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Explain solution rd sharma class 12 chapter 16 Increasing and decreasing function exercise multiple choice question, question 33

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decreasing in \left ( \frac{\pi}{2},\pi \right )

Hints: Find the derivative of given equation.

Given: f(x)=4 \sin ^{3} x-6 \sin ^{2} x+12 \sin x+100

Solution: We have,

               f(x)=4 \sin ^{3} x-6 \sin ^{2} x+12 \sin x+100

               \begin{aligned} f^{\prime}(x) &=12 \sin ^{2} x \cos x-12 \sin x \cos x+12 \cos x \\ &=12 \cos x\left[\sin ^{2} x-\sin x+1\right] \\ &=12 \cos x\left[\sin ^{2} x+(1-\sin x)\right] \end{aligned}


Now 1-\sin x \geq 0 and \sin^{2} x \geq 0

               \sin ^{2} x+1-\sin x \geq 0

Hence, f'{x}>0 when \cos x>0 , i.e., x\epsilon \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )

So, f(x) is increasing when x\epsilon \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )

f'{x}<0 , when \cos x>0 , i.e., x\epsilon \left ( \frac{\pi}{2}, \frac{3\pi}{2} \right )

Hence, f(x) is decreasing when x\epsilon \left ( \frac{\pi}{2}, \frac{3\pi}{2} \right )

f(x) is decreasing in x\epsilon \left ( \frac{\pi}{2},\pi \right )

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