JEE MAIN - Mathematics (2015 (Offline) - No. 1)
If the function.
$$g\left( x \right) = \left\{ {\matrix{ {k\sqrt {x + 1} ,} & {0 \le x \le 3} \cr {m\,x + 2,} & {3 < x \le 5} \cr } } \right.$$
is differentiable, then the value of $$k+m$$ is :
$$g\left( x \right) = \left\{ {\matrix{ {k\sqrt {x + 1} ,} & {0 \le x \le 3} \cr {m\,x + 2,} & {3 < x \le 5} \cr } } \right.$$
is differentiable, then the value of $$k+m$$ is :
$${{10} \over 3}$$
$$4$$
$$2$$
$${{16} \over 5}$$
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
$$m = {2 \over 5},\,\,k = {8 \over 5}$$
$$\therefore$$ $$k+m=2$$
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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