<s>[INST] Prove , there exists $\theta , \phi \in (\frac{\pi}{6},\frac{\pi}{3})$ such that $f'(\theta) = 0$ and $f'(\phi)\neq 0$ Let the function
$$f(\theta) = \begin{vmatrix} \sin\theta & \cos\theta & \tan\theta  \\ \sin(\frac{\pi}{6}) & \cos(\frac{\pi}{6}) & \tan(\frac{\pi}{6}) & \\ \sin(\frac{\pi}{3}) & \cos(\frac{\pi}{3}) & \tan(\frac{\pi}{3})   \end{vmatrix} $$
where 
$\theta \in \left[ \frac{\pi}{6},\frac{\pi}{3} \right]$ and $f'(\theta)$     denote the derivative of $f$ with respect to $\theta$. Which of the following statements is/are TRUE?
(I) There exists $\theta \in (\frac{\pi}{6},\frac{\pi}{3})$ such that $f'(\theta) = 0$
(II) There exists $\theta \in (\frac{\pi}{6},\frac{\pi}{3})$ such that $f'(\theta)\neq  0$


*

*I only

*II only

*Both I and II

*Neither I Nor II



I try to explain $:$

$$f(\theta) = \begin{vmatrix} \sin\theta & \cos\theta & \tan\theta  \\ \frac{1}{2} & \frac{\sqrt{3}}{2} & \frac{1}{\sqrt{3}} & \\ \frac{\sqrt{3}}{2} & \frac{1}{2} & \sqrt{3}   \end{vmatrix} $$
$f(\theta) =$$\left(\frac{3}{2}-\frac{1}{2\sqrt{3}}\right)sin\left(\theta\right)-\left(\frac{1}{2\sqrt{3}}-\frac{1}{2}\right)cos\left(\theta\right)+\left(\frac{1}{4}-\frac{3}{4}\right)tan\left(\theta\right)$
Since , 
$f'(\theta) =$$\frac{1}{6}\left(-3\sec ^2\left(\theta\right)-\left(\sqrt{3}-9\right)\cos \left(\theta\right)+\left(\sqrt{3}-3\right)\sin \left(\theta\right)\right)$
I know option (3) is true , since it is possible for both statements .

Can you prove in formal way ?

 [/INST] Without explicitly expanding the determinant we see that
$$f(\pi/6) = f(\pi/3) = 0$$
since two of the row are equal in these cases giving us a zero determinant. The mean value theorem then tells us that there exists a $\theta\in(\pi/6,\pi/3)$ where $f'(\theta) = 0$.
If II is false then $f’(\theta) \equiv 0 \implies f(\theta) \equiv 0$ on $(\pi/6,\pi/3)$. Thus a single non-zero value of $f$ on the interval is enough to show that II is true. Take for example $\theta = \pi/4$.
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