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Step-by-Step Solution
Step 1: Understand the functions f(x) and g(x)
We are given:
1. $f(x) = x \lvert x \rvert$
2. $g(x) = \sin x$
We want to analyze the composited function $g\bigl(f(x)\bigr)$ at $x = 0$, in terms of its differentiability and twice-differentiability.
Step 2: Express the composition $g \circ f$
The composition is
$$
(g \circ f)(x) = g\bigl(f(x)\bigr) = g\bigl(x \lvert x \rvert \bigr) = \sin\bigl(x \lvert x \rvert \bigr).
$$
Since $x \lvert x \rvert$ can be written as
\[
x \lvert x \rvert =
\begin{cases}
-x^2, & x < 0,\\
x^2, & x \ge 0,
\end{cases}
\]
it follows that
\[
(g \circ f)(x) =
\begin{cases}
\sin(-x^2) = -\sin(x^2), & x < 0,\\
\sin(x^2), & x \ge 0.
\end{cases}
\]
Step 3: Verify differentiability at x = 0
Let us compute the first derivative of $(g\circ f)(x)$. For $x \neq 0$:
\[
\frac{d}{dx}\Bigl[(g \circ f)(x)\Bigr]
=
\begin{cases}
\frac{d}{dx}\bigl[-\sin(x^2)\bigr] = -2x \cos(x^2), & x < 0, \\
\frac{d}{dx}\bigl[\sin(x^2)\bigr] = 2x \cos(x^2), & x \ge 0.
\end{cases}
\]
Now, to check differentiability at $x=0$, we need the left-hand derivative (LHD) and right-hand derivative (RHD) at 0:
LHD as $x \to 0^{-}$:
$$
\lim_{x \to 0^-} -2x \cos(x^2) = -2 \cdot 0 \cdot \cos(0) = 0.
$$
RHD as $x \to 0^{+}$:
$$
\lim_{x \to 0^+} 2x \cos(x^2) = 2 \cdot 0 \cdot \cos(0) = 0.
$$
Since the left-hand and right-hand derivatives both equal 0 at $x = 0$, the function is differentiable at $x=0$. Also, the resulting derivative expression on each side tends to 0 continuously at $x=0$, so the first derivative is continuous at $x=0$.
Step 4: Check twice differentiability at x = 0
Next, we find the second derivative for $x \neq 0$:
\[
\frac{d^2}{dx^2}\Bigl[(g \circ f)(x)\Bigr]
=
\begin{cases}
\frac{d}{dx}\bigl[-2x \cos(x^2)\bigr] = -2\cos(x^2) + 4x^2 \sin(x^2), & x < 0, \\
\frac{d}{dx}\bigl[2x \cos(x^2)\bigr] = 2\cos(x^2) - 4x^2 \sin(x^2), & x \ge 0.
\end{cases}
\]
We now check the left-hand limit and the right-hand limit of the second derivative at $x=0$:
Left-hand second derivative (as $x \to 0^{-}$):
$$
\lim_{x \to 0^-}\Bigl[-2 \cos(x^2) + 4x^2 \sin(x^2)\Bigr]
= -2 \cos(0) + 4 \cdot 0 \cdot \sin(0) = -2.
$$
Right-hand second derivative (as $x \to 0^{+}$):
$$
\lim_{x \to 0^+}\Bigl[2 \cos(x^2) - 4x^2 \sin(x^2)\Bigr]
= 2 \cos(0) - 4 \cdot 0 \cdot \sin(0) = 2.
$$
Because these two limits are not equal ($-2 \neq 2$), the second derivative does not exist at $x=0$. Thus the function is not twice differentiable at $x=0$.
Step 5: Match with the given statements
Statement-1: "$g \circ f$ is differentiable at $x=0$ and its derivative is continuous at that point."
We found that $(g \circ f)$ is indeed differentiable at $x=0$ and the first derivative is continuous there. So Statement-1 is true.
Statement-2: "$g \circ f$ is twice differentiable at $x=0$."
We found the left-hand and right-hand limits of the second derivative at $x=0$ are different. Hence it is not twice differentiable at $x=0$. So Statement-2 is false.
Conclusion
Therefore, the correct choice is:
Statement-1 is true, Statement-2 is false.
