Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f(x)=\{\begin{array}{l}-1,-2\le x<0\\ x^2-1,0\le x\le 2\end{array}$ and $g(x)=|f(x)|+f(|x|)$. Then, in the interval $(-2,2)$, $g$ is :

A Journey into Smoothness

Welcome, fellow traveler on the JEE Advanced path. Today, we are not just solving a problem; we are dissecting a function.

We are looking at $f(x)$ and its mysterious cousin $g(x)=|f(x)|+f(|x|)$. This is a classic JEE trap, designed to test your ability to handle absolute values and piecewise definitions with precision. Let's peel back the layers together.

The Symmetry of $f(|x|)$

First, let's look at $f(|x|)$. Many students panic here, thinking they need to define it separately for every possible value of $x$.

But look closer. The input to $f$ is $|x|$, which is always non-negative. Our original function $f(x)$ is defined as $x^2-1$ for all non-negative inputs.

Therefore, $f(|x|)$ is simply $(|x|{)}^2-1$. Since $(|x|{)}^2$ is identical to $x^2$, we find that:

This holds for the entire interval $x\in (-2,2)$. The symmetry is beautiful, isn't it? We have already simplified half of our problem.

The Absolute Value Trap

Now, let's turn our attention to $|f(x)|$. This is where the real work begins. We have two regions to consider: $x\in [-2,0)$ and $x\in [0,2]$.

For $x\in [-2,0)$, $f(x)=-1$. Thus, $|f(x)|=|-1|=1$.

For $x\in [0,2]$, $f(x)=x^2-1$. Here is the catch: $x^2-1$ is not always positive. It crosses the x-axis at $x=1$. So, we must split this interval further:

- If $0\le x<1$, $x^2-1$ is negative, so $|f(x)|=-(x^2-1)=1-x^2$. - If $1\le x\le 2$, $x^2-1$ is positive, so $|f(x)|=x^2-1$.

Synthesizing $g(x)$

Now, we combine our findings to construct $g(x)=|f(x)|+f(|x|)$.

1. For $x\in [-2,0)$:

2. For $x\in [0,1)$:

3. For $x\in [1,2]$:

Look at that! The function $g(x)$ is defined in three distinct pieces. The magic happens when we check the transition points.

The Smoothness Test

Finally, we test for differentiability at $x=0$ and $x=1$.

At $x=0$: The left-hand derivative (LHD) of $x^2$ is $2x$, which is $0$ at $x=0$. The right-hand derivative (RHD) of $0$ is $0$. Since $0=0$, the function is smooth at $x=0$.

At $x=1$: The LHD of $0$ is $0$. The RHD of $2x^2-2$ is $4x$, which is $4$ at $x=1$. Since $0 eq 4$, we have found our sharp corner!

We have navigated the piecewise landscape and found that $g(x)$ is continuous everywhere, but it fails to be differentiable at exactly one point: $x=1$.

This is the essence of JEE Advanced mathematics—not just calculating, but visualizing the behavior of functions. Keep this intuition, and no problem will ever be too daunting.