Question

The least and the greatest value of ${\left({\sin }^{-1}x\right)}^3+{\left({\cos }^{-1}x\right)}^3$

• A. $\frac{-\pi }{2},\frac{\pi }{2}$
• B. $\frac{-{\pi }^3}{8},\frac{{\pi }^3}{8}$
• C. $\frac{{\pi }^3}{32},\frac{7{\pi }^3}{8}$
• D. None of these

Answer 1

Answer: (C)

$\frac{{\pi }^3}{32},\frac{7{\pi }^3}{8}$

REF.Image.

$f\left(x\right)=(si{n}^{-1}x{)}^3+(co{s}^{-1}x{)}^3$

$=(si{n}^{-1}x{)}^3+(\frac{\pi }{2}-si{n}^{-1}x{)}^3$

$f^{\prime }\left(x\right)=3(si{n}^{-1}x{)}^2\frac{1}{\sqrt{1-x^2}}+3(\frac{\pi }{2}-si{n}^{-1}x{)}^2\times \left(\frac{-1}{\sqrt{1-x^2}}\right)$

$=\frac{3}{\sqrt{1-x^2}}[si{n}^{-1}x-\frac{\pi }{2}+si{n}^{-1}x][si{n}^{-1}x+\frac{\pi }{2}-si{n}^{-1}x]$

$=\frac{3\pi }{2\sqrt{1-x^2}}(2si{n}^{-1}x-\frac{\pi }{2})$

$=\frac{3\pi }{\sqrt{1-x^2}}(si{n}^{-1}x-\frac{\pi }{4})$

for critical point $f^{\prime }\left(x\right)=0$

$\therefore si{n}^{-1}x=\pi /4$

$\therefore x=\frac{1}{\sqrt{2}}$

$\therefore$ at $x=\frac{1}{\sqrt{2}},f^{\prime }\left(x\right)$ will be min

and min value $f\left(\frac{1}{\sqrt{2}}\right)=(si{n}^{-1}\frac{1}{\sqrt{2}}{)}^3+(co{s}^{-1}\frac{1}{\sqrt{2}}{)}^3$

$=(\frac{\pi }{4}{)}^3+(\frac{\pi }{4}{)}^3=\frac{{\pi }^3}{3^2}$

now $f(-1)=\left(si{n}^{-1}\right(-1){)}^3+(co{s}^{-1}(-1){)}^3$

$=(-\frac{\pi }{2}{)}^3+(\pi {)}^3$

$={\pi }^3-\frac{{\pi }^3}{8}=\frac{7{\pi }^3}{8}$

also check for $f\left(1\right)=\left(si{n}^{-1}\right(1){)}^3+(co{s}^{-1}1{)}^3$

$=(\frac{\pi }{2}{)}^3+0^3=\frac{{\pi }^3}{8}$

$\therefore$ max value of $f\left(x\right)$ will be $\frac{7{\pi }^3}{8}$