Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $a,b\in R$ and $f:R\to R$ be defined by $f(x)=a\cos (|x^3-x|)+b|x|\sin (|x^3+x|)$. Then $f$ is

Imagine you are standing before a massive, intimidating fortress. That is how many students feel when they see a function like $f(x)=a\cos (|x^3-x|)+b|x|\sin (|x^3+x|)$.

It looks jagged, sharp, and impossible to handle. But in the world of JEE Advanced, we do not attack the fortress head-on. We dismantle it, brick by brick.

Today, we are going to master the art of differentiability by breaking this function into two manageable pieces: $g_1(x)=a\cos (|x^3-x|)$ and $g_2(x)=b|x|\sin (|x^3+x|)$.

The Illusion of the Absolute Value

Let us look at $g_1(x)=a\cos (|x^3-x|)$. The absolute value sign here is a classic 'scarecrow'—it is designed to make you panic and start splitting the domain into cases where $x^3-x>0$ and $x^3-x<0$.

But stop! Take a deep breath and recall the fundamental property of the cosine function: $\cos (-\theta )=\cos (\theta )$. Because cosine is an even function, it does not care about the sign of its argument.

Therefore, $\cos (|u|)=\cos (u)$ for any real $u$. Suddenly, the absolute value vanishes! We are left with $g_1(x)=a\cos (x^3-x)$.

This is a composition of a polynomial and a trigonometric function, both of which are smooth and differentiable everywhere. The fortress wall has crumbled. $g_1(x)$ is differentiable for all $x\in \mathbb{R}$.

The First Principle at $x=0$

Now, we turn our attention to $g_2(x)=b|x|\sin (|x^3+x|)$. This is where the real work begins. At $x=0$, we cannot simply rely on standard differentiation rules because of the $|x|$ term.

We must return to the roots of calculus: the First Principle. We define the derivative at $x=0$ as:

Substituting our function, we get:

As $h$ approaches zero, we know that $\sin (\theta )\approx \theta$. Thus, $\sin (|h^3+h|)\approx |h^3+h|=|h(h^2+1)|=|h|(h^2+1)$.

Our limit expression transforms into:

Since $|h|\cdot |h|=h^2$, the expression becomes:

As $h\to 0$, this limit is clearly $0$. Because the limit exists and is finite, $g_2(x)$ is differentiable at $x=0$ for any real value of $b$.

The Neighborhood of $x=1$

Finally, let us examine $x=1$. We do not need the First Principle here. We simply look at the neighborhood of $x=1$.

For values of $x$ very close to $1$, $x$ is positive, so $|x|=x$. Similarly, $x^3+x$ is positive, so $|x^3+x|=x^3+x$.

The absolute value signs are effectively removed by the nature of the point we are investigating. Our function simplifies to $g_2(x)=bx\sin (x^3+x)$.

This is a product of two functions: a polynomial $x$ and a trigonometric function $\sin (x^3+x)$. Both are differentiable everywhere. By the product rule, their product is also differentiable. Thus, $g_2(x)$ is differentiable at $x=1$ for any $b$.

We have systematically dismantled the problem. We found that $g_1(x)$ is differentiable everywhere, and $g_2(x)$ is differentiable at both $x=0$ and $x=1$.

Since the sum of two differentiable functions is always differentiable, the original function $f(x)$ is differentiable at both points for all real values of $a$ and $b$.

You see? The complexity was just a mask. By staying calm and applying the fundamental definitions, you have conquered the problem. Keep this mindset for your JEE journey—never fear the function; simply understand its behavior.