Match the following. List – I List – II (A) Let y x = cos ⁡ 3 cos - 1 ⁡ x , x ϵ - 1 , 1 , x ≠ ± 3 2 . Then 1 y x x 2 - 1 d 2 y x d x 2 + x d y x d x equals (P) 1 (B) Let A 1 , A 2 , … , A n n 2 be t

Question

Match the following.

	List – I		List – II
(A)	Let $y (x) = \cos (3 \cos^{- 1} x), x ϵ [- 1, 1], x \neq \pm \frac{\sqrt{3}}{2} .$ Then $\frac{1}{y (x)} \{(x^{2} - 1) \frac{d^{2} y (x)}{d x^{2}} + x \frac{d y (x)}{d x}\}$ equals	(P)	1
(B)	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 $\vec{a_{k}}$ be the position vector of the point $A_{k}, k = 1, 2, \dots n .$ If $\|\sum_{k = 1}^{n - 1} (\vec{a_{k}} \times \vec{a_{k + 1}})\| = \|\sum_{k = 1}^{n - 1} (\vec{a_{k}} \cdot \vec{a_{k + 1}})\|$ , then the minimum value of n is	(Q)	2
(C)	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	(R)	8
(D)	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	(S)	9

TestHub · Accepted Answer

P y = cos ⁡ 3 cos - 1 ⁡ x y ′ = 3 sin ⁡ 3 cos - 1 ⁡ x 1 - x 2 1 - x 2 y ′ = 3 sin ⁡ 3 cos - 1 ⁡ x ⇒ - x 1 - x 2 y ′ + 1 - x 2 y " = 3 c o s ⁡ ( 3 c o s - 1 x ) . - 3 1 - x 2 ⇒ - x y ′ + 1 - x 2 y " = - 9 y ⇒ 1 y [ x 2 - 1 y " + x y ′ ] = 9 Q a k × a k + 1 = r 2 sin ⁡ 2 π n a k . a k + 1 = r 2 cos ⁡ 2 π n ⇒ ∑ k = 1 n - 1 a k → × a k + 1 → = ∑ k = 1 n - 1 a k . a k + 1 ⇒ r 2 n - 1 sin ⁡ 2 π n = r 2 n - 1 cos ⁡ 2 π n tan ⁡ 2 π n = 1 ⇒ n = 8 4 k + 1 ⇒ n = 8 R h 2 6 + 1 2 3 = 1 , h = ± 2 Tangent at (2, 1) is 2 x 6 + y 3 = 1 x + y = 3 S tan - 1 ⁡ 1 2 x + 1 + tan - 1 ⁡ 1 4 x + 1 = tan - 1 ⁡ 2 x 2 tan - 1 ⁡ 3 x + 1 4 x 2 + 3 x = tan - 1 ⁡ 2 x 2 ⇒ 3 x 2 - 7 x - 6 = 0 x = - 2 3 , 3

Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

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JEE Advanced (2028)