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

The correct option is B ($3$)

Given $f\left(x\right)=(kx+e^x)h\left(x\right)$ and also $h\left(0\right)=5,h^{\prime }\left(0\right)=-2$ and $f^{\prime }\left(0\right)=18$

Differentiating with respect to x, we get

$f^{\prime }\left(x\right)=(kx+e^x)h^{\prime }\left(x\right)+h\left(x\right)(k+e^x)$
Put $x=0$, we get
$\Rightarrow f^{\prime }\left(0\right)=(0+e^0)h^{\prime }\left(0\right)+h\left(0\right)(k+e^0)$
So, $f^{\prime }\left(0\right)=(0+1)h^{\prime }\left(0\right)+h\left(0\right)(k+1)$
$\Rightarrow 18=-2+5(k+1)$
$\Rightarrow 18+2=5k+5$
$\Rightarrow 5k+5=20$
$\Rightarrow 5k=15$
$\therefore k=3$