If the function g x = { k x + 1 , 0 ≤ x ≤ 3 m x + 2 , 3 < x ≤ 5 is differentiable, then the value of k + m is

LHD at x = 3 lim x ⟶ 3 - ⁡ k x + 1 - 2 k x - 3 = lim x → 3 - ⁡k x - 3 x - 3 . 1 x + 1 + 2 = k 4 RHD at x = 3 lim x → 3 + ⁡ m x + 2 - 2 k x - 3 Since this limit exists if 3 m + 2 - 2 k = 0 . . . i ( 0 0 form) So, lim x → 3 + ⁡ m x + 2 - 3 m + 2 x - 3 = m Hence m = k 4 … ii From i & ii m = 2 5 & k = 8 5 ⇒ k + m = 2

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MathematicsContinuity - DifferentiabilityDifferentiabilityEasy2 minPYQ_2015

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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

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Solution:

LHD at $x = 3$

$\lim_{x ⟶ 3^{-}} (\frac{k \sqrt{x + 1} - 2 k}{x - 3})$

$= \lim_{x \to 3^{-}} k (\frac{x - 3}{x - 3}) . \frac{1}{\sqrt{x + 1} + 2} = \frac{k}{4}$

RHD at $x = 3$

$\lim_{x \to 3^{+}} \frac{(m x + 2) - 2 k}{x - 3}$

Since this limit exists if $3 m + 2 - 2 k = 0 . . . (i)$ ( $\frac{0}{0}$ form)

So, $\lim_{x \to 3^{+}} \frac{(m x + 2) - (3 m + 2)}{x - 3} = m$

Hence $m = \frac{k}{4} \dots (ii)$

From $(i) & (ii)$

$m = \frac{2}{5} & k = \frac{8}{5} \Rightarrow k + m = 2$

Stream:JEESubject:MathematicsTopic:Continuity - DifferentiabilitySubtopic:Differentiability

⏱ 2mℹ️ Source: PYQ_2015

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