If $f(x) = x^3 - x^2 + (x-1)\sin x$, then which of the following statements is/are true?

Question

$f(x) = x^3 - x^2 + (x-1)\sin x$

$fg(x) = f(x) g(x)$

Which of the following is true?

A) If g is continuous at x = 1, then fg is differentiable at x = 1

B) If fg is differentiable at x = 1, then g is continuous at x = 1

C) If g is differentiable at x = 1, then fg is differentiable at x = 1

D) If fg is differentiable at x = 1, then g is differentiable at x = 1.

The way I solved :

(fg)'(x) = f '(x)g(x) + f(x)g'(x)

(fg)'(1) = f '(1)g(1) + f(1)g'(1)

(fg)'(1) = (1 + sin1) (g(1))

so for (fg)'(x) to be continuous at x=1, g(x) should be continuous at x=1 and vice versa.

so for (fg)(x) to be differentiable at x=1, g(x) should be continuous at x=1 and vice versa.

This way I get option A,B,C but the answer given is only A,C.

Please help me understand why the reverse is not true?

Answer 1

You have $$f(x) = x^3 - x^2 + (x-1)\sin x = (x-1)(x^2+\sin x)$$ therefore the inequality $$\lvert f(x) g(x) \vert \le 5\lvert x- 1 \rvert \lvert g(x) \rvert $$ for $\lvert x-1 \rvert \le 1$.

Now if $\lim\limits_{x \to 1} g(x) =0$, the inequality above proves that $fg$ is differentiable at $1$ with the derivative vanishing at that point. This whatever the value of $g(1)$ is. For example, if $g$ is equal to $\lvert x-1 \rvert$ for $x \neq 1$ and $g(1) = 1$, $fg$ is differentiable at $1$. However, $g$ is not continuous at $1$, proving that B doesn't hold.