Let g: R → R be a differentiable function with g(0) = 0, g'(0) = 0 and g'(1) ≠ 0. Let f(x) = {x/|x|

Q: Why do we need to check LHD and RHD separately? Can't we just differentiate the function directly?

Great question! Differentiation rules like the chain rule or product rule assume the function is already differentiable. When you have absolute values or piecewise definitions, the function might have a 'sharp corner' where the derivative is undefined. Checking LHD and RHD is the only way to be 100% sure.

Q: Why is $h(x) = e^{|x|}$ not differentiable at $x=0$?

Think of the graph. $e^{|x|}$ looks like a 'V' shape with a rounded bottom, but at $x=0$, the slope jumps from $-1$ to $1$. A function is differentiable only if the slope is continuous. Since $-1 \neq 1$, it's not differentiable.

Q: In Option D, how did you just replace $|f(x)|$ with $|g(x)|$?

Because $f(x) = \frac{x}{|x|} g(x)$, taking the absolute value gives $|f(x)| = |\frac{x}{|x|}| \cdot |g(x)|$. Since $|\frac{x}{|x|}| = 1$ for all $x \neq 0$, it simplifies perfectly to $|g(x)|$.

Question

Let $g: R 	o R$ be a differentiable function with $g(0) = 0, g'(0) = 0$ and $g'(1) 
eq 0$. Let $f(x) = \begin{cases} \frac{x}{|x|} g(x), & x 
eq 0 \ 0, & x = 0 \end{cases}$ and $h(x) = e^{|x|}$ for all $x \in R$. Let $(foh)(x)$ denote $f(h(x))$ and $(hof)(x)$ denote $h(f(x))$. Then which of the following is (are) true?

Options:

(A) $f$ is differentiable at $x=0$

(B) $h$ is differentiable at $x=0$

(C) $foh$ is differentiable at $x=0$

(D) $hof$ is differentiable at $x=0$

Sigma Percentile · Accepted Answer

### The Symphony of Differentiability Welcome, fellow traveler on the road to JEE Advanced mastery! Today, we are going to peel back the layers of a problem that tests not just your algebraic skills, but your fundamental intuition about what it means for a function to be "smooth." We are dealing with differentiability, the heartbeat of calculus. Let's embark on this journey. ### Phase 1: The Anatomy of $f(x)$ We start with the definition: $$f(x) = \frac{x}{|x|} g(x) ext{ for } x eq 0, ext{ and } f(0) = 0$$ The term $\frac{x}{|x|}$ is the classic signum function, $ ext{sgn}(x)$. It is $-1$ for negative $x$ and $+1$ for positive $x$. This is the "trap"—the function changes its definition abruptly at $x=0$. To check if $f$ is differentiable at $x=0$, we must use the first principle: $$f'(0) = \lim_{x o 0} \frac{f(x) - f(0)}{x - 0}$$ Substituting $f(0)=0$, we get: $$\lim_{x o 0} \frac{\frac{x}{|x|} g(x)}{x}$$ For $x > 0$, this simplifies to $\lim_{x o 0^+} \frac{g(x)}{x}$, which is $g'(0) = 0$. For $x < 0$, it becomes $\lim_{x o 0^-} \frac{-g(x)}{x}$, which is $-g'(0) = 0$. Since both sides yield $0$, $f$ is indeed differentiable at $x=0$. A smooth start! ### Phase 2: The Sharp Corner of $h(x)$ Next, we look at $h(x) = e^{|x|}$. Whenever you see an absolute value, your alarm bells should ring. Let's visualize the graph. It is an exponential curve that reflects across the y-axis, creating a "V" shape at the origin. At $x=0$, the slope from the right is $1$, and from the left, it is $-1$. Because $1 eq -1$, the function has a sharp corner. It is continuous, but definitely not differentiable. **Option B is false.** ### Phase 3: The Composition Dance Now, let's tackle the compositions. First, consider $(f \circ h)(x) = f(e^{|x|})$. Since $e^{|x|} \ge 1$ for all $x$, the input to $f$ is always positive. For positive inputs, $f(u) = g(u)$. So, $f(h(x)) = g(e^{|x|})$. To check differentiability at $x=0$, we look at the RHD and LHD. For $x > 0$, the derivative is $g'(e^x) \cdot e^x$, which at $x=0$ is $g'(1)$. For $x < 0$, the derivative is $g'(e^{-x}) \cdot (-e^{-x})$, which at $x=0$ is $-g'(1)$. Since $g'(1) eq 0$, these two values are not equal. Thus, $f \circ h$ is not differentiable. **Option C is false.** ### Phase 4: The Smoothing Effect Finally, we reach the climax: $(h \circ f)(x) = h(f(x)) = e^{|f(x)|}$. We need to evaluate: $$\lim_{x o 0} \frac{e^{|f(x)|} - 1}{x}$$ This looks intimidating, but let's use a clever trick. We rewrite it as: $$\lim_{x o 0} \left( \frac{e^{|f(x)|} - 1}{|f(x)|} \cdot \frac{|f(x)|}{x} ight)$$ The first part, $\frac{e^u - 1}{u}$ where $u = |f(x)|$, is a standard limit that approaches $1$. The second part, $\frac{|f(x)|}{x}$, simplifies to $\frac{|g(x)|}{x}$ because $|\frac{x}{|x|}| = 1$. As we found earlier, the limit of $\frac{|g(x)|}{x}$ as $x o 0$ is $0$. So, we have $1 \cdot 0 = 0$. The derivative exists! **Option D is true.** What a beautiful result! We've seen how composition can either preserve a sharp corner or, in the case of $h(f(x))$, smooth it out entirely. Keep this intuition close, and you'll conquer any differentiability problem the JEE throws at you!

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Visualized Solution

Analyzing the Given Functions

Setup for $f^{'} (0)$

Right Hand Derivative of $f (x)$

Left Hand Derivative of $f (x)$

Graph of $h (x) = e^{∣ x ∣}$

Checking Differentiability of $h (x)$

Setup for $f (h (x))$

Right Hand Derivative of $f (h (x))$

Left Hand Derivative of $f (h (x))$

Setup for $h (f (x))$

Algebraic Manipulation of the Limit

Evaluating the Exponential Limit

Evaluating the Modulus Limit

Final Derivative of $h (f (x))$

Final Summary

The Sigma Insight: Differentiability of a Function

The Symphony of Differentiability

Phase 1

Phase 2

Phase 3

Phase 4

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Visualized Solution

Analyzing the Given Functions

Setup for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(0)

Right Hand Derivative of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Left Hand Derivative of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Graph of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h(x)=e∣x∣

Checking Differentiability of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h(x)

Setup for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(h(x))

Right Hand Derivative of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(h(x))

Left Hand Derivative of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(h(x))

Setup for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h(f(x))