Question

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

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

Answer 1

Answer: (C)

Let, $sinx=t$
$\therefore Ifx\in \left[0,\frac{\pi }{6}\right]\Rightarrow t\in \left[0,\frac{1}{2}\right]$
Now, $f\left(t\right)=t^3-3t$
$\Rightarrow f^{{}^{\prime }}\left(t\right)=3t^2-3$ $=3\left(t-1\right)\left(t+1\right)$

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