IIT-JEE 2007 · Paper 2 · Mathematics Q65

MathematicsMatrix match+6Question 65 of 66

Let $(x,y)$ be such that
$\sin^{-1}(ax)+\cos^{-1}(y)+\cos^{-1}(bxy)=\frac\pi2.$
Match the statements in Column I with statements in Column II and indicate your answer by darkening the appropriate bubbles in the 4 × 4 matrix given in the ORS.

Column I

A
If $a=1$ and $b=0$ , then $(x,y)$
B
If $a=1$ and $b=1$ , then $(x,y)$
C
If $a=1$ and $b=2$ , then $(x,y)$
D
If $a=2$ and $b=2$ , then $(x,y)$

Column II

P
lies on the circle $x^2+y^2=1$
Q
lies on $\left(x^2-1\right)\left(y^2-1\right)=0$
R
lies on $y=x$
S
lies on $\left(4x^2-1\right)\left(y^2-1\right)=0$

Answer and worked solution

Official answer: A → P; B → Q; C → P; D → S

Rewrite. $\cos^{-1}(bxy)=\dfrac\pi2-\sin^{-1}(ax)-\cos^{-1}y=\cos^{-1}(ax)-\cos^{-1}y$ . Taking cosines,
$bxy=axy+\sqrt{1-a^2x^2}\sqrt{1-y^2}.$

A, $a=1,b=0$ : $\sqrt{1-x^2}\sqrt{1-y^2}=-xy$ ; squaring, $(1-x^2)(1-y^2)=x^2y^2$ , i.e. $x^2+y^2=1$ . → P.
B, $a=b=1$ : $\sqrt{1-x^2}\sqrt{1-y^2}=0$ : $(x^2-1)(y^2-1)=0$ . → Q.
C, $a=1,b=2$ : $\sqrt{1-x^2}\sqrt{1-y^2}=xy$ ; squaring gives $x^2+y^2=1$ . → P.
D, $a=b=2$ : $\sqrt{1-4x^2}\sqrt{1-y^2}=0$ : $(4x^2-1)(y^2-1)=0$ . → S.

Answer: A → P; B → Q; C → P; D → S

Topics:Inverse trigonometric functions

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IIT-JEE 2007 · Paper 2, Mathematics question 65