JEE Main 2015MathematicsContinuity and DifferentiabilityEasyMCQ

JEE Main 2015 — Continuity and Differentiability Question with Solution

JEE Main 2015 (04 Apr)

Question

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

Choose an option

Show full solutionCorrect option: B
Correct answer
B2

Step-by-step explanation

LHD at x=3

limx⟶3-⁡kx+1-2kx-3

= limx→3-⁡kx-3x-3.1x+1+2=k4

RHD at x=3

limx→3+⁡mx+2-2kx-3

Since this limit exists if 3m+2-2k=0 ...i (00 form)

So,limx→3+⁡mx+2-3m+2x-3=m

Hence m=k4 …ii

From i & ii

m=25 & k=85  ⇒k+m=2 

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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.