JEE Main 2015 · 4 Apr · Shift 1 · Mathematics Q71

MathematicsSingle correct+4 / -1Question 71 of 90

If the function.
$g(x)=\begin{cases}k\sqrt{x+1},&0\le x\le3\\mx+2,&3<x\le5\end{cases}$
is differentiable, then the value of $k+m$ is :

\dfrac{16}5

\dfrac{10}3

Answer and worked solution

Official answer: A

Then $2k=\frac{3k}4+2$ , so $k=\frac85$ , $m=\frac25$ and $k+m=2$ .

JEE Main 2015 · 4 Apr · Shift 1, Mathematics question 71