SATHEE: Inverse Circular Functions 2 Question 9

Inverse Circular Functions 2 Question 9

9. Match List I with List II and select the correct answer using the code given below the lists.

List I	List II
P. $\frac{1}{y^{2}} \frac{\cos (\tan^{- 1} y) + y \sin {(\tan^{- 1} y)}^{2}}{\cot (\sin^{- 1} y) + \tan (\sin^{- 1} y)} + y^{4}$ takes value	1.	$\frac{1}{2} \sqrt{\frac{5}{3}}$
Q. If $\cos x + \cos y + \cos z = 0 = \sin x + \sin y + \sin z$ , then	2.	$\sqrt{2}$
possible value of $\cos \frac{x - y}{2}$ is
R. If $\cos \frac{π}{4} - x \cos 2 x + \sin x \sin 2 x \sec x$
$= \cos x \sin 2 x \sec x + \cos \frac{π}{4} + x \cos 2 x$ , then
possible value of $\sec x$ is
S. If $\cot (\sin^{- 1} \sqrt{1 - x^{2}}) = \sin [\tan^{- 1} (x \sqrt{6})]$ ,
$x = 0$ . Then, possible value of $x$ is

Codes

	P	Q	R	S
(a)	4	3	1	2
(b)	4	3	2	1
(c)	3	4	2	1
(d)	3	4	1	2

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Answer:

Correct Answer: 9. (a)

Solution:

P. Here, innermost function is inverse.

$∴$ Put $\tan^{- 1} y = θ \Rightarrow \tan θ = y$

$\frac{1}{y^{2}} \cdot \frac{\cos (\tan^{- 1} y) + y \sin (\tan^{- 1} y)}{\cot (\sin^{- 1} y) + \tan (\sin^{- 1} y)} + y^{4}$

$= \frac{1}{y^{2}} \frac{\frac{1}{\sqrt{1 + y^{2}}} + \frac{y^{2}}{\sqrt{1 + y^{2}}}}{\frac{\sqrt{1 - y^{2}}}{y} + \frac{y}{\sqrt{1 - y^{2}}}} + y^{4}$

$= \frac{1}{y^{2}} \cdot y^{2} (1 - y^{4}) + y^{4}^{1 / 2} = 1$

Q. Given,

$\cos x + \cos y = - \cos z$

and

$\sin x + \sin y = - \sin z$

On squaring and adding, we get

$\cos^{2} x + \sin^{2} x + \cos^{2} y + \sin^{2} y + 2 \cos x \cos y$

$+ 2 \sin x \sin y = 1$

$\Rightarrow 2 + 2 [\cos (x - y)] = 1 \Rightarrow \cos (x - y) = - \frac{1}{2}$

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

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

$\Rightarrow \cos \frac{x - y}{2} = \frac{1}{2}$ R. $\cos 2 x \cdot \cos \frac{π}{4} - x - \cos \frac{π}{4} + x + 2 \sin^{2} x$

$= 2 \sin x \cdot \cos x$

$\Rightarrow \cos 2 x \cdot (\sqrt{2} \sin x) + 2 \sin^{2} x = 2 \sin x \cdot \cos x$

$\Rightarrow \sqrt{2} \sin x [\cos 2 x + \sqrt{2} \sin x - \sqrt{2} \cos x] = 0$

$\Rightarrow \sin x = 0, (\cos x - \sin x) (\cos x + \sin x - \sqrt{2}) = 0$

$\Rightarrow \sec x = 1$ or $\tan x = 1$

$\Rightarrow \sec x = 1$ or $\sqrt{2}$ S. $\cot (\sin^{- 1} \sqrt{1 - x^{2}}) = \sin (\tan^{- 1} (x \sqrt{6}))$

$\begin{aligned} \Rightarrow \frac{x}{\sqrt{1 - x^{2}}} = \frac{x \sqrt{6}}{\sqrt{1 + 6 x^{2}}} \\ \Rightarrow 1 + 6 x^{2} = 6 - 6 x^{2} \\ \Rightarrow 12 x^{2} = 5 \Rightarrow x = \sqrt{\frac{5}{12}} = \frac{\sqrt{5}}{2 \sqrt{3}} \end{aligned}$