JEE Main 2015MathematicsContinuity and DifferentiabilityMediumMCQ

JEE Main 2015 — Continuity and Differentiability Question with Solution

JEE Main 2015 (11 Apr Online)

Question

Let k be a non - zero real number. If fx=(ex−1)2sin xklog 1+x4,x ≠ 012                                       ,x = 0  is a continuous function at x=0, then the value of k is

Choose an option

Show full solutionCorrect option: C
Correct answer
C3

Step-by-step explanation

If the function is continuous then

limx→0fx=f0

limx→0(ex−1)2sinxk.log1+x4=12

⇒  limx→0ex-1x2sinxkk .  xk .  log1+x44.  x4   = 12

⇒ 11k .14=12

⇒ 4k=12

⇒ k=3

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Continuity and Differentiability chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

About this question

This is a previous-year question from JEE Main 2015, covering the Continuity and Differentiability chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.