Question

Consider the system of equations ${\cos }^{-1}x+({\sin }^{-1}y{)}^2=\frac{p{\pi }^2}{4}$ and $\left({\cos }^{-1}x\right)({\sin }^{-1}y{)}^2=\frac{{\pi }^4}{16},p\in Z$
The value of $p$ for which system has a solution is

• A. $4$
• B. $-4$
• C. $2$
• D. $-2$

Answer 1

The correct option is C $2$

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

We have $a+b^2=\frac{p{\pi }^2}{4}.........\left(\text{i}\right)$
and $a{b}^2=\frac{{\pi }^4}{16}$.............(ii)

Since $b^2\in [0,{\pi }^2/4],\text{we get}a+b^2\in [0,\pi +{\pi }^2/4]$
So, from Eq. (i) we get$0\le \frac{p{\pi }^2}{4}\le \pi +\frac{{\pi }^2}{4}$
i.e., $0\le p\le \frac{4}{\pi }+1$
Since $p\in Z,\text{so}p=0,1\text{or}2.$
Substituting the value of $b^2$ from Eq (i) in Eq (ii) we get
$a(\frac{p{\pi }^2}{4}-a)=\frac{{\pi }^4}{16}$
$\Rightarrow 16a^2-4p{\pi }^{2}a+{\pi }^4=0$.....(iii)
Since $a\in R$ we have $D\ge 0$
i.e., $16p^2{\pi }^4-64{\pi }^4\ge 0$
or$p^2\ge 4$ or$p\ge 2$ or $p=2$