Question Error Report

Question is incomplete/wrong

Question not belongs to this Chapter

Answer is wrong

Solution is wrong

Answer & Solution is not matching

Spelling mistake

Image missing

Website not working properly

Other (not listed above)

Error description

Q. Let $(x, y)$ be such that $\sin ^{-1}(a x)+\cos ^{-1}(y)+\cos ^{-1}(b x y)=\frac{\pi}{2}$ 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(4 x^{2}-1\right)\left(y^{2}-1\right)= 0 $.

JEE AdvancedJEE Advanced 2007

A

$( A ) \rightarrow( P ) ;( B ) \rightarrow( Q ) ;( C ) \rightarrow( P ) ;( D ) \rightarrow( Q )$

B

$( A ) \rightarrow( P ) ;( B ) \rightarrow( Q ) ;( C ) \rightarrow( S ) ;( D ) \rightarrow( S )$

C

$( A ) \rightarrow( Q ) ;( B ) \rightarrow( Q ) ;( C ) \rightarrow( P ) ;( D ) \rightarrow( S )$

D

$( A ) \rightarrow( P ) ;( B ) \rightarrow( Q ) ;( C ) \rightarrow( P ) ;( D ) \rightarrow( S )$

Solution:

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