Question

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

• 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)

$\left(\sin ^{-1} x\right)^{3}+\left(\cos ^{-1} x\right)^{3}-a \pi^{3}=0$
$\Rightarrow \left(\sin ^{-1} x+\cos ^{-1} x\right)\left[\left(\sin ^{-1} x\right)^{2}-\sin ^{-1} x \cos ^{-1} x+\left(\cos ^{-1} x\right)^{2}\right]=a \pi^{3}$
$\Rightarrow \frac{\pi}{2}\left[\left(\sin ^{-1} x+\cos ^{-1} x\right)^{2}-3 \sin ^{-1} x \cos ^{-1} x\right]=a \pi^{3} \pi^{2}$
$\Rightarrow \left(\frac{\pi}{2}\right)^{2}-3 \sin ^{-1} x\left(\frac{\pi}{2}-\sin ^{-1} x\right)=2 a \pi^{2}$
$\Rightarrow \frac{\frac{\pi^{2}}{4}-2 a \pi^{2}}{3}=\frac{\pi}{2}\left(\sin ^{-1} x\right)-\left(\sin ^{-1} x\right)^{2}$
$\Rightarrow \frac{\pi^{2}-8 a \pi^{2}}{12}=\frac{\pi}{2}\left(\sin ^{-1} x\right)-\left(\sin ^{-1} x\right)^{2}$
$\Rightarrow \left(\sin ^{-1} x\right)^{2}-\frac{\pi}{2} \sin ^{-1} x+\left(\frac{\pi}{4}\right)^{2}=\left(\frac{\pi}{4}\right)^{2}-\frac{\pi^{2}-8 a \pi^{2}}{12}$
$\Rightarrow \left(\sin ^{-1} x-\frac{\pi}{4}\right)^{2}=\frac{\pi^{2}}{16}-\frac{\pi^{2}}{12}+\frac{\not 8^{2} a \pi^{2}}{12_{3}}=\frac{32 a \pi^{2}-\pi^{2}}{48}$
Now
$-\frac{\pi}{2} \leq \sin ^{-1} x \leq \frac{\pi}{2}$
$ \Rightarrow \frac{-\pi}{2}-\frac{\pi}{4} \leq \sin ^{-1} x-\frac{\pi}{4} \leq \frac{\pi}{2}-\frac{\pi}{4}$
or $\frac{-3 \pi}{4} \leq \sin ^{-1} x-\frac{\pi}{4} \leq \frac{\pi}{4}$
$\Rightarrow 0 \leq\left(\sin ^{-1} x-\frac{\pi}{4}\right)^{2} \leq \frac{9 \pi^{2}}{16}$
Therefore,
$0 \leq \frac{32 a \pi^{2}-\pi^{2}}{48} \leq \frac{9 \pi^{2}}{16}$
$ \Rightarrow 0 \leq \frac{32 a-1}{48} \leq \frac{9}{16} $
$\Rightarrow 0 \leq 32 a-1 \leq 27$
$ \Rightarrow 1 \leq 32 a \leq 28 $
$\Rightarrow \frac{1}{32} \leq a \leq \frac{28}{32} $
$\Rightarrow a \in\left[\frac{1}{32}, \frac{7}{8}\right]$
Therefore, Statement I is false and II is true.