Animated Solution for Mathematics - Differentiation: The number of points in $(-\infty ,\infty )$, for which $x^2-x\sin x-\cos x=0$, is

Welcome, fellow traveler of the JEE landscape. Today, we encounter an equation that looks deceptively simple but hides a beautiful, elegant structure beneath its surface. We are tasked with finding the number of real roots for the equation $x^2-x\sin x-\cos x=0$.

When you first look at this, your instinct might be to reach for algebraic identities or perhaps try to isolate $x$. But pause—take a breath. This is a transcendental equation. It refuses to be tamed by simple algebra. Instead, we must invite Calculus to the table.

Let us define our function as $f(x)=x^2-x\sin x-\cos x$. Our goal is to find the number of times this curve kisses or crosses the x-axis.

Imagine you are standing on a graph, and $f(x)$ represents your altitude. If you start at a high altitude, descend into a valley, and then climb back up to the clouds, you must have crossed the ground level at least twice. This is the intuition we will use.

To understand the 'lay of the land,' we need to know where the function is climbing and where it is falling. We calculate the derivative $f^{\prime }(x)$.

Using the standard rules of differentiation—the power rule for $x^2$, the product rule for $x\sin x$, and the derivative of $\cos x$—we get:

Look closely at this expression. It looks a bit messy, but watch what happens when we simplify. The $-\sin x$ and $+\sin x$ terms cancel out with the grace of a perfectly choreographed dance. We are left with:

This is the moment of truth. We set $f^{\prime }(x)=0$ to find our critical points. We know that for any real $x$, the value of $\cos x$ is trapped between $-1$ and $1$.

Consequently, the term $(2-\cos x)$ is always between $1$ and $3$. It is strictly positive! This means the sign of $f^{\prime }(x)$ depends entirely on the sign of $x$.

When $x<0$, $f^{\prime }(x)$ is negative, meaning our function is strictly decreasing. When $x>0$, $f^{\prime }(x)$ is positive, meaning our function is strictly increasing.

At $x=0$, we find our global minimum. Evaluating $f(0)$, we get:

Now, consider the behavior of the function as $x$ approaches infinity. The $x^2$ term is the 'heavyweight champion' here; it grows much faster than the oscillating $\sin x$ and $\cos x$ terms.

Thus, ${\lim }_{x\to \infty }f(x)=\infty$ and ${\lim }_{x\to -\infty }f(x)=\infty$.

We have a function that starts at positive infinity, drops down to a minimum of $-1$ at $x=0$, and then climbs back up to positive infinity. Because the function is continuous, it must cross the x-axis exactly once while descending from $+\infty$ to $-1$, and exactly once while ascending from $-1$ to $+\infty$.

There are no other turns, no other wiggles, and no other surprises. The logic is ironclad. We have found exactly 2 real roots.

You see, the beauty of mathematics lies not in brute force, but in understanding the behavior of the system. You have successfully navigated the landscape of this function. Keep this clarity with you as you tackle the next challenge!