Question

If the equation $({\sin }^{-1}x{)}^3+({\cos }^{-1}x{)}^3=a{\pi }^3$, has a solution, then '$a$' lies in the interval

• A. $[-\frac{1}{3},\frac{7}{8}]$
• B. $[\frac{1}{32},\frac{27}{32}]$
• C. $[\frac{1}{32},\frac{7}{8}]$
• D. $[\frac{1}{11},\frac{9}{8}]$

Answer 1

The correct option is C $[\frac{1}{32},\frac{7}{8}]$

Let $f\left(x\right)=({\sin }^{-1}x{)}^3+({\cos }^{-1}x{)}^3$
Put ${\sin }^{-1}x=t$
Then $f\left(t\right)=t^3+{(\frac{\pi }{2}-t)}^3$
$f^{\prime }\left(t\right)=3[t^2-{(\frac{\pi }{2}-t)}^2]$
$\Rightarrow f^{\prime }\left(t\right)=3\cdot \frac{\pi }{2}(2t-\frac{\pi }{2})$

Critical point: $f^{\prime }\left(t\right)=0\Rightarrow t=\frac{\pi }{4}$
Also, $\frac{-\pi }{2}\le t\le \frac{\pi }{2}$

$f(t=\frac{\pi }{4})={\left(\frac{\pi }{4}\right)}^3+{\left(\frac{\pi }{4}\right)}^3=\frac{{\pi }^3}{32}\to \text{minimum}$
$f(t=\frac{-\pi }{2})=\frac{7{\pi }^3}{8}\to \text{maximum}$
$f(t=\frac{\pi }{2})=\frac{{\pi }^3}{8}$
$\therefore$ Range of $f$ is $[\frac{1}{32}{\pi }^3,\frac{7}{8}{\pi }^3]$