JEE Main 2021MathematicsDifferentiationHardNumerical

JEE Main 2021 — Differentiation Question with Solution

JEE Main 2021 (17 Mar Shift 1)

Question

If fx=sincos-11-22x1+22x and its first derivative with respect to x is -baloge2 when  x=1, where a and b are integers, then the minimum value of a2-b2 is _______.

Enter your answer

Show full solutionCorrect answer: 481
Correct answer
481

Step-by-step explanation

Given fx=sincos-11-22x1+22x at x=1;22x=2x2

For sincos-11-2x21+2x2

Let tan-12x=θ;θ∈-π2,π2

∴sincos-1cos2θ=sin2θ

 If x>1⇒π2>θ>π4∴ π>2θ>π2

=2sinθcosθ=2tanθ1+tan2θ

Hence, fx=2·2x1+22x

∴f'x=1+22x2.2xln2-22x·2·ln2·2·2x1+22x

∴f'1=20ln2-32ln225=-1225ln2

So, a=25, b=12.

Then, a2-b2=252-122

=625-144

=481

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Differentiation 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 2021, covering the Differentiation 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.