Question

If ${\cos }^{-1}x+({\sin }^{-1}y{)}^2=\frac{p{\pi }^2}{4}$ and
$\left({\cos }^{-1}x\right)({\sin }^{-1}y{)}^2=\frac{{\pi }^4}{16},p\in Z$, then which of the following is/are correct ?

• A. There are exactly two values of $p$ for which the system has a solution.
• B. There is exactly one value of $p$ for which the system has a solution.
• C. $x=\cos \frac{{\pi }^2}{4}$
• D. $y=0,\pm 1$

Answer 1

Answer: (B), (C)

The correct options are
B There is exactly one value of $p$ for which the system has a solution.
C $x=\cos \frac{{\pi }^2}{4}$

Let ${\cos }^{-1}x=a\Rightarrow a\in [0,\pi ]$
and ${\sin }^{-1}y=b\Rightarrow b\in [-\frac{\pi }{2},\frac{\pi }{2}]$

$\Rightarrow a+b^2=\frac{p{\pi }^2}{4}...\left(1\right)a{b}^2=\frac{{\pi }^4}{16}...\left(2\right)$

Since, $b^2\in [0,\frac{{\pi }^2}{4}]$
$a+b^2\in [0,\pi +\frac{{\pi }^2}{4}]$

From eqn $\left(1\right)$,
$0\le \frac{p{\pi }^2}{4}\le \pi +\frac{{\pi }^2}{4}$
$\Rightarrow 0\le p\le \frac{4}{\pi }+1$

Since, $p\in Z$, so $p=0,1\text{or}2$

Substitute the value of $b^2$ from eqn$\left(1\right)$ in eqn$\left(2\right)$, we get
$a(\frac{p{\pi }^2}{4}-a)=\frac{{\pi }^4}{16}$
$\Rightarrow 16a^2-4p{\pi }^{2}a+{\pi }^4=0...\left(3\right)$
Since, $a\in R$, so $D\ge 0$
$\Rightarrow 16p^2{\pi }^4-64{\pi }^4\ge 0$
$\Rightarrow p^2\ge 4$
$\Rightarrow p=2$

Put $p=2$ in eqn$\left(3\right)$, we get
$16a^2-8{\pi }^{2}a+{\pi }^4=0$
$\Rightarrow (4a-{\pi }^2{)}^2=0$
$\Rightarrow a=\frac{{\pi }^2}{4}={\cos }^{-1}x$
$\Rightarrow x=\cos \frac{{\pi }^2}{4}$

From eqn$\left(2\right)$,
$\Rightarrow b^2=\frac{{\pi }^2}{4}$
$\Rightarrow b=\pm \frac{\pi }{2}={\sin }^{-1}y$
$\Rightarrow y=\pm 1$