Let the function f: R rightarrow R be defined | vedclass

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(D) Given $f(x) = x - x^2 + (x - 1) \sin x = -(x - 1)^2 + (x - 1) \sin x = (x - 1) [-(x - 1) + \sin x]$.Note that $f(1) = 0$.Also,$f'(x) = 1 - 2x + \s

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$A, C$

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(D) Given $f(x) = x - x^2 + (x - 1) \sin x = -(x - 1)^2 + (x - 1) \sin x = (x - 1) [-(x - 1) + \sin x]$.Note that $f(1) = 0$.Also,$f'(x) = 1 - 2x + \sin x + (x - 1) \cos x$. Thus,$f'(1) = 1 - 2 + \sin 1 + 0 = \sin 1 - 1$.Let $h(x) = (fg)(x) = f(x)g(x)$.For differentiability at $x=1$,we check $h'(1) = \lim_{k 	o 0} \frac{f(1+k)g(1+k) - f(1)g(1)}{k} = \lim_{k 	o 0} \frac{f(1+k)g(1+k)}{k}$.Since $f(1+k) = k(-k + \sin(1+k))$,the limit becomes $\lim_{k 	o 0} \frac{k(-k + \sin(1+k))g(1+k)}{k} = \lim_{k 	o 0} (-k + \sin(1+k))g(1+k) = \sin(1) \cdot g(1)$.If $g$ is continuous at $x=1$,$g(1+k) 	o g(1)$,so the limit exists. Thus,$(A)$ is true.If $g$ is differentiable at $x=1$,it is continuous,so $(C)$ is true.If $fg$ is differentiable,it does not imply $g$ is continuous or differentiable (e.g.,if $f(x)=0$ at $x=1$,$fg$ could be differentiable even if $g$ is not). Thus,$(B)$ and $(D)$ are false.Therefore,the correct statements are $(A)$ and $(C)$.

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