Let ( x , y ) be such that sin - 1 ( a x ) + cos - 1 ( y ) + cos - 1 ( b x y ) = 2 . 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.

Question

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. [2007]

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

Smart Achievers · Accepted Answer

(4)

(A) → p; (B) → q; (C) → p; (D) → s

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

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

$Let \cos^{- 1} y = α, \cos^{- 1} (b x y) = β, \cos^{- 1} (a x) = γ$

$Then y = \cos α, b x y = \cos β, a x = \cos γ$

$∴$ we get $α + β = γ$ and $\cos β = b x y$

$\Rightarrow \cos (γ - α) = \cos β = b x y$

$\Rightarrow \cos γ \cos α + \sin γ \sin α = b x y$

$\Rightarrow a x y + \sin γ \sin α = b x y \Rightarrow (a - b) x y = - \sin α \sin γ$

$\Rightarrow {(a - b)}^{2} x^{2} y^{2} = \sin^{2} α \sin^{2} γ$

$= (1 - \cos^{2} α) (1 - \cos^{2} γ)$

$∴$ ${(a - b)}^{2} x^{2} y^{2} = (1 - y^{2}) (1 - a^{2} x^{2})$ $\dots (1)$

$(A)$ $For a = 1, b = 0, equation (i) reduces to$

$x^{2} y^{2} = (1 - x^{2}) (1 - y^{2}) \Rightarrow x^{2} + y^{2} = 1$

$(B)$ $For a = 1, b = 1, equation (i) becomes$

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

$(C)$ $For a = 1, b = 2, equation (i) reduces to$

$x^{2} y^{2} = (1 - x^{2}) (1 - y^{2}) \Rightarrow x^{2} + y^{2} = 1$

$(D)$ $For a = 2, b = 2, equation (i) reduces to$

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

Solution

1 (A) → s; (B) → p; (C) → q; (D) → q

2 (A) → s; (B) → p; (C) → q; (D) → p

3 (A) → s; (B) → q; (C) → p; (D) → p

4 (A) → p; (B) → q; (C) → p; (D) → s

Want Us to Email you About Special Offers & Updates?