Match List I with List II and select the correct answer using the code given below the lists : [2013] List I List II P. ( 1 y 2 ( cos ( tan - 1 y ) + y sin ( tan - 1 y ) cot ( sin - 1 y ) + tan ( sin - 1 y ) ) 2 + y 4 )

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Other Competitive Exams

NDA

NDA

MATHS

Courses

Class Room Courses
Online Live Courses
Online Recorded Courses

ONLINE COURSES

More

Solution

Q.
Match List I with List II and select the correct answer using the code given below the lists : [2013]

List I List II

P. $(\frac{1}{y^{2}} {{(\frac{\cos (\tan^{- 1} y) + y \sin (\tan^{- 1} y)}{\cot (\sin^{- 1} y) + \tan (\sin^{- 1} y)})}^{2} + y^{4})}^{1 / 2}$ takes value 1. $\frac{1}{2} \sqrt{\frac{5}{3}}$

Q. If $\cos x + \cos y + \cos z = \sin x + \sin y + \sin z,$
then possible value of $\cos (\frac{x - y}{2})$ is 2. $\sqrt{2}$

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 3. $\frac{1}{2}$

S. If $\cot (\sin^{- 1} \sqrt{1 - x^{2}}) = \sin (\tan^{- 1} (x \sqrt{6})),$ $x \neq 0,$ then possible value of $x$ is 4. $1$

Codes:

1 P - 4, Q - 3, R - 1, S - 2

2 P - 4, Q - 3, R - 2, S - 1

3 P - 3, Q - 4, R - 2, S - 1

4 P - 3, Q - 4, R - 1, S - 2

Ans.
(2)

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

   $= {[\frac{1}{y^{2}} {(\frac{\cos (\cos^{- 1} \frac{1}{\sqrt{1 + y^{2}}}) + y \sin (\sin^{- 1} \frac{y}{\sqrt{1 + y^{2}}})}{\cot (\cot^{- 1} \frac{\sqrt{1 - y^{2}}}{y}) + \tan (\tan^{- 1} \frac{y}{\sqrt{1 - y^{2}}})})}^{2} + y^{4}]}^{\frac{1}{2}}$

   $= {[\frac{1}{y^{2}} {(\frac{\sqrt{1 + y^{2}}}{\frac{1}{\frac{1}{y \sqrt{1 - y^{2}}}}})}^{2} + y^{4}]}^{\frac{1}{2}}$

   $= {(1 - y^{4} + y^{4})}^{\frac{1}{2}} = 1$    $∴$ $(P) \to (4)$

$(Q) \cos x + \cos y = - \cos z$    $\dots (i)$

   $and \sin x + \sin y = - \sin z$ $\dots (i i)$

On squaring (i) and (ii) and then adding, we get

${(\cos x + \cos y)}^{2} + {(\sin x + \sin y)}^{2} = \cos^{2} z + \sin^{2} z$

   $\Rightarrow 2 + 2 \cos (x - y) = 1$

$\Rightarrow 4 \cos^{2} \frac{x - y}{2} = 1$ $\Rightarrow \cos \frac{x - y}{2} = \pm \frac{1}{2}$

$∴$ $(Q) \to (3)$

$(R) \cos (\frac{π}{4} - x) \cos 2 x + \sin x \sin 2 x \sec x$

   $= \cos x \sin 2 x \sec x + \cos (\frac{π}{4} + x) \cos 2 x$

$\Rightarrow \cos 2 x [\cos (\frac{π}{4} - x) - \cos (\frac{π}{4} + x)] = \sin 2 x \sec x (\cos x - \sin x)$

$\Rightarrow 2 \sin \frac{π}{4} \sin x \cos 2 x = 2 \sin x (\cos x - \sin x)$

   $\Rightarrow 2 \sin x [\frac{1}{\sqrt{2}} (\cos^{2} x - \sin^{2} x) - (\cos x - \sin x)] = 0$

$\Rightarrow 2 \sin x (\cos x - \sin x) (\frac{\cos x + \sin x}{\sqrt{2}} - 1) = 0$

   $\Rightarrow \sin x = 0 or \tan x = 1 or \cos (x - \frac{π}{4}) = 1$

$\Rightarrow x = 0 or \frac{π}{4} \Rightarrow \sec x = 1 or \sqrt{2}$

   $∴$ $(R) \to (2, 4)$

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

   $\Rightarrow \frac{x}{\sqrt{1 - x^{2}}} = \frac{x \sqrt{6}}{\sqrt{1 + 6 x^{2}}}$ $\Rightarrow x = \pm \frac{1}{2} \sqrt{\frac{5}{3}}$

$∴$ $(S) \to (1)$

Hence,   $(P) \to (4), (Q) \to (3), (R) \to (2, 4), (S) \to (1)$

Want Us to Email you About Special Offers & Updates?