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Question: Answered & Verified by Expert
If fx=cosx.cos2x.cos4x .cos8x .cos16x, then f'π4 is (where  f'(x)=ddxf(x))
MathematicsDifferentiationJEE Main
Options:
  • A 2
  • B 12
  • C 1
  • D None of these
Solution:
1755 Upvotes Verified Answer
The correct answer is: 2
fx=2sinx.cosx.cos2x.cos4x.cos8x.cos16x2sinx

=sin32x25sinx

f'x=132.32cos32x.sinx-cosx.sin32xsin2x

f'π4=32.12-12.032.122=2

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