SATHEE: Inverse Circular Functions 2 Question 10

Inverse Circular Functions 2 Question 10

10. Let $(x, y)$ be such that

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

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

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Answer:

Correct Answer: 10. $A \to p; B \to q; C \to p; D \to s$

Solution:

A. If $a = 1, b = 0$ , then $\sin^{- 1} x + \cos^{- 1} y = 0$

$\Rightarrow \sin^{- 1} x = - \cos^{- 1} y \Rightarrow x^{2} + y^{2} = 1$

B. If $a = 1$ and $b = 1$ , then

$\begin{aligned} \sin^{- 1} x + \cos^{- 1} y + \cos^{- 1} x y = \frac{π}{2} \\ \Rightarrow \cos^{- 1} x - \cos^{- 1} y = \cos^{- 1} x y \\ \Rightarrow x y + \sqrt{1 - x^{2}} \sqrt{1 - y^{2}} = x y \Rightarrow (x^{2} - 1) (y^{2} - 1) = 0 \end{aligned}$

C. If $a = 1, b = 2$ , then

$\begin{aligned} \sin^{- 1} x + \cos^{- 1} y + \cos^{- 1} (2 x y) = \frac{π}{2} \\ \Rightarrow \cos^{- 1} x - \cos^{- 1} y = \cos^{- 1} (2 x y) \\ \Rightarrow x y + \sqrt{1 - x^{2}} \sqrt{1 - y^{2}} = 2 x y \Rightarrow x^{2} + y^{2} = 1 \end{aligned}$

D. If $a = 2$ and $b = 2$ , then

$\begin{aligned} \sin^{- 1} (2 x) + \cos^{- 1} (y) + \cos^{- 1} (2 x y) = \frac{π}{2} \\ \Rightarrow \cos^{- 1} (2 x) - \cos^{- 1} (y) = \cos^{- 1} (2 x y) \\ \Rightarrow 2 x y + \sqrt{1 - 4 x^{2}} \sqrt{1 - y^{2}} = 2 x y \\ \Rightarrow (4 x^{2} - 1) (y^{2} - 1) = 0 \end{aligned}$

Inverse Circular Functions 2 Question 10

10. Let $(x, y)$ be such that

Column I

Answer:

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Inverse Circular Functions 2 Question 10

10. Let (x,y) be such that

Column I

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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10. Let $(x, y)$ be such that