JEE MAIN - Mathematics (2021 - 25th July Morning Shift - No. 1)
Let $$f(x) = 3{\sin ^4}x + 10{\sin ^3}x + 6{\sin ^2}x - 3$$, $$x \in \left[ { - {\pi \over 6},{\pi \over 2}} \right]$$. Then, f is :
increasing in $$\left( { - {\pi \over 6},{\pi \over 2}} \right)$$
decreasing in $$\left( {0,{\pi \over 2}} \right)$$
increasing in $$\left( { - {\pi \over 6},0} \right)$$
decreasing in $$\left( { - {\pi \over 6},0} \right)$$
Explanation
$$f(x) = 3{\sin ^4}x + 10{\sin ^3}x + 6{\sin ^2}x - 3,x \in \left[ { - {\pi \over 6},{\pi \over 2}} \right]$$
$$f'(x) = 12{\sin ^3}x\cos x + 30{\sin ^2}x\cos x + 12\sin x\cos x$$
$$ = 6\sin x\cos x(2{\sin ^2}x + 5\sin x + 2)$$
$$ = 6\sin x\cos x(2\sin x + 1)(\sin + 2)$$
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Decreasing in $$\left( { - {\pi \over 6},0} \right)$$
$$f'(x) = 12{\sin ^3}x\cos x + 30{\sin ^2}x\cos x + 12\sin x\cos x$$
$$ = 6\sin x\cos x(2{\sin ^2}x + 5\sin x + 2)$$
$$ = 6\sin x\cos x(2\sin x + 1)(\sin + 2)$$
Decreasing in $$\left( { - {\pi \over 6},0} \right)$$
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