpha+\beta+\gamma$$, then the ordered pair $$(\mathrm{V}, \mathrm{p})$$ is equal to :</p> | [{"identifier": "A", "content": "$$(48,-13)$$"}, {"identifier": "B", "content": "$$(24,-13)$$"}, {"identifier": "C", "content": "$$(48,11)$$"}, {"identifier": "D", "content": "$$(24,-5)$$"}] | ["B"] | null | <p>Let $${\overrightarrow a _1} = ( - 2,1, - 3)$$ and $${\overrightarrow a _2} = ( - 1,2, - 2)$$</p>
<p>Vector normal to plane $$\overline n = {\overrightarrow a _1} \times {\overrightarrow a _2}$$</p>
<p>$$\overline n = (4, - 1, - 3)$$</p>
<p>Plane through $$(2,2, - 2)$$ and normal to $$\overline n $$</p>
<p>$$(x - 2,y - 2,z + 2)\,.\,(4, - 1, - 3) = 0$$</p>
<p>$$ \Rightarrow 4x - y - 3z = 12$$</p>
<p>$$ \Rightarrow {x \over 3} + {y \over { - 12}} + {z \over { - 4}} = 1$$</p>
<p>Intercepts $$\alpha$$, $$\beta$$, $$\gamma$$ are $$3, - 12, - 4$$</p>
<p>$$P = \alpha + \beta + \gamma = - 13$$</p>
<p>$$V = {1 \over 6} \times 3 \times 12 \times 4 = 24$$</p> | mcq | jee-main-2022-online-28th-july-evening-shift |
1l6reqbxr | maths | 3d-geometry | plane-in-space | <p>Let $$Q$$ be the foot of perpendicular drawn from the point $$P(1,2,3)$$ to the plane $$x+2 y+z=14$$. If $$R$$ is a point on the plane such that $$\angle P R Q=60^{\circ}$$, then the area of $$\triangle P Q R$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{\\sqrt{3}}{2}$$"}, {"identifier": "B", "content": "$$ \\sqrt{3}$$"}, {"identifier": "C", "content": "$$2 \\sqrt{3}$$"}, {"identifier": "D", "content": "3"}] | ["B"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7xrxbtg/d0e01ee5-c717-48d4-b80a-6a1b9ae27182/345bdd40-320e-11ed-bab2-5faeb0367b9d/file-1l7xrxbth.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7xrxbtg/d0e01ee5-c717-48d4-b80a-6a1b9ae27182/345bdd40-320e-11ed-bab2-5faeb0367b9d/file-1l7xrxbth.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 29th July Evening Shift Mathematics - 3D Geometry Question 116 English Explanation">
<br><br>$$
P Q=\left|\frac{1+4+3-14}{\sqrt{6}}\right|=\sqrt{6}
$$
<br><br>$$
Q R=\frac{P Q}{\tan 60^{\circ}}=\frac{\sqrt{6}}{\sqrt{3}}=\sqrt{2}
$$
<br><br>$$
\text { Area }(\triangle P Q R)=\frac{1}{2} \cdot P Q \cdot Q R=\sqrt{3}
$$ | mcq | jee-main-2022-online-29th-july-evening-shift |
1l6resdnr | maths | 3d-geometry | plane-in-space | <p>If $$(2,3,9),(5,2,1),(1, \lambda, 8)$$ and $$(\lambda, 2,3)$$ are coplanar, then the product of all possible values of $$\lambda$$ is:</p> | [{"identifier": "A", "content": "$$\\frac{21}{2}$$"}, {"identifier": "B", "content": "$$\\frac{59}{8}$$"}, {"identifier": "C", "content": "$$\\frac{57}{8}$$"}, {"identifier": "D", "content": "$$\\frac{95}{8}$$"}] | ["D"] | null | $\because A(2,3,9), B(5,2,1), C(1, \lambda, 8)$ and D$(\lambda, 2,3)$ are coplanar.
<br/><br/>$$ \therefore $$ $[\overrightarrow{\mathrm{AB}} \,\,\,\,\overrightarrow{\mathrm{AC}} \,\,\,\, \overrightarrow{\mathrm{AD}}]=0$
<br/><br/>$\left|\begin{array}{ccc}3 & -1 & -8 \\ -1 & \lambda-3 & -1 \\ \lambda-2 & -1 & -6\end{array}\right|=0$
<br/><br/>$\Rightarrow[-6(\lambda-3)-1]-8(1-(\lambda-3)(\lambda-2))+(6+(\lambda$
$-2)=0$
<br/><br/>$$ \Rightarrow $$ $3(-6 \lambda+17)-8\left(-\lambda^2+5 \lambda-5\right)+(\lambda+4)=8$
<br/><br/>$$ \Rightarrow $$ $8 \lambda^2-57 \lambda+95=0$
<br/><br/>$$ \therefore $$ $\lambda_1 \lambda_2=\frac{95}{8}$ | mcq | jee-main-2022-online-29th-july-evening-shift |
ldo9tm7i | maths | 3d-geometry | plane-in-space | If a point $\mathrm{P}(\alpha, \beta, \gamma)$ satisfying
<br/><br/>$$\left( {\matrix{
\alpha & \beta & \gamma \cr
} } \right)\left( {\matrix{
2 & {10} & 8 \cr
9 & 3 & 8 \cr
8 & 4 & 8 \cr
} } \right) = \left( {\matrix{
0 & 0 & 0 \cr
} } \right)$$
<br/><br/>lies on the plane $2 x+4 y+3 z=5$, then $6 \alpha+9 \beta+7 \gamma$ is equal to : | [{"identifier": "A", "content": "$\\frac{11}{5}$"}, {"identifier": "B", "content": "11"}, {"identifier": "C", "content": "$-1$"}, {"identifier": "D", "content": "$\\frac{5}{4}$"}] | ["B"] | null | Point $\mathrm{P}(\alpha, \beta, \gamma)$ lies on the plane $2 x+4 y+3 z=5$,
<br/><br/>$$ \therefore $$ $2 \alpha+4 \beta+3 \gamma=5$ ........(1)
<br/><br/>Given, $$\left( {\matrix{
\alpha & \beta & \gamma \cr
} } \right)\left( {\matrix{
2 & {10} & 8 \cr
9 & 3 & 8 \cr
8 & 4 & 8 \cr
} } \right) = \left( {\matrix{
0 & 0 & 0 \cr
} } \right)$$
<br/><br/>$$ \therefore $$ $2 \alpha+9 \beta+8 \gamma=0$ .......(2)
<br/><br/>and $10 \alpha+3 \beta+4 \gamma=0$ ........(3)
<br/><br/>and $8 \alpha+8 \beta+8 \gamma=0$ ..........(4)
<br/><br/>Subtract (4) from (2)
<br/><br/>$-6 \alpha+\beta=0$
$\beta=6 \alpha$
<br/><br/>From equation (4)
<br/><br/>$8 \alpha+48 \alpha+8 \gamma=0$
<br/><br/>$\gamma=-7 \alpha$
<br/><br/>From equation (1)
<br/><br/>$2 \alpha+24 \alpha-21 \alpha=5$
<br/><br/>$5 \alpha=5$
<br/><br/>$\alpha=1$
<br/><br/>$\beta=+6, \quad \gamma=-7$
<br/><br/>$\therefore 6 \alpha+9 \beta+7 \gamma$
<br/><br/>$=6+54-49$
<br/><br/>$=11$ | mcq | jee-main-2023-online-31st-january-evening-shift |
1ldonegez | maths | 3d-geometry | plane-in-space | <p>Let the image of the point $$P(2,-1,3)$$ in the plane $$x+2 y-z=0$$ be $$Q$$. <br/><br/>Then the distance of the plane $$3 x+2 y+z+29=0$$ from the point $$Q$$ is :</p> | [{"identifier": "A", "content": "$$2\\sqrt{14}$$"}, {"identifier": "B", "content": "$$\\frac{22\\sqrt2}{7}$$"}, {"identifier": "C", "content": "$$\\frac{24\\sqrt2}{7}$$"}, {"identifier": "D", "content": "$$3\\sqrt{14}$$"}] | ["D"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lea9v38f/9887ff8f-08e7-4f18-9711-257b854f4854/0e8003e0-afb7-11ed-a5f0-99851f9df37c/file-1lea9v38g.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lea9v38f/9887ff8f-08e7-4f18-9711-257b854f4854/0e8003e0-afb7-11ed-a5f0-99851f9df37c/file-1lea9v38g.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 1st February Morning Shift Mathematics - 3D Geometry Question 107 English Explanation">
<br><br>Equation of line PM $\frac{x-2}{1}=\frac{y+1}{2}=\frac{z-3}{-1}=\lambda$
<br><br>Any point on line $=(\lambda+2,2 \lambda-1,-\lambda+3)$
<br><br>For point ' $m$ ' $(\lambda+2)+2(2 \lambda-1)-(3-\lambda)=0$
<br><br>$$ \Rightarrow $$ $$
\lambda=\frac{1}{2}
$$
<br><br>Point $\mathrm{m}\left(\frac{1}{2}+2,2 \times \frac{1}{2}-1, \frac{-1}{2}+3\right)$
<br><br>$$
=\left(\frac{5}{2}, 0, \frac{5}{2}\right)
$$
<br><br>For Image $\mathrm{Q}(\alpha, \beta, \gamma)$
<br><br>$$
\begin{aligned}
& \frac{\alpha+2}{2}=\frac{5}{2}, \frac{\beta-1}{2}=0, \\\\
& \frac{\gamma+3}{2}=\frac{5}{2}
\end{aligned}
$$
<br><br>$$
\begin{aligned}
& Q:(3,1,2) \\\\
& d=\left|\frac{3(3)+2(1)+2+29}{\sqrt{3^2+2^2+1^2}}\right| \\\\
& \Rightarrow d=\frac{42}{\sqrt{14}}=3 \sqrt{14}
\end{aligned}
$$ | mcq | jee-main-2023-online-1st-february-morning-shift |
1ldr83jqr | maths | 3d-geometry | plane-in-space | <p>If $$\lambda_{1} < \lambda_{2}$$ are two values of $$\lambda$$ such that the angle between the planes $$P_{1}: \vec{r}(3 \hat{i}-5 \hat{j}+\hat{k})=7$$ and
$$P_{2}: \vec{r} \cdot(\lambda \hat{i}+\hat{j}-3 \hat{k})=9$$ is $$\sin ^{-1}\left(\frac{2 \sqrt{6}}{5}\right)$$, then the square of the length of perpendicular from the point $$\left(38 \lambda_{1}, 10 \lambda_{2}, 2\right)$$ to the plane $$P_{1}$$ is ______________.</p> | [] | null | 315 | <p>$${P_1}:\overrightarrow r \,.\,(3\widehat i - 5\widehat j + \widehat k) = 7$$</p>
<p>$${P_2}:\overrightarrow r \,.\,(\lambda \widehat i + \widehat j - 3\widehat k) = 9$$</p>
<p>Let angle between P<sub>1</sub> and P<sub>2</sub> is $$\theta$$</p>
<p>Then $$\cos \theta = {{3\lambda - 5 - 3} \over {\sqrt {35} \sqrt {{\lambda ^2} + 10} }}$$</p>
<p>But $$\sin \theta = {{2\sqrt 6 } \over 5}$$</p>
<p>$$\therefore$$ $${{{{(3\lambda - 8)}^2}} \over {35({\lambda ^2} + 10)}} = 1 - {{24} \over {25}}$$</p>
<p>$$ \Rightarrow 5(9{\lambda ^2} + 64 - 48\lambda ) = 7{\lambda ^2} + 70$$</p>
<p>$$ \Rightarrow 38{\lambda ^2} - 240\lambda + 250 = 0$$</p>
<p>$$ \Rightarrow 19{\lambda ^2} - 120\lambda + 125 = 0$$</p>
<p>$$ \Rightarrow (19\lambda - 25)(\lambda - 5) = 0$$</p>
<p>$$\therefore$$ $${\lambda _1} = {{25} \over {19}},{\lambda _2} = 5$$</p>
<p>So, point (50, 50, 2)</p>
<p>$$\therefore$$ $$d = {{|150 - 250 + 2 - 7|} \over {\sqrt {35} }} = 315$$</p> | integer | jee-main-2023-online-30th-january-morning-shift |
1ldyb64xl | maths | 3d-geometry | plane-in-space | <p>The distance of the point (7, $$-$$3, $$-$$4) from the plane passing through the points (2, $$-$$3, 1), ($$-$$1, 1, $$-$$2) and (3, $$-$$4, 2) is :</p> | [{"identifier": "A", "content": "$$4\\sqrt2$$"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "$$5\\sqrt2$$"}] | ["D"] | null | $A(2,-3,1), B(-1,1,-2), C(3,-4,2)$
<br/><br/>
$$
\begin{aligned}
& \overrightarrow{A B}=-3 \hat{i}+4 \hat{j}-3 \hat{k} \quad \overrightarrow{A C}=\hat{i}-\hat{j}+\hat{k} \\\\
& \vec{n}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\\\
-3 & 4 & -3 \\\\
1 & -1 & 1
\end{array}\right|=\hat{i}-\hat{k}
\end{aligned}
$$
<br/><br/>
Let equation of plane is $x-z+\lambda=0$ passes through point $A(2,-3,1) \Rightarrow \lambda=-1$
<br/><br/>
Equation of plane is $x-z-1=0$
<br/><br/>
Distance of point $(7,-3,-4)$ from the plane $x-z-$ $1=0$ is $5 \sqrt{2}$ | mcq | jee-main-2023-online-24th-january-morning-shift |
lgnwzi53 | maths | 3d-geometry | plane-in-space | Let the foot of perpendicular of the point $P(3,-2,-9)$ on the plane passing through the points $(-1,-2,-3),(9,3,4),(9,-2,1)$ be $Q(\alpha, \beta, \gamma)$. Then the distance of $Q$ from the origin is : | [{"identifier": "A", "content": "$\\sqrt{38}$"}, {"identifier": "B", "content": "$\\sqrt{29}$"}, {"identifier": "C", "content": "$\\sqrt{42}$"}, {"identifier": "D", "content": "$\\sqrt{35}$"}] | ["C"] | null | <p>The equation of the plane passing through points $A(-1, -2, -3)$, $B(9, 3, 4)$, and $C(9, -2, 1)$ can be written using the determinant :</p>
<p>$$
\left|\begin{array}{ccc}
x+1 & y+2 & z+3 \\
10 & 5 & 7 \\
10 & 0 & 4
\end{array}\right|=0
$$</p>
<p>Expanding the determinant, we get :</p>
<p>$$
2x + 3y - 5z - 7 = 0
$$</p>
<p>Next, we find the foot of the perpendicular from point $P(3, -2, -9)$ to the plane. Using the coordinates of $P$ and the equation of the plane, we can find the ratio of the perpendicular distance to the sum of the squares of the coefficients of $x$, $y$, and $z$ :</p>
<p>$$
\frac{2(3) + 3(-2) - 5(-9) - 7}{2^2 + 3^2 + (-5)^2} = \frac{-38}{38}
$$</p>
<p>We can now find the coordinates of the foot of the perpendicular, $Q(\alpha, \beta, \gamma)$ :</p>
<p>$$
\frac{\alpha - 3}{2} = \frac{\beta + 2}{3} = \frac{\gamma + 9}{-5} = -\frac{38}{38}
$$</p>
<p>Solving for $\alpha$, $\beta$, and $\gamma$ :</p>
<p>$$
\alpha = 3 - 2\left(-\frac{38}{38}\right) = 1
$$</p>
<p>$$
\beta = -2 + 3\left(-\frac{38}{38}\right) = -5
$$</p>
<p>$$
\gamma = -9 - 5\left(-\frac{38}{38}\right) = -4
$$</p>
<p>So, the coordinates of the foot of the perpendicular are $Q(1, -5, -4)$. Now, we can find the distance of point $Q$ from the origin :</p>
<p>$$
OQ = \sqrt{\alpha^2 + \beta^2 + \gamma^2} = \sqrt{1^2 + (-5)^2 + (-4)^2} = \sqrt{42}
$$</p> | mcq | jee-main-2023-online-15th-april-morning-shift |
lgny51ak | maths | 3d-geometry | plane-in-space | Let the system of linear equations
<br/><br/>$-x+2 y-9 z=7$
<br/><br/>$-x+3 y+7 z=9$
<br/><br/>$-2 x+y+5 z=8$
<br/><br/>$-3 x+y+13 z=\lambda$
<br/><br/>has a unique solution $x=\alpha, y=\beta, z=\gamma$. Then the distance of the point
<br/><br/>$(\alpha, \beta, \gamma)$ from the plane $2 x-2 y+z=\lambda$ is : | [{"identifier": "A", "content": "11"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "13"}, {"identifier": "D", "content": "9"}] | ["B"] | null | $$
\begin{aligned}
& -x+2 y-9 z=7-(1) \\\\
& -x+3 y-7 z=9-(2) \\\\
& -2 x+y+5 z=8-(3) \\\\
& (2)-(1) \\\\
& y+16 z=2 \quad(4) \\\\
& (3)-2 \times(1) \\\\
& -3 y+23 z=-6-(5) \\\\
& 3 \times(4)+(5) \\\\
& 71 z=0 \Rightarrow z=0 \\\\
& \quad y=2 \\\\
& (-3,2,0) \rightarrow(\alpha, \beta, \gamma) \\\\
& \text { Put in }-3 x+y+13 z=\lambda \\\\
& \lambda=9+2=11
\end{aligned}
$$
<br/><br/>The equation of the plane is:
<br/><br/>$$
2x - 2y + z = \lambda
$$
<br/><br/>We already know that $\lambda = 11$. So the equation of the plane becomes:
<br/><br/>$$
2x - 2y + z = 11
$$
<br/><br/>Now, let's find the distance $d$ of the point $(\alpha, \beta, \gamma) = (-3, 2, 0)$ from the plane using the formula :
<br/><br/>$$
d = \frac{|Ax + By + Cz + D|}{\sqrt{A^2 + B^2 + C^2}}
$$
<br/><br/>Plugging in the values for the point and the plane equation, we get:
<br/><br/>$$
d = \frac{|2(-3) - 2(2) + 0 - 11|}{\sqrt{2^2 + (-2)^2 + 1^2}} = \frac{|-6 - 4 - 11|}{\sqrt{4 + 4 + 1}} = \frac{21}{\sqrt{9}} = 7
$$
<br/><br/>So, the distance of the point $(-3, 2, 0)$ from the plane $2x - 2y + z = 11$ is 7. | mcq | jee-main-2023-online-15th-april-morning-shift |
1lgoxhquj | maths | 3d-geometry | plane-in-space | <p>Let $$\mathrm{N}$$ be the foot of perpendicular from the point $$\mathrm{P}(1,-2,3)$$ on the line passing through the points $$(4,5,8)$$ and $$(1,-7,5)$$. Then the distance of $$N$$ from the plane $$2 x-2 y+z+5=0$$ is :</p> | [{"identifier": "A", "content": "7"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "9"}, {"identifier": "D", "content": "8"}] | ["A"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lh2hc8pe/d4ba8260-fd3b-48a4-aed6-cb4a8b6ea85d/ac53ff20-e6d2-11ed-b683-39a79c7e644a/file-1lh2hc8pf.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lh2hc8pe/d4ba8260-fd3b-48a4-aed6-cb4a8b6ea85d/ac53ff20-e6d2-11ed-b683-39a79c7e644a/file-1lh2hc8pf.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 13th April Evening Shift Mathematics - 3D Geometry Question 74 English Explanation">
<br><br>Equation of line
$$
\frac{x-4}{4-1}=\frac{y-5}{5-(-7)}=\frac{z-8}{8-5}
$$
<br><br>$$
\frac{x-4}{3}=\frac{y-5}{12}=\frac{z-8}{3}
$$
<br><br>Let point $\mathrm{N}(3 \lambda+4,12 \lambda+5,3 \lambda+8)$
<br><br>$$
\overrightarrow{\mathrm{PN}}=(3 \lambda+4-1) \hat{i}+(12 \lambda+5-(-2)) \hat{j}+(3 \lambda+8-3) \hat{\mathrm{k}}
$$
<br><br>$$
\overrightarrow{\mathrm{PN}}=(3 \lambda+3) \hat{i}+(12 \lambda+7) \hat{\mathrm{j}}+(3 \lambda+5) \hat{\mathrm{k}}
$$
<br><br>And parallel vector to line (say $\vec{a}=3 \hat{i}+12 \hat{j}+3 \hat{k}$ )
<br><br>Now, $\overrightarrow{\mathrm{PN}} \cdot \overrightarrow{\mathrm{a}}=0$
<br><br>$$
\begin{aligned}
& (3 \lambda+3) 3+(12 \lambda+7) 12+(3 \lambda+5) 3=0 \\\\
& 162 \lambda+108=0 \Rightarrow \lambda=\frac{-108}{162}=\frac{-2}{3}
\end{aligned}
$$
<br><br>So point $\mathrm{N}$ is $(2,-3,6)$
<br><br>Now distance of
N(2, -3, 6) from
$$2 x-2 y+z+5=0$$ is
<br><br> $=\left|\frac{2(2)-2(-3)+6+5}{\sqrt{4+4+1}}\right|=7$ | mcq | jee-main-2023-online-13th-april-evening-shift |
1lgpxn2hg | maths | 3d-geometry | plane-in-space | <p>Let the equation of plane passing through the line of intersection of the planes $$x+2 y+a z=2$$ and $$x-y+z=3$$ be $$5 x-11 y+b z=6 a-1$$. For $$c \in \mathbb{Z}$$, if the distance of this plane from the point $$(a,-c, c)$$ is $$\frac{2}{\sqrt{a}}$$, then $$\frac{a+b}{c}$$ is equal to :</p> | [{"identifier": "A", "content": "$$-$$2"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "$$-$$4"}] | ["D"] | null | Given the equation of the plane passing through the intersection of the two given planes:
<br/><br/>$$P: (x + 2y + az - 2) + \lambda(x - y + z - 3) = 0$$
<br/><br/>$$\Rightarrow x(\lambda+1)+y(2-\lambda)+z(a+\lambda)-2-3 \lambda=0$$
<br/><br/>This is the same as the given equation $$5x - 11y + bz = 6a - 1$$.
<br/><br/>Now, comparing the coefficients of the corresponding variables in both equations:
<br/><br/>$$\frac{\lambda+1}{5} = \frac{2-\lambda}{-11} = \frac{a+\lambda}{b} = \frac{2+3\lambda}{6a-1}$$
<br/><br/>Solving for $$\lambda$$:
<br/><br/>$$-11\lambda -11 = 10 - 5\lambda$$
<br/><br/>$$6\lambda = -21 \Rightarrow \lambda = -\frac{7}{2}$$
<br/><br/>Now, substituting the value of $$\lambda$$ back into the equations:
<br/><br/>$$\frac{2-\lambda}{-11} = \frac{2+3\lambda}{6a-1} \Rightarrow \frac{2+\frac{7}{2}}{-11} = \frac{2-\frac{21}{2}}{6a-1}$$
<br/><br/>From this equation, we find the value of a :
<br/><br/>$$6a - 1 = 17 \Rightarrow a = 3$$
<br/><br/>Now, substituting the value of $$a$$ and $$\lambda$$ into the equation:
<br/><br/>$$\frac{2-\lambda}{-11} = \frac{a+\lambda}{b} \Rightarrow -\frac{1}{2} = \frac{3 - \frac{7}{2}}{b}$$
<br/><br/>$$\Rightarrow -\frac{b}{2} = -\frac{1}{2} \Rightarrow b = 1$$
<br/><br/>Therefore, the point $$(a, -c, c) \equiv (3, -c, c)$$.
<br/><br/>The given distance is $$\frac{2}{\sqrt{a}} = \frac{2}{\sqrt{3}}$$.
<br/><br/>The plane is: $$5x - 11y + z = 17$$.
<br/><br/>Now, let's find the distance:
<br/><br/>$$\left|\frac{15 + 11c + c - 17}{\sqrt{147}}\right| = \frac{2}{\sqrt{3}}$$
<br/><br/>$$\Rightarrow c = -1, \frac{4}{3}$$
<br/><br/>Since $$c \in \mathbb{Z}$$, we have $$c = -1$$.
<br/><br/>Therefore, $$\frac{a+b}{c} = \frac{3+1}{-1} = -4$$. | mcq | jee-main-2023-online-13th-april-morning-shift |
1lgq12wu6 | maths | 3d-geometry | plane-in-space | <p>Let the image of the point $$\left(\frac{5}{3}, \frac{5}{3}, \frac{8}{3}\right)$$ in the plane $$x-2 y+z-2=0$$ be P. If the distance of the point $$Q(6,-2, \alpha), \alpha > 0$$, from $$\mathrm{P}$$ is 13 , then $$\alpha$$ is equal to ___________.</p> | [] | null | 15 | Image of point $\left(\frac{5}{3}, \frac{5}{3}, \frac{8}{3}\right)$
<br/><br/>$$
\begin{aligned}
& \frac{x-\frac{5}{3}}{1}=\frac{y-\frac{5}{3}}{-2}=\frac{\mathrm{z}-\frac{8}{3}}{1}=\frac{-2\left(1 \times \frac{5}{3}+(-2) \times \frac{8}{3}+1 \times \frac{8}{3}-2\right)}{1^2+2^2+1^2} =\frac{1}{3} \\\\
& \therefore x=2, y=1, \mathrm{z}=3 \\\\
&PQ^2 = 13^2=(6-2)^2+(-2-1)^2+(\alpha-3)^2 \\\\
& \Rightarrow(\alpha-3)^2=144 \Rightarrow \alpha=15(\because \alpha>0)
\end{aligned}
$$ | integer | jee-main-2023-online-13th-april-morning-shift |
1lgrgo0q2 | maths | 3d-geometry | plane-in-space | <p>Let the plane $$x+3 y-2 z+6=0$$ meet the co-ordinate axes at the points A, B, C. If the orthocenter of the triangle $$\mathrm{ABC}$$ is $$\left(\alpha, \beta, \frac{6}{7}\right)$$, then $$98(\alpha+\beta)^{2}$$ is equal to ___________.</p> | [] | null | 288 | $$
\begin{aligned}
& \mathrm{A}(-6,0,0) \quad \mathrm{B}(0,-2,0) \mathrm{C}=(0,0,3) \\\\
& \overrightarrow{\mathrm{AB}}=6 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}, \quad \overrightarrow{\mathrm{BC}}=2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}, \\\\
& \overrightarrow{\mathrm{AC}}=6 \hat{\mathrm{i}}+3 \hat{\mathrm{k}}
\end{aligned}
$$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1li9wiey7/d8727013-7167-49ce-a48e-7f381925a62f/b4de6bf0-feb3-11ed-9a3f-cb8a261721bd/file-1li9wiey8.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1li9wiey7/d8727013-7167-49ce-a48e-7f381925a62f/b4de6bf0-feb3-11ed-9a3f-cb8a261721bd/file-1li9wiey8.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 12th April Morning Shift Mathematics - 3D Geometry Question 69 English Explanation">
<br><br>$$
\begin{aligned}
& \overrightarrow{\mathrm{AH}} \cdot \overrightarrow{\mathrm{BC}}=0 \\\\
& \left(\alpha+6, \beta, \frac{6}{7}\right) \cdot(0,2,3)=0 \\\\
& \beta=\frac{-9}{7} \\\\
& \overrightarrow{\mathrm{CH}} \cdot \overrightarrow{\mathrm{AB}}=0 \\\\
& \left(\alpha, \beta, \frac{-15}{7}\right) \cdot(6,-2,0)=0 \\\\
& 6 \alpha-2 \beta=0 \\\\
& \alpha=\frac{-3}{7}
\end{aligned}
$$
<br><br>$$
98(\alpha+\beta)^2=(98) \frac{(144)}{49}=288
$$ | integer | jee-main-2023-online-12th-april-morning-shift |
1lgsuokhr | maths | 3d-geometry | plane-in-space | <p>Let the line passing through the points $$\mathrm{P}(2,-1,2)$$ and $$\mathrm{Q}(5,3,4)$$ meet the plane $$x-y+z=4$$ at the point $$\mathrm{R}$$. Then the distance of the point $$\mathrm{R}$$ from the plane $$x+2 y+3 z+2=0$$ measured parallel to the line $$\frac{x-7}{2}=\frac{y+3}{2}=\frac{z-2}{1}$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\sqrt{31}$$"}, {"identifier": "B", "content": "$$\\sqrt{189}$$"}, {"identifier": "C", "content": "$$\\sqrt{61}$$"}, {"identifier": "D", "content": "3"}] | ["D"] | null | Equation of line $P Q$ :
<br><br>$$
\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}=\lambda
$$
<br><br>Let $R$ be $(3 \lambda+2,4 \lambda-1,2 \lambda+2)$
<br><br>$\mathrm{R}$ lies on plane $x-y+z=4$
<br><br>$$
\begin{aligned}
& \therefore \quad 3 \lambda+2-4 \lambda+1+2 \lambda+2=4 \\\\
& \Rightarrow \quad \lambda=-1 \\\\
& \therefore \quad R(-1,-5,0)
\end{aligned}
$$
<br><br>$$
\text { Let } S R \text { be }: \frac{x+1}{2}=\frac{y+5}{2}=\frac{z - 0}{1}=k
$$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lifho571/8169bced-736c-4022-8d65-aab144d24688/5f25cfd0-01c6-11ee-a619-03f3ffceb323/file-1lifho572.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lifho571/8169bced-736c-4022-8d65-aab144d24688/5f25cfd0-01c6-11ee-a619-03f3ffceb323/file-1lifho572.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 11th April Evening Shift Mathematics - 3D Geometry Question 65 English Explanation">
<br><br>Point $$
S:(2 k-1,2 k-5, k)
$$
<br><br>$S$ lies on plane: $x+2 y+3 z+2=0$
<br><br>$$
\begin{aligned}
& \Rightarrow(2 k-1)+(4 k-10)+3 k+2=0 \\\\
& \Rightarrow 9 k-9=0 \Rightarrow k=1
\end{aligned}
$$
<br><br>$$
\therefore \mathrm{S}=(1,-3,1)
$$
<br><br>$$
\therefore S R=\sqrt{4+4+1}=3
$$ | mcq | jee-main-2023-online-11th-april-evening-shift |
1lgsvumqy | maths | 3d-geometry | plane-in-space | <p>Let P be the plane passing through the points $$(5,3,0),(13,3,-2)$$ and $$(1,6,2)$$.
For $$\alpha \in \mathbb{N}$$, if the distances of the points $$\mathrm{A}(3,4, \alpha)$$ and $$\mathrm{B}(2, \alpha, a)$$ from the plane P are 2 and 3 respectively, then the positive value of a is :</p> | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "3"}] | ["B"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lig7g1t0/e607e161-c3d1-45fc-96f7-d8420574551c/2b69fe40-022b-11ee-8eea-09e354b91938/file-1lig7g1t1.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lig7g1t0/e607e161-c3d1-45fc-96f7-d8420574551c/2b69fe40-022b-11ee-8eea-09e354b91938/file-1lig7g1t1.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 11th April Evening Shift Mathematics - 3D Geometry Question 67 English Explanation">
<br><br>$$
\begin{aligned}
& \overrightarrow{A B}=8 \hat{i}-2 \hat{k} \\\\
& \overrightarrow{A C}=-4 \hat{i}+3 \hat{j}+2 \hat{k}
\end{aligned}
$$
<br><br>$$
\begin{aligned}
& \overrightarrow{A B} \times \overrightarrow{A C}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
8 & 0 & -2 \\
-4 & 3 & 2
\end{array}\right| \\\\
& =6 \hat{i}-8 \hat{j}+24 \hat{k}
\end{aligned}
$$
<br><br>Equation of plane : $6 x-8 y+24 z=d$ passes through $(5,3,0)$
<br><br>$$
\begin{aligned}
& 6 \times 5-8 \times 3+24 \times 0=d \\\\
& \Rightarrow d=6 \\\\
& 6 x-8 y+24 z=6 \\\\
& \Rightarrow 3 x-4 y+12 z=3
\end{aligned}
$$
<br><br>Distance of point $(3,4, \alpha)$
<br><br>$$
\frac{9-16+12 \alpha-3}{\sqrt{9+16+144}}=2 \Rightarrow \alpha=3
$$
<br><br>Distance of point $(2, \alpha, a)$
<br><br>$$
\begin{aligned}
& \frac{3 \times 2-4 \times 3+12 \times a-3}{13}=3 \\\\
& \Rightarrow 12 a-9=39 \\\\
& \Rightarrow 12 a=48 \\\\
& \Rightarrow a=4
\end{aligned}
$$ | mcq | jee-main-2023-online-11th-april-evening-shift |
1lguu3law | maths | 3d-geometry | plane-in-space | <p>Let $$(\alpha, \beta, \gamma)$$ be the image of the point $$\mathrm{P}(2,3,5)$$ in the plane $$2 x+y-3 z=6$$. Then $$\alpha+\beta+\gamma$$ is equal to :</p> | [{"identifier": "A", "content": "10"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "12"}] | ["A"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1ln74ubxe/c878026c-5267-42b0-b10f-7150edf4c2a2/04392c20-602b-11ee-84d4-f1c1aec911f3/file-6y3zli1ln74ubxf.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1ln74ubxe/c878026c-5267-42b0-b10f-7150edf4c2a2/04392c20-602b-11ee-84d4-f1c1aec911f3/file-6y3zli1ln74ubxf.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh; vertical-align: baseline;" alt="JEE Main 2023 (Online) 11th April Morning Shift Mathematics - 3D Geometry Question 62 English Explanation">
<br><br>Co-ordinate of image of point $P(2,3,5)$ in the plane $2 x+y-3 z=6$ is
<br><br>$$
\begin{aligned}
\frac{\alpha-2}{2} & =\frac{\beta-3}{1}=\frac{\gamma-5}{-3}=\frac{-2(2 \times 2+3-3 \times 5-6)}{2^2+1^2+(-3)^2} \\\\
\frac{\alpha-2}{2} & =\frac{\beta-3}{1}=\frac{\gamma-5}{-3}=2 \\\\
\frac{\alpha-2}{2} & =2, \frac{\beta-3}{1}=2, \frac{\gamma-5}{-3}=2 \\\\
\alpha & =6, \beta=5, \gamma=-1
\end{aligned}
$$
<br><br>Hence, $\alpha+\beta+\gamma=6+5-1=10$ | mcq | jee-main-2023-online-11th-april-morning-shift |
1lguuhrui | maths | 3d-geometry | plane-in-space | <p>If equation of the plane that contains the point $$(-2,3,5)$$ and is perpendicular to each of the planes $$2 x+4 y+5 z=8$$ and $$3 x-2 y+3 z=5$$ is $$\alpha x+\beta y+\gamma z+97=0$$ then $$\alpha+\beta+\gamma=$$</p> | [{"identifier": "A", "content": "15"}, {"identifier": "B", "content": "16"}, {"identifier": "C", "content": "17"}, {"identifier": "D", "content": "18"}] | ["A"] | null | The equation of plane that passes through the point $(-2,3,5)$ is
<br/><br/>$$
a(x+2)+b(y-3)+c(z-5)=0
$$ ..........(i)
<br/><br/>The plane is perpendicular to
<br/><br/>$$
\begin{array}{ll}
2 x+4 y+5 z =8 \text { and } 3 x-2 y+3 z=5 \\\\
\end{array}
$$
<br/><br/>$$
\begin{array}{ll}
\therefore 2 a+4 b+5 c=0 ..........(ii)\\\\
\text { and } 3 a-2 b+3 c=0 ..........(iii)
\end{array}
$$
<br/><br/>$$
\begin{aligned}
& \therefore \quad \frac{\mathrm{a}}{\left|\begin{array}{cc}
4 & 5 \\
-2 & 3
\end{array}\right|}=\frac{-\mathrm{b}}{\left|\begin{array}{ll}
2 & 5 \\
3 & 3
\end{array}\right|}=\frac{\mathrm{c}}{\left|\begin{array}{cc}
2 & 4 \\
3 & -2
\end{array}\right|} \\\\
& \Rightarrow \frac{\mathrm{a}}{22}=\frac{\mathrm{b}}{9}=\frac{\mathrm{c}}{-16}
\end{aligned}
$$
<br/><br/>$\therefore$ From Equation (i)
equation of plane
<br/><br/>$$
\begin{aligned}
&22(x+2)+9(y-3)-16(z-5)=0 \\\\
&\Rightarrow 22 x+9 y-16 z+97=0
\end{aligned}
$$
<br/><br/>Here, $\alpha=22, \beta=9, \gamma=-16$
<br/><br/>$$
\therefore \alpha+\beta+\gamma=22+9-16=15
$$ | mcq | jee-main-2023-online-11th-april-morning-shift |
1lgvpmh4j | maths | 3d-geometry | plane-in-space | <p>Let the image of the point $$\mathrm{P}(1,2,6)$$ in the plane passing through the points $$\mathrm{A}(1,2,0), \mathrm{B}(1,4,1)$$ and $$\mathrm{C}(0,5,1)$$ be $$\mathrm{Q}(\alpha, \beta, \gamma)$$. Then $$\left(\alpha^{2}+\beta^{2}+\gamma^{2}\right)$$ is equal to :</p> | [{"identifier": "A", "content": "76"}, {"identifier": "B", "content": "62"}, {"identifier": "C", "content": "70"}, {"identifier": "D", "content": "65"}] | ["D"] | null | Equation of plane passing through the points $A(1,2$, $0), B(1,4,1)$ and $C(0,5,1)$ is
<br/><br/>$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-1 & y-2 & z-0 \\
0 & 2 & 1 \\
-1 & 3 & 1
\end{array}\right|=0 \\\\
& \Rightarrow x+y-2 z=3
\end{aligned}
$$
<br/><br/>Now $Q(\alpha, \beta, \gamma)$ is the image of the point $P(1,2,6)$ in the plane $x+y-2 z-3=0$
<br/><br/>$$
\begin{array}{ll}
&\therefore \frac{\alpha-1}{1}=\frac{\beta-2}{1}=\frac{\gamma-6}{-2}=\frac{-2[1+2-2(6)-3]}{1^2+1^2+(-2)^2} \\\\
&\Rightarrow \frac{\alpha-1}{1}=\frac{\beta-2}{1}=\frac{\gamma-6}{-2}=4 \\\\
&\Rightarrow \alpha=5, \beta=6, \gamma=-2
\end{array}
$$
<br/><br/>$$
\text { Hence, } \alpha^2+\beta^2+\gamma^2=5^2+6^2+(-2)^2=65
$$ | mcq | jee-main-2023-online-10th-april-evening-shift |
1lgvpsiky | maths | 3d-geometry | plane-in-space | <p>Let the line $$\frac{x}{1}=\frac{6-y}{2}=\frac{z+8}{5}$$ intersect the lines $$\frac{x-5}{4}=\frac{y-7}{3}=\frac{z+2}{1}$$ and $$\frac{x+3}{6}=\frac{3-y}{3}=\frac{z-6}{1}$$ at the points $$\mathrm{A}$$ and $$\mathrm{B}$$ respectively. Then the distance of the mid-point of the line segment $$\mathrm{AB}$$ from the plane $$2 x-2 y+z=14$$ is :</p> | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "$$\\frac{10}{3}$$"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "$$\\frac{11}{3}$$"}] | ["C"] | null | We have, $\frac{x}{1}=\frac{6-y}{2}=\frac{z+8}{5}$ intersect the line
<br/><br/>$\frac{x-5}{4}=\frac{y-7}{3}=\frac{z+2}{1}$ and $\frac{x+3}{6}=\frac{3-y}{3}=\frac{z-6}{1}$
<br/><br/>$$
\begin{array}{ll}
& \text { Now, } \frac{x}{1}=\frac{6-y}{2}=\frac{z+8}{5}=\lambda ...........(i)\\\\
& \Rightarrow x=\lambda, y=6-2 \lambda, z=5 \lambda-8
\end{array}
$$
<br/><br/>$$
\begin{array}{ll}
&\text { Also, } \frac{x-5}{4} =\frac{y-7}{3}=\frac{z+2}{1}=k ...........(ii)\\\\
&\Rightarrow x =4 k+5, y=3 k+7, z=k-2
\end{array}
$$
<br/><br/>$$
\begin{array}{rlrl}
& \frac{x+3}{6}=\frac{3-y}{3}=\frac{z-6}{1}=\mu ..........(iii)\\\\
&\Rightarrow x = 6 \mu-3, y=3-3 \mu, z=\mu+6
\end{array}
$$
<br/><br/>On solving Eqs. (i) and (ii), we get $\lambda=1, k=-1$
<br/><br/>$\therefore$ Co-ordinate of $A$ is $(1,4,-3)$
<br/><br/>On solving Eqs. (i) and (iii), we get $\lambda=3, \mu=1$
<br/><br/>$\therefore$ Co-ordinate of $\beta$ is $(3,0,7)$
<br/><br/>Co-ordinate of mid-point of $A B$ is $\left(\frac{1+3}{2}, \frac{4+0}{2}, \frac{-3+7}{2}\right)$ or $(2,2,2)$
<br/><br/>Perpendicular distance of mid-point of $A B$ from the plane $2 x-2 y+z=14$ is
<br/><br/>$$
\frac{|2(2)-2(2)+2-14|}{\sqrt{2^2+(-2)^2+1^2}}=4
$$ | mcq | jee-main-2023-online-10th-april-evening-shift |
1lgyop0gk | maths | 3d-geometry | plane-in-space | <p>Let $$\mathrm{P}_{1}$$ be the plane $$3 x-y-7 z=11$$ and $$\mathrm{P}_{2}$$ be the plane passing through the points $$(2,-1,0),(2,0,-1)$$, and $$(5,1,1)$$. If the foot of the perpendicular drawn from the point $$(7,4,-1)$$ on the line of intersection of the planes $$P_{1}$$ and $$P_{2}$$ is $$(\alpha, \beta, \gamma)$$, then $$\alpha+\beta+\gamma$$ is equal to ___________.</p> | [] | null | 11 | Equation of plane $\mathrm{P}_2$ passing through $(2,-1,0),(2,0$, $-1)$ and $(5,1,1)$ is
<br/><br/>$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-5 & y-1 & z-1 \\
3 & 2 & 1 \\
3 & 1 & 2
\end{array}\right|=0 \\\\
& \Rightarrow(x-5)(4-1)-(y-1)(6-3)+(z-1)(3-6)=0 \\\\
& \Rightarrow 3 x-15-3 y+3-3 z+3=0 \\\\
& \Rightarrow 3 x-3 y-3 z-9=0 \\\\
& \Rightarrow x-y-z=3 .......(i)
\end{aligned}
$$
<br/><br/>Now, direction ratios of line of intersection of $P_1$ and $\mathrm{P}_2$ is
<br/><br/>$$
\begin{aligned}
& \left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & -1 & -1 \\
3 & -1 & -7
\end{array}\right| \\\\
& =\hat{i}(7-1)-\hat{j}(-7+3)+\hat{k}(-1+3) \\\\
& =6 \hat{i}+4 \hat{j}+2 \hat{k}
\end{aligned}
$$
<br/><br/>At $z=0, x-y=3$
[from (i)]
<br/><br/>$$
3 x-y=11
$$
<br/><br/>On solving, we get
<br/><br/>$$
x=4 \text { and } y=1
$$
<br/><br/>So, equation of line is
<br/><br/>$$
\frac{x-4}{6}=\frac{y-1}{4}=\frac{z-2}{6}=k
$$
<br/><br/>$$
\begin{aligned}
& \therefore(\alpha, \beta, \gamma)=(6 k+4,4 k+1,2 k) \\\\
& \Rightarrow(6)(\alpha-7)+4(\beta-4)+2(\gamma+1)=0 \\\\
& \Rightarrow 6(6 k+4-7)+4(4 k+1-4)+2(2 k+1)=0 \\\\
& \Rightarrow 36 k-18+16 k-12+4 k+4=0 \\\\
& \Rightarrow 56 k=26 \Rightarrow k=\frac{1}{2} \\\\
& \text { So, } \alpha=7, \beta=3 \text { and } \gamma=1 \\\\
& \therefore \alpha+\beta+\gamma=7+3+1=11
\end{aligned}
$$ | integer | jee-main-2023-online-8th-april-evening-shift |
1lgzzukx0 | maths | 3d-geometry | plane-in-space | <p>If the equation of the plane containing the line <br/><br/>$$x+2 y+3 z-4=0=2 x+y-z+5$$ and perpendicular to the plane <br/><br/>$\vec{r}=(\hat{i}-\hat{j})+\lambda(\hat{i}+\hat{j}+\hat{k})+\mu(\hat{i}-2 \hat{j}+3 \hat{k})$<br/><br/> is $a x+b y+c z=4$, then $$(a-b+c)$$ is equal to :</p> | [{"identifier": "A", "content": "18"}, {"identifier": "B", "content": "22"}, {"identifier": "C", "content": "20"}, {"identifier": "D", "content": "24"}] | ["B"] | null | Equation of plane $\mathrm{P}$ containing the given lines is
<br/><br/>$$
\begin{aligned}
& (x+2 y+3 z-4)+\lambda(2 x+y-z+5)=0 \\\\
& \Rightarrow(1+2 \lambda) x+(2+\lambda) y+(3-\lambda) z+(-4+5 \lambda)=0
\end{aligned}
$$
<br/><br/>Now, plane $\mathrm{P}$ is perpendicular to plane $\mathrm{P}^{\prime}$
<br/><br/>$$
\vec{r}=(\hat{i}-\hat{j})+\lambda(\hat{i}+\hat{j}+\hat{k})+\mu(\hat{i}-2 \hat{j}+3 \hat{k})
$$
<br/><br/>So, normal to plane $\mathrm{P}^{\prime}$ is
<br/><br/>$$
\begin{aligned}
& \vec{n}=(\hat{i}+\hat{j}+\hat{k}) \times(\hat{i}-2 \hat{j}+3 \hat{k}) \\\\
& \Rightarrow \vec{n}=5 \hat{i}-2 \hat{j}-3 \hat{k}
\end{aligned}
$$
<br/><br/>$\therefore \mathrm{P}$ and $\mathrm{P}^{\prime}$ are perpendicular
<br/><br/>$$
\begin{aligned}
& \therefore 5(1+2 \lambda)-2(2+\lambda)-3(3-\lambda)=0 \\\\
& \Rightarrow 5+10 \lambda-4-2 \lambda-9+3 \lambda=0 \\\\
& \Rightarrow 11 \lambda=8 \Rightarrow \lambda=\frac{8}{11}
\end{aligned}
$$
<br/><br/>$$
\begin{array}{r}
\therefore P:\left(1+\frac{16}{11}\right) x+\left(2+\frac{8}{11}\right) y+\left(3-\frac{8}{11}\right) z+\left(5 \times \frac{8}{11}-4\right)
=0
\end{array}
$$
<br/><br/>i.e., $27 x+30 y+25 z=4$
<br/><br/>which is same as $a x+b y+c z=4$
<br/><br/>$$
\begin{aligned}
& \therefore a=27, b=30 \text { and } c=25 \\\\
& \Rightarrow a-b+c=27-30+25=22
\end{aligned}
$$ | mcq | jee-main-2023-online-8th-april-morning-shift |
1lh244z4l | maths | 3d-geometry | plane-in-space | <p>Let the image of the point $$\mathrm{P}(1,2,3)$$ in the plane $$2 x-y+z=9$$ be $$\mathrm{Q}$$. If the coordinates of the point $$\mathrm{R}$$ are $$(6,10,7)$$, then the square of the area of the triangle $$\mathrm{PQR}$$ is _____________.</p> | [] | null | 594 | Let $Q(x, y, z)$ be the image of $P(1,2,3)$ in the plane
<br><br>$$
\begin{aligned}
& 2 x-y+z=9 \\\\
& \therefore \frac{x-1}{2}=\frac{y-2}{-1}=\frac{z-3}{1}
=\frac{-2(2 \times 1+(-1)(2)+(1)(3)(-9)}{(2)^2+(-1)^2+(1)^2}
\end{aligned}
$$
<br><br>$$
\begin{aligned}
& \Rightarrow \frac{x-1}{2}=\frac{y-2}{-1}=\frac{z-3}{1}=\frac{-2(-6)}{6}=2 \\\\
& \Rightarrow x=5, y=0, z=5
\end{aligned}
$$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lo34mk9p/c7b53e64-218a-4775-ad23-59117bd05abd/40f9acd0-71c3-11ee-84b8-cf7c0a25cec4/file-6y3zli1lo34mk9q.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lo34mk9p/c7b53e64-218a-4775-ad23-59117bd05abd/40f9acd0-71c3-11ee-84b8-cf7c0a25cec4/file-6y3zli1lo34mk9q.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh; vertical-align: baseline;" alt="JEE Main 2023 (Online) 6th April Morning Shift Mathematics - 3D Geometry Question 48 English Explanation">
<br><br>Now, $\overrightarrow {PQ} =4 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and <br><br>$\overrightarrow{P R}=5 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$
<br><br>$\therefore$ Area of the $\triangle P Q R=\frac{1}{2}|\overrightarrow{P Q} \times \overrightarrow{P R}|$
<br><br>Now, $\overrightarrow{P Q} \times \overrightarrow{P R}$
<br><br>$$
\begin{aligned}
& =\left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
4 & -2 & 2 \\
5 & 8 & 4
\end{array}\right| \\\\
& =\hat{\mathbf{i}}(-8-16)-\hat{\mathbf{j}}(16-10)+\hat{\mathbf{k}}(32+10) \\\\
& =-24 \hat{\mathbf{i}}-6 \hat{\mathbf{j}}+42 \hat{\mathbf{k}}
\end{aligned}
$$
<br><br>$$
\begin{aligned}
\therefore \frac{1}{2} \mid(-24 \hat{\mathbf{i}} & -6 \hat{\mathbf{j}}+42 \hat{\mathbf{k}})|=|-12 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+21 \hat{\mathbf{k}} \mid \\\\
& =\sqrt{(-12)^2+(-3)^2+(21)^2} \\\\
& =\sqrt{144+9+441}=\sqrt{594}
\end{aligned}
$$
<br><br>$\therefore$ The square of the area of the $\triangle P Q R=594$ | integer | jee-main-2023-online-6th-april-morning-shift |
1lh2xv9fi | maths | 3d-geometry | plane-in-space | <p>A plane P contains the line of intersection of the plane $$\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=6$$ and $$\vec{r} \cdot(2 \hat{i}+3 \hat{j}+4 \hat{k})=-5$$. If $$\mathrm{P}$$ passes through the point $$(0,2,-2)$$, then the square of distance of the point $$(12,12,18)$$ from the plane $$\mathrm{P}$$ is :</p> | [{"identifier": "A", "content": "310"}, {"identifier": "B", "content": "620"}, {"identifier": "C", "content": "1240"}, {"identifier": "D", "content": "155"}] | ["B"] | null | Given plane $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=6$ and
<br/><br/>$$
\vec{r} \cdot(2 \hat{i}+3 \hat{j}+4 \hat{k})=-5
$$
<br/><br/>Equation of plane passing through both plane
<br/><br/>$$
\begin{aligned}
& \mathrm{P}_1 \rightarrow(x \hat{i}+y \hat{j}+2 \hat{k})(\hat{i}+\hat{j}+\hat{k})=6 \\\\
& \mathrm{P}_1=x+y+z=6 \\\\
& \mathrm{P}_2 \rightarrow(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(2 \hat{i}+3 \hat{j}+4 \hat{k})=-5 \\\\
& \mathrm{P}_2 \rightarrow=2 x+3 y+4 z=-5
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \mathrm{P}_1+\lambda \mathrm{P}_2=0 \\\\
& \Rightarrow(x+y+z-6)+\lambda(2 x+3 y+4 z+5)=0
\end{aligned}
$$
<br/><br/>passes through $(0,2,-2)$
<br/><br/>$$
\begin{aligned}
& \Rightarrow(0+2-2-6)+\lambda(2 \times 0+3 \times 2+4 \times(-2)+5)=0 \\\\
& \Rightarrow \lambda=2
\end{aligned}
$$
<br/><br/>Equation of plane
<br/><br/>$$5 x+7 y+9 z+4=0$$
<br/><br/>$\therefore$ Distance of the point $(12,12,18)$ from the plane
<br/><br/>$$
\begin{aligned}
d & =\frac{5 \times 12+7 \times 12+9 \times 18+4}{\sqrt{5^2+7^2+9^2}} \\\\
& =\frac{60+84+162+4}{\sqrt{155}}=\frac{310}{\sqrt{155}} \\\\
\therefore d^2 & =\frac{310 \times 310}{155}=620
\end{aligned}
$$ | mcq | jee-main-2023-online-6th-april-evening-shift |
i0c95WVQI0WQ06dx | maths | application-of-derivatives | maxima-and-minima | The maximum distance from origin of a point on the curve
<br/>$$x = a\sin t - b\sin \left( {{{at} \over b}} \right)$$
<br/>$$y = a\cos t - b\cos \left( {{{at} \over b}} \right),$$ both $$a,b > 0$$ is | [{"identifier": "A", "content": "$$a-b$$ "}, {"identifier": "B", "content": "$$a+b$$ "}, {"identifier": "C", "content": "$$\\sqrt {{a^2} + {b^2}} $$ "}, {"identifier": "D", "content": "$$\\sqrt {{a^2} - {b^2}} $$ "}] | ["B"] | null | Distance of origin from $$\left( {x,y} \right) = \sqrt {{x^2} + {y^2}} $$
<br><br>$$ = \sqrt {{a^2} + {b^2} - 2ab\cos \left( {t - {{at} \over b}} \right)} ;$$
<br><br>$$ \le \sqrt {{a^2} + {b^2} + 2ab} $$ $$\left[ {{{\left\{ {\cos \left( {t - {{at} \over b}} \right)} \right\}}_{\min }} = - 1} \right]$$
<br><br>$$=a+b$$
<br><br>$$\therefore$$ Maximum distance from origin $$=a+b$$ | mcq | aieee-2002 |
mM353u3qL1EToDRE | maths | application-of-derivatives | maxima-and-minima | The real number $$x$$ when added to its inverse gives the minimum sum at $$x$$ equal : | [{"identifier": "A", "content": "-2 "}, {"identifier": "B", "content": "2 "}, {"identifier": "C", "content": "1 "}, {"identifier": "D", "content": "-1"}] | ["C"] | null | $$f(x) = y = x + {1 \over x}$$ or $${{dy} \over {dx}} = 1 - {1 \over {{x^2}}}$$
<br><br>For max. or min, $$1 - {1 \over {{x^2}}} = 0 \Rightarrow x = \pm 1$$
<br><br>$${{{d^2}y} \over {d{x^2}}} = {2 \over {{x^3}}} $$
<br/><br/>$$ {\left( {{{{d^2}y} \over {d{x^2}}}} \right)_{x = 1}} = 2$$ ($$+ve$$)
<br><br>$$\therefore$$ f(x) will be minima at $$x=1$$. | mcq | aieee-2003 |
xIrzYEUCC29YS3QC | maths | application-of-derivatives | maxima-and-minima | If the function $$f\left( x \right) = 2{x^3} - 9a{x^2} + 12{a^2}x + 1,$$ where $$a>0,$$ attains its maximum and minimum at $$p$$ and $$q$$ respectively such that $${p^2} = q$$ , then $$a$$ equals | [{"identifier": "A", "content": "$${1 \\over 2}$$ "}, {"identifier": "B", "content": "$$3$$"}, {"identifier": "C", "content": "$$1$$ "}, {"identifier": "D", "content": "$$2$$ "}] | ["D"] | null | $$f\left( x \right) = 2{x^3} - 9a{x^2} + 12{a^2}x + 1$$
<br><br>$$f'\left( x \right) = 6{x^2} - 18ax + 12{a^2};$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,f''\left( x \right) = 12x - 18a$$
<br><br>For max. or min.
<br><br>$$6{x^2} - 18ax + 12{a^2} = 0$$
<br><br>$$ \Rightarrow {x^2} - 3ax + 2{a^2} = 0$$
<br><br>$$ \Rightarrow x = a$$ or $$x=2a.$$
<br><br>At $$x=a$$
<br><br>$$f''(a) = 12a - 18a = -6a < 0$$
<br><br>At $$x=a$$ maximum As f''($$a$$) < 0.
<br><br>At $$x=2a$$
<br><br>$$f''(a) = 24a - 18a = 6a > 0$$
<br><br>At $$x=2a$$ minimum As f''(2$$a$$) > 0.
<br><br>$$\therefore$$ $$p=a$$ and $$q=2a$$
<br><br>As per question $${p^2} = q$$
<br><br>$$\therefore$$ $${a^2} = 2a $$
<br><br>$$ \Rightarrow $$ $$a(a - 2) = 0$$
<br><br>$$ \therefore $$ $$a$$ = 0, 2
<br><br>but $$a > 0,$$ therefore, $$a=2.$$ | mcq | aieee-2003 |
2tGr74lgODUL42wb | maths | application-of-derivatives | maxima-and-minima | Area of the greatest rectangle that can be inscribed in the
<br/>ellipse $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$ | [{"identifier": "A", "content": "$$2ab$$ "}, {"identifier": "B", "content": "$$ab$$ "}, {"identifier": "C", "content": "$$\\sqrt {ab} $$ "}, {"identifier": "D", "content": "$${a \\over b}$$ "}] | ["A"] | null | Area of rectangle $$ABCD$$ $$ = 2a\,\cos \,\theta $$
<br><br>$$\left( {2b\,\sin \,\theta } \right) = 2ab\,\sin \,2\theta $$
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263702/exam_images/e7w2mcfgqs0qwtwa1jkf.webp" loading="lazy" alt="AIEEE 2005 Mathematics - Application of Derivatives Question 197 English Explanation">
<br><br>$$ \Rightarrow $$ Area of greatest rectangle is equal to $$2ab$$
<br><br>When $$\sin \,2\theta = 1.$$ | mcq | aieee-2005 |
BPl54gj84e83x2jD | maths | application-of-derivatives | maxima-and-minima | The function $$f\left( x \right) = {x \over 2} + {2 \over x}$$ has a local minimum at | [{"identifier": "A", "content": "$$x=2$$ "}, {"identifier": "B", "content": "$$x=-2$$ "}, {"identifier": "C", "content": "$$x=0$$ "}, {"identifier": "D", "content": "$$x=1$$ "}] | ["A"] | null | $$f\left( x \right) = {x \over 2} + {2 \over x} \Rightarrow f'\left( x \right) = {1 \over 2} - {2 \over {{x^2}}} = 0$$
<br><br>$$ \Rightarrow {x^2} = 4$$ or $$x=2,-2;$$ $$\,\,\,\,\,f''\left( x \right) = {4 \over {{x^3}}}$$
<br><br>$$f''{\left. {\left( x \right)} \right]_{x = 2}} = + ve \Rightarrow f\left( x \right)$$
<br><br>has local min at $$x=2.$$ | mcq | aieee-2006 |
AEdU5WTln3rLzwom | maths | application-of-derivatives | maxima-and-minima | A triangular park is enclosed on two sides by a fence and on the third side by a straight river bank. The two sides having fence are of same length $$x$$. The maximum area enclosed by the park is | [{"identifier": "A", "content": "$${3 \\over 2}{x^2}$$ "}, {"identifier": "B", "content": "$$\\sqrt {{{{x^3}} \\over 8}} $$ "}, {"identifier": "C", "content": "$${1 \\over 2}{x^2}$$ "}, {"identifier": "D", "content": "$$\\pi {x^2}$$ "}] | ["C"] | null | Area $$ = {1 \over 2}{x^2}\,\sin \,\theta $$
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265305/exam_images/jl4uz1aheqoemq1myxxa.webp" loading="lazy" alt="AIEEE 2006 Mathematics - Application of Derivatives Question 192 English Explanation">
<br><br>Maximum value of $$\sin \theta $$ is $$1$$ at $$\theta = {\pi \over 2}$$
<br><br>$${A_{\max }} = {1 \over 2}{x^2}$$ | mcq | aieee-2006 |
SOofWOrrmqiZbPr1 | maths | application-of-derivatives | maxima-and-minima | If $$p$$ and $$q$$ are positive real numbers such that $${p^2} + {q^2} = 1$$, then the maximum value of $$(p+q)$$ is | [{"identifier": "A", "content": "$${1 \\over 2}$$ "}, {"identifier": "B", "content": "$${1 \\over {\\sqrt 2 }}$$ "}, {"identifier": "C", "content": "$${\\sqrt 2 }$$ "}, {"identifier": "D", "content": "$$2$$"}] | ["C"] | null | Given that $${p^2} + {q^2} = 1$$
<br><br>$$\therefore$$ $$p = \cos \theta $$ and $$q = \sin \theta $$
<br><br>Then $$p+q$$ $$ = \cos \theta + \sin \theta $$
<br><br>We know that
<br><br>$$ - \sqrt {{a^2} + {b^2}} \le a\cos \theta + b\sin \theta \le \sqrt {{a^2} + {b^2}} $$
<br><br>$$\therefore$$ $$ - \sqrt 2 \le \cos \theta + \sin \theta \le \sqrt 2 $$
<br><br>Hence max. value of $$p + q$$ is $$\sqrt 2 $$ | mcq | aieee-2007 |
U4dcftyIinHNo7sg | maths | application-of-derivatives | maxima-and-minima | Suppose the cubic $${x^3} - px + q$$ has three distinct real roots
<br/>where $$p>0$$ and $$q>0$$. Then which one of the following holds? | [{"identifier": "A", "content": "The cubic has minima at $$\\sqrt {{p \\over 3}} $$ and maxima at $$-\\sqrt {{p \\over 3}} $$"}, {"identifier": "B", "content": "The cubic has minima at $$-\\sqrt {{p \\over 3}} $$ and maxima at $$\\sqrt {{p \\over 3}} $$"}, {"identifier": "C", "content": "The cubic has minima at both $$\\sqrt {{p \\over 3}} $$ and $$-\\sqrt {{p \\over 3}} $$"}, {"identifier": "D", "content": "The cubic has maxima at both $$\\sqrt {{p \\over 3}} $$ and $$-\\sqrt {{p \\over 3}} $$"}] | ["A"] | null | Let $$y = {x^3} - px + q$$
<br><br>$$ \Rightarrow {{dy} \over {dx}} = 3{x^2} - p$$
<br><br>For $${{dy} \over {dx}} = 0 \Rightarrow 3{x^2} - p = 0$$
<br><br>$$ \Rightarrow x = \pm \sqrt {{p \over 3}} $$
<br><br>$${{{d^2}y} \over {d{x^2}}} = 6x$$
<br><br>$${\left. {{{{d^2}y} \over {d{x^2}}}} \right|_{x = \sqrt {{p \over 3}} }} = + ve\,\,\,\,$$ and
<br><br>$$\,\,\,\,\,\,\,\,\,\,$$ $${\left. {\,\,\,{{{d^2}y} \over {d{x^2}}}} \right|_{x = - \sqrt {{p \over 3}} }} = - ve$$
<br><br>$$\therefore$$ $$y$$ has ninima at $$x = \sqrt {{p \over 3}} $$
<br><br>and maxima at $$x = - \sqrt {{p \over 3}} $$ | mcq | aieee-2008 |
h0h4CCP2OjJaHXK7 | maths | application-of-derivatives | maxima-and-minima | Given $$P\left( x \right) = {x^4} + a{x^3} + b{x^2} + cx + d$$ such that $$x=0$$ is the only
<br/>real root of $$P'\,\left( x \right) = 0.$$ If $$P\left( { - 1} \right) < P\left( 1 \right),$$ then in the interval $$\left[ { - 1,1} \right]:$$ | [{"identifier": "A", "content": "$$P(-1)$$ is not minimum but $$P(1)$$ is the maximum of $$P$$"}, {"identifier": "B", "content": "$$P(-1)$$ is the minimum but $$P(1)$$ is not the maximum of $$P$$"}, {"identifier": "C", "content": "Neither $$P(-1)$$ is the minimum nor $$P(1)$$ is the maximum of $$P$$"}, {"identifier": "D", "content": "$$P(-1)$$ is the minimum and $$P(1)$$ is the maximum of $$P$$"}] | ["A"] | null | We have $$P\left( x \right) = {x^4} + a{x^3} + b{x^2} + cx + d$$
<br><br>$$ \Rightarrow P'\left( x \right) = 4\,{x^3} + 3a{x^2} + 2bx + c$$
<br><br>But $$P'\left( 0 \right) = 0 \Rightarrow c = 0$$
<br><br>$$\therefore$$ $$P\left( x \right) = {x^4} + a{x^3} + b{x^2} + d$$
<br><br>As given that $$P\left( { - 1} \right) < P\left( a \right)$$
<br><br>$$ \Rightarrow 1 - a + b + d\,\, < \,\,1 + a + b + d \Rightarrow a > 0$$
<br><br>Now $$P'\left( x \right) = 4{x^3} + 3a{x^2} + 2bx = x\left( {4{x^2} + 3ax + 2b} \right)$$
<br><br>As $$P'\left( x \right) = 0,$$ there is only one solution $$x = 0,$$
<br><br>therefore $$4{x^2} + 3ax + 2b = 0$$ should not have any real roots i.e. $$D < 0$$
<br><br>$$ \Rightarrow 9{a^2} - 32b < 0$$
<br><br>$$ \Rightarrow b > {{9{a^2}} \over {32}} > 0$$
<br><br>Hence $$a,b > 0 \Rightarrow P'\left( x \right) = 4{x^3} + 3a{x^2} + 2bx > 0$$
<br><br>$$\forall x > 0$$
<br><br>$$\therefore$$ $$P(x)$$ is an increasing function on $$\left( {0,1} \right)$$
<br><br>$$\therefore$$ $$P\left( 0 \right) < P\left( a \right)$$
<br><br>Similarly we can prove $$P\left( x \right)$$ is decreasing on $$\left( { - 1,0} \right)$$
<br><br>$$\therefore$$ $$P\left( { - 1} \right) > P\left( 0 \right)$$
<br><br>So we can conclude that
<br><br>Max $$P\left( x \right) = P\left( 1 \right)$$ and Min $$P\left( x \right) = P\left( 0 \right)$$
<br><br>$$ \Rightarrow P\left( { - 1} \right)$$ is not minimum but $$P\left( 1 \right)$$ is the maximum of $$P.$$ | mcq | aieee-2009 |
20SUnhwmW0otvw8z | maths | application-of-derivatives | maxima-and-minima | Let $$f:R \to R$$ be defined by
$$$f\left( x \right) = \left\{ {\matrix{
{k - 2x,\,\,if} & {x \le - 1} \cr
{2x + 3,\,\,if} & {x > - 1} \cr
} } \right.$$$
<p>If $$f$$has a local minimum at $$x=-1$$, then a possible value of $$k$$ is </p> | [{"identifier": "A", "content": "$$0$$ "}, {"identifier": "B", "content": "$$ - {1 \\over 2}$$ "}, {"identifier": "C", "content": "$$-1$$ "}, {"identifier": "D", "content": "$$1$$"}] | ["C"] | null | $$f\left( x \right) = \left\{ {\matrix{
{k - 2x,\,\,\,\,if\,\,\,\,x \le - 1} \cr
{2x + 3,\,\,\,\,if\,\,\,\,x > - 1} \cr
} } \right.$$
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264221/exam_images/up9xtippomu7kefhskum.webp" loading="lazy" alt="AIEEE 2010 Mathematics - Application of Derivatives Question 184 English Explanation">
<br><br>This is true where $$k=-1$$ | mcq | aieee-2010 |
WAccCVwQ4VUOLsfn | maths | application-of-derivatives | maxima-and-minima | Let $$f:R \to R$$ be a continuous function defined by
$$$f\left( x \right) = {1 \over {{e^x} + 2{e^{ - x}}}}$$$
<p><b>Statement - 1 :</b> $$f\left( c \right) = {1 \over 3},$$ for some $$c \in R$$.</p>
<p><b>Statement - 2 :</b> $$0 < f\left( x \right) \le {1 \over {2\sqrt 2 }},$$ for all $$x \in R$$</p> | [{"identifier": "A", "content": "Statement - 1 is true, Statement -2 is true; Statement - 2 is <b>not</b> a correct explanation for Statement - 1."}, {"identifier": "B", "content": "Statement - 1 is true, Statement - 2 is false."}, {"identifier": "C", "content": "Statement - 1 is false, Statement - 2 is true."}, {"identifier": "D", "content": "Statement - 1 is true, Statement -2 is true; Statement -2 is a correct explanation for Statement - 1."}] | ["D"] | null | $$f\left( x \right) = {1 \over {{e^x} + 2{e^{ - x}}}} = {{{e^x}} \over {{e^{2x}} + 2}}$$
<br><br>$$f'\left( x \right) = {{\left( {{e^{2x}} + 2} \right)e{}^x - 2{e^{2x}}.{e^x}} \over {{{\left( {{e^{2x}} + 2} \right)}^2}}}$$
<br><br>$$f'\left( x \right) = 0 \Rightarrow {e^{2x}} + 2 = 2{e^{2x}}$$
<br><br>$${e^{2x}} = 2 \Rightarrow {e^x} = \sqrt 2 $$
<br><br>maximum $$f\left( x \right) = {{\sqrt 2 } \over 4} = {1 \over {2\sqrt 2 }}$$
<br><br>$$0 < f\left( x \right) \le {1 \over {2\sqrt 2 }}\,\,\forall x \in R$$
<br><br>Since $$0 < {1 \over 3} < {1 \over {2\sqrt 2 }} \Rightarrow $$ for some $$c \in R$$
<br><br>$$f\left( c \right) = {1 \over 3}$$ | mcq | aieee-2010 |
Lcf95duImInmLqTv | maths | application-of-derivatives | maxima-and-minima | For $$x \in \left( {0,{{5\pi } \over 2}} \right),$$ define $$f\left( x \right) = \int\limits_0^x {\sqrt t \sin t\,dt.} $$ Then $$f$$ has | [{"identifier": "A", "content": "local minimum at $$\\pi $$ and $$2\\pi $$"}, {"identifier": "B", "content": "local minimum at $$\\pi $$ and local maximum at $$2\\pi $$"}, {"identifier": "C", "content": "local maximum at $$\\pi $$ and local minimum at $$2\\pi $$"}, {"identifier": "D", "content": "local maximum at $$\\pi $$ and $$2\\pi $$"}] | ["C"] | null | $$f'\left( x \right) = \sqrt x \sin x$$
<br><br>At local maxima or minima, $$f'\left( x \right) = 0$$
<br><br>$$ \Rightarrow x = 0$$ or $$sin$$ $$x=0$$
<br><br>$$ \Rightarrow x = 2\pi ,\,\,\pi \in \left( {0,{{5\pi } \over 2}} \right)$$
<br><br>$$f''\left( x \right) = \sqrt x \cos \,x + {1 \over {2\sqrt x }}\sin \,x$$
<br><br>$$ = {1 \over {2\sqrt x }}\left( {2x\,\cos \,x + \sin \,x} \right)$$
<br><br>At $$x = \pi ,$$ $$f''\left( x \right) < 0$$
<br><br>Hence, local maxima at $$x = \pi $$
<br><br>At $$x = 2\pi ,\,\,\,f''\left( x \right) > 0$$
<br><br>Hence local minima at $$x = 2\pi $$ | mcq | aieee-2011 |
iXvsDjZDvJ3Y3hNO | maths | application-of-derivatives | maxima-and-minima | Let $$a,b \in R$$ be such that the function $$f$$ given by $$f\left( x \right) = In\left| x \right| + b{x^2} + ax,\,x \ne 0$$ has extreme values at $$x=-1$$ and $$x=2$$
<p><b>Statement-1 :</b> $$f$$ has local maximum at $$x=-1$$ and at $$x=2$$.</p>
<p><b>Statement-2 :</b> $$a = {1 \over 2}$$ and $$b = {-1 \over 4}$$</p> | [{"identifier": "A", "content": "Statement - 1 is false, Statement - 2 is true."}, {"identifier": "B", "content": "Statement - 1 is true , Statement - 2 is true; Statement - 2 is a correct explanation for Statement - 1."}, {"identifier": "C", "content": "Statement - 1 is true, Statement - 2 is true; Statement - 2 is <b>not</b> a correct explanation for Statement - 1. "}, {"identifier": "D", "content": "Statement - 1 is true, Statement - 2 is false."}] | ["B"] | null | Given, $$f\left( x \right) = \ln \left| x \right| + b{x^2} + ax$$
<br><br>$$\therefore$$ $$f'\left( x \right) = {1 \over x} + 2bx + a$$
<br><br>At $$x=-1,$$ $$f'\left( x \right) = - 1 - 2b + a = 0$$
<br><br>$$ \Rightarrow a - 2b = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
<br><br>At $$x=2,$$ $$\,\,f'\left( x \right) = {1 \over 2} + 4b + a = 0$$
<br><br>$$ \Rightarrow a + 4b = - {1 \over 2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$
<br><br>On solving $$(i)$$ and $$(ii)$$ we get $$a = {1 \over 2},b = - {1 \over 4}$$
<br><br>Thus, $$f'\left( x \right) = {1 \over x} - {x \over 2} + {1 \over 2} = {{2 - {x^2} + x} \over {2x}}$$
<br><br>$$ = {{ - {x^2} + x + 2} \over {2x}} = {{ - \left( {{x^2} - x - 2} \right)} \over {2x}} = {{ - \left( {x + 1} \right)\left( {x - 2} \right)} \over {2x}}$$
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265826/exam_images/evqbnlvd5jv0hmjutzfj.webp" loading="lazy" alt="AIEEE 2012 Mathematics - Application of Derivatives Question 179 English Explanation">
<br><br>So maximum at $$x=-1,2$$
<br><br>Hence both the statements are true and statement $$2$$ is a correct explanation for $$1.$$ | mcq | aieee-2012 |
wyiDjCvI5NPRr2DZ | maths | application-of-derivatives | maxima-and-minima | A line is drawn through the point $$(1, 2)$$ to meet the coordinate axes at $$P$$ and $$Q$$ such that it forms a triangle $$OPQ,$$ where $$O$$ is the origin. If the area of the triangle $$OPQ$$ is least, then the slope of the line $$PQ$$ is : | [{"identifier": "A", "content": "$$-{1 \\over 4}$$"}, {"identifier": "B", "content": "$$-4$$ "}, {"identifier": "C", "content": "$$-2$$ "}, {"identifier": "D", "content": "$$-{1 \\over 2}$$"}] | ["C"] | null | Equation of a line passing through $$\left( {{x_1},{y_1}} \right)$$ having
<br><br>slope $$m$$ is given by $$y - {y_1} = m\left( {x - {x_1}} \right)$$
<br><br>Since the line $$PQ$$ is passing through $$(1,2)$$ therefore its
<br><br>equation is
<br><br>$$\left( {y - 2} \right) = m\left( {x - 1} \right)$$
<br><br>where $$m$$ is the slope of the line $$PQ$$.
<br><br>Now, point $$P\left( {x,0} \right)$$ will also satisfy the equation of $$PQ$$
<br><br>$$\therefore$$ $$y - 2 = m\left( {x - 1} \right)$$
<br><br>$$ \Rightarrow 0 - 2 = m\left( {x - 1} \right)$$
<br><br>$$ \Rightarrow - 2 = m\left( {x - 1} \right)$$
<br><br>$$ \Rightarrow x - 1 = {{ - 2} \over m}$$
<br><br>$$ \Rightarrow x = {{ - 2} \over m} + 1$$
<br><br>Also, $$OP = \sqrt {\left( {x - 0} \right){}^2 + {{\left( {0 - 0} \right)}^2}} = x = {{ - 2} \over m} + 1$$
<br><br>Similarly, point $$Q\left( {0,y} \right)$$ will satisfy equation of $$PQ$$
<br><br>$$\therefore$$ $$y - 2 = m\left( {x - 1} \right)$$
<br><br>$$ \Rightarrow y - 2 = m\left( { - 1} \right) \Rightarrow y = 2 - m$$b
<br><br>and $$OQ = y = 2 - m$$
<br><br>Area of $$\Delta POQ = {1 \over 2}\left( {OP} \right)\left( {OQ} \right) = {1 \over 2}\left( {1 - {2 \over m}} \right)\left( {2 - m} \right)$$
<br><br>( As Area of $$\Delta = {1 \over 2} \times $$ base $$\,\, \times \,\,$$ height )
<br><br>$$ = {1 \over 2}\left[ {2 - m - {4 \over m} + 2} \right] = {1 \over 2}\left[ {4 - \left( {m + {4 \over m}} \right)} \right]$$
<br><br>$$ = 2 - {m \over 2} - {2 \over m}$$
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265674/exam_images/oasqyywx4wcmnissw76e.webp" loading="lazy" alt="AIEEE 2012 Mathematics - Application of Derivatives Question 178 English Explanation">
<br><br>Let Area $$ = f\left( m \right) = 2 - {m \over 2} - {2 \over m}$$
<br><br>Now, $$f'\left( m \right) = {{ - 1} \over 2} + {2 \over {{m^2}}}$$
<br><br>Put $$f'\left( m \right) = 0$$
<br><br>$$ \Rightarrow {m^2} = 4 \Rightarrow m = \pm 2$$
<br><br>Now, $$f''\left( m \right) = {{ - 4} \over {{m^3}}}$$
<br><br>$${\left. {f''\left( m \right)} \right|_{m = 2}} = - {1 \over 2} < 0$$
<br><br>$${\left. {f''\left( m \right)} \right|_{m = - 2}} = {1 \over 2} > 0$$
<br><br>Area will be least at $$m=-2$$
<br><br>Hence, slope of $$PQ$$ is $$-2.$$ | mcq | aieee-2012 |
uhpBt8tZl7jVNmvk | maths | application-of-derivatives | maxima-and-minima | If $$x=-1$$ and $$x=2$$ are extreme points of $$f\left( x \right) = \alpha \,\log \left| x \right|+\beta {x^2} + x$$ then | [{"identifier": "A", "content": "$$\\alpha = 2,\\beta = - {1 \\over 2}$$ "}, {"identifier": "B", "content": "$$\\alpha = 2,\\beta = {1 \\over 2}$$ "}, {"identifier": "C", "content": "$$\\alpha = - 6,\\beta = {1 \\over 2}$$ "}, {"identifier": "D", "content": "$$\\alpha = - 6,\\beta = -{1 \\over 2}$$ "}] | ["A"] | null | Let $$f\left( x \right) = \alpha \log \left| x \right| + \beta {x^2} + x$$
<br><br>Differentiating both sides,
<br><br>$$f'\left( x \right) = {\alpha \over x} + 2\beta x + 1$$
<br><br>Since $$x=-1$$ and $$x=2$$ are extreme points therefore
<br><br>$$f'\left( x \right) = 0$$ at these points.
<br><br>Put $$x = - 1$$ and $$x = 2$$ in $$f'\left( x \right),$$
<br><br>we get $$ - \alpha - 2\beta + 1 = 0$$
<br><br>$$ \Rightarrow \alpha + 2\beta = 1\,\,...\left( i \right)$$
<br><br>$${\alpha \over 2} + 4\beta + 1 = 0$$
<br><br>$$ \Rightarrow \alpha + 8\beta = - 2\,\,...\left( {ii} \right)$$
<br><br>On solving $$(i)$$ and $$(ii)$$, we get
<br><br>$$6\beta = - 3 \Rightarrow \beta = - {1 \over 2}$$
<br><br>$$\therefore$$ $$\,\,\,\,\alpha = 2$$ | mcq | jee-main-2014-offline |
JAEcF3gxW0HGEZRy | maths | application-of-derivatives | maxima-and-minima | Let $$f(x)$$ be a polynomial of degree four having extreme values
<br/>at $$x=1$$ and $$x=2$$. If $$\mathop {\lim }\limits_{x \to 0} \left[ {1 + {{f\left( x \right)} \over {{x^2}}}} \right] = 3$$, then f$$(2)$$ is equal to : | [{"identifier": "A", "content": "$$0$$ "}, {"identifier": "B", "content": "$$4$$ "}, {"identifier": "C", "content": "$$-8$$ "}, {"identifier": "D", "content": "$$-4$$ "}] | ["A"] | null | $$\mathop {\lim }\limits_{x \to 0} \left[ {1 + {{f\left( x \right)} \over {{x^2}}}} \right] = 3 \Rightarrow \mathop {Lim}\limits_{x \to 0} {{f\left( x \right)} \over {{x^2}}} = 2$$
<br><br>So, $$f(x)$$ contains terms in $$x{}^2,{x^3}$$ and $${x^4}$$
<br><br>Let $$f\left( x \right) = {a_1}{x^2} + {a_2}{x^3} + {a_3}{x^4}$$
<br><br>Since $$\mathop {\lim }\limits_{x \to 0} {{f\left( x \right)} \over {{x^2}}} = 2 \Rightarrow {a_1} = 2$$
<br><br>Hence, $$f\left( x \right) = 2{x^2} + {a_2}{x^3} + {a_3}{x^4}$$
<br><br>$$f'\left( x \right) = 4x + 3{a_2}{x^2} + 4{a_3}{x^3}$$
<br><br>As given: $$f'\left( 1 \right) = 0$$ and $$f'\left( 2 \right) = 0$$
<br><br>Hence, $$4 + 3{a_2} + 4{a_3} = 0\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
<br><br>and $$8 + 12{a_2} + 32{a_3} = 0\,\,\,\,\,...\left( 1 \right)$$
<br><br>By $$4x\left( {eq1} \right) - eq\left( 2 \right),$$ we get
<br><br>$$16 + 12{a_2} + 16{a_3} - \left( {8 + 12{a_2} + 32{a_3}} \right) = 0$$
<br><br>$$ \Rightarrow 8 - 16{a_3} = 0 \Rightarrow {a_3} = 1/2$$
<br><br>and by eqn. $$\left( 1 \right),4 + 3{a_2} + 4/2 = 0 \Rightarrow {a_2} = - 2$$
<br><br>$$ \Rightarrow f\left( x \right) = 2{x^2} - 2{x^3} + {1 \over 2}{x^4}$$
<br><br>$$f\left( 2 \right) = 2 \times 4 - 2 \times 8 + {1 \over 2} \times 16 = 0$$ | mcq | jee-main-2015-offline |
jrSbPCcMERZCP1ui | maths | application-of-derivatives | maxima-and-minima | A wire of length $$2$$ units is cut into two parts which are bent respectively to form a square of side $$=x$$ units and a circle of radius $$=r$$ units. If the sum of the areas of the square and the circle so formed is minimum, then: | [{"identifier": "A", "content": "$$x=2r$$ "}, {"identifier": "B", "content": "$$2x=r$$ "}, {"identifier": "C", "content": "$$2x = \\left( {\\pi + 4} \\right)r$$ "}, {"identifier": "D", "content": "$$\\left( {4 - \\pi } \\right)x = \\pi \\,\\, r$$ "}] | ["A"] | null | $$4x + 2\pi r = 2$$ $$\,\,\,$$ $$ \Rightarrow 2x + \pi r = 1$$
<br><br>$$S = {x^2} + \pi {r^2}$$
<br><br>$$S = {\left( {{{1 - \pi r} \over 2}} \right)^2} + \pi {r^2}$$
<br><br>$${{dS} \over {dr}} = 2\left( {{{1 - \pi r} \over 2}} \right)\left( {{{ - \pi } \over 2}} \right) + 2\pi r$$
<br><br>$$ \Rightarrow {{ - \pi } \over 2} + {{{\pi ^2}r} \over 2} + 2\pi r = 0$$
<br><br>$$ \Rightarrow r = {1 \over {\pi + 4}}$$
<br><br>$$ \Rightarrow x = {2 \over {\pi + 4}}\,$$
<br><br>$$ \Rightarrow x = 2r$$ | mcq | jee-main-2016-offline |
byueGoFmiFA0tNnAffoxb | maths | application-of-derivatives | maxima-and-minima | The minimum distance of a point on the curve y = x<sup>2</sup>−4 from the origin is : | [{"identifier": "A", "content": "$${{\\sqrt {19} } \\over 2}$$"}, {"identifier": "B", "content": "$$\\sqrt {{{15} \\over 2}} $$"}, {"identifier": "C", "content": "$${{\\sqrt {15} } \\over 2}$$"}, {"identifier": "D", "content": "$$\\sqrt {{{19} \\over 2}} $$"}] | ["C"] | null | Let point on the curve
<br><br>y = x<sup>2</sup> $$-$$ 4 is ($$\alpha $$<sup>2</sup>, $$\alpha $$<sup>2</sup> $$-$$ 4)
<br><br>$$ \therefore $$ Distance of the point ($$\alpha $$<sup>2</sup>, $$\alpha $$<sup>2</sup> $$-$$ 4) from origin,
<br><br>D = $$\sqrt {{\alpha ^2} + {{\left( {{\alpha ^2} - 4} \right)}^2}} $$
<br><br>$$ \Rightarrow $$ D<sup>2</sup> = $$\alpha $$<sup>2</sup> + $$\alpha $$<sup>4</sup> + 16 $$-$$ 8$$\alpha $$<sup>2</sup>
<br><br>$$=$$ $$\alpha $$<sup>4</sup> $$-$$ 7$$\alpha $$<sup>2</sup> + 16
<br><br>$$ \therefore $$ $${{d{D^2}} \over {d\alpha }}$$ = 4$$\alpha $$<sup>3</sup> $$-$$ 14$$\alpha $$
<br><br>Now, $${{d{D^2}} \over {d\alpha }}$$ = 0
<br><br>$$ \Rightarrow $$ 4$$\alpha $$<sup>3</sup> $$-$$ 14$$\alpha $$ = 0
<br><br>$$ \Rightarrow $$ 2$$\alpha $$ (2$$\alpha $$<sup>2</sup> $$-$$ 7) = 0
<br><br>$$\alpha $$ = 0 or $$\alpha $$<sup>2</sup> = $${7 \over 2}$$
<br><br>$${{{d^2}{D^2}} \over {d{\alpha ^2}}} = 12{\alpha ^2} - 14$$
<br><br>$$ \therefore $$ $${\left( {{{{d^2}{D^2}} \over {d{\alpha ^2}}}} \right)_{at\,\,\alpha = 0}} = - 14 < 0$$
<br><br>$${\left( {{{{d^2}{D^2}} \over {d{\alpha ^2}}}} \right)_{at\,\,{\alpha ^2} = {7 \over 2}}} = 28 > 0$$
<br><br>$$\therefore\,\,\,$$ Distance is minimum at $$\alpha $$<sup>2</sup> = $${7 \over 2}$$
<br><br>$$ \therefore $$ Minimum distance
<br><br>D = $$\sqrt {{{49} \over 4} - {{49} \over 4} + 16} $$
<br><br>= $${{\sqrt {15} } \over 2}$$ | mcq | jee-main-2016-online-9th-april-morning-slot |
JRHASG5PZuMAMaza | maths | application-of-derivatives | maxima-and-minima | Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector. Then the
maximum area (in sq. m) of the flower-bed, is : | [{"identifier": "A", "content": "10"}, {"identifier": "B", "content": "25"}, {"identifier": "C", "content": "30"}, {"identifier": "D", "content": "12.5"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265411/exam_images/o27yks7ncu1vj6sgzmlx.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2017 (Offline) Mathematics - Application of Derivatives Question 167 English Explanation">
We have
<br><br>Total length = r + r + r$$\theta $$ = 20
<br><br>$$ \Rightarrow $$ 2r + r$$\theta $$ = 20
<br><br>$$ \Rightarrow $$ $$\theta = {{20 - 2r} \over r}$$ .......(1)
<br><br>A = Area = $${\theta \over {2\pi }} \times \pi {r^2}$$ = $${1 \over 2}{r^2}\theta $$ = $${1 \over 2}{r^2}\left( {{{20 - 2r} \over r}} \right)$$
<br><br>$$ \Rightarrow $$ A = 10r – r<sup>2</sup>
<br><br>For A to be maximum
<br><br>$${{dA} \over {dr}} = 0$$
<br><br>$$ \Rightarrow $$ 10 – 2r = 0
<br><br>$$ \Rightarrow $$ r = 5
<br><br>$${{{d^2}A} \over {d{r^2}}} = - 2 < 0$$
<br><br>$$ \therefore $$ For r = 5, A is maximum
From (1)
<br><br>$${\theta = {{20 - 2\left( 5 \right)} \over r}}$$ = 2
<br><br>A = $${2 \over {2\pi }} \times \pi {\left( 5 \right)^2}$$ = 25 sq. m | mcq | jee-main-2017-offline |
YesRI3v7YkrCJy2r | maths | application-of-derivatives | maxima-and-minima | Let $$f\left( x \right) = {x^2} + {1 \over {{x^2}}}$$ and $$g\left( x \right) = x - {1 \over x}$$,
<br/>$$x \in R - \left\{ { - 1,0,1} \right\}$$.
<br/>If $$h\left( x \right) = {{f\left( x \right)} \over {g\left( x \right)}}$$, then the local minimum value of h(x) is | [{"identifier": "A", "content": "$$2\\sqrt 2 $$"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "-3"}, {"identifier": "D", "content": "$$-2\\sqrt 2 $$"}] | ["A"] | null | Given $$f\left( x \right) = {x^2} + {1 \over {{x^2}}}$$ and $$g\left( x \right) = x - {1 \over x}$$
<br><br>As $$h\left( x \right) = {{f\left( x \right)} \over {g\left( x \right)}}$$
<br><br>= $${{{x^2} + {1 \over {{x^2}}}} \over {x - {1 \over x}}}$$
<br><br>= $${{{{\left( {x - {1 \over x}} \right)}^2} + 2.x.{1 \over x}} \over {x - {1 \over x}}}$$
<br><br>= $${{{{\left( {x - {1 \over x}} \right)}^2} + 2} \over {x - {1 \over x}}}$$
<br><br>Let $${x - {1 \over x} = t}$$
<br><br>So $$f\left( t \right) = {{{t^2} + 2} \over t}$$ = $$t + {2 \over t}$$
<br><br>then $$f'\left( t \right) = 1 - {2 \over {{t^2}}}$$
<br><br>At maximum or minimum $$f'\left( t \right) = 0$$.
<br><br>$$\therefore$$ $$1 - {2 \over {{t^2}}} = 0$$
<br><br>$$ \Rightarrow t = \pm \sqrt 2 $$
<br><br>Now we need to find $$f''\left( t \right)$$ which will tell at which point among $$ + \sqrt 2 $$ and $$ - \sqrt 2 $$ will be maximum and minimum.
<br><br>$$f''\left( t \right) = + {4 \over {{t^3}}}$$
<br><br>So when $$t = + \sqrt 2 $$, then $$f''\left( t \right) = + {4 \over {{{\left( {\sqrt 2 } \right)}^3}}}$$ = $$ + \sqrt 2 $$
<br><br>As $$f''\left( t \right) > 0$$ when $$t = + \sqrt 2 $$ then at $$t = + \sqrt 2 $$ the function $$f\left( t \right)$$ will be minimum.
<br><br>Min of $$f\left( t \right) = {{{{\left( {\sqrt 2 } \right)}^2} + 2} \over {\sqrt 2 }}$$ = $${4 \over {\sqrt 2 }}$$ = $$2\sqrt 2 $$
<br><br>So local minimum of $$f\left( t \right)$$ = $$2\sqrt 2 $$
| mcq | jee-main-2018-offline |
45nNp849vryhuiO8gPGhJ | maths | application-of-derivatives | maxima-and-minima | If a right circular cone, having maximum volume, is inscribed in a sphere of radius 3 cm, then the curved surface area (in cm<sup>2</sup>) of this cone is : | [{"identifier": "A", "content": "$$6\\sqrt 2 \\pi $$"}, {"identifier": "B", "content": "$$6\\sqrt 3 \\pi $$"}, {"identifier": "C", "content": "$$8\\sqrt 2 \\pi $$"}, {"identifier": "D", "content": "$$8\\sqrt 3 \\pi $$"}] | ["D"] | null | Sphere of radius r = 3 cm
<br><br>Let b, h be base radius and height of cone respectively.
<br><br>So, volume of cone = $${1 \over 2}$$ $$\pi $$b<sup>2</sup>h
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265946/exam_images/x9rxyugujvfepiew1pgy.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2018 (Online) 15th April Morning Slot Mathematics - Application of Derivatives Question 166 English Explanation">
<br><br>In right $$\Delta $$ABC by Pythagoras theorem
<br><br>(h $$-$$ r)<sup>2</sup> + b<sup>2</sup> = r<sup>2</sup>
<br><br>$$ \Rightarrow $$ b<sup>2</sup> = r<sup>2</sup> $$-$$ (h $$-$$ r)<sup>2</sup> = r<sup>2</sup> $$-$$ (h<sup>2</sup> $$-$$ 2hr + r<sup>2</sup>) = 2hr $$-$$ h<sup>2</sup>
<br><br>$$\therefore\,\,\,$$ Volume (v) = $${1 \over 3}$$ $$\pi $$h[2hr $$-$$ h<sup>2</sup>] = $${1 \over 3}$$ [ 2h<sup>2</sup>r $$-$$ h<sup>3</sup>]
<br><br>$${{dv} \over {dh}}$$ = $${1 \over 3}$$ [4hr $$-$$ 3h<sup>2</sup>] = 0
<br><br>$$ \Rightarrow $$ h (4r $$-$$ 3h) = 0
<br><br>$${{{d^2}v} \over {d{h^2}}}$$ = $${1 \over 3}$$ [4r $$-$$ 6h]
<br><br>At h = $${{4r} \over 3}$$, $${{{d^2}y} \over {d{h^2}}}$$ = $${1 \over 3}$$ $$\left[ {4r - {{4r} \over 3} \times 6} \right] = {1 \over 3}\left[ {4r - 8r} \right] < 0$$
<br><br>$$ \Rightarrow $$ maximum volume cours at h = $${{4r} \over 3}$$ = $${4 \over 3}$$ $$ \times $$ 3 = 4 cm
<br><br>As from (1),
<br><br>(h $$-$$ r)<sup>2</sup> + b<sup>2</sup> = r<sup>2</sup>
<br><br>$$ \Rightarrow $$ b<sup>2</sup> = 2hr $$-$$ h<sup>2</sup> = 2.$${{4r} \over 3}$$ r $$-$$ $${{16{r^2}} \over 9}$$ = $${{8{r^2}} \over 3}$$ $$-$$ $${{16{r^2}} \over 9}$$
<br><br>= $${{\left( {24 - 16} \right){r^2}} \over 9}$$ = $${{8{r^2}} \over 9}$$
<br><br>$$ \Rightarrow $$ b = $${{2\sqrt 2 } \over 3}$$ r = 2 $$\sqrt 2 \,\,m$$
<br><br>Therefore curved surface area = $$\pi bl$$
<br><br>= $$\pi $$b$$\sqrt {{h^2} + {r^2}} $$ = $$\pi $$2$$\sqrt 2 $$ $$\sqrt {{4^2} + 8} $$ = 8$$\sqrt 3 $$$$\pi $$ cm<sup>2</sup> | mcq | jee-main-2018-online-15th-april-morning-slot |
rkBBF51Mq0IdrNDlgz0jp | maths | application-of-derivatives | maxima-and-minima | Let M and m be respectively the absolute maximum and the absolute minimum values of the function, f(x) = 2x<sup>3</sup> $$-$$ 9x<sup>2</sup> + 12x + 5 in the interval [0, 3]. Then M $$-$$m is equal to : | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "1"}] | ["B"] | null | <p>To determine the absolute maximum (M) and absolute minimum (m) of the function $ f(x) = 2x^3 - 9x^2 + 12x + 5 $ over the interval $[0, 3]$, we need to examine its critical points and endpoints.</p>
<p>First, we find the derivative of the function, $ f'(x) $, to locate the critical points:</p>
<p>
<p>$$ f'(x) = \frac{d}{dx} (2x^3 - 9x^2 + 12x + 5) = 6x^2 - 18x + 12 $$</p>
</p>
<p>Next, we set the derivative equal to zero to find the critical points:</p>
<p>
<p>$$ 6x^2 - 18x + 12 = 0 $$</p>
</p>
<p>Simplifying this equation by dividing by 6:</p>
<p>
<p>$$ x^2 - 3x + 2 = 0 $$</p>
</p>
<p>We solve this quadratic equation using the factorization method:</p>
<p>
<p>$$ (x - 1)(x - 2) = 0 $$</p>
</p>
<p>So, the critical points are:</p>
<p>
<p>$$ x = 1 \quad \text{and} \quad x = 2 $$</p>
</p>
<p>We now evaluate the function $ f(x) $ at the critical points and at the endpoints of the interval [0, 3]:</p>
<p>1. At $ x = 0 $:</p>
<p>
<p>$$ f(0) = 2(0)^3 - 9(0)^2 + 12(0) + 5 = 5 $$</p>
</p>
<p>2. At $ x = 1 $:</p>
<p>
<p>$$ f(1) = 2(1)^3 - 9(1)^2 + 12(1) + 5 = 2 - 9 + 12 + 5 = 10 $$</p>
</p>
<p>3. At $ x = 2 $:</p>
<p>
<p>$$ f(2) = 2(2)^3 - 9(2)^2 + 12(2) + 5 = 16 - 36 + 24 + 5 = 9 $$</p>
</p>
<p>4. At $ x = 3 $:</p>
<p>
<p>$$ f(3) = 2(3)^3 - 9(3)^2 + 12(3) + 5 = 54 - 81 + 36 + 5 = 14 $$</p>
</p>
<p>We now identify the absolute maximum (M) and absolute minimum (m) from the above values:</p>
<ul>
<li>Maximum value $ M = 14 $ at $ x = 3 $</li>
<li>Minimum value $ m = 5 $ at $ x = 0 $</li>
</ul>
<p>Thus, the difference $ M - m $ is:</p>
<p>
<p>$$ M - m = 14 - 5 = 9 $$</p>
</p>
<p>Therefore, the correct answer is:</p>
<p><strong>Option B: 9</strong></p> | mcq | jee-main-2018-online-16th-april-morning-slot |
R9TaH8Cmtts4XSPllBy4F | maths | application-of-derivatives | maxima-and-minima | A helicopter is flying along the curve given by y – x<sup>3/2</sup> = 7, (x $$ \ge $$ 0). A soldier positioned at the point $$\left( {{1 \over 2},7} \right)$$ wants to shoot down the helicopter when it is nearest to him. Then this nearest distance is - | [{"identifier": "A", "content": "$${1 \\over 6}\\sqrt {{7 \\over 3}} $$"}, {"identifier": "B", "content": "$${{\\sqrt 5 } \\over 6}$$"}, {"identifier": "C", "content": "$${1 \\over 2}$$"}, {"identifier": "D", "content": "$${1 \\over 3}$$$$\\sqrt {{7 \\over 3}} $$"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264139/exam_images/woynacqz8rh7gnukj3c5.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th January Evening Slot Mathematics - Application of Derivatives Question 153 English Explanation">
<br>$$y - {x^{3/2}} = 7\left( {x \ge 0} \right)$$
<br><br>$${{dy} \over {dx}} = {3 \over 2}{x^{1/2}}$$
<br><br>$$\left( {{3 \over 2}\sqrt x } \right)\left( {{{7 - y} \over {{1 \over 2} - x}}} \right) = - 1$$
<br><br>$$\left( {{3 \over 2}\sqrt x } \right)\left( {{{ - {x^{3/2}}} \over {{1 \over 2} - x}}} \right) = - 1$$
<br><br>$${3 \over 2}.{x^2} = {1 \over 2} - x$$
<br><br>$$3{x^2} = 1 - 2x$$
<br><br>$$3{x^2} + 2x - 1 = 0$$
<br><br>$$3{x^2} + 3x - x - 1 = 0$$
<br><br>$$\left( {x + 1} \right)\left( {3x - 1} \right) = 0$$
<br><br>$$ \therefore $$ $$x = - 1$$ (rejected)
<br><br>$$x = {1 \over 3}$$
<br><br>$$y = 7 + {x^{3/2}} = 7 + {\left( {{1 \over 3}} \right)^{3/2}}$$
<br><br>$${\ell _{AB}} = \sqrt {{{\left( {{1 \over 2} - {1 \over 3}} \right)}^2} + {{\left( {{1 \over 3}} \right)}^3}} = \sqrt {{1 \over {36}} + {1 \over {27}}} $$
<br><br>$$ = \sqrt {{{3 + 4} \over {9 \times 12}}} $$
<br><br>$$ = \sqrt {{7 \over {108}}} = {1 \over 6}\sqrt {{7 \over 3}} $$ | mcq | jee-main-2019-online-10th-january-evening-slot |
H7ek3BudnKTxOUvEFo54m | maths | application-of-derivatives | maxima-and-minima | If ƒ(x) is a non-zero polynomial of degree four,
having local extreme points at x = –1, 0, 1; then
the set
<br/>S = {x $$ \in $$ R : ƒ(x) = ƒ(0)}<br/>
Contains exactly : | [{"identifier": "A", "content": "four rational numbers."}, {"identifier": "B", "content": "four irrational numbers."}, {"identifier": "C", "content": "two irrational and one rational number."}, {"identifier": "D", "content": "two irrational and two rational numbes."}] | ["C"] | null | Local extreme points of f(x) is at x = –1, 0, 1.
<br><br>$$ \therefore $$ f'(x) = 0 has three solutions x = –1, 0, 1.
<br><br>$$ \therefore $$ f'(x) = k(x + 1)x(x - 1)
<br><br>$$\int {f'(x)dx} = \int {k(x + 1)x(x - 1)dx} $$
<br><br>f(x) = $$k\left[ {{{{x^4}} \over 4} - {{{x^2}} \over 2}} \right] + C$$
<br><br>Also given that
<br><br>ƒ(x) = ƒ(0)
<br><br>$$ \therefore $$ $$k\left[ {{{{x^4}} \over 4} - {{{x^2}} \over 2}} \right] + C$$ = C
<br><br>$$ \Rightarrow $$ $${{{x^2}} \over 2}\left( {{{{x^2}} \over 2} - 1} \right)$$ = 0
<br><br>$$ \Rightarrow $$ x = 0, $$ - \sqrt 2 $$, $$ \sqrt 2 $$
<br><br>$$ \therefore $$ x has two irrational and one rational value. | mcq | jee-main-2019-online-9th-april-morning-slot |
2wavbu9DmtWZrZAiqfr4E | maths | application-of-derivatives | maxima-and-minima | The height of a right circular cylinder of maximum
volume inscribed in a sphere of radius 3 is | [{"identifier": "A", "content": "$$\\sqrt 3 $$"}, {"identifier": "B", "content": "$$2\\sqrt 3 $$"}, {"identifier": "C", "content": "$$\\sqrt 6 $$"}, {"identifier": "D", "content": "$${2 \\over 3} {\\sqrt 3} $$"}] | ["B"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265888/exam_images/flpfpj2fdznelp4hlns0.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265702/exam_images/kgucy7y7y3d2pmdomcey.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 8th April Evening Slot Mathematics - Application of Derivatives Question 145 English Explanation"></picture>
<br>Here r = 3 cos$$\theta $$
<br><br>and $${h \over 2}$$ = 3 sin$$\theta $$
<br><br>$$ \Rightarrow $$ h = 6 sin$$\theta $$
<br><br>We know, Volume of cylinder
<br><br>V = $$\pi $$ r<sup>2</sup> h
<br><br>$$ \Rightarrow $$ V = $$\pi $$ (9cos<sup>2</sup>$$\theta $$)(6 sin$$\theta $$)
<br><br>$$ \Rightarrow $$ V = 54$$\pi $$ (sin$$\theta $$ - sin<sup>3</sup>$$\theta $$)
<br><br>For maxima/minima of volume,
<br><br>$${{dV} \over {dh}} = 0$$
<br><br>$$ \Rightarrow $$ cos$$\theta $$ - 3sin<sup>2</sup>$$\theta $$ cos$$\theta $$ = 0
<br><br>$$ \Rightarrow $$ cos$$\theta $$(1 - 3sin<sup>2</sup>$$\theta $$) = 0
<br><br>$$ \Rightarrow $$ cos$$\theta $$ = 0
<br><br>$$ \Rightarrow $$ $$\theta $$ = $${\pi \over 2}$$ (not possible)
<br><br>or 1 - 3sin<sup>2</sup>$$\theta $$ = 0
<br><br>$$ \Rightarrow $$ sin$$\theta $$ = $${1 \over {\sqrt 3 }}$$
<br><br>$$ \therefore $$ h = 6 sin$$\theta $$ = $${6 \over {\sqrt 3 }}$$ = $$2\sqrt 3 $$ | mcq | jee-main-2019-online-8th-april-evening-slot |
bsO11rG2h8VURkIcQKDH2 | maths | application-of-derivatives | maxima-and-minima | If S<sub>1</sub> and S<sub>2</sub> are respectively the sets of local
minimum and local maximum points of the function,
<br/><br/>ƒ(x) = 9x<sup>4</sup> + 12x<sup>3</sup> – 36x<sup>2</sup> + 25, x $$ \in $$ R,
then : | [{"identifier": "A", "content": "S<sub>1</sub> = {\u20131}; S<sub>2</sub> = {0, 2}"}, {"identifier": "B", "content": "S<sub>1</sub> = {\u20132}; S<sub>2</sub> = {0, 1}"}, {"identifier": "C", "content": "S<sub>1</sub> = {\u20132, 0}; S<sub>2</sub> = {1}"}, {"identifier": "D", "content": "S<sub>1</sub> = {\u20132, 1}; S<sub>2</sub> = {0}"}] | ["D"] | null | ƒ(x) = 9x<sup>4</sup> + 12x<sup>3</sup> – 36x<sup>2</sup> + 25
<br><br>ƒ'(x) = 36x<sup>3</sup> + 36x<sup>2</sup> – 72x
<br><br>ƒ'(x) = 36x(x<sup>2</sup> + x – 2)
<br><br>ƒ'(x) = 36x(x + 2)(x - 1)
<picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263290/exam_images/wchaj60kqtal4nuywplf.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267823/exam_images/n3w3sauf5usnpuniw1s3.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 8th April Morning Slot Mathematics - Application of Derivatives Question 148 English Explanation"></picture>
<br><br>While moving
left to right on x-axis whenever derivative changes sign from
negative to positive, we get local minima, and
whenever derivative changes sign from positive
to negative, we get local maxima.
<br><br>$$ \therefore $$ S<sub>1</sub> = {–2, 1}
<br><br>S<sub>2</sub> = {0} | mcq | jee-main-2019-online-8th-april-morning-slot |
tOS3ma3QAHGsURo5qS5KT | maths | application-of-derivatives | maxima-and-minima | The maximum value of the function f(x) = 3x<sup>3</sup> – 18x<sup>2</sup> + 27x – 40 on the set S = $$\left\{ {x\, \in R:{x^2} + 30 \le 11x} \right\}$$ is : | [{"identifier": "A", "content": "$$-$$ 222"}, {"identifier": "B", "content": "$$-$$ 122"}, {"identifier": "C", "content": "$$122$$ "}, {"identifier": "D", "content": "222"}] | ["C"] | null | S = {x $$ \in $$ R, x<sup>2</sup> + 30 $$-$$ 11x $$ \le $$ 0}
<br><br>= {x $$ \in $$ R, 5 $$ \le $$ x $$ \le $$ 6}
<br><br>Now f(x) = 3x<sup>3</sup> $$-$$ 18x<sup>2</sup> + 27x $$-$$ 40
<br><br>$$ \Rightarrow $$ f '(x) = 9(x $$-$$ 1)(x $$-$$ 3),
<br><br>which is positive in [5, 6]
<br><br>$$ \Rightarrow $$ f(x) increasing in [5, 6]
<br><br>Hence maximum value = f(6) = 122 | mcq | jee-main-2019-online-11th-january-morning-slot |
wQ3wgXwJYh85rfjWcJkJl | maths | application-of-derivatives | maxima-and-minima | The shortest distance between the point $$\left( {{3 \over 2},0} \right)$$ and the curve y = $$\sqrt x $$, (x > 0), is - | [{"identifier": "A", "content": "$${{\\sqrt 3 } \\over 2}$$"}, {"identifier": "B", "content": "$${5 \\over 4}$$"}, {"identifier": "C", "content": "$${3 \\over 2}$$"}, {"identifier": "D", "content": "$${{\\sqrt 5 } \\over 2}$$"}] | ["D"] | null | Let points $$\left( {{3 \over 2},0} \right),\left( {{t^2},t} \right),t > 0$$
<br><br>Distance = $$\sqrt {{t^2} + {{\left( {{t^2} - {3 \over 2}} \right)}^2}} $$
<br><br>= $$\sqrt {{t^4} - 2{t^2} + {9 \over 4}} = \sqrt {{{\left( {{t^2} - 1} \right)}^2} + {5 \over 4}} $$
<br><br>So minimum distance is $$\sqrt {{5 \over 4}} = {{\sqrt 5 } \over 2}$$ | mcq | jee-main-2019-online-10th-january-morning-slot |
U1DtFitPjt3mM5fLmUIkD | maths | application-of-derivatives | maxima-and-minima | The maximum volume (in cu.m) of the right circular cone having slant height 3 m is : | [{"identifier": "A", "content": "2$$\\sqrt3$$$$\\pi $$"}, {"identifier": "B", "content": "3$$\\sqrt3$$$$\\pi $$"}, {"identifier": "C", "content": "6$$\\pi $$"}, {"identifier": "D", "content": "$${4 \\over 3}\\pi $$"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263605/exam_images/srfihf2ilic7xfbenv64.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th January Morning Slot Mathematics - Application of Derivatives Question 156 English Explanation">
<br><br>$$ \therefore $$ h = 3 cos$$\theta $$
<br><br>r = 3 sin$$\theta $$
<br><br>We know volume of right circular cone,
<br><br>V = $${1 \over 3}\pi {r^2}h$$
<br><br>= $${1 \over 3}\pi $$(3 sin$$\theta $$)<sup>2</sup> 3 cos$$\theta $$
<br><br>= 9 $$\pi $$ sin<sup>2</sup>$$\theta $$ cos$$\theta $$
<br><br>For maximum or minimum value of volume,
<br><br>$${{dv} \over {d\theta }}$$ = 0
<br><br>$$ \therefore $$ (2sin$$\theta $$ cos$$\theta $$) cos$$\theta $$ + 3sin<sup>2</sup>$$\theta $$ .($$-$$ sin$$\theta $$) = 0
<br><br>$$ \Rightarrow $$ 2 sin$$\theta $$ cos<sup>2</sup>$$\theta $$ $$-$$ sin<sup>3</sup>$$\theta $$ = 0
<br><br>$$ \Rightarrow $$ 2 sin$$\theta $$(1 $$-$$ sin<sup>2</sup>$$\theta $$) $$-$$ sin<sup>3</sup> $$\theta $$ = 0
<br><br>$$ \Rightarrow $$ 2 sin$$\theta $$ $$-$$ 2 sin<sup>3</sup>$$\theta $$ $$-$$ sin<sup>3</sup>$$\theta $$ = 0
<br><br>$$ \Rightarrow $$ 3 sin<sup>3</sup>$$\theta $$ = 2 sin$$\theta $$
<br><br>$$ \Rightarrow $$ sin<sup>2</sup>$$\theta $$ = $${2 \over 3}$$
<br><br>$$ \Rightarrow $$ sin$$\theta $$ = $$\sqrt {{2 \over 3}} $$
<br><br>$${{{d^2}v} \over {d{\theta ^2}}}$$ = 2cos$$\theta $$ $$-$$ 3(3sin$$\theta $$ cos$$\theta $$)
<br><br>= 2 cos$$\theta $$ $$-$$ 9 sin$$\theta $$ cos$$\theta $$
<br><br>= 2 $$ \times $$ $${1 \over {\sqrt 3 }}$$ $$-$$ 9 $$ \times $$ $${{\sqrt 2 } \over {\sqrt 3 }}$$ $$ \times $$ $${1 \over {\sqrt 3 }}$$
<br><br>= $${2 \over {\sqrt 3 }} - 3\sqrt 2 \, < 0$$
<br><br>$$ \therefore $$ Volume is maximum
<br><br>when sin$$\theta $$ = $$\sqrt {{2 \over 3}} $$
<br><br>$$ \therefore $$ Maximum volume is
<br><br>= 9 $$\pi $$ $${\left( {\sqrt {{2 \over 3}} } \right)^2} \times {1 \over {\sqrt 3 }}$$
<br><br>= 9 $$\pi $$ $$ \times $$ $${2 \over 3} \times {1 \over {\sqrt 3 }}$$
<br><br>= $$2\sqrt 3 \,\pi $$ | mcq | jee-main-2019-online-9th-january-morning-slot |
1lBNcIAtmid94BAZmo7k9k2k5hkbtc7 | maths | application-of-derivatives | maxima-and-minima | Let ƒ(x) be a polynomial of degree 3 such that
ƒ(–1) = 10, ƒ(1) = –6, ƒ(x) has a critical point
at x = –1 and ƒ'(x) has a critical point at x = 1.
Then ƒ(x) has a local minima at x = _______. | [] | null | 3 | Let f(x) = ax<sup>3</sup>
+ bx<sup>2</sup>
+ cx + d
<br><br>Given f(-1) = 10, f(1) = -6
<br><br>$$ \therefore $$ -a + b - c + d = 10 ....(i)
<br><br>and a + b + c + d = -6 ......(ii)
<br><br>adding (i) + (ii)
<br><br>2(b + d) = 4
<br><br>$$ \Rightarrow $$ b + d = 2 ....(iii)
<br><br>f'(x) = 3ax<sup>2</sup> + 2bx + c
<br><br>Given f'(-1) = 0
<br><br>$$ \Rightarrow $$ 3a - 2b + c = 0 .....(iv)
<br><br>f"(x) = 6ax + 2b
<br><br>Given f"(1) = 0
<br><br>$$ \therefore $$ 6a + 2b = 0 ....(v)
<br><br>$$ \Rightarrow $$ b = -3a
<br><br>adding (iv) + (v), we get
<br><br>9a + c = 0 ....(vi)
<br><br>$$ \Rightarrow $$ $$9\left( {{{ - b} \over 3}} \right)$$ + c = 0
<br><br>$$ \Rightarrow $$ c = 3b
<br><br>f(x) =
$${{{ - b} \over 3}{x^3}}$$ + bx<sup>2</sup> + 3bx + (2 - b)
<br><br>$$ \Rightarrow $$ f'(x) = -bx<sup>2</sup>
+ 2bx + 3b
<br><br> = -b(x<sup>2</sup>
- 2x - 3)
<br><br>At maxima and minima f'(x) = 0
<br><br>$$ \therefore $$ (x<sup>2</sup>
- 2x - 3) = 0
<br><br>$$ \Rightarrow $$ (x - 3) (x + 1) = 0
<br><br>x = 3, -1
<br><br>As a + b + c + d = -6
<br><br>$$ \Rightarrow $$ $${{{ - b} \over 3}}$$ + b + 3b + 2 - b = -6
<br><br>$$ \Rightarrow $$ b = -3
<br><br>$$ \therefore $$ f'(x) = 3(x<sup>2</sup>
- 2x - 3)
<br><br>$$ \Rightarrow $$ f''(x) = 3(2x
- 2)
<br><br>At x = 3, f''(x) = 3(2.3
- 2) = 12 > 0
<br><br>$$ \therefore $$ Minima
at x = 3. | integer | jee-main-2020-online-8th-january-evening-slot |
bWXZpKvMYXQpzsJPLljgy2xukg38i6xk | maths | application-of-derivatives | maxima-and-minima | The set of all real values of $$\lambda $$ for which the
function<br/><br/>
$$f(x) = \left( {1 - {{\cos }^2}x} \right)\left( {\lambda + \sin x} \right),x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$$<br/><br/>has exactly one maxima and exactly one
minima, is :
| [{"identifier": "A", "content": "$$\\left( { - {3 \\over 2},{3 \\over 2}} \\right) - \\left\\{ 0 \\right\\}$$"}, {"identifier": "B", "content": "$$\\left( { - {3 \\over 2},{3 \\over 2}} \\right)$$"}, {"identifier": "C", "content": "$$\\left( { - {1 \\over 2},{1 \\over 2}} \\right) - \\left\\{ 0 \\right\\}$$"}, {"identifier": "D", "content": "$$\\left( { - {1 \\over 2},{1 \\over 2}} \\right)$$"}] | ["A"] | null | $$f(x) = \left( {1 - {{\cos }^2}x} \right)\left( {\lambda + \sin x} \right)$$
<br><br>$$ \Rightarrow $$ f(x) = sin<sup>2</sup> x($$\lambda $$ + sinx) ....(1)
<br><br>$$ \therefore $$ f'(x) = 2sinx cosx ($$\lambda $$ +sinx) + sin<sup>2</sup>x (cosx)
<br><br>$$ \Rightarrow $$ f'(x) = sin2x($${{2\lambda + 3\sin x} \over 2}$$)
<br><br>For maixma and minima, f'(x) = 0
<br><br>$$ \therefore $$ sin2x = 0 or 2$$\lambda $$ + 3sinx = 0
<br><br>when sin2x = 0 $$ \Rightarrow $$ x = 0
<br><br>or when 2$$\lambda $$ + 3sinx = 0
<br><br>$$ \Rightarrow $$ sin x = $$ - {{2\lambda } \over 3}$$
<br><br>As $$x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$$
<br><br>$$ \therefore $$ -1 < sinx < 1
<br><br>$$ \Rightarrow $$ -1 < $$ - {{2\lambda } \over 3}$$ < 1
<br><br>$$ \Rightarrow $$ $$ - {3 \over 2}$$ < $$\lambda $$ < $$ {3 \over 2}$$
<br><br>$$ \therefore $$ $$\lambda $$ $$ \in $$ $$\left( { - {3 \over 2},{3 \over 2}} \right)$$ - {0}
<br><br><b>Note :</b> If $$\lambda $$ = 0 $$ \Rightarrow $$ f(x) = sin<sup>3</sup>x [from (1)]
<br><br>Which is monotonic. so no maxima/minima.
| mcq | jee-main-2020-online-6th-september-evening-slot |
cMYWcwhlFTJFfWbA6Ejgy2xukfw0x33e | maths | application-of-derivatives | maxima-and-minima | The position of a moving car at time t is <br/>given by f(t) = at<sup>2</sup> + bt + c, t > 0, where a, b and c are real
numbers greater than 1. Then the average speed of the car over the time interval [t<sub>1</sub>
, t<sub>2</sub>
] is
attained at the point :
| [{"identifier": "A", "content": "$${{\\left( {{t_1} + {t_2}} \\right)} \\over 2}$$"}, {"identifier": "B", "content": "$${{\\left( {{t_2} - {t_1}} \\right)} \\over 2}$$"}, {"identifier": "C", "content": "2a(t<sub>1</sub>\n + t<sub>2</sub>) + b"}, {"identifier": "D", "content": "a(t<sub>2</sub>\n\u2013 t<sub>1</sub>) + b"}] | ["A"] | null | V<sub>av</sub> = $${{f\left( {{t_2}} \right) - f\left( {{t_1}} \right)} \over {{t_2} - {t_1}}}$$ = f'(t)
<br><br>$$ \Rightarrow $$ $${{a\left( {t_2^2 - t_1^2} \right) - b\left( {{t_2} - {t_1}} \right)} \over {{t_2} - {t_1}}}$$ = 2$$a$$t + b
<br><br>$$ \Rightarrow $$ a(t<sub>2</sub> + t<sub>1</sub>) + b = 2at + b
<br><br>$$ \Rightarrow $$ t = $${{{t_1} + {t_2}} \over 2}$$ | mcq | jee-main-2020-online-6th-september-morning-slot |
ZG73oFifL0rz2mngI3jgy2xukfqc6v02 | maths | application-of-derivatives | maxima-and-minima | If x = 1 is a critical point of the function
<br/>f(x) = (3x<sup>2</sup>
+ ax – 2 – a)e<sup>x</sup>
, then : | [{"identifier": "A", "content": "x = 1 is a local maxima and x = $$ - {2 \\over 3}$$ is a\nlocal minima of f."}, {"identifier": "B", "content": "x = 1 and x = $$ - {2 \\over 3}$$ are local maxima of f."}, {"identifier": "C", "content": "x = 1 and x = $$ - {2 \\over 3}$$ are local minima of f."}, {"identifier": "D", "content": "x = 1 is a local minima and x = $$ - {2 \\over 3}$$ is a local\nmaxima of f."}] | ["D"] | null | f(x) = (3x<sup>2</sup>
+ ax – 2 – a)e<sup>x</sup>
<br><br>$$ \therefore $$ f'(x) = e<sup>x</sup>(6x + a) + (3x<sup>2</sup>
+ ax – 2 – a)e<sup>x</sup>
<br><br> = e<sup>x</sup>(3x<sup>2</sup> + x(6 + a) – 2)
<br><br>f '(x) = 0 at x = 1
<br><br>$$ \Rightarrow $$ 3 + (6 + a) – 2 = 0
<br><br>$$ \Rightarrow $$ a = -7
<br><br>$$ \therefore $$ f'(x) = e<sup>x</sup>(3x<sup>2</sup> – x – 2)
<br><br>= e<sup>x</sup>(x – 1) (3x + 2)
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266001/exam_images/fwsvcudxolhotbhb0swn.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 5th September Evening Slot Mathematics - Application of Derivatives Question 116 English Explanation">
<br>x = 1 is a local minima and x = $$ - {2 \over 3}$$ is a local
maxima of f. | mcq | jee-main-2020-online-5th-september-evening-slot |
hOgXPc5U67INRRRyO9jgy2xukfg6uw39 | maths | application-of-derivatives | maxima-and-minima | If the point P on the curve, 4x<sup>2</sup> + 5y<sup>2</sup> = 20 is <br/>farthest from the point Q(0, -4), then PQ<sup>2</sup> is equal to:
| [{"identifier": "A", "content": "36"}, {"identifier": "B", "content": "48"}, {"identifier": "C", "content": "21"}, {"identifier": "D", "content": "29"}] | ["A"] | null | Given ellipse is $${{{x^2}} \over 5} + {{{y^2}} \over 4} = 1$$<br><br>Let point P is $$(\sqrt 5 \cos \theta ,\,2\sin \theta )$$<br><br>$${(PQ)^2}=5{\cos ^2}\theta + {(2\sin \theta + 4)^2}$$
<br><br>$$ \Rightarrow $$ (PQ)<sup>2</sup> = $$5{\cos ^2}\theta + 4{\sin ^2}\theta + 16\sin \theta + 16$$
<br><br>$$ \Rightarrow $$ (PQ)<sup>2</sup> = $${\cos ^2}\theta + 4{\cos ^2}\theta + 4{\sin ^2}\theta + 16\sin \theta + 16$$
<br><br>$$ \Rightarrow $$ (PQ)<sup>2</sup> = $${\cos ^2}\theta + 4 + 16\sin \theta + 16$$
<br><br>$$ \Rightarrow $$ $${(PQ)^2} = {\cos ^2}\theta + 16\sin \theta + 20$$
<br><br>$$ \Rightarrow $$ $${(PQ)^2} = - {\sin ^2}\theta + 16\sin \theta + 21$$
<br><br>= $$ - \left( {{{\sin }^2}\theta - 2.8\sin \theta + 64} \right) + 64 + 21$$
<br><br>= $$85 - {(\sin \theta - 8)^2}$$<br><br>will be maximum when sin $$\theta $$ = 1<br><br>$$ \Rightarrow {(PQ)^2}_{\max } = 85 - 49 = 36$$ | mcq | jee-main-2020-online-5th-september-morning-slot |
HBKNx1NQFioVfmrFstjgy2xukfahe3zt | maths | application-of-derivatives | maxima-and-minima | The area (in sq. units) of the largest rectangle ABCD whose vertices A and B lie on the x-axis and vertices C and D lie on the parabola, y = x<sup>2</sup>–1 below the x-axis, is : | [{"identifier": "A", "content": "$${1 \\over {3\\sqrt 3 }}$$"}, {"identifier": "B", "content": "$${2 \\over {3\\sqrt 3 }}$$"}, {"identifier": "C", "content": "$${4 \\over {3\\sqrt 3 }}$$"}, {"identifier": "D", "content": "$${4 \\over 3}$$"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264037/exam_images/uk75haohyuzity3go0py.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 4th September Evening Slot Mathematics - Application of Derivatives Question 119 English Explanation">
<br><br>Area (A) = 2t. (1$$ - $$t<sup>2</sup>)<br><br>(0 < t < 1)<br><br>A = 2t $$ - $$ 2t<sup>3</sup><br><br>$${{dA} \over {dt}} = 2 - 6{t^2}$$ = 0<br><br>$$ \Rightarrow $$ $$t = $$$$ \pm $$$${1 \over {\sqrt 3 }}$$<br><br>$$ \therefore $$ $$ {A_{\max }} = |{2 \over {\sqrt 3 }}\left( {1 - {1 \over 3}} \right)| = {4 \over {3\sqrt 3 }}$$ | mcq | jee-main-2020-online-4th-september-evening-slot |
Bss5ARx998ScyyRnSIjgy2xukewssjfx | maths | application-of-derivatives | maxima-and-minima | If p(x) be a polynomial of degree three that has
a local maximum value 8 at x = 1 and a local
minimum value 4 at x = 2; then p(0) is equal to : | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "12"}, {"identifier": "C", "content": "-12"}, {"identifier": "D", "content": "-24"}] | ["C"] | null | Since p(x) has relative extreme at
<br><br>x = 1 & 2
<br><br>so p'(x) = 0 at x = 1 & 2
<br><br>$$ \therefore $$ Let p'(x) = A(x – 1) (x – 2)
<br><br>$$ \Rightarrow $$ p(x) = $$\int {A\left( {{x^2} - 3x + 2} \right)dx} $$
<br><br>$$ \Rightarrow $$ p(x) = $${A\left( {{{{x^2}} \over 3} - 3{x^2} + 2x} \right)}$$ + C ...(1)
<br><br>As P(1) = 8
<br><br>From (1)
<br><br>$$8 = A\left( {{1 \over 3} - {3 \over 2} + 2} \right) + C$$
<br><br>$$ \Rightarrow $$ 8 = $${{5A} \over 6} + C$$
<br><br>$$ \Rightarrow $$ 48 = 5A + 5C ...(2)
<br><br>Also P(2) = 4
<br><br>From (1)
<br><br>$$4 = A\left( {{8 \over 3} - 6 + 4} \right) + C$$
<br><br>$$ \Rightarrow $$ 4 = $${{2A} \over 3} + C$$
<br><br>$$ \Rightarrow $$ 12 = 2A + 3C ......(3)
<br><br>Form (3) & (4), C = –12, A = 24
<br><br>Now p(0) = C = -12 | mcq | jee-main-2020-online-2nd-september-morning-slot |
mmwaTQKip2yYQJU89z7k9k2k5fo9w02 | maths | application-of-derivatives | maxima-and-minima | Let ƒ(x) be a polynomial of degree 5 such that x = ±1 are its critical points.
<br/><br/>If $$\mathop {\lim }\limits_{x \to 0} \left( {2 + {{f\left( x \right)} \over {{x^3}}}} \right) = 4$$, then which one of the following is not true? | [{"identifier": "A", "content": "\u0192(1) - 4\u0192(-1) = 4."}, {"identifier": "B", "content": "x = 1 is a point of minima and x = -1 is a point of maxima of \u0192."}, {"identifier": "C", "content": "x = 1 is a point of maxima and x = -1 is a point of minimum of \u0192."}, {"identifier": "D", "content": "\u0192 is an odd function."}] | ["B"] | null | let f(x) = ax<sup>5</sup> + bx<sup>4</sup> + cx<sup>3</sup> + dx<sup>2</sup> + ex + f
<br><br>Given $$\mathop {\lim }\limits_{x \to 0} \left( {2 + {{f\left( x \right)} \over {{x^3}}}} \right) = 4$$
<br><br>$$ \Rightarrow $$ $$\mathop {\lim }\limits_{x \to 0} \left( {2 + {{a{x^5} + b{x^4} + c{x^3} + d{x^2} + ex + f} \over {{x^3}}}} \right)$$ = 4
<br><br>As this limit exists so d = e = f = 0
<br><br>$$ \therefore $$ f(x) = ax<sup>5</sup> + bx<sup>4</sup> + cx<sup>3</sup>
<br><br>$$\mathop {\lim }\limits_{x \to 0} \left( {2 + {{a{x^5} + b{x^4} + c{x^3}} \over {{x^3}}}} \right)$$ = 4
<br><br>$$ \Rightarrow $$ 2 + c = 4
<br><br>$$ \Rightarrow $$ c = 2
<br><br>$$ \therefore $$ f(x) = ax<sup>5</sup> + bx<sup>4</sup> + 2x<sup>3</sup>
<br><br>$$ \Rightarrow $$ f'(x) = 5ax<sup>4</sup> + 4bx<sup>3</sup> + 6x<sup>2</sup>
<br><br>As x = ±1 are its critical points so f'(x) = 0 at x = ±1.
<br><br>f'(1) = 5a + 4b + 6 = 0 ....(1)
<br><br>and f'(-1) = 5a - 4b + 6 = 0 .....(2)
<br><br>Solving (1) and (2),
<br><br>a = $$ - {6 \over 5}$$ and b = 0
<br><br>$$ \therefore $$ f(x) = $$ - {6 \over 5}{x^5} + 2{x^3}$$
<br><br>So f'(x) = -6x<sup>4</sup> + 6x<sup>2</sup> = 6x<sup>2</sup>(1 + x)(1 - x)
<br><br> Sign scheme for f'(x)
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265993/exam_images/ignyryih3n3jwrth8owo.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 7th January Evening Slot Mathematics - Application of Derivatives Question 134 English Explanation">
<br><br>It is clear that maxima at x = 1 and minima at x
= –1. | mcq | jee-main-2020-online-7th-january-evening-slot |
XCWdLnNu3Bl74gtzgZ1klri0zmo | maths | application-of-derivatives | maxima-and-minima | The minimum value of $$\alpha $$ for which the <br/>equation $${4 \over {\sin x}} + {1 \over {1 - \sin x}} = \alpha $$
has at least one
solution in $$\left( {0,{\pi \over 2}} \right)$$ is ....... | [] | null | 9 | $$f(x) = {4 \over {\sin x}} + {1 \over {1 - \sin x}}$$<br><br>Let sinx = t $$ \because $$ $$x \in \left( {0,{\pi \over 2}} \right) \Rightarrow 0 < t < 1$$<br><br>$$f(t) = {4 \over t} + {1 \over {1 - t}}$$<br><br>$$f'(t) = {{ - 4} \over {{t^2}}} + {1 \over {{{(1 - t)}^2}}}$$<br><br>$$ = {{{t^2} - 4{{(1 - t)}^2}} \over {{t^2}{{(1 - t)}^2}}}$$<br><br>$$ = {{(t - 2(1 - t))(t + 2(1 - t))} \over {{t^2}{{(1 - t)}^2}}}$$<br><br>$$ = {{(3t - 2)(2 - t)} \over {{t^2}{{(1 - t)}^2}}}$$<br><br>$${f_{\min }}$$ at $$t = {2 \over 3}$$<br><br>$${\alpha _{\min }} = f\left( {{2 \over 3}} \right) = {4 \over {{2 \over 3}}} + {1 \over {1 - {2 \over 3}}}$$<br><br>$$ = 6 + 3$$<br><br>$$ = 9$$ | integer | jee-main-2021-online-24th-february-morning-slot |
8ssmAtzXibYG3PMlMr1kls5x94e | maths | application-of-derivatives | maxima-and-minima | Let f(x) be a polynomial of degree 6 in x, in which the coefficient of x<sup>6</sup> is unity and it has extrema at x = $$-$$1 and x = 1. If $$\mathop {\lim }\limits_{x \to 0} {{f(x)} \over {{x^3}}} = 1$$, then $$5.f(2)$$ is equal to _________. | [] | null | 144 | $$f(x) = {x^6} + a{x^5} + b{x^4} + {x^3}$$<br><br>$$\therefore$$ $$f'(x) = 6{x^5} + 5a{x^4} + 4b{x^3} + 3{x^2}$$<br><br>Roots 1 & $$-$$1<br><br>$$ \therefore $$ $$6 + 5z + 4b + 3 = 0$$ & $$ - 6 + 5a - 4b + 3 = 0$$ solving<br><br>$$a = - {3 \over 5}$$<br><br>$$b = - {3 \over 2}$$<br><br>$$ \therefore $$ $$f(x) = {x^6} - {3 \over 5}{x^5} - {3 \over 2}{x^4} + {x^3}$$<br><br>$$ \therefore $$ $$5.f(2) = 5\left[ {64 - {{96} \over 5} - 24 + 8} \right] = 144$$ | integer | jee-main-2021-online-25th-february-morning-slot |
cPe9u55dm7F2RKDxWx1kluh1ma0 | maths | application-of-derivatives | maxima-and-minima | The maximum slope of the curve $$y = {1 \over 2}{x^4} - 5{x^3} + 18{x^2} - 19x$$ occurs at the point : | [{"identifier": "A", "content": "$$\\left( {3,{{21} \\over 2}} \\right)$$"}, {"identifier": "B", "content": "(0, 0)"}, {"identifier": "C", "content": "(2, 9)"}, {"identifier": "D", "content": "(2, 2)"}] | ["D"] | null | Given, $$y = {1 \over 2}{x^4} - 5{x^3} + 18{x^2} - 19x$$
<br><br>$${{dy} \over {dx}} = {1 \over 2} \times 4{x^3} - 15{x^2} + 36x - 19$$<br><br>$$ \Rightarrow $$ Slope M = $$2{x^3} - 15{x^2} + 36x - 19$$<br><br>At max of slope $${{dM} \over {dx}} = 0$$<br><br>$$ \therefore $$ $${{dM} \over {dx}} = 6{x^2} - 30x + 36 = 0$$<br><br>$$ \Rightarrow 6({x^2} - 5x + 6) = 0$$<br><br>$$ \Rightarrow 6(x - 2)(x - 3) = 0$$<br><br>$$ \therefore $$ $$x = 2,3$$<br><br>Now, $${{{d^2}M} \over {d{x^2}}} = 6(2x - 5)$$<br><br>at $$x = 2,{{{d^2}M} \over {d{x^2}}} = 6(4 - 5) = - 6 < 0$$<br><br>$$ \therefore $$ at x = 2 slope is maximum.
<br><br>At x = 2,
<br><br>y = 8 - 40 + 72 - 38 = 2
<br><br>$$ \therefore $$ Required point = (2, 2) | mcq | jee-main-2021-online-26th-february-morning-slot |
MNEisjfUr6CRd4ILoq1kmiz0uri | maths | application-of-derivatives | maxima-and-minima | The maximum value of <br/><br/>$$f(x) = \left| {\matrix{
{{{\sin }^2}x} & {1 + {{\cos }^2}x} & {\cos 2x} \cr
{1 + {{\sin }^2}x} & {{{\cos }^2}x} & {\cos 2x} \cr
{{{\sin }^2}x} & {{{\cos }^2}x} & {\sin 2x} \cr
} } \right|,x \in R$$ is : | [{"identifier": "A", "content": "$$\\sqrt 5 $$"}, {"identifier": "B", "content": "$${3 \\over 4}$$"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "$$\\sqrt 7 $$"}] | ["A"] | null | $$f(x) = \left| {\matrix{
{{{\sin }^2}x} & {1 + {{\cos }^2}x} & {\cos 2x} \cr
{1 + {{\sin }^2}x} & {{{\cos }^2}x} & {\cos 2x} \cr
{{{\sin }^2}x} & {{{\cos }^2}x} & {\sin 2x} \cr
} } \right|$$
<br><br>$${C_1} \to {C_1} + {C_2}$$<br><br>= $$\left| {\matrix{
2 & {1 + {{\cos }^2}x} & {\cos 2x} \cr
2 & {{{\cos }^2}x} & {\cos 2x} \cr
1 & {{{\cos }^2}x} & {\sin 2x} \cr
} } \right|$$<br><br>$${R_1} \to {R_1} - {R_2}$$<br><br>= $$\left| {\matrix{
0 & 1 & 0 \cr
2 & {{{\cos }^2}x} & {\cos 2x} \cr
1 & {{{\cos }^2}x} & {\sin 2x} \cr
} } \right|$$<br><br>$$ = ( - 1)[2\sin 2x - \cos 2x] = \cos 2x - 2\sin 2x$$
<br><br>We know, maximum value of acosx $$ \pm $$ bsinx
<br><br>= $$\sqrt {{a^2} + {b^2}} $$
<br><br>$$ \therefore $$ Here maximum value = $$\sqrt {{1^2} + {{\left( { - 2} \right)}^2}} $$$$ = \sqrt 5 $$ | mcq | jee-main-2021-online-16th-march-evening-shift |
axb7TtMHVL2HxjaxWs1kmko64sd | maths | application-of-derivatives | maxima-and-minima | Let f : [$$-$$1, 1] $$ \to $$ R be defined as f(x) = ax<sup>2</sup> + bx + c for all x$$\in$$[$$-$$1, 1], where a, b, c$$\in$$R such that f($$-$$1) = 2, f'($$-$$1) = 1 for x$$\in$$($$-$$1, 1) the maximum value of f ''(x) is $${{1 \over 2}}$$. If f(x) $$ \le $$ $$\alpha$$, x$$\in$$[$$-$$1, 1], then the least value of $$\alpha$$ is equal to _________. | [] | null | 5 | $$f(x) = a{x^2} + bx + c$$<br><br>$$f'(x) = 2ax + b,$$<br><br>$$f''(x) = 2a$$<br><br>Given, $$f''( - 1) = {1 \over 2}$$<br><br>$$ \Rightarrow a = {1 \over 4}$$<br><br>$$f'( - 1) = 1 \Rightarrow b - 2a = 1$$<br><br>$$ \Rightarrow b = {3 \over 2}$$<br><br>$$f( - 1) = a - b + c = 2$$<br><br>$$ \Rightarrow c = {{13} \over 4}$$<br><br>Now, $$f(x) = {1 \over 4}({x^2} + 6x + 13),x \in [ - 1,1]$$<br><br>$$f'(x) = {1 \over 4}(2x + 6) = 0$$<br><br>$$ \Rightarrow x = - 3 \notin [ - 1,1]$$<br><br>$$f(1) = 5,f( - 1) = 2$$<br><br>$$f(x) \le 5$$<br><br>So, $$\alpha$$<sub>minimum</sub> = 5 | integer | jee-main-2021-online-17th-march-evening-shift |
1krpwaqea | maths | application-of-derivatives | maxima-and-minima | Let $$A = [{a_{ij}}]$$ be a 3 $$\times$$ 3 matrix, where $${a_{ij}} = \left\{ {\matrix{
1 & , & {if\,i = j} \cr
{ - x} & , & {if\,\left| {i - j} \right| = 1} \cr
{2x + 1} & , & {otherwise.} \cr
} } \right.$$<br/><br/>Let a function f : R $$\to$$ R be defined as f(x) = det(A). Then the sum of maximum and minimum values of f on R is equal to: | [{"identifier": "A", "content": "$$ - {{20} \\over {27}}$$"}, {"identifier": "B", "content": "$${{88} \\over {27}}$$"}, {"identifier": "C", "content": "$${{20} \\over {27}}$$"}, {"identifier": "D", "content": "$$ - {{88} \\over {27}}$$"}] | ["D"] | null | $$A = \left[ {\matrix{
1 & { - x} & {2x + 1} \cr
{ - x} & 1 & { - x} \cr
{2x + 1} & { - x} & 1 \cr
} } \right]$$<br><br>$$\left| A \right| = 4{x^3} - 4{x^2} - 4x = f(x)$$<br><br>$$f'(x) = 4(3{x^2} - 2x - 1) = 0$$<br><br>$$ \Rightarrow x = 1;x = {{ - 1} \over 3}$$<br><br>$$\therefore$$ $$f(1) = - 4;f\left( { - {1 \over 3}} \right) = {{20} \over {27}}$$<br><br>Sum $$ = - 4 + {{20} \over 7} = - {{88} \over {27}}$$ | mcq | jee-main-2021-online-20th-july-morning-shift |
1krpzb31f | maths | application-of-derivatives | maxima-and-minima | Let 'a' be a real number such that the function f(x) = ax<sup>2</sup> + 6x $$-$$ 15, x $$\in$$ R is increasing in $$\left( { - \infty ,{3 \over 4}} \right)$$ and decreasing in $$\left( {{3 \over 4},\infty } \right)$$. Then the function g(x) = ax<sup>2</sup> $$-$$ 6x + 15, x$$\in$$R has a : | [{"identifier": "A", "content": "local maximum at x = $$-$$ $${{3 \\over 4}}$$"}, {"identifier": "B", "content": "local minimum at x = $$-$$$${{3 \\over 4}}$$"}, {"identifier": "C", "content": "local maximum at x = $${{3 \\over 4}}$$"}, {"identifier": "D", "content": "local minimum at x = $${{3 \\over 4}}$$"}] | ["A"] | null | $${{ - B} \over {2A}} = {3 \over 4}$$<br><br>$$ \Rightarrow {{ - (6)} \over {2a}} = {3 \over 4}$$<br><br>$$ \Rightarrow a = {{ - 6 \times 4} \over 6} \Rightarrow a = - 4$$<br><br>$$\therefore$$ $$g(x) = 4{x^2} - 6x + 15$$<br><br>Local max. at $$x = {{ - B} \over {2A}} = - {{( - 6)} \over {2( - 4)}}$$<br><br>$$ = {{ - 3} \over 4}$$ | mcq | jee-main-2021-online-20th-july-morning-shift |
1krrsddj5 | maths | application-of-derivatives | maxima-and-minima | The sum of all the local minimum values of the twice differentiable function f : R $$\to$$ R defined by $$f(x) = {x^3} - 3{x^2} - {{3f''(2)} \over 2}x + f''(1)$$ is : | [{"identifier": "A", "content": "$$-$$22"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "$$-$$27"}, {"identifier": "D", "content": "0"}] | ["C"] | null | $$f(x) = {x^3} - 3{x^2} - {{3f''(2)} \over 2}x + f''(1)$$ ..... (i)<br><br>$$f(x) = 3{x^2} - 6x - {3 \over 2}f''(2)$$ ..... (ii)<br><br>$$f''(x) = 6x - 6$$ ..... (iii)<br><br>Now, is 3<sup>rd</sup> equation<br><br>$$f''(2) = 12 - 6 = 6$$<br><br>$$f''(11 = 0)$$<br><br>Use (ii)<br><br>$$f'(x) = 3{x^2} - 6x - {3 \over 2}f''(2)$$<br><br>$$f'(x) = 3{x^2} - 6x - {3 \over 2} \times 6$$<br><br>$$f'(x) = 3{x^2} - 6x - 9$$<br><br>$$f'(x) = 0$$<br><br>$$3{x^2} - 6x - 9 = 0$$<br><br>$$\Rightarrow$$ x = $$-$$1 & 3<br><br>Use (iii)<br><br>$$f''(x) = 6x - 6$$<br><br>$$f''( - 1) = - 12 < 0$$ maxima<br><br>$$f''(3) = 12 > 0$$ minima.<br><br>Use (i)<br><br>$$f(x) = {x^3} - 3{x^2} - {3 \over 2}f''(2)x + f''(1)$$<br><br>$$f(x) = {x^3} - 3{x^2} - {3 \over 2} \times 6 \times x + 0$$<br><br>$$f(x) = {x^3} - 3{x^2} - 9x$$<br><br>$$f(3) = 27 - 27 - 9 \times 3 = - 27$$ | mcq | jee-main-2021-online-20th-july-evening-shift |
1ktbihwcy | maths | application-of-derivatives | maxima-and-minima | A wire of length 36 m is cut into two pieces, one of the pieces is bent to form a square and the other is bent to form a circle. If the sum of the areas of the two figures is minimum, and the circumference of the circle is k (meter), then $$\left( {{4 \over \pi } + 1} \right)k$$ is equal to _____________. | [] | null | 36 | Let x + y = 36<br><br>x is perimeter of square and y is perimeter of circle side of square = x/4<br><br>radius of circle = $${y \over {2\pi }}$$<br><br>Sum Areas = $${\left( {{x \over 4}} \right)^2} + \pi {\left( {{y \over {2\pi }}} \right)^2}$$<br><br>$$ = {{{x^2}} \over {16}} + {{{{(36 - x)}^2}} \over {4\pi }}$$<br><br>For min Area :<br><br>$$x = {{144} \over {\pi + 4}}$$<br><br>$$\Rightarrow$$ Radius = y = 36 $$-$$ $${{144} \over {\pi + 4}}$$<br><br>$$\Rightarrow$$ k = $${{36\pi } \over {\pi + 4}}$$<br><br>$$\left( {{4 \over \pi } + 1} \right)k$$ = 36 | integer | jee-main-2021-online-26th-august-morning-shift |
1ktcydszh | maths | application-of-derivatives | maxima-and-minima | The local maximum value of the function $$f(x) = {\left( {{2 \over x}} \right)^{{x^2}}}$$, x > 0, is | [{"identifier": "A", "content": "$${\\left( {2\\sqrt e } \\right)^{{1 \\over e}}}$$"}, {"identifier": "B", "content": "$${\\left( {{4 \\over {\\sqrt e }}} \\right)^{{e \\over 4}}}$$"}, {"identifier": "C", "content": "$${(e)^{{2 \\over e}}}$$"}, {"identifier": "D", "content": "1"}] | ["C"] | null | $$f(x) = {\left( {{2 \over x}} \right)^{{x^2}}}$$ ; x > 0<br><br>$$\ln f(x) = {x^2}(\ln 2 - \ln x)$$<br><br>$$f'(x) = f(x)\{ - x + (\ln 2 - \ln x)2x\} $$<br><br>$$f'(x) = \underbrace {f(x)}_ + \,.\,\underbrace x_ + \underbrace {(2\ln 2 - 2\ln x - 1)}_{g(x)}$$<br><br>$$g(x) = 2{\ln ^2} - 2\ln x - 1$$<br><br>$$ = \ln {4 \over {{x^2}}} - 1 = 0 \Rightarrow x = {2 \over {\sqrt e }}$$<br><br> <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264656/exam_images/ss5f5bqsk3a61xv45vq9.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267368/exam_images/ulmjxqnxjjh3icn3gqsy.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266510/exam_images/wec0srmmwngoa5mbqhd5.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264724/exam_images/xhueaftoffkovgut0qew.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 26th August Evening Shift Mathematics - Application of Derivatives Question 86 English Explanation"></picture> <br><br>$$LM = {2 \over {\sqrt e }}$$<br><br>Local maximum value = $${\left( {{2 \over {2/\sqrt e }}} \right)^{{4 \over e}}} = {e^{{2 \over e}}}$$ | mcq | jee-main-2021-online-26th-august-evening-shift |
1kteodvui | maths | application-of-derivatives | maxima-and-minima | A wire of length 20 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a regular hexagon. Then the length of the side (in meters) of the hexagon, so that the combined area of the square and the hexagon is minimum, is : | [{"identifier": "A", "content": "$${5 \\over {2 + \\sqrt 3 }}$$"}, {"identifier": "B", "content": "$${{10} \\over {2 + 3\\sqrt 3 }}$$"}, {"identifier": "C", "content": "$${5 \\over {3 + \\sqrt 3 }}$$"}, {"identifier": "D", "content": "$${{10} \\over {3 + 2\\sqrt 3 }}$$"}] | ["D"] | null | Let the wire is cut into two pieces of length x and 20 $$-$$ x.<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263573/exam_images/qgzb6vevzrjgj6qtkfoq.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th August Morning Shift Mathematics - Application of Derivatives Question 85 English Explanation"><br><br>Area of square = $${\left( {{x \over 4}} \right)^2}$$ <br><br>Area of regular hexagon = $$6 \times {{\sqrt 3 } \over 4}{\left( {{{20 - x} \over 6}} \right)^2}$$<br><br>Total area = $$A(x) = {{{x^2}} \over {16}} + {{3\sqrt 3 } \over 2}{{{{(20 - x)}^2}} \over {36}}$$<br><br>$$A'(x) = {{2x} \over {16}} + {{3\sqrt 3 \times 2} \over {2 \times 36}}(20 - x)( - 1)$$<br><br>A'(x) = 0 at $$x = {{40\sqrt 3 } \over {3 + 2\sqrt 3 }}$$<br><br>Length of side of regular Hexagon $$ = {1 \over 6}(20 - x)$$<br><br>$$ = {1 \over 6}\left( {20 - {{4.\sqrt 3 } \over {3 + 2\sqrt 3 }}} \right)$$<br><br>$$ = {{10} \over {2 + 2\sqrt 3 }}$$ | mcq | jee-main-2021-online-27th-august-morning-shift |
1kteoog7u | maths | application-of-derivatives | maxima-and-minima | The number of distinct real roots of the equation 3x<sup>4</sup> + 4x<sup>3</sup> $$-$$ 12x<sup>2</sup> + 4 = 0 is _____________. | [] | null | 4 | 3x<sup>4</sup> + 4x<sup>3</sup> $$-$$ 12x<sup>2</sup> + 4 = 0<br><br>So, let f(x) = 3x<sup>4</sup> + 4x<sup>3</sup> $$-$$ 12x<sup>2</sup> + 4<br><br>$$\therefore$$ f'(x) = 12x(x<sup>2</sup> + x $$-$$ 2)<br><br>= 12x (x + 2) (x $$-$$ 1)<br><br>$$ \therefore $$ f'(x) = 12x<sup>3</sup> + 12x<sup>2</sup> – 24x = 12x(x + 2) (x – 1)
<br><br>Points of extrema are at x = 0, –2, 1
<br><br>f(0) = 4
<br><br>f(–2) = –28
<br><br>f(1) = –1
<br><br>So, 4 Real Roots
<br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266991/exam_images/mswypgvszyrm0wd5qq5t.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th August Morning Shift Mathematics - Application of Derivatives Question 84 English Explanation"> | integer | jee-main-2021-online-27th-august-morning-shift |
1ktg2hnod | maths | application-of-derivatives | maxima-and-minima | A box open from top is made from a rectangular sheet of dimension a $$\times$$ b by cutting squares each of side x from each of the four corners and folding up the flaps. If the volume of the box is maximum, then x is equal to : | [{"identifier": "A", "content": "$${{a + b - \\sqrt {{a^2} + {b^2} - ab} } \\over {12}}$$"}, {"identifier": "B", "content": "$${{a + b - \\sqrt {{a^2} + {b^2} + ab} } \\over 6}$$"}, {"identifier": "C", "content": "$${{a + b - \\sqrt {{a^2} + {b^2} - ab} } \\over 6}$$"}, {"identifier": "D", "content": "$${{a + b + \\sqrt {{a^2} + {b^2} + ab} } \\over 6}$$"}] | ["C"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266717/exam_images/wemjsa96uqnlahmrdgwc.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264426/exam_images/yprti8h0wfn66ietnppr.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266134/exam_images/u0nyg29w60wvazbqlpbt.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th August Evening Shift Mathematics - Application of Derivatives Question 83 English Explanation 1"></picture> <br><br>V = l . b . h = (a $$-$$ 2x)(b $$-$$ 2x) x<br><br>$$\Rightarrow$$ V(x) = (2x $$-$$ a)(2x $$-$$ b) x<br><br>$$\Rightarrow$$ V(x) = 4x<sup>3</sup> $$-$$ 2(a + b)x<sup>2</sup> + abx<br><br>$$ \Rightarrow {d \over {dx}}v(x) = 12{x^2} - 4(a + b)x + ab$$<br><br>$${d \over {dx}}(v(x)) = 0 \Rightarrow 12{x^2} - 4(a + b)x + ab = 0 < _\beta ^\alpha $$<br><br>$$ \Rightarrow x = {{4(a + b) \pm \sqrt {16{{(a + b)}^2} - 48ab} } \over {2(12)}}$$<br><br>$$ = {{(a + b) \pm \sqrt {{a^2} + {b^2} - ab} } \over 6}$$<br><br>Let $$x = \alpha = {{(a + b) + \sqrt {{a^2} + {b^2} - ab} } \over 6}$$<br><br>$$\beta = {{(a + b) - \sqrt {{a^2} + {b^2} - ab} } \over 6}$$<br><br>Now, $$12(x - \alpha )(x - \beta ) = 0$$<br><br> <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265372/exam_images/uhxfq60ihh4ljz6tust2.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th August Evening Shift Mathematics - Application of Derivatives Question 83 English Explanation 2"> <br><br>$$\therefore$$ x = $$\beta$$ $$ = {{a + b - \sqrt {{a^2} + {b^2} - ab} } \over b}$$ | mcq | jee-main-2021-online-27th-august-evening-shift |
1ktio607h | maths | application-of-derivatives | maxima-and-minima | The number of real roots of the equation <br/><br/>$${e^{4x}} + 2{e^{3x}} - {e^x} - 6 = 0$$ is : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "0"}] | ["C"] | null | Let $${e^x} = t > 0$$<br><br>$$f(t) = {t^4} + 2{t^3} - t - 6 = 0$$<br><br>$$f'(t) = 4{t^3} + 6{t^2} - 1$$<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266142/exam_images/xzvjyoinhhpqhttdoa57.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 31st August Morning Shift Mathematics - Application of Derivatives Question 82 English Explanation 1"><br><br>$$f''(t) = 12{t^2} + 12t > 0$$<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264918/exam_images/d9q64amnu43sedw4rgz4.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 31st August Morning Shift Mathematics - Application of Derivatives Question 82 English Explanation 2"><br><br>$$f(0) = - 6,f(1) = - 4,f(2) = 24$$<br><br>$$\Rightarrow$$ Number of real roots = 1 | mcq | jee-main-2021-online-31st-august-morning-shift |
1ktkevgzw | maths | application-of-derivatives | maxima-and-minima | Let f(x) be a cubic polynomial with f(1) = $$-$$10, f($$-$$1) = 6, and has a local minima at x = 1, and f'(x) has a local minima at x = $$-$$1. Then f(3) is equal to ____________. | [] | null | 22 | <p>Let f(x) = ax<sup>3</sup> + bx<sup>2</sup> + cx + d</p>
<p>f'(x) = 3ax<sup>2</sup> + 2bx + c $$\Rightarrow$$ f''(x) = 6ax + 2b</p>
<p>f'(x) has local minima at x = $$-$$1, so</p>
<p>$$\because$$ f''($$-$$1) = 0 $$\Rightarrow$$ $$-$$6a + 2b = 0 $$\Rightarrow$$ b = 3a ..... (i)</p>
<p>f(x) has local minima at x = 1</p>
<p>f'(1) = 0</p>
<p>$$\Rightarrow$$ 3a + 6a + c = 0</p>
<p>$$\Rightarrow$$ c = $$-$$9a ..... (ii)</p>
<p>f(1) = $$-$$10</p>
<p>$$\Rightarrow$$ $$-$$5a + d = $$-$$10 ..... (iii)</p>
<p>f($$-$$1) = 6</p>
<p>$$\Rightarrow$$ 11a + d = 6 ..... (iv)</p>
<p>Solving Eqs. (iii) and (iv)</p>
<p>a = 1, d = $$-$$5</p>
<p>From Eqs. (i) and (ii),</p>
<p>b = 3, c = $$-$$9</p>
<p>$$\therefore$$ f(x) = x<sup>3</sup> + 3x<sup>2</sup> $$-$$ 9x $$-$$ 5</p>
<p>So, f(3) = 27 + 27 $$-$$ 27 $$-$$ 5 = 22</p> | integer | jee-main-2021-online-31st-august-evening-shift |
1l545dst9 | maths | application-of-derivatives | maxima-and-minima | <p>A wire of length 22 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into an equilateral triangle. Then, the length of the side of the equilateral triangle, so that the combined area of the square and the equilateral triangle is minimum, is :</p> | [{"identifier": "A", "content": "$${{22} \\over {9 + 4\\sqrt 3 }}$$"}, {"identifier": "B", "content": "$${{66} \\over {9 + 4\\sqrt 3 }}$$"}, {"identifier": "C", "content": "$${{22} \\over {4 + 9\\sqrt 3 }}$$"}, {"identifier": "D", "content": "$${{66} \\over {4 + 9\\sqrt 3 }}$$"}] | ["B"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5nil7kj/2ee06cba-de30-46ff-9d0f-1870cf0f243a/6bf44540-04d1-11ed-93b8-936002ac8631/file-1l5nil7kk.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5nil7kj/2ee06cba-de30-46ff-9d0f-1870cf0f243a/6bf44540-04d1-11ed-93b8-936002ac8631/file-1l5nil7kk.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 29th June Morning Shift Mathematics - Application of Derivatives Question 78 English Explanation"></p>
<p>$$4a + 3b = 22$$</p>
<p>Total area $$ = A = {a^2} + {{\sqrt 3 } \over 4}{b^2}$$</p>
<p>$$A = {\left( {{{22 - 3b} \over 4}} \right)^2} + {{\sqrt 3 } \over 4}{b^2}$$</p>
<p>$${{dA} \over {dB}} = 2\left( {{{22 - 3b} \over 4}} \right)\left( {{{ - 3} \over 4}} \right) + {{\sqrt 3 } \over 4}\,.\,2b = 0$$</p>
<p>$$ \Rightarrow {{\sqrt {3b} } \over 2} = {3 \over 8}(22 - 3b)$$</p>
<p>$$ \Rightarrow 4\sqrt 3 b = 66 - 9b$$</p>
<p>$$ \therefore b = {{66} \over {9 + 4\sqrt 3 }}$$</p> | mcq | jee-main-2022-online-29th-june-morning-shift |
1l589itwu | maths | application-of-derivatives | maxima-and-minima | <p>The sum of the absolute minimum and the absolute maximum values of the <br/><br/>function f(x) = |3x $$-$$ x<sup>2</sup> + 2| $$-$$ x in the interval [$$-$$1, 2] is :</p> | [{"identifier": "A", "content": "$${{\\sqrt {17} + 3} \\over 2}$$"}, {"identifier": "B", "content": "$${{\\sqrt {17} + 5} \\over 2}$$"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "$${{9 - \\sqrt {17} } \\over 2}$$"}] | ["A"] | null | $f(x)=\left|x^2-3 x-2\right|-x \forall x \in[-1,2]$<br><br>
$\Rightarrow f(x)=\left\{\begin{array}{l}x^2-4 x-2 \text { if }-1 \leq x<\frac{3-\sqrt{17}}{2} \\ -x^2+2 x+2 \text { if } \frac{3-\sqrt{17}}{2} \leq x \leq 2\end{array}\right.$<br><br>
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lc6nga5p/f6bd0c59-facc-4024-b756-658e8619a898/a12a41d0-8620-11ed-ba1b-3b1de775931b/file-1lc6nga5q.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lc6nga5p/f6bd0c59-facc-4024-b756-658e8619a898/a12a41d0-8620-11ed-ba1b-3b1de775931b/file-1lc6nga5q.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 26th June Morning Shift Mathematics - Application of Derivatives Question 73 English Explanation"><br>
$f(x)_{\max }=3$<br><br>
$f(x)_{\min }=f\left(\frac{3-\sqrt{17}}{2}\right)$
$=\frac{\sqrt{17}-3}{2}$ | mcq | jee-main-2022-online-26th-june-morning-shift |
1l58f6wgh | maths | application-of-derivatives | maxima-and-minima | <p>Consider a cuboid of sides 2x, 4x and 5x and a closed hemisphere of radius r. If the sum of their surface areas is a constant k, then the ratio x : r, for which the sum of their volumes is maximum, is :</p> | [{"identifier": "A", "content": "2 : 5"}, {"identifier": "B", "content": "19 : 45"}, {"identifier": "C", "content": "3 : 8"}, {"identifier": "D", "content": "19 : 15"}] | ["B"] | null | <p>$$\because$$ $${s_1} + {s_2} = k$$</p>
<p>$$76{x^2} + 3\pi {r^2} = k$$</p>
<p>$$\therefore$$ $$152x{{dx} \over {dr}} + 6\pi r = 0$$</p>
<p>$$\therefore$$ $${{dx} \over {dr}} = {{ - 6\pi r} \over {152x}}$$</p>
<p>Now $$V = 40{x^3} + {2 \over 3}\pi {r^3}$$</p>
<p>$$\therefore$$ $${{dv} \over {dr}} = 120{x^2}\,.\,{{dx} \over {dr}} + 2\pi {r^2} = 0$$</p>
<p>$$ \Rightarrow 120{x^2}\,.\,\left( {{{ - 6\pi } \over {152}}{r \over x}} \right) + 2\pi {r^2} = 0$$</p>
<p>$$ \Rightarrow 120\left( {{x \over r}} \right) = 2\pi \left( {{{152} \over {6\pi }}} \right)$$</p>
<p>$$ \Rightarrow \left( {{x \over r}} \right) = {{152} \over 3}{1 \over {120}} = {{19} \over {45}}$$</p> | mcq | jee-main-2022-online-26th-june-evening-shift |
1l59lf1ub | maths | application-of-derivatives | maxima-and-minima | <p>Let $$f(x) = |(x - 1)({x^2} - 2x - 3)| + x - 3,\,x \in R$$. If m and M are respectively the number of points of local minimum and local maximum of f in the interval (0, 4), then m + M is equal to ____________.</p> | [] | null | 3 | <p>$$f(x) = \left| {(x - 1)(x + 1)(x - 3)} \right| + (x - 3)$$</p>
<p>$$f(x) = \left\{ {\matrix{
{(x - 3)({x^2})} & {3 \le x \le 4} \cr
{(x - 3)(2 - {x^2})} & {1 \le x < 3} \cr
{(x - 3)({x^2})} & {0 < x < 1} \cr
} } \right.$$</p>
<p>$$f'(x) = \left\{ {\matrix{
{3{x^2} - 6x} & {3 < x < 4} \cr
{ - 3{x^2} + 6x + 2} & {1 < x < 3} \cr
{3{x^2} - 6x} & {0 < x < 1} \cr
} } \right.$$</p>
<p>$$f'({3^ + }) > 0\,\,\,f'({3^ - }) < 0 \to $$ Minimum</p>
<p>$$f'({1^ + }) > 0\,\,\,f'({1^ - }) < 0 \to $$ Minimum</p>
<p>$$x \in (1,3)\,\,f'(x) = 0$$ at one point $$\to$$ Maximum</p>
<p>$$x \in (3,4)\,\,f'(x) \ne 0$$</p>
<p>$$x \in (0,1)\,\,f'(x) \ne 0$$</p>
<p>So, 3 points.</p> | integer | jee-main-2022-online-25th-june-evening-shift |
1l5c1bifq | maths | application-of-derivatives | maxima-and-minima | <p>The sum of absolute maximum and absolute minimum values of the function $$f(x) = |2{x^2} + 3x - 2| + \sin x\cos x$$ in the interval [0, 1] is :</p> | [{"identifier": "A", "content": "$$3 + {{\\sin (1){{\\cos }^2}\\left( {{1 \\over 2}} \\right)} \\over 2}$$"}, {"identifier": "B", "content": "$$3 + {1 \\over 2}(1 + 2\\cos (1))\\sin (1)$$"}, {"identifier": "C", "content": "$$5 + {1 \\over 2}(\\sin (1) + \\sin (2))$$"}, {"identifier": "D", "content": "$$2 + \\sin \\left( {{1 \\over 2}} \\right)\\cos \\left( {{1 \\over 2}} \\right)$$"}] | ["B"] | null | $f(x)=|(2 x-1)(x+2)|+\frac{\sin 2 x}{2}$
<br/><br/>
$0 \leq x<\frac{1}{2} \quad f(x)=(1-2 x)(x+2)+\frac{\sin 2 x}{2}$
<br/><br/>
$f^{\prime}(x)=-4 x-3+\cos 2 x<0$
<br/><br/>
For $x \geq \frac{1}{2}: \quad f^{\prime}(x)=4 x+3+\cos 2 x>0$
<br/><br/>
So, minima occurs at $x=\frac{1}{2}$
<br/><br/>
$$
\begin{aligned}
\left.f(x)\right|_{\min } &=\left|2\left(\frac{1}{2}\right)^{2}+\frac{3}{2}-2\right|+\sin \left(\frac{1}{2}\right) \cdot \cos \left(\frac{1}{2}\right) \\\\
&=\frac{1}{2} \sin 1
\end{aligned}
$$
<br/><br/>
So, maxima is possible at $x=0$ or $x=1$
<br/><br/>
Now checking for $x=0$ and $x=1$, we can see it attains its maximum value at $x=1$
<br/><br/>
$$
\begin{aligned}
\left.f(x)\right|_{\max }=&|2+3-2|+\frac{\sin 2}{2} \\\\
&=3+\frac{1}{2} \sin 2
\end{aligned}
$$
<br/><br/>
Sum of absolute maximum and minimum value
<br/><br/>$=3+\frac{1}{2}(\sin 1+\sin 2)$
<br/><br/>$=3+\frac{1}{2}(\sin 1+2\sin 1\cos 1)$
<br/><br/>= $$3 + {1 \over 2}(1 + 2\cos (1))\sin (1)$$ | mcq | jee-main-2022-online-24th-june-morning-shift |
1l5vzjzdv | maths | application-of-derivatives | maxima-and-minima | <p>If xy<sup>4</sup> attains maximum value at the point (x, y) on the line passing through the points (50 + $$\alpha$$, 0) and (0, 50 + $$\alpha$$), $$\alpha$$ > 0, then (x, y) also lies on the line :</p> | [{"identifier": "A", "content": "y = 4x"}, {"identifier": "B", "content": "x = 4y"}, {"identifier": "C", "content": "y = 4x + $$\\alpha$$"}, {"identifier": "D", "content": "x = 4y $$-$$ $$\\alpha$$"}] | ["A"] | null | <p>Equation of line passing through the point (50 + $$\alpha$$, 0) and (0, 50 + $$\alpha$$) is</p>
<p>$$y - 0 = {{50 + \alpha - 0} \over {0 - (50 + \alpha )}}\left( {x - (50 + \alpha )} \right)$$</p>
<p>$$ \Rightarrow y = - 1\left( {x - (50 + \alpha )} \right)$$</p>
<p>$$ \Rightarrow y = (50 + \alpha ) - x$$</p>
<p>$$ \Rightarrow x + y = 50 + \alpha $$</p>
<p>Let $$p = x{y^4}$$</p>
<p>$$ = (50 + \alpha - y){y^4}$$</p>
<p>$$ = (50 + \alpha ){y^4} - {y^5}$$</p>
<p>For maximum or minimum value of p,</p>
<p>$${{dp} \over {dy}} = 0$$</p>
<p>$$ \Rightarrow 4{y^3}(50 + \alpha ) - 5{y^4} = 0$$</p>
<p>$$ \Rightarrow {y^3}[200 + 4\alpha - 5y] = 0$$</p>
<p>$$\therefore$$ $$y = 0$$</p>
<p>or</p>
<p>$$200 + 4\alpha - 5y = 0$$</p>
<p>$$ \Rightarrow y = {4 \over 5}(50 + \alpha )$$</p>
<p>$$\therefore$$ $$x = 50 + \alpha - y$$</p>
<p>$$ = 50 + \alpha - {4 \over 5}(50 + \alpha )$$</p>
<p>$$ = 50 + \alpha \left( {1 - {4 \over 5}} \right)$$</p>
<p>$$ = {1 \over 5}(50 + \alpha )$$</p>
<p>$$ \Rightarrow 4x = {4 \over 5}(50 + \alpha ) = y$$</p>
<p>$$ \Rightarrow y = 4x$$</p>
<p>$$\therefore$$ (x, y) lies on the line y = 4x</p> | mcq | jee-main-2022-online-30th-june-morning-shift |
1l5vzpx90 | maths | application-of-derivatives | maxima-and-minima | <p>Let $$f(x) = 4{x^3} - 11{x^2} + 8x - 5,\,x \in R$$. Then f :</p> | [{"identifier": "A", "content": "has a local minina at $$x = {1 \\over 2}$$"}, {"identifier": "B", "content": "has a local minima at $$x = {3 \\over 4}$$"}, {"identifier": "C", "content": "is increasing in $$\\left( {{1 \\over 2},{3 \\over 4}} \\right)$$"}, {"identifier": "D", "content": "is decreasing in $$\\left( {{1 \\over 2},{4 \\over 3}} \\right)$$"}] | ["D"] | null | <p>Given,</p>
<p>$$f(x) = 4{x^3} - 11{x^2} + 8x - 5$$</p>
<p>$$\therefore$$ $$f'(x) = 12{x^2} - 22x + 8$$</p>
<p>$$ = 2(6{x^2} - 11x + 4)$$</p>
<p>$$ = 2(6{x^2} - 8x - 3x + 4)$$</p>
<p>$$ = 2\left[ {2x(3x - 4) - 1(3x - 4)} \right]$$</p>
<p>$$ = 2\left[ {(3x - 4)(2x - 1)} \right]$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5ykgpbg/402463f1-cb39-4303-b051-0d9df56ac983/6004f6c0-0ae5-11ed-a51c-73986e88f75f/file-1l5ykgpbh.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5ykgpbg/402463f1-cb39-4303-b051-0d9df56ac983/6004f6c0-0ae5-11ed-a51c-73986e88f75f/file-1l5ykgpbh.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 30th June Morning Shift Mathematics - Application of Derivatives Question 59 English Explanation 1"></p>
<p>$$f'(x)$$ is positive before $${1 \over 2}$$ means slope of f(x) is positive before $${1 \over 2}$$ and f'(x) is negative after $${1 \over 2}$$ means slope of f(x) is negative after $${1 \over 2}$$.</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5yki2b2/272d4c44-a92f-4bfd-a59e-312affd66d01/85dcc4e0-0ae5-11ed-a51c-73986e88f75f/file-1l5yki2b3.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5yki2b2/272d4c44-a92f-4bfd-a59e-312affd66d01/85dcc4e0-0ae5-11ed-a51c-73986e88f75f/file-1l5yki2b3.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 30th June Morning Shift Mathematics - Application of Derivatives Question 59 English Explanation 2"></p>
<p>$$\therefore$$ At $${1 \over 2}$$, f(x) is local maximum</p>
<p>Also, $$f'(x)$$ is negative before $${4 \over 3}$$ means slope of f(x) is negative before $${4 \over 3}$$ and f'(x) is positive after $${4 \over 3}$$ means slope of f(x) is positive after $${4 \over 3}$$.</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5ykjlo9/ac69243c-25e4-42c2-af51-acef287e6cb5/b0a1b690-0ae5-11ed-a51c-73986e88f75f/file-1l5ykjloa.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5ykjlo9/ac69243c-25e4-42c2-af51-acef287e6cb5/b0a1b690-0ae5-11ed-a51c-73986e88f75f/file-1l5ykjloa.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 30th June Morning Shift Mathematics - Application of Derivatives Question 59 English Explanation 3"></p>
<p>$$\therefore$$ At $${4 \over 3},\,f(x)$$ is local minimum</p>
<p>From wavy curve method,</p>
<p>$$f'(x) > 0,\,\forall x \in \left( { - \alpha ,\,{1 \over 2}} \right) \cup \left( {{4 \over 3},\,\alpha } \right)$$</p>
<p>$$\therefore$$ f(x) increasing in $$x \in \left( { - \alpha ,\,{1 \over 2}} \right) \cup \left( {{4 \over 3},\,\alpha } \right)$$</p>
<p>$$f'(x) < 0\,\forall x \in \left( {{1 \over 2},{4 \over 3}} \right)$$</p>
<p>$$\therefore$$ f(x) decreasing in $$x \in \left( {{1 \over 2},{4 \over 3}} \right)$$</p> | mcq | jee-main-2022-online-30th-june-morning-shift |
1l6dv7vz2 | maths | application-of-derivatives | maxima-and-minima | <p>If the absolute maximum value of the function $$f(x)=\left(x^{2}-2 x+7\right) \mathrm{e}^{\left(4 x^{3}-12 x^{2}-180 x+31\right)}$$ in the interval $$[-3,0]$$ is $$f(\alpha)$$, then :
</p> | [{"identifier": "A", "content": "$$\\alpha=0$$"}, {"identifier": "B", "content": "$$ \\alpha=-3$$"}, {"identifier": "C", "content": "$$\\alpha \\in(-1,0)$$"}, {"identifier": "D", "content": "$$\\alpha \\in(-3,-1]$$"}] | ["B"] | null | Given, $f(x)=\underbrace{\left(x^{2}-2 x+7\right)}_{f_{1}(x)} \underbrace{e^{\left(4 x^{3}-12 x^{2}-180 x+31\right)}}_{f_{2}(x)}$
<br/><br/>
$f_{1}(x)=x^{2}-2 x+7$
<br/><br/>
$f_{1}^{\prime}(x)=2 x-2$, so $f(x)$ is decreasing in $[-3,0]$ and positive also
<br/><br/>
$f_{2}(x)=e^{4 x^{3}-12 x^{2}-180 x+31}$
<br/><br/>
$f_{2}^{\prime}(x)=e^{4 x^{3}-12 x^{2}-180 x+31} \cdot 12 x^{2}-24 x-180$
<br/><br/>
$=12(x-5)(x+3) e^{4 x^{3}-12 x^{2}-180 x+31}$
<br/><br/>
So, $f_{2}(x)$ is also decreasing and positive in $\{-3,0\}$
<br/><br/>
$\therefore$ absolute maximum value of $f(x)$ occurs at $x=-3$
<br/><br/>
$\therefore \quad \alpha=-3$ | mcq | jee-main-2022-online-25th-july-morning-shift |
1l6dvag5k | maths | application-of-derivatives | maxima-and-minima | <p>The curve $$y(x)=a x^{3}+b x^{2}+c x+5$$ touches the $$x$$-axis at the point $$\mathrm{P}(-2,0)$$ and cuts the $$y$$-axis at the point $$Q$$, where $$y^{\prime}$$ is equal to 3 . Then the local maximum value of $$y(x)$$ is:
</p> | [{"identifier": "A", "content": "$$\\frac{27}{4}$$"}, {"identifier": "B", "content": "$$\\frac{29}{4}$$"}, {"identifier": "C", "content": "$$\\frac{37}{4}$$"}, {"identifier": "D", "content": "$$\\frac{9}{2}$$"}] | ["A"] | null | $f(x)=y=a x^{3}+b x^{2}+c x+5 \quad \ldots$ (i)
<br/><br/>
$$
\frac{d y}{d x}=3 a x^{2}+2 b x+c \quad \ldots (ii)
$$
<br/><br/>
Touches $x$-axis at $P(-2,0)$
<br/><br/>
$\left.\Rightarrow y\right|_{x=-2}=0 \Rightarrow-8 a+4 b-2 c+5=0 \quad \ldots ...(iii)$
<br/><br/>
Touches $x$-axis at $P(-2,0)$ also implies
<br/><br/>
$$
\left.\frac{d y}{d x}\right|_{x=-2}=0 \Rightarrow 12 a-4 b+c=0 \quad \ldots...(iv)
$$
<br/><br/>
$y=f(x)$ cuts $y$-axis at $(0,5)$
<br/><br/>
Given, $\left.\frac{d y}{d x}\right|_{x=0}=c=3$
<br/><br/>
From (iii), (iv) and (v)
<br/><br/>
$$
\begin{aligned}
&a=-\frac{1}{2}, b=-\frac{3}{4}, c=3 \\\\
&\Rightarrow f(x)=\frac{-x^{2}}{2}-\frac{3}{4} x^{2}+3 x+5 \\\\
&f^{\prime}(x)=\frac{-3}{2} x^{2}-\frac{3}{2} x+3 \\\\
&=\frac{-3}{2}(x+2)(x-1) \\\\
&f^{\prime}(x)=0 \text { at } x=-2 \text { and } x=1
\end{aligned}
$$
<br/><br/>
By first derivative test $x=1$ in point of local maximum Hence local maximum value of $f(x)$ is $f(1)$
<br/><br/>
i.e., $\frac{27}{4}$ | mcq | jee-main-2022-online-25th-july-morning-shift |
1l6f3c9hf | maths | application-of-derivatives | maxima-and-minima | <p>The sum of the maximum and minimum values of the function $$f(x)=|5 x-7|+\left[x^{2}+2 x\right]$$ in the interval $$\left[\frac{5}{4}, 2\right]$$, where $$[t]$$ is the greatest integer $$\leq t$$, is ______________.</p> | [] | null | 15 | <p>$$f(x) = |5x - 7| + [{x^2} + 2x]$$</p>
<p>$$ = |5x - 7| + [{(x + 1)^2}] - 1$$</p>
<p>Critical points of</p>
<p>$$f(x) = {7 \over 5},\sqrt 5 - 1,\,\sqrt 6 - 1,\,\sqrt 7 - 1,\,\sqrt 8 - 1,\,2$$</p>
<p>$$\therefore$$ Maximum or minimum value of $$f(x)$$ occur at critical points or boundary points</p>
<p>$$\therefore$$ $$f\left( {{5 \over 4}} \right) = {3 \over 4} + 4 = {{19} \over 4}$$</p>
<p>$$f\left( {{7 \over 5}} \right) = 0 + 4 = 4$$</p>
<p>as both $$|5x - 7|$$ and $${x^2} + 2x$$ are increasing in nature after $$x = {7 \over 5}$$</p>
<p>$$\therefore$$ $$f(2) = 3 + 8 = 11$$</p>
<p>$$\therefore$$ $$f{\left( {{7 \over 5}} \right)_{\min }} = 4$$ and $$f{(2)_{\max }} = 11$$</p>
<p>Sum is $$4 + 11 = 15$$</p> | integer | jee-main-2022-online-25th-july-evening-shift |
1l6hy2szh | maths | application-of-derivatives | maxima-and-minima | <p>If the maximum value of $$a$$, for which the function $$f_{a}(x)=\tan ^{-1} 2 x-3 a x+7$$ is non-decreasing in $$\left(-\frac{\pi}{6}, \frac{\pi}{6}\right)$$, is $$\bar{a}$$, then $$f_{\bar{a}}\left(\frac{\pi}{8}\right)$$ is equal to :</p> | [{"identifier": "A", "content": "$$\n8-\\frac{9 \\pi}{4\\left(9+\\pi^{2}\\right)}\n$$"}, {"identifier": "B", "content": "$$8-\\frac{4 \\pi}{9\\left(4+\\pi^{2}\\right)}$$"}, {"identifier": "C", "content": "$$8\\left(\\frac{1+\\pi^{2}}{9+\\pi^{2}}\\right)$$"}, {"identifier": "D", "content": "$$8-\\frac{\\pi}{4}$$"}] | ["A"] | null | $\text {Given, }$
<br/><br/>$$
\begin{aligned}
f_a(x) & =\tan ^{-1} 2 x-3 a x+7 \\\\
f_a^{\prime}(x) & =\frac{2}{1+4 x^2}-3 a
\end{aligned}
$$
<br/><br/>As the function $f_a^{\prime}(x)$ is non-decreasing
<br/><br/>$$
\begin{aligned}
& \text { in }\left(-\frac{\pi}{6}, \frac{\pi}{6}\right), \\\\
& f_a^{\prime}(x) \geq 0
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \Rightarrow \frac{2}{1+4 x^2}-3 a \geq 0 \Rightarrow \frac{2}{1+4 x^2} \geq 3 a \\\\
& \Rightarrow a \leq \frac{2}{3\left(1+4 x^2\right)}, \text { when } x \in\left(-\frac{\pi}{6}, \frac{\pi}{6}\right) \\\\
& \because a \text { is maximum when } x^2=\frac{\pi^2}{36},
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
a_{\max } & =\frac{2}{3\left(1+\frac{4 \pi^2}{36}\right)}=\frac{2 \times 12}{36+4 \pi^2} \\\\
& =\frac{6}{9+\pi^2} \\\\
\therefore \bar{a} & =\frac{6}{9+\pi^2}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \therefore f_a(x)=\tan ^{-1} 2 x-\frac{18}{9+\pi^2} x+7 \\\\
& f_\pi\left(\frac{\pi}{8}\right)=\tan ^{-1} 2\left(\frac{\pi}{8}\right)-\frac{18}{9+\pi^2} \times \frac{\pi}{8}+7 \\\\
& =\tan ^{-1} \frac{\pi}{4}-\frac{9 \pi}{36+4 \pi^2}+7 \\\\
& =8-\frac{9 \pi}{36+4 \pi^2}=8-\frac{9 \pi}{4\left(9+\pi^2\right)}
\end{aligned}
$$ | mcq | jee-main-2022-online-26th-july-evening-shift |
1l6m63rp3 | maths | application-of-derivatives | maxima-and-minima | <p>If the minimum value of $$f(x)=\frac{5 x^{2}}{2}+\frac{\alpha}{x^{5}}, x>0$$, is 14 , then the value of $$\alpha$$ is equal to :</p> | [{"identifier": "A", "content": "32"}, {"identifier": "B", "content": "64"}, {"identifier": "C", "content": "128"}, {"identifier": "D", "content": "256"}] | ["C"] | null | <p>$$f(x) = {{5{x^2}} \over 2} + {\alpha \over {{x^5}}}\,\,\{ x > 0\} $$</p>
<p>$$f'(x) = 5x - {{5\alpha } \over {{x^6}}} = 0$$</p>
<p>$$ \Rightarrow x = {(\alpha )^{{1 \over 7}}}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7rb2qy2/d1f9b6e9-ed06-4643-80c4-d9d8e53ea216/9bffa4a0-2e7f-11ed-8702-156c00ced081/file-1l7rb2qy3.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7rb2qy2/d1f9b6e9-ed06-4643-80c4-d9d8e53ea216/9bffa4a0-2e7f-11ed-8702-156c00ced081/file-1l7rb2qy3.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 28th July Morning Shift Mathematics - Application of Derivatives Question 50 English Explanation"></p>
<p>$$f{(x)_{\min }} = {{5{{(\alpha )}^{{2 \over 7}}}} \over 2} + {\alpha \over {{\alpha ^{{5 \over 7}}}}} = 14$$</p>
<p>$${5 \over 2}{\alpha ^{{2 \over 7}}} + {\alpha ^{{2 \over 7}}} = 14$$</p>
<p>$${7 \over 2}{\alpha ^{{2 \over 7}}} = 14$$</p>
<p>$$\alpha = 128$$</p> | mcq | jee-main-2022-online-28th-july-morning-shift |
1l6p31rwv | maths | application-of-derivatives | maxima-and-minima | <p>Let $$f(x)=3^{\left(x^{2}-2\right)^{3}+4}, x \in \mathrm{R}$$. Then which of the following statements are true?</p>
<p>$$\mathrm{P}: x=0$$ is a point of local minima of $$f$$</p>
<p>$$\mathrm{Q}: x=\sqrt{2}$$ is a point of inflection of $$f$$</p>
<p>$$R: f^{\prime}$$ is increasing for $$x>\sqrt{2}$$</p> | [{"identifier": "A", "content": "Only P and Q"}, {"identifier": "B", "content": "Only P and R"}, {"identifier": "C", "content": "Only Q and R"}, {"identifier": "D", "content": "All P, Q and R"}] | ["D"] | null | <p>$$f(x) = {3^{{{({x^2} - 2)}^3} + 4}},\,x \in R$$</p>
<p>$$f(x) = {81.3^{{{({x^2} - 2)}^3}}}$$</p>
<p>$$f'(x) = {81.3^{{{({x^2} - 2)}^3}}}\ln 2.3({x^2} - 2)2x$$</p>
<p>$$ = (486\ln 2)\left( {{3^{{{({x^2} - 2)}^3}}}({x^2} - 2)x} \right)$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7sso9mz/26e78098-e521-4cb7-8650-9f6cbc3d58bd/364659b0-2f51-11ed-85dd-19dc023e9ad1/file-1l7sso9n0.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7sso9mz/26e78098-e521-4cb7-8650-9f6cbc3d58bd/364659b0-2f51-11ed-85dd-19dc023e9ad1/file-1l7sso9n0.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 29th July Morning Shift Mathematics - Application of Derivatives Question 48 English Explanation"></p>
<p>$$ \Rightarrow x = 0$$ is the local minima.</p>
<p>$$f''(x) = (486\ln 2)\left( {{3^{{{({x^2} - 2)}^3}}}\,.\,({x^2} - 2)(5{x^2} - 2 + 6{x^2}\ln 3({x^2} - 2))} \right)$$</p>
<p>$$f''(x) = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = \sqrt 2 $$</p>
<p>$$f''\left( {{{\sqrt 2 }^ + }} \right) > 0$$</p>
<p>$$f''\left( {{{\sqrt 2 }^ - }} \right) < 0$$</p>
<p>$$ \Rightarrow x = \sqrt 2 $$ is point of inflection</p>
<p>$$f''(x) > 0\,\forall x > \sqrt 2 $$</p>
<p>$$ \Rightarrow f'(x)$$ is increasing for $$x > \sqrt 2 $$</p> | mcq | jee-main-2022-online-29th-july-morning-shift |
1ldo4vr0v | maths | application-of-derivatives | maxima-and-minima | <p>The sum of the absolute maximum and minimum values of the function $$f(x)=\left|x^{2}-5 x+6\right|-3 x+2$$ in the interval $$[-1,3]$$ is equal to :</p> | [{"identifier": "A", "content": "13"}, {"identifier": "B", "content": "24"}, {"identifier": "C", "content": "10"}, {"identifier": "D", "content": "12"}] | ["C"] | null | The sum of the absolute maximum and minimum values of the function $$f(x) = \left|x^2 - 5x + 6\right| - 3x + 2$$ in the interval $$[-1, 3]$$ can be found by finding the maximum and minimum values of $$f(x)$$ in this interval and then adding them.
<br/><br/>
First, let's find the critical points of $$f(x)$$. To do this, we will find the zeros of the expression inside the absolute value:
<br/><br/>
$$x^2 - 5x + 6 = 0$$
<br/><br/>
Solving this quadratic equation, we find that the zeros are $$x = 2$$ and $$x = 3$$. These are the critical points of $$f(x)$$.
<br/><br/>
Next, we evaluate $$f(x)$$ at these critical points and at the endpoints of the interval $$[-1, 3]$$:
<br/><br/>
$$f(-1) = \left|(-1)^2 - 5(-1) + 6\right| - 3(-1) + 2 = 17$$
<br/><br/>
$$f(2) = \left|(2)^2 - 5(2) + 6\right| - 3(2) + 2 = -4$$
<br/><br/>
$$f(3) = \left|(3)^2 - 5(3) + 6\right| - 3(3) + 2 = -7$$
<br/><br/>
So, the minimum value of $$f(x)$$ is $$-7$$ and the maximum value is $$17$$.
<br/><br/>
So, the sum of the absolute maximum and minimum values of the function is $$-7 + 17 = 10$$.
<br/><br/><b>Note :</b> The absolute maximum and minimum values of a function are the largest and smallest values that the function takes on a given interval, respectively. These values are also called the "extrema" of the function. The absolute maximum value is the highest point on the graph of the function, and the absolute minimum value is the lowest point. | mcq | jee-main-2023-online-1st-february-evening-shift |
1ldpsc9p1 | maths | application-of-derivatives | maxima-and-minima | <p>A wire of length $$20 \mathrm{~m}$$ is to be cut into two pieces. A piece of length $$l_{1}$$ is bent to make a square of area $$A_{1}$$ and the other piece of length $$l_{2}$$ is made into a circle of area $$A_{2}$$. If $$2 A_{1}+3 A_{2}$$ is minimum then $$\left(\pi l_{1}\right): l_{2}$$ is equal to :</p> | [{"identifier": "A", "content": "6 : 1"}, {"identifier": "B", "content": "1 : 6"}, {"identifier": "C", "content": "4 : 1"}, {"identifier": "D", "content": "3 : 1"}] | ["A"] | null | $ \ell_{1}+\ell_{2}=20 \Rightarrow \frac{\mathrm{d} \ell_{2}}{\mathrm{~d} \ell_{1}}=-1$
<br/><br/>$\mathrm{A}_{1}=\left(\frac{\ell_{1}}{4}\right)^{2}$ and $\mathrm{A}_{2}=\pi\left(\frac{\ell_{2}}{2 \pi}\right)^{2}$
<br/><br/>Let $\mathrm{S}=2 \mathrm{~A}_{1}+3 \mathrm{~A}_{2}=\frac{\ell_{1}^{2}}{8}+\frac{3 \ell_{2}^{2}}{4 \pi}$
<br/><br/>$\frac{\mathrm{ds}}{\mathrm{d} \ell_1}=0 \Rightarrow \frac{2 \ell_{1}}{8}+\frac{6 \ell_{2}}{4 \pi} \cdot \frac{\mathrm{d} \ell_{2}}{\mathrm{~d} \ell_{1}}=0$
<br/><br/>$\Rightarrow \frac{\ell_{1}}{4}=\frac{6 \ell_{2}}{4 \pi} $
<br/><br/>$\Rightarrow \frac{\pi \ell_{1}}{\ell_{2}}=6$ | mcq | jee-main-2023-online-31st-january-morning-shift |
ldqwo36h | maths | application-of-derivatives | maxima-and-minima | If the functions $f(x)=\frac{x^3}{3}+2 b x+\frac{a x^2}{2}$
<br/><br/>and $g(x)=\frac{x^3}{3}+a x+b x^2, a \neq 2 b$ <br/><br/>have a common extreme point, then $a+2 b+7$ is equal to : | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "$\\frac{3}{2}$"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "4"}] | ["A"] | null | <p>$$f'(x)=x^2+2b+ax$$</p>
<p>$$g'(x)=x^2+a+2bx$$</p>
<p>$$\Rightarrow x=1$$ is common root</p>
<p>$$a+2b+1=0$$</p> | mcq | jee-main-2023-online-30th-january-evening-shift |
1ldu57hdx | maths | application-of-derivatives | maxima-and-minima | <p>Let the function $$f(x) = 2{x^3} + (2p - 7){x^2} + 3(2p - 9)x - 6$$ have a maxima for some value of $$x < 0$$ and a minima for some value of $$x > 0$$. Then, the set of all values of p is</p> | [{"identifier": "A", "content": "$$\\left( { - {9 \\over 2},{9 \\over 2}} \\right)$$"}, {"identifier": "B", "content": "$$\\left( {{9 \\over 2},\\infty } \\right)$$"}, {"identifier": "C", "content": "$$\\left( {0,{9 \\over 2}} \\right)$$"}, {"identifier": "D", "content": "$$\\left( { - \\infty ,{9 \\over 2}} \\right)$$"}] | ["D"] | null | $f^{\prime}(x)=6 x^{2}+2 x(2 p-7)+3(2 p-9)$
<br/><br/>
$x_{1}<0, x_{2}>0$
<br/><br/>
$\Rightarrow f^{\prime}(0)<0$
<br/><br/>
$\Rightarrow p<\frac{9}{2}$<br/><br/>
$$p \in \left( { - \infty ,{9 \over 2}} \right)$$ | mcq | jee-main-2023-online-25th-january-evening-shift |
1ldv1bkyt | maths | application-of-derivatives | maxima-and-minima | <p>Let $$x=2$$ be a local minima of the function $$f(x)=2x^4-18x^2+8x+12,x\in(-4,4)$$. If M is local maximum value of the function $$f$$ in ($$-4,4)$$, then M =</p> | [{"identifier": "A", "content": "$$18\\sqrt6-\\frac{33}{2}$$"}, {"identifier": "B", "content": "$$18\\sqrt6-\\frac{31}{2}$$"}, {"identifier": "C", "content": "$$12\\sqrt6-\\frac{33}{2}$$"}, {"identifier": "D", "content": "$$12\\sqrt6-\\frac{31}{2}$$"}] | ["C"] | null | $$
\begin{aligned}
& f(x)=8 x^3-36 x+8 \\\\
& =4\left(2 x^3-9 x+2\right) \\\\
& =4(x-2)\left(2 x^2+4 x-1\right) \\\\
& =4(x-2)\left(x-\frac{-2+\sqrt{6}}{2}\right)\left(x-\frac{-2 \sqrt{6}}{2}\right)
\end{aligned}
$$<br/><br/>
Local maxima occurs at $x=\frac{-2+\sqrt{6}}{2}=x_0$<br/><br/>
$$
f\left(x_0\right)=12 \sqrt{6}-\frac{33}{2}
$$ | mcq | jee-main-2023-online-25th-january-morning-shift |
lgnz80z6 | maths | application-of-derivatives | maxima-and-minima | Consider the triangles with vertices $A(2,1), B(0,0)$ and $C(t, 4), t \in[0,4]$.
<br/><br/>If the maximum and the minimum perimeters of such triangles are obtained at <br/><br/>$t=\alpha$ and $t=\beta$ respectively, then $6 \alpha+21 \beta$ is equal to ___________. | [] | null | 48 | We have a triangle with vertices $A(2,1)$, $B(0,0)$, and $C(t, 4)$, where $t$ belongs to the interval $[0,4]$.
<br><br>We want to find the maximum and minimum perimeters of such triangles, which occur at $t=\alpha$ and $t=\beta$, respectively.
<br><br>To find the minimum perimeter, we use a geometric approach. Reflect point $B$ over the line $y=4$ to get $B'(0,8)$. The line segment $AB'$ intersects the line $y=4$ (which is the $y$-coordinate of point $C$) at the point which gives the minimum perimeter.
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1li61mesu/46698a3c-f1f2-4c38-a7f4-cdbe7d5c6371/8f09b1d0-fc94-11ed-8452-d7b44c8805e4/file-1li61mesv.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1li61mesu/46698a3c-f1f2-4c38-a7f4-cdbe7d5c6371/8f09b1d0-fc94-11ed-8452-d7b44c8805e4/file-1li61mesv.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 15th April Morning Shift Mathematics - Application of Derivatives Question 38 English Explanation">
<br>The slope of $AB'$ is :
<br><br>$$m_{AB'} = \frac{8 - 1}{0 - 2} = \frac{-7}{2}$$
<br><br>The equation of the line $AB'$ is then $y - 1 = m_{AB'}(x - 2)$, or $7x + 2y = 16$.
<br><br>Solving this equation for $y = 4$ yields $x = \frac{8}{7}$. So, the minimum perimeter is achieved at point $C\left(\frac{8}{7}, 4\right)$, so $\beta = \frac{8}{7}$.
<br><br>For the maximum perimeter, we notice that it will be achieved when point $C$ is either at $(0,4)$ or at $(4,4)$, since these are the furthest points from $A(2,1)$ within the range of $t$. By calculating the perimeters at these points, we find that the maximum perimeter is achieved at $\alpha = 4$.
<br><br>Finally, we calculate $6\alpha + 21\beta = 6 \cdot 4 + 21 \cdot \frac{8}{7} = 24 + 24 = 48$.
<br><br>Therefore, $6\alpha + 21\beta = 48$. | integer | jee-main-2023-online-15th-april-morning-shift |
1lgq0kk6g | maths | application-of-derivatives | maxima-and-minima | <p>$$\max _\limits{0 \leq x \leq \pi}\left\{x-2 \sin x \cos x+\frac{1}{3} \sin 3 x\right\}=$$</p> | [{"identifier": "A", "content": "$$\\frac{5 \\pi+2+3 \\sqrt{3}}{6}$$"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "$$\\frac{\\pi+2-3 \\sqrt{3}}{6}$$"}, {"identifier": "D", "content": "$$\\pi$$"}] | ["A"] | null | <p>Given the function:</p>
<p>$$
f(x) = x - 2\sin{x}\cos{x} + \frac{1}{3}\sin{3x}
$$</p>
<p>We want to find the maximum value of this expression for $$0 \leq x \leq \pi$$.</p>
<p>Step 1: Rewrite the expression
<br/><br/>Notice that we can rewrite the expression as:
<br/><br/>$$
f(x) = x - \sin{2x} + \frac{1}{3}\sin{3x}
$$</p>
<p>Step 2: Find the first derivative
<br/><br/>Now, let's find the derivative of this expression with respect to $$x$$:
<br/><br/>$$
f'(x) = 1 - 2\cos{2x} + \cos{3x}
$$</p>
<p>Step 3: Find the critical points
<br/><br/>To find the critical points, we set $$f'(x) = 0$$:
<br/><br/>$$
1 - 2\cos{2x} + \cos{3x} = 0
$$</p>
<p>Step 4: Solve for x
<br/><br/>This equation can be rewritten as:
<br/><br/>$$
(2\cos{x} + \sqrt{3})(2\cos{x} - \sqrt{3})(\cos{x} - 1) = 0
$$</p>
<p>We get three possible solutions for $$x$$:
<br/><br/>$$
\cos{x} = \frac{-\sqrt{3}}{2}, \frac{\sqrt{3}}{2}, 1
$$</p>
<p>Which gives us:
$$
x = \frac{5\pi}{6}, \frac{\pi}{6}, 0
$$</p>
<p>Step 5: Find the second derivative
<br/><br/>Now, let's find the second derivative of the expression:
<br/><br/>$$
f''(x) = 4\sin{2x} - 3\sin{3x}
$$</p>
<p>Step 6: Analyze the critical points
Evaluate the second derivative at the critical points:</p>
<ol>
<li>$$f''\left(\frac{5\pi}{6}\right) = -2\sqrt{3} - \sqrt{3} < 0$$, indicating a local maximum.</li>
<li>$$f''\left(\frac{\pi}{6}\right) = 2\sqrt{3} - \sqrt{3} > 0$$, indicating a local minimum.</li>
<li>$$f''(0) = 0$$, inconclusive.</li>
</ol>
<p>Step 7: Evaluate the function at the local maximum point and the boundary points
Since we are looking for the maximum value of $$f(x)$$, we can now evaluate the function at the local maximum point and the boundary points:</p>
<ol>
<li>$$f\left(\frac{5\pi}{6}\right) = \frac{5\pi}{6} + \frac{\sqrt{3}}{2} + \frac{1}{3} = \frac{5\pi + 2 + 3\sqrt{3}}{6}$$</li>
<li>$$f(0) = 0$$</li>
<li>$$f(\pi) = \pi$$</li>
</ol>
<p>Step 8: Compare the values
The maximum value of $$f(x)$$ is given by $$f\left(\frac{5\pi}{6}\right) = \frac{5\pi + 2 + 3\sqrt{3}}{6}$$.</p> | mcq | jee-main-2023-online-13th-april-morning-shift |
1lgrg9l7x | maths | application-of-derivatives | maxima-and-minima | <p>If the local maximum value of the function $$f(x)=\left(\frac{\sqrt{3 e}}{2 \sin x}\right)^{\sin ^{2} x}, x \in\left(0, \frac{\pi}{2}\right)$$ , is $$\frac{k}{e}$$, then $$\left(\frac{k}{e}\right)^{8}+\frac{k^{8}}{e^{5}}+k^{8}$$ is equal to</p> | [{"identifier": "A", "content": "$$e^{3}+e^{6}+e^{10}$$"}, {"identifier": "B", "content": "$$e^{3}+e^{5}+e^{11}$$"}, {"identifier": "C", "content": "$$e^{3}+e^{6}+e^{11}$$"}, {"identifier": "D", "content": "$$e^{5}+e^{6}+e^{11}$$"}] | ["C"] | null | $$
\text { Let } y=\left(\frac{\sqrt{3 e}}{2 \sin x}\right)^{\sin ^2 x}
$$
<br/><br/>$$
\ln \mathrm{y}=\sin ^2 \mathrm{x} \cdot \ln \left(\frac{\sqrt{3 \mathrm{e}}}{2 \sin \mathrm{x}}\right)
$$
<br/><br/>$$
\frac{1}{y} y^{\prime}=\ln \left(\frac{\sqrt{3 e}}{2 \sin x}\right) 2 \sin x \cos x+\sin ^2 x \frac{2 \sin x}{\sqrt{3 e}} \frac{\sqrt{3 e}}{2}(-\operatorname{cosec} x \cot x)
$$
<br/><br/>For maxima or minima, $$
\frac{d y}{d x}=0
$$
<br/><br/>$$
\Rightarrow \ln \left(\frac{\sqrt{3 e}}{2 \sin x}\right) 2 \sin x \cos x-\sin x \cos x=0
$$
<br/><br/>$$
\Rightarrow \sin x \cos x\left[2 \ln \left(\frac{\sqrt{3 e}}{2 \sin x}\right)-1\right]=0
$$
<br/><br/>$$
\Rightarrow \ln \left(\frac{3 \mathrm{e}}{4 \sin ^2 \mathrm{x}}\right)=1
$$
<br/><br/>$$
\Rightarrow \frac{3 e}{4 \sin ^2 x}=e
$$
<br/><br/>$$
\Rightarrow \sin ^2 x=\frac{3}{4}
$$
<br/><br/>$$
\Rightarrow \sin \mathrm{x}=\frac{\sqrt{3}}{2} \quad\left(\text { as } \mathrm{x} \in\left(0, \frac{\pi}{2}\right)\right)
$$
<br/><br/>$$
\Rightarrow \text { Local max value }=\left(\frac{\sqrt{3 \mathrm{e}}}{\sqrt{3}}\right)^{3 / 4}=\mathrm{e}^{3 / 8}=\frac{\mathrm{k}}{\mathrm{e}}
$$
<br/><br/>$$
\Rightarrow \mathrm{k}^8=\mathrm{e}^{11}
$$
<br/><br/>$$ \therefore $$ $$
\left(\frac{k}{e}\right)^8+\frac{k^8}{e^5}+k^8=e^3+e^6+e^{11}
$$ | mcq | jee-main-2023-online-12th-april-morning-shift |
1lgxt9s39 | maths | application-of-derivatives | maxima-and-minima | <p>A square piece of tin of side 30 cm is to be made into a box without top by cutting a square from each corner and folding up the flaps to form a box. If the volume of the box is maximum, then its surface area (in cm$$^2$$) is equal to :</p> | [{"identifier": "A", "content": "1025"}, {"identifier": "B", "content": "900"}, {"identifier": "C", "content": "800"}, {"identifier": "D", "content": "675"}] | ["C"] | null | Given, length of square sheet of side is $30 \mathrm{~cm}$.
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnbdiknq/edf7bbaa-df8b-4c8e-bede-daa23c7634e0/12049450-6280-11ee-95f7-df7702d42956/file-6y3zli1lnbdiknr.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lnbdiknq/edf7bbaa-df8b-4c8e-bede-daa23c7634e0/12049450-6280-11ee-95f7-df7702d42956/file-6y3zli1lnbdiknr.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh; vertical-align: baseline;" alt="JEE Main 2023 (Online) 10th April Morning Shift Mathematics - Application of Derivatives Question 32 English Explanation">
<br><br>Let, side of small square be $x \mathrm{~cm}$.
<br><br>Since, a square piece of tin is to be made into a box without top by cutting square from each corner and folding up the flaps to from a box. Then, this shape formed a cuboidal shape.
<br><br>Thus, volume of the box $(V)$
<br><br>$$
\begin{aligned}
& =(30-2 x)(30-2 x) x \\\\
&\Rightarrow V =x(30-2 x)^2
\end{aligned}
$$
<br><br>On differentiating both side with respect to $x$, we get
<br><br>$$
\begin{aligned}
\frac{d V}{d x} & =(30-2 x)^2+x\{2(30-2 x) \cdot(-2)\} \\\\
& =(30-2 x)^2-4 x(30-2 x) \\\\
& =(30-2 x)\{30-6 x\}
\end{aligned}
$$
<br><br>Therefore, maximum of $V, \frac{d V}{d x}=0$
<br><br>$$
\begin{array}{ll}
&\Rightarrow (30-2 x)(30-6 x)=0 \\\\
&\Rightarrow x=15 \text { or } x=5 \quad[x=15 \text { (not taken) }]
\end{array}
$$
<br><br>Thus, $x=5$
<br><br>So, the surface area
<br><br>$$
\begin{array}{ll}
=4 \times(30-2 x) \times x+(30-2 x)^2 & \\\\
=4 \times(30-10) \times 5+(30-10)^2 & (\text { Put } x=5) \\\\
=400+400=800 &
\end{array}
$$ | mcq | jee-main-2023-online-10th-april-morning-shift |
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