JEE Main 2016MathematicsContinuity and DifferentiabilityEasyMCQ

JEE Main 2016 — Continuity and Differentiability Question with Solution

JEE Main 2016 (09 Apr Online)

Question

If the function fx= -x,x<1a+cos-1⁡x+b,1 ≤x ≤2 is differentiable at x=1, then ab is equal to 

Choose an option

Show full solutionCorrect option: A
Correct answer
Aπ+22

Step-by-step explanation

fx= -xx<1a+cos-1⁡x+b1 ≤x ≤2

fx is continuous at x=1 as fx is differentiable at x=1.

⇒ lim x→1 f( x )= lim x→ 1 + ( a+ cos −1 ( x+b ) )=f( 1 )

⇒ -1=a+cos-1⁡1+b

cos-1⁡1+b=-1-a ........i

fx is differentiable at x=1.

⇒ LHD=RHD

⇒ -1=-11-1+b2

⇒ 1-1+b2=1

⇒ b= -1 ......ii

From i ⇒ cos-1⁡(0)= -1-a

∴ -1-a=π2

a= -1-π2

a=-π-22 ......iii

∴ ab=π+22

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About this question

This is a previous-year question from JEE Main 2016, 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.