| List I | | List II | | --- | --- | --- | | P. | Let $y(x)=\cos \left(3 \cos ^{-1} x\right), x \in[-1,1], x \neq \pm \frac{\sqrt{3}}{2}$. Then $\frac{1}{y(x)}\left\{\left(x^{2}-1\right) \frac{d^{2} y(x)}{d x^{2}}+x \frac{d y(x)}{d x}\right\}$ equals | 1. 1 | | 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 $\overrightarrow{a_{k}}$ be the position vector of the point $A_{k}, k=1,2, \ldots, n$. If $\left|\sum_{k=1}^{n-1}\left(\overrightarrow{a_{k}} \times \overrightarrow{a_{k+1}}\right)\right|=\left|\sum_{k=1}^{n-1}\left(\overrightarrow{a_{k}} \cdot \overrightarrow{a_{k+1}}\right)\right|$, then the minimum value of $n$ is | 2. 2 | | R. | If the normal from the point $P(h, 1)$ on the ellipse $\frac{x^{2}}{6}+\frac{y^{2}}{3}=1$ is perpendicular to the line $x+y=8$, then the value of $h$ is | 3. 8 | | S. | Number of positive solutions satisfying the equation $\tan ^{-1}\left(\frac{1}{2 x+1}\right)+\tan ^{-1}\left(\frac{1}{4 x+1}\right)=\tan ^{-1}\left(\frac{2}{x^{2}}\right)$ is | 4. 9 | P Q R S

Chapter_pending — JEE Advanced · 2014 · 25 May 2014 · Paper 2 Question with Solution

List I		List II
P.	Let $y (x) = cos (3 cos^{- 1} x), x \in [- 1, 1], x \neq = \pm \frac{3}{2}$ . Then $\frac{1}{y ( x )} {(x^{2} - 1) \frac{d ^{2} y ( x )}{d x ^{2}} + x \frac{d y ( x )}{d x}}$ equals	1. 1
Q.	Let $A_{1}, A_{2}, \dots, A_{n} (n > 2)$ be the vertices of a regular polygon of $n$ sides with its centre at the origin. Let $a_{k}$ be the position vector of the point $A_{k}, k = 1, 2, \dots, n$ . If $\sum_{k = 1}^{n - 1} (a_{k} \times a_{k + 1}) = \sum_{k = 1}^{n - 1} (a_{k} \cdot a_{k + 1})$ , then the minimum value of $n$ is	2. 2
R.	If the normal from the point $P (h, 1)$ on the ellipse $\frac{x ^{2}}{6} + \frac{y ^{2}}{3} = 1$ is perpendicular to the line $x + y = 8$ , then the value of $h$ is	3. 8
S.	Number of positive solutions satisfying the equation $tan^{- 1} (\frac{1}{2 x + 1}) + tan^{- 1} (\frac{1}{4 x + 1}) = tan^{- 1} (\frac{2}{x ^{2}})$ is	4. 9

P Q R S

(A)P-4, Q-3, R-2, S-1✓ Correct
(B)P-2, Q-4, R-3, S-1
(C)P-4, Q-3, R-1, S-2
(D)P-2, Q-4, R-1, S-3

Answer: (A)

Solution

P → 4 (equals 9): Using

y = cos (3 cos^{- 1} x) = 4 x^{3} - 3 x

, we get

\frac{d y}{d x} = 12 x^{2} - 3

and

\frac{d ^{2} y}{d x ^{2}} = 24 x

. Substituting:

\frac{1}{4 x ^{3} - 3 x} [(x^{2} - 1) (24 x) + x (12 x^{2} - 3)] = \frac{36 x ^{3} - 27 x}{4 x ^{3} - 3 x} = 9

.

Q → 3 (minimum n = 8): The condition relates the sum of cross products to dot products of consecutive vertices. For a regular n-gon, this equality holds when

n = 8

.

R → 2 (h = 2): The slope of the tangent to the ellipse at a point is

- \frac{x _{1}}{2 y _{1}}

. For the normal to be perpendicular to

x + y = 8

(slope -1), the tangent must have slope -1, giving

x_{1} = 2 y_{1}

. With the point

P (h, 1)

on the ellipse,

h = 2

.

S → 1 (one positive solution): Using the addition formula for inverse tangent and simplifying:

3 x^{2} - 7 x - 6 = 0

, giving

x = 3

x = - \frac{2}{3}

. Only

x = 3

is positive.

Answer: P→4, Q→3, R→2, S→1, which is 4321.

JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Single correct option

JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Chapter_pending

Solution

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