Question

If the function.
$g(x) = \begin{cases} k\sqrt{x+1} , & \text{$0 \le x \le 3$} \\[2ex] mx+2, & \text{3 < x $\le$ 5} \end{cases}$
is differentiable, the value of $k + m$ is

• A. 2
• B. $\frac{16}{5}$
• C. $\frac{10}{3}$
• D. 4

Answer 1

Answer: (A)

$g(x)=\begin{cases}k \sqrt{x+1} & x \in[0,3] \\ m x+2 & x \in(3,5]\end{cases}$
$g(x)$ diff $\Rightarrow g(x)$ continuous
$\therefore g\left(3^{-}\right)=g\left(3^{+}\right)$
$\Rightarrow k \sqrt{4}=3 m+2$
$\Rightarrow 2 k=3 m+2 \ldots \ldots(1)$
Again
$g'\left(3^{+}\right)=g'\left(3^{-}\right)$
$\Rightarrow m =\left(\frac{ k }{2 \sqrt{ X +1}}\right)_{ x =3}=\frac{ k }{4}$
$\Rightarrow 4 m = k \ldots . .(2)$
from $(1) \&(2)$
$2 k =3 m +2 \Rightarrow 8 m =3 m +2$
$5 m =2$
$\& k =4 m =\frac{8}{5}$
$\Rightarrow k + m =\frac{10}{5}=2$