QJEEM15Ma68e4: JEE Main 2015 Mathematics - Class 12 JEEM15Ma68e4 Solution: If the function. g ( x ) = { k x + 1 , 0 ≤ x ≤ 3 m x + 2 , 3 <…

Question

Verified Solution • 2025–26 Session moderate

If the function.

g (x) = {\begin{matrix} k \sqrt{x + 1}, & 0 \leq x \leq 3 \\ m x + 2, & 3 < x \leq 5 \end{matrix}

is differentiable, then the value of

k+m

is :

Accepted Answer

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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