Discussion Forum

If  the function   $g(x)=\begin{cases}k\sqrt{x+1}, & 0 \leq x \leq3\\mx+2, &3<x \leq 5\end{cases}$ differentiable , then value of k+m is

Answer:

Explanation:

 Since , g(x) is differentiable  $\Rightarrow$ g(x)  must be continuous.

$\therefore$    $g(x)=\begin{cases}k\sqrt{x+1}, & 0 \leq x \leq3\\mx+2, &3<x \leq 5\end{cases}$

 At x=3,   RHL=3m+2

   and at x=3,   LHL=2k

   $\therefore$               2k=m+2     .........(i)

 Also, 

              $g'(x)=\begin{cases}\frac{k}{2\sqrt{k+1}}, & 0\leq x < 3\\m, & 3<x \leq5 \end{cases}$

   $\therefore$   $L\left\{g'(3)\right\}=\frac{k}{4}$   and   $R\left\{g'(3)\right\}=m$

   $\Rightarrow$  $\frac{k}{4}=m$    i.e, k=4m

 On , solving Eqs(i)   and (ii) , we get

   $k=\frac{8}{5},m=\frac{2}{5}$

  $\Rightarrow$             k+m=2