If the function g (x) = cases k x + 1 , & 0 ≤ x ≤ 3 mx + 2, & 3 < x ≤ 5 cases is differentiable, then the value of k + m is

Correct answer: B. 2 LHD at x = 3 _ x ⟶ 3 ^ - ( k x + 1 - 2k x - 3 ) = _ x → 3 ^ - k ((x - 3)/(x - 3)) . 1 x + 1 + 2 = (k)/(4) RHD at x = 3 _ x → 3 ^ + ((mx + 2) - 2k)/(x - 3) Since this limit exists if 3m + 2 - 2k = 0 . . . (i) ( (0)/(0) form) So, _ x → 3 ^ + ((mx + 2) - (3m + 2))/(x - 3) = m Hence m = (k)/(4) (ii) From (i) and (ii) m = (2)/(5) and k = (8)/(5) ⇒ k + m = 2

JEE Main 20154 April

Exam: JEE Main
Year: 2015
Date: 4 April
Subject: Mathematics
Chapter: Continuity and Differentiability

JEE Main 2015 · 4 April · Mathematics · Continuity and Differentiability

If the function g (x) = cases k x + 1 , & 0 ≤ x ≤ 3 mx + 2, & 3 < x...

If the function

g \left(x\right) = {

\begin{cases}k\sqrt{x + 1}\text{,} & 0 \le x \le 3 \\ mx + 2\text{,} & 3 < x \le 5\end{cases}$

is differentiable, then the value of

k + m$ is

Options

A $4$
B $2$
C $\dfrac{16}{5}$
D $\dfrac{10}{3}$

Correct answer

B. $2$

Step-by-step solution

LHD at

x = 3

\lim_{x ⟶ {3}^{-}} \left(\dfrac{k\sqrt{x + 1} - 2k}{x - 3}\right)

= \lim_{x \to {3}^{-}} k \left(\dfrac{x - 3}{x - 3}\right) . \dfrac{1}{\sqrt{x + 1} + 2} = \dfrac{k}{4}

RHD at

x = 3

\lim_{x \to {3}^{+}} \dfrac{\left(mx + 2\right) - 2k}{x - 3}

Since this limit exists if

3m + 2 - 2k = 0 . . . \left(i\right)

(

\dfrac{0}{0}

form)

So,

\lim_{x \to {3}^{+}} \dfrac{\left(mx + 2\right) - \left(3m + 2\right)}{x - 3} = m

Hence

m = \dfrac{k}{4} \ldots \left(ii\right)

From \left(i\right) \text{ and } \left(ii\right)

<br /><br />

m = \dfrac{2}{5} \text{ and } k = \dfrac{8}{5} ⇒ k + m = 2

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