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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Differentiation

NTA Abhyas 2020 · Mathematics · Differentiation

If f(x) = cos x. 2x. 4x. 8x. 16x , then f^ ' ((π)/(4)) is ( wheref^...

If f(x)=cosx.cos⁡2x.cos⁡4x.cos⁡8x.cos⁡16xf\left(x\right) = cos x. \cos 2x. \cos 4x. \cos 8x. \cos 16x, then f′(π4)f^{'} \left(\dfrac{\pi}{4}\right) is (wheref′(x)=ddx( wheref^{'}( x ) = \dfrac{d}{dx} f( x ) )$$

Options

  1. A2\sqrt{2}
  2. B12\dfrac{1}{\sqrt{2}}
  3. C11
  4. DNone of these

Correct answer

A. 2\sqrt{2}

Step-by-step solution

f(x)=2sinx.cosx.cos⁡2x.cos⁡4x.cos⁡8x.cos⁡16x2sinxf\left(x\right) = \dfrac{2sin x. cos x. \cos 2x. \cos 4x. \cos 8x. \cos 16x}{2sin x}

=sin⁡32x25sinx= \dfrac{\sin 32x}{{2}^{5}sin x}

∴f′(x)=132.32cos⁡32x.sinx−cosx.sin⁡32xsin⁡2x∴ f^{'}\left(x\right) = \dfrac{1}{32}. \dfrac{32\cos 32x. sin x - cos x. \sin 32x}{\sin^{2} x}

∴f′(π4)=32.12−12.032.(12)2=2∴ f^{'}\left(\dfrac{\pi}{4}\right) = \dfrac{32. \dfrac{1}{2} - \dfrac{1}{2}. 0}{32. \left(\dfrac{1}{2}\right)^{2}} = \sqrt{2}
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