JEE Main 2021MathematicsApplication of DerivativesHardMCQ

JEE Main 2021 — Application of Derivatives Question with Solution

JEE Main 2021 (27 Aug Shift 2)

Question

Let M and m respectively be the maximum and minimum values of the function f(x)=tan-1(sinx+cosx) in 0,π2. Then the value of tan(M-m) is equal to:

Choose an option

Show full solutionCorrect option: B
Correct answer
B3-22

Step-by-step explanation

Given, the function, f(x)=tan-1(sinx+cosx) in 0,π2

f'x=cosx-sinx1+sinx+cosx2

=cosx-sinx1+sin2x

f'x=0

cosx-sinx=0

tanx=1

x=π4
Then, 1≤sinx+cosx≤2         x∈π4,tan-12
M=tan-1(2)

m=tan-1(1)

M-m=tan-12-tan-11

M-m=tan-12-11+2

M-m=tan-1(3-22)

tan(M-m)=3-22

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Application of Derivatives 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 Application of Derivatives 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.