Question

Let $h\left(x\right)$ be differentiable for all $x$ and let $f\left(x\right)=(kx+e^x)h\left(x\right)$ where $k$ is some constant. If $h\left(0\right)=5,h^{\prime }\left(0\right)=-2$ and $f^{\prime }\left(0\right)=18$, then the value of $k$ is equal to

• A. $3$
• B. $4$
• C. $5$
• D. $1$

Answer 1

Answer: (A)

$3$

$f\left(x\right)=(kx+e^x).h\left(x\right)$
$h\left(0\right)=5,h^{\prime }\left(0\right)=-2,f^{\prime }\left(0\right)=18$
$f^{\prime }\left(x\right)=\frac{d}{dx}\left[(kx+e^x)h\right(x\left)\right]$
$\Rightarrow f^{\prime }\left(x\right)=(kx+e^x)\frac{d}{dx}h\left(x\right)+h\left(x\right)\frac{d}{dx}(kx+e^x)$
$\Rightarrow f^{\prime }\left(x\right)=(kx+e^x)h^{\prime }\left(x\right)+h\left(x\right)(k+e^x)$
put $x=0$, we get
$f^{\prime }\left(0\right)=(0+1)h^{\prime }\left(0\right)+h\left(0\right)(k+e^0)$
$f^{\prime }\left(0\right)=h^{\prime }\left(0\right)+h\left(0\right)(k+1)$
$18=-2+5(k+1)$
$20=5(k+1)$
$4=k+1$
$k=3$