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

NTA Abhyas 2020 · Mathematics · Differentiation

If y = cos xcos 2xcos 4xcos 8x , then (dy)/(dx) at x = (π)/(2) is

If y=cosxcos2xcos4xcos8xy = cos xcos 2xcos 4xcos 8x, then dydx\dfrac{dy}{dx} at x=π2x = \dfrac{\pi}{2} is

Options

  1. A11
  2. B00
  3. C44
  4. D1616

Correct answer

A. 11

Step-by-step solution

y = \dfrac{\sin 16x}{16sin x}(∵cosAcos2A.........cos2n−1A=sin⁡(2nA)2nsinA)\left(∵ cos Acos 2A . . . . . . . . . cos {2}^{n - 1}A = \dfrac{\sin \left({2}^{n}A\right)}{{2}^{n}sin A}\right)
So, dydx=116sinx⋅16cos⁡16x−sin⁡16xcosx(sinx)2\dfrac{dy}{dx} = \dfrac{1}{16}\dfrac{sin x ⋅ 16\cos 16x - \sin 16xcos x}{{\left(sin x\right)}^{2}}
Putting x=π2x = \dfrac{\pi}{2}, we get,
dydx=116⋅{1.16(1)−0(1)2}=1\dfrac{dy}{dx} = \dfrac{1}{16} ⋅ \left\{\dfrac{1.16\left(1\right) - 0}{{\left(1\right)}^{2}}\right\} = 1
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