Let f(x)=3 sin ^4 x+10 sin ^3 x+6 sin ^2 x-3
Question:

Let $f(x)=3 \sin ^{4} x+10 \sin ^{3} x+6 \sin ^{2} x-3$

$x \in\left[-\frac{\pi}{6}, \frac{\pi}{2}\right] .$ Then, $f$ is :

  1. increasing in $\left(-\frac{\pi}{6}, \frac{\pi}{2}\right)$

  2. decreasing in $\left(0, \frac{\pi}{2}\right)$

  3. increasing in $\left(-\frac{\pi}{6}, 0\right)$

  4. decreasing in $\left(-\frac{\pi}{6}, 0\right)$


Correct Option: , 4

Solution:

$f(x)=3 \sin ^{4} x+10 \sin ^{3} x+6 \sin ^{2} x-3, x \in\left[-\frac{\pi}{6}, \frac{\pi}{2}\right]$

$f^{\prime}(x)=12 \sin ^{3} x \cos x+30 \sin ^{2} x \cos x+12 \sin x \cos x$

$=6 \sin x \cos x\left(2 \sin ^{2} x+5 \sin x+2\right)$

$=6 \sin x \cos x(2 \sin x+1)(\sin +2)$

Decreasing in $\left(-\frac{\pi}{6}, 0\right)$

Administrator

Leave a comment

Please enter comment.
Please enter your name.