Inverse Circular Functions 2 Question 10
10. Let $(x, y)$ be such that
$\sin ^{-1}(a x)+\cos ^{-1}(y)+\cos ^{-1}(b x y)=\frac{\pi}{2}$.
(2007)
Column I
Column II
A. If $a=1$ and $b=0$, then $(x, y) \quad$ p. lies on the circle $x^{2}+y^{2}=1$
B. If $a=1$ and $b=1$, then $(x, y) \quad$ q. lies on $\left(x^{2}-1\right)\left(y^{2}-1\right)=0$
C. If $a=1$ and $b=2$, then $(x, y) \quad$ r. lies on $y=x$
D. If $a=2$ and $b=2$, then $(x, y) \quad$ s. lies on $\left(4 x^{2}-1\right)\left(y^{2}-1\right)=0$
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Answer:
Correct Answer: 10. $A \rightarrow p ; B \rightarrow q ; C \rightarrow p ; D \rightarrow s$
Solution:
- A. If $a=1, b=0$, then $\sin ^{-1} x+\cos ^{-1} y=0$
$ \Rightarrow \quad \sin ^{-1} x=-\cos ^{-1} y \Rightarrow x^{2}+y^{2}=1 $
B. If $a=1$ and $b=1$, then
$ \begin{aligned} & \sin ^{-1} x+\cos ^{-1} y+\cos ^{-1} x y=\frac{\pi}{2} \\ & \Rightarrow \quad \cos ^{-1} x-\cos ^{-1} y=\cos ^{-1} x y \\ & \Rightarrow \quad x y+\sqrt{1-x^{2}} \sqrt{1-y^{2}}=x y \Rightarrow\left(x^{2}-1\right)\left(y^{2}-1\right)=0 \end{aligned} $
C. If $a=1, b=2$, then
$ \begin{aligned} & \sin ^{-1} x+\cos ^{-1} y+\cos ^{-1}(2 x y)=\frac{\pi}{2} \\ & \Rightarrow \quad \cos ^{-1} x-\cos ^{-1} y=\cos ^{-1}(2 x y) \\ & \Rightarrow \quad x y+\sqrt{1-x^{2}} \sqrt{1-y^{2}}=2 x y \quad \Rightarrow \quad x^{2}+y^{2}=1 \end{aligned} $
D. If $a=2$ and $b=2$, then
$ \begin{aligned} & \sin ^{-1}(2 x)+\cos ^{-1}(y)+\cos ^{-1}(2 x y)=\frac{\pi}{2} \\ & \Rightarrow \quad \cos ^{-1}(2 x)-\cos ^{-1}(y)=\cos ^{-1}(2 x y) \\ & \Rightarrow \quad 2 x y+\sqrt{1-4 x^{2}} \sqrt{1-y^{2}}=2 x y \\ & \Rightarrow \quad\left(4 x^{2}-1\right)\left(y^{2}-1\right)=0 \end{aligned} $
