JEE Main 2024MathematicsDifferentiationHardMCQ

JEE Main 2024 — Differentiation Question with Solution

JEE Main 2024 (29 Jan Shift 1)

Question

Suppose fx=2x+2-xtanxtan-1x2-x+17x2+3x+13. Then the value of f'0 is equal to

Choose an option

Show full solutionCorrect option: C
Correct answer
Cπ

Step-by-step explanation

Given: fx=2x+2-xtanxtan-1x2-x+17x2+3x+13

We know that, f'0=limh→0fh-f0h

⇒f'0=limh→02h+2-htanhtan-1h2-h+17h2+3h+13-20+20tan0tan-10-0+10+0+13h

⇒f'0=limh→02h+2-htanhtan-1h2-h+1h7h2+3h+13

⇒f'0=limh→02h+2-htan-1h2-h+17h2+3h+13

⇒f'0=20+20tan-110+13

⇒f'0=2π4

⇒f'0=π

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About this question

This is a previous-year question from JEE Main 2024, 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.