Let k be a non - zero real number. If f(x) = \ array ll ( e ^ x - 1 ) ^ 2 ((x)/(k)) (1 + (x)/(4)) , x ≠ 0 array cases 12, & x = 0 cases ) is a continuous function at x = 0 , then the value of k is

Correct answer: C. 3 If the function is continuous then _ x → 0 f(x) = f(0) _ x → 0 ( e ^ x - 1 ) ^ 2 sin(x)/(k) . (1 + (x)/(4)) = 12 ⇒ _ x → 0 ( e ^ x - 1 x ) ^ 2 (sin(x)/(k))/(k . (x)/(k)) . ( (1 + (x)/(4)))/(4 . (x)/(4)) = 12 ⇒(1)/((1)/(k) . (1)/(4)) = 12 ⇒4k = 12 ⇒ k = 3

JEE Main 201511 AprilOnline

Exam: JEE Main
Year: 2015
Date: 11 April
Mode: Online
Subject: Mathematics
Chapter: Continuity and Differentiability

JEE Main 2015 · 11 April · Online · Mathematics · Continuity and Differentiability

Let k be a non - zero real number. If f(x) = \ array ll ( e ^ x - 1...

Let

k be a non - zero real number. If

f\left(x\right) = \left\{

\begin{array}{ll}\dfrac{{( {e}^{x} - 1 )}^{2}}{\sin \left(\dfrac{x}{k}\right)\log \left(1 + \dfrac{x}{4}\right)} , x \ne 0\end{array}

\begin{cases}12, & x = 0\end{cases}

\right) is a continuous function at

x = 0

, then the value of

k

Options

A $2$
B $4$
C $3$
D $1$

Correct answer

C. $3$

Step-by-step solution

If the function is continuous then

\lim_{x \to 0}f\left(x\right) = f\left(0\right)

\lim_{x \to 0}\dfrac{{( {e}^{x} - 1 )}^{2}}{sin\dfrac{x}{k} . \log \left(1 + \dfrac{x}{4}\right)} = 12

⇒\lim_{x \to 0}\dfrac{{\left(\dfrac{{e}^{x} - 1}{x}\right)}^{2}}{\dfrac{sin\dfrac{x}{k}}{k . \dfrac{x}{k}} . \dfrac{\log \left(1 + \dfrac{x}{4}\right)}{4 . \dfrac{x}{4}}}

= 12

⇒\dfrac{1}{\dfrac{1}{k} . \dfrac{1}{4}} = 12

⇒4k = 12

⇒

k = 3

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