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 and indicate your answer by darkening the appropriate bubbles in the $4 \times 4$ matrix given in the ORS. | Column I | Column II | |----------|----------| | (A) If $a=1$ and $b=0$, then $(x, y)$ | (p) lies on the circle $x^{2}+y^{2}=1$ | | (B) If $a=1$ and $b=1$, then $(x, y)$ | (q) lies on $\left(x^{2}-1\right)\left(y^{2}-1\right)=0$ | | (C) If $a=1$ and $b=2$, then $(x, y)$ | (r) lies on $y=x$ | | (D) If $a=2$ and $b=2$, then $(x, y)$ | (s) lies on $\left(4 x^{2}-1\right)\left(y^{2}-1\right)=0$ |

(A) If : , which gives . Matches (p). (B) If : . This simplifies to , giving . Matches (q). (C) If : . This simplifies to . Matches (p). (D) If : . This simplifies to . Matches (s).

JEE Advanced · 2007 · 08 Apr 2007 · Paper 2 · Inverse Trigonometric Functions

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Let

(x, y)

be such that

sin^{- 1} (a x) + cos^{- 1} (y) + cos^{- 1} (b x y) = \frac{π}{2} .

Match the statements in Column I with statements in Column II and indicate your answer by darkening the appropriate bubbles in the

4 \times 4

matrix given in the ORS.

Column I	Column II
(A) If $a = 1$ and $b = 0$ , then $(x, y)$	(p) lies on the circle $x^{2} + y^{2} = 1$
(B) If $a = 1$ and $b = 1$ , then $(x, y)$	(q) lies on $(x^{2} - 1) (y^{2} - 1) = 0$
(C) If $a = 1$ and $b = 2$ , then $(x, y)$	(r) lies on $y = x$
(D) If $a = 2$ and $b = 2$ , then $(x, y)$	(s) lies on $(4 x^{2} - 1) (y^{2} - 1) = 0$

Answer: A → I, B → II, C → I, D → IV

Solution

(A) If

a = 1, b = 0

sin^{- 1} x + cos^{- 1} y = \frac{π}{2}

, which gives

x^{2} + y^{2} = 1

. Matches (p).

(B) If

a = 1, b = 1

sin^{- 1} x + cos^{- 1} y + cos^{- 1} (x y) = \frac{π}{2}

. This simplifies to

1 - x^{2} 1 - y^{2} = 0

, giving

(x^{2} - 1) (y^{2} - 1) = 0

. Matches (q).

(C) If

a = 1, b = 2

sin^{- 1} x + cos^{- 1} y + cos^{- 1} (2 x y) = \frac{π}{2}

. This simplifies to

x^{2} + y^{2} = 1

. Matches (p).

(D) If

a = 2, b = 2

sin^{- 1} (2 x) + cos^{- 1} y + cos^{- 1} (2 x y) = \frac{π}{2}

. This simplifies to

(4 x^{2} - 1) (y^{2} - 1) = 0

. Matches (s).

JEE Advanced · 2007 · 08 Apr 2007 · Paper 2 · Match the columns

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JEE Advanced · 2007 · 08 Apr 2007 · Paper 2 · Inverse Trigonometric Functions

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