Question

Let $(x,y)$ be such that ${\sin }^{-1}(ax)+{\cos }^{-1}(y)+{\cos }^{-1}(bxy)=\frac{\pi }{2}$ Match the statements in Column I with statements in Column II:

Column I		Column II
A	If $\alpha =1$ and $b=0$, then $(x,y)$	P	lies on the circle $x^2+y^2=1$
B	If $\alpha =1$ and $b=1$, then $(x,y)$	Q	lies on $\left(x^2-1\right)\left(y^2-1\right)=0.$
C	If $\alpha =1$ and $b=2$, then $(x,y)$	R	lies on $y=x$
D	If $\alpha =2$ and $b=2$, then $(x,y)$	S	lies on $\left(4x^2-1\right)\left(y^2-1\right)=0$.

• A. $(A)\to (P);(B)\to (Q);(C)\to (P);(D)\to (Q)$
• B. $(A)\to (P);(B)\to (Q);(C)\to (S);(D)\to (S)$
• C. $(A)\to (Q);(B)\to (Q);(C)\to (P);(D)\to (S)$
• D. $(A)\to (P);(B)\to (Q);(C)\to (P);(D)\to (S)$

Answer 1

Answer: (D)

(A) $\to (P)$
If $a=1$ and $b=0$, then
${\sin }^{-1}x+{\cos }^{-1}y+\frac{\pi }{2}=\frac{\pi }{2}$
$\Rightarrow {\sin }^{-1}y=-{\cos }^{-1}y$
$\Rightarrow {\sin }^{-1}x=-{\sin }^{-1}\sqrt{1-y^2}$
$\Rightarrow -x=\sqrt{1-y^2}$
$\Rightarrow x^2+y^2=1$
$(B)\to (Q)$
If $a=1$ and $b=1$, then
${\sin }^{-1}x+{\cos }^{-1}y+{\cos }^{-1}xy=\frac{\pi }{2}$
${\cos }^{-1}x-{\cos }^{-1}y={\cos }^{-1}xy$
$\Rightarrow xy+\sqrt{1-y^2}\sqrt{1-x^2}=xy$
$\Rightarrow \left(x^2-1\right)\left(y^2-1\right)=0$
$(C)\to (P)$
If $a=1$ and $b=2$, then
${\sin }^{-1}x+{\cos }^{-1}y+{\cos }^{-1}(2xy)=\frac{\pi }{2}$
${\cos }^{-1}x-{\cos }^{-1}y={\cos }^{-1}(2xy)$
$\Rightarrow xy+\sqrt{1-x^2}\sqrt{1-y^2}=2xy$
$\Rightarrow \sqrt{1-x^2}\sqrt{1-y^2}=xy$
$\Rightarrow 1-x^2-y^2+x^2{y}^2=x^2{y}^2$
$\Rightarrow x^2+y^2=1$
(D) $\to (S)$
If $a=2$ and $b=2$, we get
${\sin }^{-2x}+{\cos }^{-1}y+{\cos }^{-1}(2xy)=\frac{\pi }{2}$
${\cos }^{-1}2x-{\cos }^{-1}y={\cos }^{-1}(2xy)$
$2xy+\sqrt{1+4x^2}\sqrt{1-y^2}=2xy$
$\Rightarrow \left(4x^2-1\right)\left(y^2-1\right)=0$