Differentiability of Product Function with Given Quadratic and Sine Term

To determine which of the statements are TRUE, let us analyze the differentiability of the product function $h(x) = (fg)(x) = f(x)g(x)$ at $x = 1$.

The function $f: \mathbb{R} \to \mathbb{R}$ is given by: $f(x) = x^3 - x^2 + (x - 1)\sin x$

We can factor out $(x - 1)$ from the first two terms: $f(x) = x^2(x - 1) + (x - 1)\sin x = (x - 1)(x^2 + \sin x)$

Evaluating $f(x)$ at $x = 1$: $f(1) = (1 - 1)(1^2 + \sin 1) = 0$

For any function $g: \mathbb{R} \to \mathbb{R}$, the value of the product function $h(x) = (fg)(x)$ at $x = 1$ is: $h(1) = f(1)g(1) = 0 \cdot g(1) = 0$

By definition, $h(x)$ is differentiable at $x = 1$ if and only if the following limit exists and is finite: $h'(1) = \lim_{x \to 1} \frac{h(x) - h(1)}{x - 1} = \lim_{x \to 1} \frac{f(x)g(x) - 0}{x - 1}$

Substituting $f(x) = (x - 1)(x^2 + \sin x)$: $h'(1) = \lim_{x \to 1} \frac{(x - 1)(x^2 + \sin x)g(x)}{x - 1} = \lim_{x \to 1} (x^2 + \sin x)g(x)$

Since $\lim_{x \to 1} (x^2 + \sin x) = 1 + \sin 1 \neq 0$, the limit $h'(1)$ exists if and only if $\lim_{x \to 1} g(x)$ exists and is finite.

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Analysis of the Options:

Statement (A): Provided $g$ is continuous at $x = 1$, the function $fg$ is differentiable at $x = 1$ If $g$ is continuous at $x = 1$, then: $\lim_{x \to 1} g(x) = g(1)$ Thus, the limit for $h'(1)$ evaluates to: $h'(1) = \lim_{x \to 1} (x^2 + \sin x)g(x) = (1 + \sin 1)g(1)$ Since $g(1)$ is a finite real number, $h'(1)$ exists and is finite. Therefore, $fg$ is differentiable at $x = 1$. Option A is TRUE.

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Statement (C): If $g$ is differentiable at $x = 1$, then $fg$ is also differentiable at $x = 1$ If $g$ is differentiable at $x = 1$, it implies that $g$ is continuous at $x = 1$. By statement (A), continuity of $g$ at $x = 1$ guarantees that $fg$ is differentiable at $x = 1$. Option C is TRUE.

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Statement (B): If $fg$ is differentiable at $x = 1$, then $g$ must be continuous at $x = 1$ Consider the function $g: \mathbb{R} \to \mathbb{R}$ defined by:

$$g(x) = \begin{cases} 1, & x \neq 1 \\ 2, & x = 1 \end{cases}$$

• Here, $\lim_{x \to 1} g(x) = 1 \neq g(1) = 2$, so $g$ is not continuous at $x = 1$.
• However, $h(1) = f(1)g(1) = 0 \cdot 2 = 0$, and: $h'(1) = \lim_{x \to 1} (x^2 + \sin x)g(x) = (1 + \sin 1) \cdot 1 = 1 + \sin 1$ Since $h'(1)$ exists, $fg$ is differentiable at $x = 1$. Thus, differentiability of $fg$ at $x = 1$ does not imply continuity of $g$ at $x = 1$. Option B is FALSE.

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Statement (D): If $fg$ is differentiable at $x = 1$, then $g$ is necessarily differentiable at $x = 1$ Using the counterexample $g(x) = |x - 1|$:

• $g(x)$ is continuous at $x = 1$, so by Statement (A), $fg$ is differentiable at $x = 1$.
• However, $g(x) = |x - 1|$ is not differentiable at $x = 1$. Thus, differentiability of $fg$ at $x = 1$ does not require $g$ to be differentiable at $x = 1$. Option D is FALSE.

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Conclusion:

The correct options are A and C.