Question

Let $(x,y)\in Q_1,2x\le y$ be such that
${\sin }^{-1}\left(ax\right)+{\cos }^{-1}\left(y\right)+{\cos }^{-1}\left(bxy\right)=\frac{\pi }{2}.$
Match the statements in Column I with the locus in Column II.
$\begin{array}{cccc}& \text{Column I}& & \text{Column II}\\ \left(A\right)& \text{If}a=1\text{and}b=0,\text{then}(x,y)& \left(p\right)& \text{lies on part of the circle}{x}^2+y^2=1\\ \left(B\right)& \text{If}a=1\text{and}b=1,\text{then}(x,y)& \left(q\right)& \text{lies on part of the curve represented by}(x^2-1)(y^2-1)=0\\ \left(C\right)& \text{If}a=1\text{and}b=2,\text{then}(x,y)& \left(r\right)& \text{lies on part of the line}y=x\\ \left(D\right)& \text{If}a=2\text{and}b=2,\text{then}(x,y)& \left(s\right)& \text{lies on part of the curve represented by}(4x^2-1)(y^2-1)=0\end{array}$

• A. $A\to p,B\to q,C\to p,D\to s$
• B. $A\to p,B\to q,C\to p,D\to r$
• C. $A\to p,B\to q,C\to r,D\to s$
• D. $A\to r,B\to q,C\to p,D\to s$

Answer 1

The correct option is A $A\to p,B\to q,C\to p,D\to s$

(A) If $a=1,b=0$, then ${\sin }^{-1}x+{\cos }^{-1}y=0$
$\Rightarrow {\sin }^{-1}x=-{\cos }^{-1}y\Rightarrow {\sin }^{-1}x=-{\sin }^{-1}\left(\frac{\sqrt{1-y^2}}{1}\right)\Rightarrow x^2+y^2=1$

(B) If $a=1$ and $b=1$, then
${\sin }^{-1}x+{\cos }^{-1}y+{\cos }^{-1}xy=\frac{\pi }{2}$
$\Rightarrow \frac{\pi }{2}-{\sin }^{-1}x-{\cos }^{-1}y={\cos }^{-1}xy$
$\Rightarrow {\cos }^{-1}x-{\cos }^{-1}y={\cos }^{-1}xy$
$\Rightarrow xy+\sqrt{1-x^2}\sqrt{1-y^2}=xy$
$\Rightarrow (x^2-1)(y^2-1)=0$

(C) If $a=1,b=2$, then
${\sin }^{-1}x+{\cos }^{-1}y+{\cos }^{-1}\left(2xy\right)=\frac{\pi }{2}$
$\Rightarrow {\cos }^{-1}x-{\cos }^{-1}y={\cos }^{-1}\left(2xy\right)$
$\Rightarrow xy+\sqrt{1-x^2}\sqrt{1-y^2}=2xy$
$\Rightarrow x^2+y^2=1$ (on squaring)

(D) If $a=2$ and $b=2$, then
${\sin }^{-1}\left(2x\right)+{\cos }^{-1}\left(y\right)+{\cos }^{-1}\left(2xy\right)=\frac{\pi }{2}$
$\Rightarrow {\cos }^{-1}\left(2x\right)-{\cos }^{-1}\left(y\right)={\cos }^{-1}\left(2xy\right)$
$\Rightarrow 2xy+\sqrt{1-4x^2}\sqrt{1-y^2}=2xy$
$\Rightarrow (4x^2-1)(y^2-1)=0$