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

JEE · Math · previous-year question

  1. A.$${{10} \over 3}$$
  2. B.$$4$$
  3. C.$$2$$correct
  4. D.$${{16} \over 5}$$

Answer

C. $$2$$

Explanation

Since $$g(x)$$ is differentiable, - it will be continuous at $$x=3$$ $$\therefore$$ $$\mathop {\lim }\limits_{x \to {3^ - }} g\left( x \right) = \mathop {\lim }\limits_{x \to {3^ + }} g\left( x \right)$$ $$2k = 3m + 2\,\,\,\,\,...\left( 1 \right)$$ Also $$g(x)$$ is differentiable at $$x=0$$ $$\therefore$$ $$\mathop {\lim }\limits_{x \to {3^ - }} g'\left( x \right) = \mathop {\lim }\limits_{x \to {3^ + }} g'\left( x \right)$$ $${K \over {2\sqrt {3 + 1} }} = m$$ $$k=4m$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$ Solving $$(1)$$ and $$(2)$$, we get $$m = {2 \over 5},\,\,k = {8 \over 5}$$ $$\therefore$$ $$k+m=2$$

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