JEE-Mains 2015 (11/30)

Calculus Level 3

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

\frac{10}{3}

\frac{16}{5}

1 solution

Chew-Seong Cheong
Jun 14, 2015

If $g(x)$ is differentiable or continuous throughout, then:

$\begin{cases} \displaystyle \lim_{x \to 3^-} g(x) = \lim_{x \to 3^+} g(x) & \Rightarrow k\sqrt{3+1} = m(3) + 2 & \Rightarrow 2k = 3m + 2 \\ \displaystyle \lim_{x \to 3^-} g'(x) = \lim_{x \to 3^+} g'(x) & \Rightarrow \dfrac{k}{2\sqrt{3+1}} = m & \Rightarrow k = 4m \end{cases}$

$\Rightarrow 2(4m) = 3m + 2 \quad \Rightarrow m = \frac{2}{5} \quad \Rightarrow k = \frac{8}{5} \quad \Rightarrow k + m = \boxed{2}$

Moderator note:

Great! A differentiable function must be continuous. If you only tried evaluating the left and right limits, then you will still be missing an equation.

JEE-Mains 2015 (11/30)

1 solution

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