JEE Advanced 2014 · Paper 2 · Mathematics Q59

MathematicsList match+3 / -1Question 59 of 60

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

List I

P
Let $y(x)=\cos(3\cos^{-1}x)$ , $x\in[-1,1]$ , $x\ne\pm\frac{\sqrt3}2$ . Then $\dfrac1{y(x)}\left\{(x^2-1)\dfrac{d^2y(x)}{dx^2}+x\dfrac{dy(x)}{dx}\right\}$ equals
Q
Let $A_1,A_2,\ldots,A_n$ $(n>2)$ be the vertices of a regular polygon of $n$ sides with its centre at the origin. Let $\vec{a_k}$ be the position vector of the point $A_k$ , $k=1,2,\ldots,n$ . If $\left|\sum_{k=1}^{n-1}(\vec{a_k}\times\vec{a_{k+1}})\right|=\left|\sum_{k=1}^{n-1}(\vec{a_k}\cdot\vec{a_{k+1}})\right|$ , then the minimum value of $n$ is
R
If the normal from the point $P(h,1)$ on the ellipse $\dfrac{x^2}6+\dfrac{y^2}3=1$ is perpendicular to the line $x+y=8$ , then the value of $h$ is
S
Number of positive solutions satisfying the equation $\tan^{-1}\left(\frac1{2x+1}\right)+\tan^{-1}\left(\frac1{4x+1}\right)=\tan^{-1}\left(\frac2{x^2}\right)$ is

List II

P → 4; Q → 3; R → 2; S → 1

P → 2; Q → 4; R → 3; S → 1

P → 4; Q → 3; R → 1; S → 2

P → 2; Q → 4; R → 1; S → 3

Answer and worked solution

Official answer: A

P: $y=\cos(3\theta)$ with $x=\cos\theta$ . Then $\frac{dy}{dx}=\frac{3\sin3\theta}{\sin\theta}$ and the Chebyshev equation $(1-x^2)y''-xy'+9y=0$ holds, so $(x^2-1)y''+xy'=9y$ . The expression is 9 → 4.
Q: Each cross product has magnitude $r^2\sin\frac{2\pi}n$ (all in the same direction) and each dot product is $r^2\cos\frac{2\pi}n$ . Equality of the sums needs $\tan\frac{2\pi}n=1$ , so $n=8$ → 3.
R: P lies on the ellipse: $\frac{h^2}6+\frac13=1$ , so $h=\pm2$ . The normal at $(x_0,y_0)$ has slope $\frac{a^2y_0}{b^2x_0}=\frac{6}{3h}=\frac2h$ , which must be 1 (perpendicular to $x+y=8$ ). So $h=2$ → 2.
S: For $x>0$ , $\tan^{-1}\frac1{2x+1}+\tan^{-1}\frac1{4x+1}=\tan^{-1}\frac{6x+2}{8x^2+6x}$ . Setting $\frac{3x+1}{x(4x+3)}=\frac2{x^2}$ gives $3x^2-7x-6=0$ , so $x=3$ or $-\frac23$ . One positive solution → 1.

(A) P → 4, Q → 3, R → 2, S → 1

JEE Advanced 2014 · Paper 2, Mathematics question 59