Animated Solution for Mathematics - Functions: Let $f_1:\mathbb{R}\to \mathbb{R},f_2:[0,\infty )\to \mathbb{R},f_3:\mathbb{R}\to \mathbb{R}$ and $f_4:\mathbb{R}\to [0,\infty )$ be defined by $f_1(x)=\{\begin{array}{ll}|x|& \text{if }x<0,\\ e^x& \text{if }x\ge 0;\end{array}$ $f_2(x)=x^2$; $f_3(x)=\{\begin{array}{ll}\sin x& \text{if }x<0,\\ x& \text{if }x\ge 0;\end{array}$ and $f_4(x)=\{\begin{array}{ll}{f}_2(f_1(x))& \text{if }x<0,\\ f_2(f_1(x))-1& \text{if }x\ge 0.\end{array}$

A Journey Through Analysis

Welcome, aspiring mathematician. Today, we are not just solving a problem; we are dissecting the DNA of calculus.

Functions are the building blocks of the universe, and understanding their behavior—whether they are continuous, differentiable, one-one, or onto—is the hallmark of a true JEE Advanced scholar. Let us walk through this analysis together, step by step.

The Warm-up with $f_2(x)$

We begin with $f_2(x)=x^2$, defined on the domain $[0,\infty )$. This is our anchor. It is a simple, elegant parabola.

Because it is a polynomial, continuity is guaranteed. To check if it is one-one, we look at its derivative:

For all $x\ge 0$, $2x\ge 0$. Since the derivative is non-negative (and only zero at a single point), the function is strictly increasing.

A strictly increasing function is the definition of one-one. It never looks back. Thus, $f_2$ is continuous and one-one. A solid start.

The 'Sharp Corner' Myth with $f_3(x)$

Now, we encounter $f_3(x)$, a piecewise function defined as $\sin x$ for $x<0$ and $x$ for $x\ge 0$. Many students see the 'piecewise' label and immediately assume it is non-differentiable. Do not fall for this trap!

Let us check the seam at $x=0$. The Left-Hand Limit (LHL) is:

The Right-Hand Limit (RHL) is:

Since LHL = RHL = $f_3(0)$, the function is continuous.

Now, for differentiability: the Left-Hand Derivative (LHD) is $\frac{d}{dx}(\sin x)=\cos x$, which at $x=0$ is $1$. The Right-Hand Derivative (RHD) is $\frac{d}{dx}(x)=1$.

Because the LHD equals the RHD, the transition is perfectly smooth. It is differentiable!

However, is it one-one? Consider $f_3(-\pi )=\sin (-\pi )=0$ and $f_3(0)=0$. Two different inputs, same output. It is many-one. Thus, $f_3$ is differentiable but not one-one.

The Composition Trap with $f_2\circ f_1$

Composition is where things get interesting. We are looking at $f_2(f_1(x))$. We know $f_1(x)$ is $|x|$ for $x<0$ and $e^x$ for $x\ge 0$.

When we plug this into $f_2(x)=x^2$, we get:

For $x<0$:

For $x\ge 0$:

Now, check the seam at $x=0$. The LHL is ${\lim }_{x\to 0^{-}}{x}^2=0$. The RHL is ${\lim }_{x\to 0^{+}}{e}^{2x}=e^0=1$.

The gap is undeniable! The function is discontinuous.

And for one-one? We found that $f_2(f_1(-2))=4$ and $f_2(f_1(\ln 2))=4$. It fails the one-one test spectacularly. This function is neither continuous nor one-one.

The Final Boss, $f_4(x)$

Finally, we arrive at $f_4(x)$. It is built upon our previous composition but with a twist: we subtract $1$ for the positive domain. This shifts the right side of the graph down, potentially closing the gap.

Let us check: LHL is $0^2=0$. RHL is $e^0-1=1-1=0$. The gap is closed! It is continuous.

Is it onto? The range for $x<0$ is $(0,\infty )$, and for $x\ge 0$ it is $[0,\infty )$. The union is $[0,\infty )$, which matches the codomain. It is onto.

But is it one-one? We tested $f_4(-1)=1$ and $f_4(\frac{1}{2}\ln 2)=1$. Again, two inputs, one output. It is not one-one.

We have navigated the landscape of these functions. We saw that continuity and differentiability are not just algebraic checks; they are geometric realities.

We saw that composition can create discontinuities or destroy one-one properties. Keep this analytical mindset, and no function will ever intimidate you again. You are ready for the exam.