<s>[INST] If $f(x)=x^{2}$ and $g(x)=x \sin x+\cos x$; Find out the no of intersecting point Question from an entrance paper
If $f(x)=x^{2}$ and $g(x)=x \sin x+\cos x$ then
A. $f$ and $g$ agree at no point
B. $f$ and $g$ agree at exactly one point
C. $f$ and $g$ agree at exactly two point
D. $f$ and $g$ agree at more then two point

My approach
$f(x)=x^{2}$
$g(x)=x \sin x+\cos x$
$\Rightarrow f(x)=g(x)$
$\Rightarrow x^{2}=x \sin x+\cos x$
$\Rightarrow x^{2}-x \sin x-\cos x=0$
$\Rightarrow F(x)=0$
$F(0)=-1$ & $F(\pi)=\pi^{2}+1 >0 $
$F^{\prime}(x)=2 x-x \cos x$
$F^{\prime}(x)=x(2-\cos x)$
Here $2-\cos x$ is greater than $0$ and $x(2-\cos x)$ will be also greater than 0 for $x>0$
$\Rightarrow F(x)$ is strictly increasing function.
So I can conclude there will be exactly two point on which they will meet.

$\boxed{\text{*My question is how to do this without calculus*}}$
Edit $1$:- I had thought like this. This a  eq of power $2$,so It must have two roots. These two roots may be real or complex conjugate.Now how to distinguish in general whether it would have real or imaginary roots for all even powered polynomials
ps Dont give solutions which is similar to mine.
 [/INST] Solution without calculus
If $x \ge \frac{\pi}{2}$, we have $g(x) \le \sqrt{x^2 + 1} < x^2$,
and thus $f(x) = g(x)$ has no real solution on $[\frac{\pi}{2}, \infty)$.
Also, $0$ is not a solution of $f(x) = g(x)$.
Since $f(0) < g(0)$ and $f(\frac{\pi}{2}) > g(\frac{\pi}{2})$, by continuity,
$f(x)=g(x)$ has at least one real solution on $(0, \frac{\pi}{2})$.
Let us prove that $f(x)=g(x)$ has exactly one real solution on $(0, \frac{\pi}{2})$.
Suppose there exist $0 < s < t < \frac{\pi}{2}$ such that $f(s) = g(s)$ and $f(t) = g(t)$. We have
\begin{align}
\frac{s^2}{\sqrt{s^2+1}} &=  \frac{s\sin s + \cos s}{\sqrt{s^2+1}} = \cos (s - \arctan s), \\
\frac{t^2}{\sqrt{t^2+1}} &=  \frac{t\sin t + \cos t}{\sqrt{t^2+1}} = \cos (t - \arctan t).
\end{align}
Since $\frac{t^2}{\sqrt{t^2+1}} - \frac{s^2}{\sqrt{s^2+1}} = 
\sqrt{t^2+1} - \frac{1}{\sqrt{t^2 + 1}} - 
\sqrt{s^2+1} + \frac{1}{\sqrt{s^2 + 1}} > 0$, we have
$$\cos (s - \arctan s) < \cos (t - \arctan t)$$
which, when combined with
$t - \arctan t \in (0, \frac{\pi}{2})$ and $s - \arctan s \in (0, \frac{\pi}{2})$, results in
$$s - \arctan s > t - \arctan t.$$
Thus, we have $\arctan t - \arctan s > t - s$ and $\tan (\arctan t - \arctan s) > \tan (t - s)$
that is $\frac{t - s}{1 + ts} > \tan (t-s)$ which, when combined with $\tan (t-s) > t-s$, results in
$\frac{t - s}{1 + ts} > t-s$ which is impossible. The desired result follows.
Thus, $f(x) = g(x)$ has exactly one real solution on $[0, \infty)$.
Since $f(x)$ and $g(x)$ are both even, $f(x) = g(x)$ has exactly two real solutions on $(-\infty, \infty)$.
We are done.
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