Let f(x) = x|x| and g(x) = sin x. Statement-1: gof is differentiable at x = 0 and its derivative is

Q: Wait, why do we reject the negative root here? Or why is the second derivative failing when the first one was fine?

This is the classic 'smoothness' trap! A function can be differentiable (having a tangent) but not twice differentiable (having a smooth curvature). The first derivative $h'(x)$ exists at $x=0$ because the slopes from the left and right match at $0$. However, the second derivative $h''(x)$ represents the rate of change of that slope. If the slope changes abruptly (a jump discontinuity in the derivative), the second derivative cannot exist. Always check the limits of the derivative itself!

Q: Why can't we just use the chain rule blindly on $\sin(x|x|)$?

The chain rule is a powerful tool, but it assumes the function is differentiable at the point of evaluation. At $x=0$, the absolute value function $|x|$ is not differentiable. Therefore, we cannot blindly apply the chain rule across the entire domain. We must use the piecewise definition to handle the transition at the origin safely. Rigor is your best friend in JEE Advanced.

Q: Is it possible for a function to be differentiable but not twice differentiable?

Absolutely! This is a fundamental concept in calculus. Think of a function like $f(x) = x|x|$. It is continuous and differentiable at $x=0$, but its derivative is $|x|$ (scaled), which has a 'kink' at the origin. When you compose this with $\sin(x)$, you inherit that behavior. It is a perfect example of how 'smoothness' can break down as you take higher-order derivatives.

Question

Let $f(x) = x|x|$ and $g(x) = \sin x$. \ 	extbf{Statement-1:} $g \circ f$ is differentiable at $x = 0$ and its derivative is continuous at that point. \ 	extbf{Statement-2:} $g \circ f$ is twice differentiable at $x = 0$.

Options:

(A) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

(B) Statement-1 is true, Statement-2 is false.

(C) Statement-1 is false, Statement-2 is true.

(D) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

Sigma Percentile · Accepted Answer

### The Anatomy of Smoothness: A Journey into $h(x) = \sin(x|x|)$ Welcome, future engineer. Today, we are not just solving a problem; we are dissecting the very nature of 'smoothness' in calculus. In the JEE Advanced arena, you will often encounter functions that look innocent but hide complex behaviors at the origin. Our function today, $h(x) = \sin(x|x|)$, is a perfect case study. Let us peel back the layers. ### Phase 1: Deconstructing the Beast We start with $f(x) = x|x|$. This is our foundation. Many students rush to differentiate this, but pause. The absolute value $|x|$ is a switch. It changes its nature at $x=0$. So, we define $f(x)$ piecewise: $$f(x) = \begin{cases} x^2 & x \ge 0 \ -x^2 & x < 0 \end{cases}$$ When we compose this with $g(x) = \sin(x)$, we get $h(x) = \sin(f(x))$. Substituting our piecewise definition, we obtain: $$h(x) = \begin{cases} \sin(x^2) & x \ge 0 \ -\sin(x^2) & x < 0 \end{cases}$$ Imagine the graph. For $x \ge 0$, it is a sine wave compressed by the square. For $x < 0$, it is the reflection of that wave. It passes through the origin smoothly. ### Phase 2: The First Derivative - The Smooth Transition To check differentiability at $x=0$, we need the first derivative, $h'(x)$. We differentiate our piecewise function using the chain rule: For $x > 0$, $$h'(x) = \frac{d}{dx}(\sin(x^2)) = 2x \cos(x^2)$$ For $x < 0$, $$h'(x) = \frac{d}{dx}(-\sin(x^2)) = -2x \cos(x^2)$$ Now, let us test the limit at $x=0$. The Left-Hand Derivative (LHD) is $\lim_{x o 0^-} (-2x \cos(x^2)) = 0$. The Right-Hand Derivative (RHD) is $\lim_{x o 0^+} (2x \cos(x^2)) = 0$. Since LHD = RHD = 0, the derivative $h'(0)$ exists and is equal to $0$. Furthermore, because the limit of $h'(x)$ as $x o 0$ is $0$, which equals $h'(0)$, the derivative is continuous. Statement-1 is undeniably true! ### Phase 3: The Second Derivative - The Hidden Trap Now, we test Statement-2. Is it twice differentiable? We need $h''(x)$. We differentiate $h'(x)$ again: For $x > 0$, $$h''(x) = \frac{d}{dx}(2x \cos(x^2)) = 2 \cos(x^2) - 4x^2 \sin(x^2)$$ For $x < 0$, $$h''(x) = \frac{d}{dx}(-2x \cos(x^2)) = -2 \cos(x^2) + 4x^2 \sin(x^2)$$ Here is where the magic—and the trap—lies. Let us find the limits of $h''(x)$ as $x$ approaches $0$: As $x o 0^+$, $h''(x) o 2 \cos(0) - 0 = 2$. As $x o 0^-$, $h''(x) o -2 \cos(0) + 0 = -2$. Look at that! The limit from the right is $2$, and the limit from the left is $-2$. There is a jump discontinuity in the second derivative at the origin. Because the limits do not match, the second derivative $h''(0)$ does not exist. Statement-2 is **false**. ### Conclusion This problem teaches us a vital lesson: differentiability is a hierarchy. A function can be continuous, then differentiable, but fail to be twice differentiable. Never assume smoothness just because the first derivative exists. Always check the limits of the derivative itself. You have navigated this complexity with precision. Keep this rigor in your toolkit, and no JEE problem will ever intimidate you again.

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Visualized Solution

Defining $f (x)$ and $g (x)$

Constructing $h (x) = g (f (x))$

Finding the First Derivative $h^{'} (x)$

Differentiability at $x = 0$

Continuity of $h^{'} (x)$

Finding the Second Derivative $h^{''} (x)$

Checking Existence of $h^{''} (0)$

Final Conclusion

The Sigma Insight: Relationship Between Continuity and Differentiability

The Anatomy of Smoothness

Phase 1

Phase 2

Phase 3

Conclusion

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Visualized Solution

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

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

Finding the First Derivative .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h′(x)

Differentiability at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=0

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

Finding the Second Derivative .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h′′(x)

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