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

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

Q55 / 90 · JEE Main 2015 (04 Apr) · Mathematics

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

A
$4$
Correct answer
B
$2$
C
$\frac{16}{5}$
D
$\frac{10}{3}$

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

Hence, the correct answer is option: 'B'.

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

Solution

Relevant Questions

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