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Question

The function f (x)=4 sin3x6 sin2x+12 sinx+100 is strictly

A
decreasing in (π2,3π2)
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B
decreasing in (π2,π)
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C
increasing in (π2,π2)
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D
increasing in (π2,3π2)
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Solution

The correct option is C increasing in (π2,π2)
We have, f (x)=4 sin3 x6 sin2 x+12 sin x+100f'(x)=12 sin2 x cos x12 sin x cos x +12 cos xf'(x)=12 cos x (sin2 xsin x+1)f'(x)=12 cos x (sin2 x+(1sin x)]Now, sin2 x0 and 1sin x0sin2 x+(1sin x)0Sign of f'(x) depends upon cos x.cos x is positive when x(π2,π2) and negative when x (π2,π) or x (π2,3π2).f(x) is increasing function when x(π2,π2) and decreasing function when x (π2,π) or x (π2,3π2).

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