Question

The difference between the maximum and minimum values of the function $f\left(x\right)=\left(sin\right)^{3} x-3sin⁡x, \, \forall x\in \left[0 , \frac{\pi }{6}\right]$ is

• A. $2$
• B. $\frac{1}{2}$
• C. $\frac{11}{8}$
• D. $\frac{7}{6}$

Answer 1

Answer: (C)

Let, $sin x=t$
$\therefore Ifx\in \left[0 , \frac{\pi }{6}\right]\Rightarrow t\in \left[0 , \frac{1}{2}\right]$
Now, $f\left(t\right)=t^{3}-3t$
$\Rightarrow f^{'}\left(t\right)=3t^{2}-3$ $=3\left(t - 1\right)\left(t + 1\right)$

Thus, $f\left(t\right)$ is decreasing $\forall t\in \left[0 , \frac{1}{2}\right]$
$\therefore \left(f \left(t\right)\right)_{m a x}=f\left(0\right)=0$
$\left(f \left(t\right)\right)_{m i n}=f\left(\frac{1}{2}\right)=\frac{1}{8}-\frac{3}{2}=-\frac{11}{8}$
$\therefore f_{m a x}-f_{m i n}=\frac{11}{8}$