Question:
Write the maximum value of 12 sin x − 9 sin2 x.
Solution:
Let $f(x)=12 \sin x-9 \sin ^{2} x$
$=-\left(9 \sin ^{2} x-12 \sin x\right)$
$=-\left[(3 \sin x)^{2}-2.3 \sin x .2+2^{2}-4\right]$
$=-\left[(3 \sin x-2)^{2}-4\right]$
$=4-(3 \sin x-2)^{2}$
Minimum value of $(3 \sin x-2)^{2}$ is 0 .
Therefore, maximum value of $4-(3 \sin x-2)^{2}$ would be 4 .
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.