( ^ - 1 x)^3 + ( ^ - 1 x)^3 ની મહતમ અને ન્યૂનતમ કિમત મેળવો.

MCQ

${({\sin ^{ - 1}}x)^3} + {({\cos ^{ - 1}}x)^3}$ ની મહતમ અને ન્યૂનતમ કિમત મેળવો.

A
$ - \frac{\pi }{2},\,\frac{\pi }{2}$
B
$ - \frac{{{\pi ^3}}}{8},\,\frac{{{\pi ^3}}}{8}$
✓
$\frac{{7{\pi ^3}}}{8},\,\,\frac{{{\pi ^3}}}{{32}}$
D
એકપણ નહીં.

✓

Answer

Correct option: C.

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

We have ${({\sin ^{ - 1}}x)^3} + {({\cos ^{ - 1}}x)^3}$
$ = {({\sin ^{ - 1}}x + {\cos ^{ - 1}}x)^3}$$ - 3{\sin ^{ - 1}}x{\cos ^{ - 1}}x({\sin ^{ - 1}}x + {\cos ^{ - 1}}x)$
$= \frac{{{\pi ^3}}}{8} - 3({\sin ^{ - 1}}x{\cos ^{ - 1}}x)\frac{\pi }{2}$
$= \frac{{{\pi ^3}}}{8} - \frac{{3\pi }}{2}{\sin ^{ - 1}}x\left( {\frac{\pi }{2} - {{\sin }^{ - 1}}x} \right)$
$= \frac{{{\pi ^3}}}{8} - \frac{{3{\pi ^2}}}{4}{\sin ^{ - 1}}x + \frac{{3\pi }}{2}{({\sin ^{ - 1}}x)^2}$
$=\frac{{{\pi ^3}}}{8} + \frac{{3\pi }}{2}\left[ {{{({{\sin }^{ - 1}}x)}^2} - \frac{\pi }{2}{{\sin }^{ - 1}}x} \right]$
$ = \frac{{{\pi ^3}}}{8} + \frac{{3\pi }}{2}\left[ {{{\left( {{{\sin }^{ - 1}}x - \frac{\pi }{4}} \right)}^2}} \right] - \frac{{3{\pi ^3}}}{{32}}$
$ = \frac{{{\pi ^3}}}{{32}} + \frac{{3\pi }}{2}{\left( {{{\sin }^{ - 1}}x - \frac{\pi }{4}} \right)^2}$
$\therefore$ The least value is $\frac{{{\pi ^3}}}{{32}}$
and since ${\left( {{{\sin }^{ - 1}}x - \frac{\pi }{4}} \right)^2} \le {\left( {\frac{{3\pi }}{4}} \right)^2}$
$\therefore$ The greatest value is $\frac{{{\pi ^3}}}{{32}} + \frac{{9{\pi ^2}}}{{16}} \times \frac{{3\pi }}{2} = \frac{{7{\pi ^3}}}{8}$.

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