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, $f(0)=f(\pi )=6$
$f\left(\frac{\pi }{2}\right)=4+6-3=7$
$f\left(\frac{\pi }{6}\right)=\frac{4}{8}+9\left(\frac{3}{4}\right)+3-3=\frac{1}{2}+\frac{27}{4}=\frac{29}{4}$
$f\left(\frac{5\pi }{6}\right)=4\left(\frac{1}{8}\right)+9\left(\frac{3}{4}\right)+3-3=\frac{29}{4}$
$\therefore f_{\max }\left(x=\frac{\pi }{6}\text{ or }\frac{5\pi }{6}\right)=\frac{29}{4}$