Key Concepts and Formulas

• Definition of Differentiability: A function $h(x)$ is differentiable at a point $x=a$ if the limit $\lim_{k \to 0} \frac{h(a+k)-h(a)}{k}$ exists. This limit is the derivative $h'(a)$.
• Continuity of a Derivative: A derivative $h'(x)$ is continuous at $x=a$ if $\lim_{x \to a} h'(x) = h'(a)$.
• Differentiability of Composite Functions: If $f$ is differentiable at $x=a$ and $g$ is differentiable at $f(a)$, then the composite function $g \circ f$ is differentiable at $x=a$, and $(g \circ f)'(a) = g'(f(a)) \cdot f'(a)$.
• Absolute Value Function: $|x|$ is not differentiable at $x=0$.
• Piecewise Functions and Differentiability: For a piecewise function to be differentiable at the point where the definition changes, the left-hand derivative and the right-hand derivative must be equal.

Step-by-Step Solution

Step 1: Define the composite function $g \circ f(x)$. We are given $f(x) = x|x|$ and $g(x) = \sin x$. The composite function $(g \circ f)(x)$ is defined as $g(f(x))$. Substituting $f(x)$ into $g(x)$: $(g \circ f)(x) = g(x|x|) = \sin(x|x|)$.

Step 2: Express the composite function as a piecewise function. The absolute value function $|x|$ is defined as: $|x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}$

Using this definition, we can write $x|x|$ as: $x|x| = \begin{cases} x \cdot x = x^2 & \text{if } x \ge 0 \\ x \cdot (-x) = -x^2 & \text{if } x < 0 \end{cases}$

Now, we can express $(g \circ f)(x) = \sin(x|x|)$ as a piecewise function: $(g \circ f)(x) = \begin{cases} \sin(-x^2) & \text{if } x < 0 \\ \sin(x^2) & \text{if } x \ge 0 \end{cases}$ Since $\sin(-u) = -\sin(u)$, we have: $(g \circ f)(x) = \begin{cases} -\sin(x^2) & \text{if } x < 0 \\ \sin(x^2) & \text{if } x \ge 0 \end{cases}$

Step 3: Check for differentiability of $g \circ f(x)$ at $x=0$. To check for differentiability at $x=0$, we need to find the left-hand derivative and the right-hand derivative at $x=0$. The derivative of $(g \circ f)(x)$ is found by differentiating each piece: For $x < 0$, $\frac{d}{dx}(-\sin(x^2)) = -\cos(x^2) \cdot (2x) = -2x\cos(x^2)$. For $x > 0$, $\frac{d}{dx}(\sin(x^2)) = \cos(x^2) \cdot (2x) = 2x\cos(x^2)$.

Now, let's evaluate the limits of these derivatives as $x$ approaches 0 from the left and right. Left-hand derivative at $x=0$: $L(g \circ f)'(0) = \lim_{x \to 0^-} (-2x\cos(x^2))$ As $x \to 0^-$, $-2x \to 0$ and $\cos(x^2) \to \cos(0) = 1$. So, $L(g \circ f)'(0) = 0 \cdot 1 = 0$.

Right-hand derivative at $x=0$: $R(g \circ f)'(0) = \lim_{x \to 0^+} (2x\cos(x^2))$ As $x \to 0^+$, $2x \to 0$ and $\cos(x^2) \to \cos(0) = 1$. So, $R(g \circ f)'(0) = 0 \cdot 1 = 0$.

Since $L(g \circ f)'(0) = R(g \circ f)'(0) = 0$, the function $(g \circ f)(x)$ is differentiable at $x=0$, and its derivative at $x=0$ is 0. This means Statement-1 is true regarding differentiability at $x=0$.

Step 4: Check for continuity of the derivative of $g \circ f(x)$ at $x=0$. The derivative of $(g \circ f)(x)$ can be written as: $(g \circ f)'(x) = \begin{cases} -2x\cos(x^2) & \text{if } x < 0 \\ 2x\cos(x^2) & \text{if } x \ge 0 \end{cases}$

For the derivative to be continuous at $x=0$, the limit of the derivative as $x$ approaches 0 must equal the value of the derivative at $x=0$. We found $(g \circ f)'(0) = 0$.

Let's check the limit of $(g \circ f)'(x)$ as $x \to 0$: $\lim_{x \to 0} (g \circ f)'(x) = \lim_{x \to 0} \begin{cases} -2x\cos(x^2) & \text{if } x < 0 \\ 2x\cos(x^2) & \text{if } x \ge 0 \end{cases}$

Left-hand limit: $\lim_{x \to 0^-} (-2x\cos(x^2)) = 0$. Right-hand limit: $\lim_{x \to 0^+} (2x\cos(x^2)) = 0$.

Since the left-hand limit and the right-hand limit are equal to 0, and $(g \circ f)'(0) = 0$, the derivative $(g \circ f)'(x)$ is continuous at $x=0$. Therefore, Statement-1 is true.

Step 5: Check for twice differentiability of $g \circ f(x)$ at $x=0$. To check for twice differentiability, we need to find the second derivative $(g \circ f)''(x)$ and see if it is defined at $x=0$. We differentiate $(g \circ f)'(x)$:

For $x < 0$, $(g \circ f)''(x) = \frac{d}{dx}(-2x\cos(x^2))$. Using the product rule and chain rule: $\frac{d}{dx}(-2x\cos(x^2)) = (-2)\cos(x^2) + (-2x)(-\sin(x^2) \cdot 2x)$ $= -2\cos(x^2) + 4x^2\sin(x^2)$.

For $x > 0$, $(g \circ f)''(x) = \frac{d}{dx}(2x\cos(x^2))$. Using the product rule and chain rule: $\frac{d}{dx}(2x\cos(x^2)) = (2)\cos(x^2) + (2x)(-\sin(x^2) \cdot 2x)$ $= 2\cos(x^2) - 4x^2\sin(x^2)$.

Now, we need to check if the left-hand second derivative and the right-hand second derivative at $x=0$ are equal. Left-hand second derivative at $x=0$: $L(g \circ f)''(0) = \lim_{x \to 0^-} (-2\cos(x^2) + 4x^2\sin(x^2))$ As $x \to 0^-$, $\cos(x^2) \to \cos(0) = 1$, and $4x^2\sin(x^2) \to 4(0)\sin(0) = 0$. So, $L(g \circ f)''(0) = -2(1) + 0 = -2$.

Right-hand second derivative at $x=0$: $R(g \circ f)''(0) = \lim_{x \to 0^+} (2\cos(x^2) - 4x^2\sin(x^2))$ As $x \to 0^+$, $\cos(x^2) \to \cos(0) = 1$, and $4x^2\sin(x^2) \to 4(0)\sin(0) = 0$. So, $R(g \circ f)''(0) = 2(1) - 0 = 2$.

Since $L(g \circ f)''(0) = -2$ and $R(g \circ f)''(0) = 2$, the left-hand second derivative is not equal to the right-hand second derivative at $x=0$. Therefore, $(g \circ f)(x)$ is not twice differentiable at $x=0$. This means Statement-2 is false.

Step 6: Evaluate the relationship between Statement-1 and Statement-2. Statement-1: $g \circ f$ is differentiable at $x=0$ and its derivative is continuous at $x=0$. We found this to be true. Statement-2: $g \circ f$ is twice differentiable at $x=0$. We found this to be false.

Since Statement-1 is true and Statement-2 is false, Statement-2 cannot be a correct explanation for Statement-1.

Common Mistakes & Tips

• Differentiability of $|x|$: Remember that $|x|$ is not differentiable at $x=0$. This is why we need to use piecewise definitions and check left/right derivatives for composite functions involving $|x|$.
• Chain Rule Application: Be careful when applying the chain rule, especially when differentiating terms like $\sin(x^2)$ or $\cos(x^2)$. The derivative of $x^2$ is $2x$.
• Continuity of Derivative: To check if the derivative is continuous, you must evaluate the limit of the derivative function as $x$ approaches the point and compare it to the derivative's value at that point.

Summary

We analyzed the composite function $(g \circ f)(x) = \sin(x|x|)$ by expressing it as a piecewise function. We then proceeded to check the differentiability of this function at $x=0$ by calculating the left-hand and right-hand derivatives. We found that the function is differentiable at $x=0$ and its derivative is also continuous at $x=0$, confirming Statement-1. Subsequently, we calculated the second derivative and found that the left-hand and right-hand second derivatives at $x=0$ are not equal, indicating that the function is not twice differentiable at $x=0$, thus disproving Statement-2. Since Statement-1 is true and Statement-2 is false, Statement-2 cannot be a correct explanation for Statement-1.

The final answer is \boxed{A}.