Question

Statement I : The equation ${\left(si{n}^{-1}x\right)}^3+\left(co{s}^{-1}{x}^3\right)-a{\pi }^3=0$ has a solution for all $a\ge \frac{1}{32}.$
Statement II : For any $xϵR,$ $si{n}^{-1}x+co{s}^{-1}x=\frac{\pi }{2}$ and $0\le {\left(si{n}^{-1}x-\frac{\pi }{4}\right)}^2\le \frac{9{\pi }^2}{16}.$

• A. Both statements I and II are true
• B. Both statements I and II are false
• C. Statement I is true and statement II is false
• D. Statement I is false and statement II is true

Answer 1

Answer: (A)

${\left({\sin }^{-1}x\right)}^3+{\left({\cos }^{-1}x\right)}^3-a{\pi }^3=0$
$\Rightarrow \left({\sin }^{-1}x+{\cos }^{-1}x\right)\left[{\left({\sin }^{-1}x\right)}^2-{\sin }^{-1}x{\cos }^{-1}x+{\left({\cos }^{-1}x\right)}^2\right]=a{\pi }^3$
$\Rightarrow \frac{\pi }{2}\left[{\left({\sin }^{-1}x+{\cos }^{-1}x\right)}^2-3{\sin }^{-1}x{\cos }^{-1}x\right]=a{\pi }^3{\pi }^2$
$\Rightarrow {\left(\frac{\pi }{2}\right)}^2-3{\sin }^{-1}x\left(\frac{\pi }{2}-{\sin }^{-1}x\right)=2a{\pi }^2$
$\Rightarrow \frac{\frac{{\pi }^2}{4}-2a{\pi }^2}{3}=\frac{\pi }{2}\left({\sin }^{-1}x\right)-{\left({\sin }^{-1}x\right)}^2$
$\Rightarrow \frac{{\pi }^2-8a{\pi }^2}{12}=\frac{\pi }{2}\left({\sin }^{-1}x\right)-{\left({\sin }^{-1}x\right)}^2$
$\Rightarrow {\left({\sin }^{-1}x\right)}^2-\frac{\pi }{2}{\sin }^{-1}x+{\left(\frac{\pi }{4}\right)}^2={\left(\frac{\pi }{4}\right)}^2-\frac{{\pi }^2-8a{\pi }^2}{12}$
$\Rightarrow {\left({\sin }^{-1}x-\frac{\pi }{4}\right)}^2=\frac{{\pi }^2}{16}-\frac{{\pi }^2}{12}+\frac{/8^{2}a{\pi }^2}{12_3}=\frac{32a{\pi }^2-{\pi }^2}{48}$
Now
$-\frac{\pi }{2}\le {\sin }^{-1}x\le \frac{\pi }{2}$
$\Rightarrow \frac{-\pi }{2}-\frac{\pi }{4}\le {\sin }^{-1}x-\frac{\pi }{4}\le \frac{\pi }{2}-\frac{\pi }{4}$
or $\frac{-3\pi }{4}\le {\sin }^{-1}x-\frac{\pi }{4}\le \frac{\pi }{4}$
$\Rightarrow 0\le {\left({\sin }^{-1}x-\frac{\pi }{4}\right)}^2\le \frac{9{\pi }^2}{16}$
Therefore,
$0\le \frac{32a{\pi }^2-{\pi }^2}{48}\le \frac{9{\pi }^2}{16}$
$\Rightarrow 0\le \frac{32a-1}{48}\le \frac{9}{16}$
$\Rightarrow 0\le 32a-1\le 27$
$\Rightarrow 1\le 32a\le 28$
$\Rightarrow \frac{1}{32}\le a\le \frac{28}{32}$
$\Rightarrow a\in \left[\frac{1}{32},\frac{7}{8}\right]$
Therefore, Statement I is false and II is true.