Find no. of points where $f$ and $g$ meet. 
If $f(x)=x^2$ and $g(x)=x \sin x+ \cos x$ then
(A) $f$ and $g$ agree at no points
(B) $f$ and $g$ agree at exactly one point
(C) $f$ and $g$ agree at exactly two points
(A) $f$ and $g$ agree at more than two points.

Trial: I think I need to find the values of $x$ for which $f(x)=g(x)$. So, I have $x^2-x \sin x- \cos x=0$. How can I solve this ? Please help

Note that $x^2$ and $x\sin x+\cos x$ are both even functions, so the graphs of $y=x^2$ and $y=x\sin x+\cos x$ are symmetric about the $y$ axis. So if we kow the story for $x\ge 0$, we know the story everywhere.
At $x=0$, the curve $y=x^2$ is below the curve $y=x\sin x+\cos x$.
But $x^2$, in the long run, is far bigger than $x\sin x+\cos x$. This is clear, but if we want formal proof, observe that for positive $x$, we have $|x\sin x+\cos x|\le x+1$. Thus if $x\gt 2$, then $x^2\gt x+1$.
So the two curves meet at some positive $x$. If you want to be formal, use the Intermediate Value Theorem.
Do they meet at more than one positive $x$? The derivative of $x^2$ is $2x$, while the derivative of $x\sin x+\cos x$ is $x\cos x$. Since $x\cos x\lt x$, at any positive $x$ the function $x^2$ grows faster than the function $x\sin x+\cos x$. So once $x^2$ gets above $x\sin x+\cos x$, the function $x\sin x+\cos x$ can never again catch up to $x^2$. A formal proof here would involve the Mean Value Theorem. 
Thus the two functions are equal at exactly one positive $x$, and, by symmetry, at exactly one negative $x$.
