JEE Advanced 2013 · Paper 2 · Mathematics Q58

MathematicsList match+3 / -1Question 58 of 60

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

List I

P
$\left(\dfrac1{y^2}\left(\dfrac{\cos(\tan^{-1}y)+y\sin(\tan^{-1}y)}{\cot(\sin^{-1}y)+\tan(\sin^{-1}y)}\right)^2+y^4\right)^{1/2}$ takes value
Q
If $\cos x+\cos y+\cos z=0=\sin x+\sin y+\sin z$ then possible value of $\cos\dfrac{x-y}2$ is
R
If $\cos\left(\dfrac\pi4-x\right)\cos2x+\sin x\sin2x\sec x=\cos x\sin2x\sec x+\cos\left(\dfrac\pi4+x\right)\cos2x$ then possible value of $\sec x$ is
S
If $\cot\left(\sin^{-1}\sqrt{1-x^2}\right)=\sin\left(\tan^{-1}\left(x\sqrt6\right)\right)$ , $x\ne0$ , then possible value of $x$ is

List II

P → 4; Q → 3; R → 1; S → 2

P → 4; Q → 3; R → 2; S → 1

P → 3; Q → 4; R → 2; S → 1

P → 3; Q → 4; R → 1; S → 2

Answer and worked solution

Official answer: B

P: $\cos(\tan^{-1}y)+y\sin(\tan^{-1}y)=\sqrt{1+y^2}$ and $\cot(\sin^{-1}y)+\tan(\sin^{-1}y)=\dfrac1{y\sqrt{1-y^2}}$ . Their ratio squared over $y^2$ is $1-y^4$ ; adding $y^4$ gives 1. → 4
Q: the three unit vectors sum to zero, so they are 120° apart: $x-y=\pm\frac{2\pi}3$ and $\cos\frac{x-y}2=\frac12$ . → 3
R: $\cos\left(\frac\pi4-x\right)-\cos\left(\frac\pi4+x\right)=\sqrt2\sin x$ , so $\sqrt2\sin x\cos2x=2\sin x(\cos x-\sin x)$ . For $\sin x\ne0$ : $(\cos x-\sin x)\left[\sqrt2(\cos x+\sin x)-2\right]=0$ , giving $x=\frac\pi4$ and $\sec x=\sqrt2$ . → 2
S: $\dfrac x{\sqrt{1-x^2}}=\dfrac{x\sqrt6}{\sqrt{1+6x^2}}$ gives $1+6x^2=6-6x^2$ , $x=\sqrt{\frac5{12}}=\frac12\sqrt{\frac53}$ . → 1

Code 4 3 2 1.

(B)

JEE Advanced 2013 · Paper 2, Mathematics question 58