Question

The least integral value of $a$ satisfying the system of equations ${\cos }^{-1}x+({\sin }^{-1}y{)}^2=\frac{a{\pi }^2}{4},({\cos }^{-1}x\left)\right({\sin }^{-1}y{)}^2=\frac{{\pi }^4}{16}$ is

Answer 1

Given:
${\cos }^{-1}x+({\sin }^{-1}y{)}^2=\frac{a{\pi }^2}{4}\cdots (i)$
$\left({\cos }^{-1}x\right)({\sin }^{-1}y{)}^2=\frac{{\pi }^4}{16}\cdots (ii)$
Let ${\cos }^{-1}x=t_1$ and $({\sin }^{-1}y{)}^2=t_2$
$0\le t_1\le \pi$ and $0\le t_2\le \frac{{\pi }^2}{4}$
From $\left(i\right)$
$0\le \frac{a{\pi }^2}{4}\le \pi +\frac{{\pi }^2}{4}$
$\Rightarrow 0\le a\le \frac{4}{\pi }+1\cdots \left(iii\right)$
From $\left(i\right)$ and $\left(ii\right)$
$t_1+t_2=\frac{a{\pi }^2}{4}$ and $t_1{t}_2=\frac{{\pi }^4}{16}$
$\Rightarrow t_1(\frac{a{\pi }^2}{4}-t_1)=\frac{{\pi }^4}{16}$
$\Rightarrow t_1^2-\frac{a{\pi }^2}{4}{t}_1+\frac{{\pi }^4}{16}=0$
Since $t_1$ is real, $D\ge 0$
$\Rightarrow \frac{a^2{\pi }^4}{16}\ge \frac{4{\pi }^4}{16}$
$\Rightarrow a^2\ge 4\Rightarrow (a+2)(a-2)\ge 0\cdots \left(iv\right)$
From $\left(iii\right)$ and $\left(iv\right)$
$2\le a\le \frac{4}{\pi }+1$
$\therefore$ The least integral value of $a$ is $2.$