Question

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

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

Answer 1

Answer: (A)

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