Question

Maximum value of $f ( x )=4 \sin ^3 x +9 \cos ^2 x +6 \sin x -3$ on the interval $[0, \pi]$ is equal to

• A. 6
• B. 7
• C. $\frac{29}{4}$
• D. $\frac{27}{4}$

Answer 1

Answer: (C)

We have $f^{\prime}(x)=\left(12 \sin ^2 x-18 \sin x+6\right) \cos x=0$
$\therefore$ Either $\cos x=0 \Rightarrow x=\frac{\pi}{2}$
or $6\left(2 \sin ^2 x-3 \sin x+1\right)=0$ or $(\sin x-1)(2 \sin x-1)=0$ or $\sin x=1$ or $\sin x=\frac{1}{2}$
$\Rightarrow x =\frac{\pi}{2}$ or $x =\frac{\pi}{6}$ or $x =\frac{5 \pi}{6}$
Now, $f (0)= f (\pi)=6$
$f\left(\frac{\pi}{2}\right)=4+6-3=7 $
$f\left(\frac{\pi}{6}\right)=\frac{4}{8}+9\left(\frac{3}{4}\right)+3-3=\frac{1}{2}+\frac{27}{4}=\frac{29}{4}$
$f\left(\frac{5 \pi}{6}\right)=4\left(\frac{1}{8}\right)+9\left(\frac{3}{4}\right)+3-3=\frac{29}{4} $
$ \therefore f_{\max }\left(x=\frac{\pi}{6} \text { or } \frac{5 \pi}{6}\right)=\frac{29}{4} $