If the function g ( x ) = * 20 c kx + 1 , ; ;0 x 3 mx + 2, ; ;3 <… - Vidyadip

Question

If the function $g\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{k\sqrt {x + 1} ,\;\;0 \le x \le 3}\\{mx + 2,\;\;3 < x \le 5}\end{array}} \right.$ is differentiable, then the value of $k + m$ is :

✓

Answer

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

$\therefore g\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
{k\sqrt {k + 1} \,\,\,,\,\,\,\,0 \le x \le 3}\\
{mx + 2\,\,\,\,\,,\,\,\,\,3 < x \le 5}
\end{array}} \right.$

At $x = 3,\,\,\,\,\,\,\,\,\,RHL = 3m + 2$

and at $x = 3,LHL = 2k$

$\therefore \,\,\,\,\,2k = 3m + 2\,\,\,\,.....\left( i \right)$

Also, $g'\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
{\frac{k}{{2\sqrt {x + 1} }}\,\,\,\,\,\,,\,\,\,\,\,\,\,\,\,\,\,0 \le x < 3}\\
{m\,\,\,\,\,\,\,\,\,\,,\,\,\,\,\,\,\,\,\,\,\,\,3 < x \le 5\,}
\end{array}} \right.$

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

$ \Rightarrow $ $\frac{k}{4} = m$ i.e. $k=4m$ ........$(ii)$

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

$k = \frac{8}{5},m = \frac{2}{5}$

$ \Rightarrow $ $k + m = 2$

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