Question
Mathematics Question on Trigonometric Functions
The value of loge2dxd(logcosxcosecx) at x=4π is
A
−22
B
22
C
−4
D
4
Answer
4
Explanation
Solution
The correct answer is (D) : 4
Let
f(x)=logcosxcosecx
=logcosxlogcosecx/
f′(x)=(logcosx)2logcosx.sinx.(−cosecxcotx−logcosecx.cosx1−sinx)
at x =4π
f′(4π)=(log21)2−log(21)+log2
=log22
∴loge2f′(x) at x=4π=4