Engineering
Mathematics
Introduction to Determinants
Question

f(x)=2cosx10xπ22cosx1012cosxf'(x)=

0

2

π2

π – 6

JEE Advance
College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

Solution
Verified BY
Verified by Zigyan

Given : f(x)=2cosx10xπ22cosx1012cosx

⇒ f(x) = 2cos⁡x(4cos2– 1) – 1(2xcos⁡– πcos⁡x)

f(x) = 8cos3– 2cos⁡– 2xcos⁡πcos⁡x

∴ '(x) = – 8 × 3cos2xsin⁡+ 2sin⁡– 2cos⁡+ 2xsin⁡π(– sin⁡x)

'(x) = – 24cos2xsin⁡+ 2sin⁡+ 2xsin⁡– 2cos⁡– πsin⁡x

fπ2=0+2+2π20π.

fπ2=2+ππ

fπ2=2

Lock Image

Please subscribe our Youtube channel to unlock this solution.