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(A) Let $a = \cos^{-1} x$ and $b^2 = (\sin^{-1} y)^2$. The given equations are $a + b^2 = \frac{n \pi^2}{4}$ and $a \cdot b^2 = \frac{\pi^4}{16}$.Thes

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$(\cos(\frac{\pi^2}{4}), \pm 1)$

Vedclass · Answer

(A) Let $a = \cos^{-1} x$ and $b^2 = (\sin^{-1} y)^2$. The given equations are $a + b^2 = \frac{n \pi^2}{4}$ and $a \cdot b^2 = \frac{\pi^4}{16}$.These are roots of the quadratic equation $t^2 - (\frac{n \pi^2}{4})t + \frac{\pi^4}{16} = 0$.For real roots, the discriminant $D \ge 0$, so $(\frac{n \pi^2}{4})^2 - 4(\frac{\pi^4}{16}) \ge 0$.$\frac{n^2 \pi^4}{16} - \frac{\pi^4}{4} \ge 0 \Rightarrow n^2 \ge 4 \Rightarrow n \ge 2$ (since $n \in Z$ and $n > 0$ for real $x, y$).For the least value $n = 2$, the quadratic becomes $t^2 - \frac{2 \pi^2}{4}t + \frac{\pi^4}{16} = 0$, which is $(t - \frac{\pi^2}{4})^2 = 0$.Thus, $a = \frac{\pi^2}{4}$ and $b^2 = \frac{\pi^2}{4}$.Since $a = \cos^{-1} x = \frac{\pi^2}{4}$, we have $x = \cos(\frac{\pi^2}{4})$.Since $b^2 = (\sin^{-1} y)^2 = \frac{\pi^2}{4}$, we have $\sin^{-1} y = \pm \frac{\pi}{2}$, so $y = \sin(\pm \frac{\pi}{2}) = \pm 1$.The solution is $(\cos(\frac{\pi^2}{4}), \pm 1)$.

Column $I$	Column $II$
$(A)$ If $a=1$ and $b=0$,then $(x, y)$	$(p)$ lies on the circle $x^2+y^2=1$
$(B)$ If $a=1$ and $b=1$,then $(x, y)$	$(q)$ lies on $(x^2-1)(y^2-1)=0$
$(C)$ If $a=1$ and $b=2$,then $(x, y)$	$(r)$ lies on $y=x$
$(D)$ If $a=2$ and $b=2$,then $(x, y)$	$(s)$ lies on $(4x^2-1)(y^2-1)=0$

For the least possible value of $n \in Z$, the solution $(x, y)$ of the equations $\cos ^{-1} x + (\sin ^{-1} y)^2 = \frac{n \pi^2}{4}$ and $(\cos ^{-1} x)(\sin ^{-1} y)^2 = \frac{\pi^4}{16}$ is

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