Question

Let $g(x)=f(x) \sin x$, where $f(x)$ is a twice differentiable function on $(-\infty, \infty)$ such that $f^{\prime}(-\pi)=1$. The value of $g^{\prime \prime}(-\pi)$ equals

• A. 1
• B. 2
• C. -2
• D. 0

Answer 1

Answer: (C)

We have $g ( x )= f ( x ) \sin x$ ....(1)
On differentiating equation (1) w.r.t. $x$, we get
$g ^{\prime}( x )= f ( x ) \cos x + f ^{\prime}( x ) \sin x$ ....(2)
Again differentiating equation (2) w.r.t. $x$, we get
$g^{\prime \prime}(x)=f(x)(-\sin x)+f^{\prime}(x) \cos x+f^{\prime}(x) \cos x+f^{\prime \prime}(x) \sin x $ ......(3)
$\Rightarrow g^{\prime \prime}(-\pi)=2 f^{\prime}(-\pi) \cos (-\pi)=2 \times 1 \times-1=-2$
$\text { Hence } g^{\prime \prime}(-\pi)=-2 $