Match List I with List II:
List I
P. Let y(x) = cos(3 cos⁻¹ x), x ∈ [-1, 1], x ≠ ±√3/2. Then (1/y(x)

Q: Why do we reject the negative root in the inverse trigonometry problem?

The equation involves $\tan^{-1}$ terms. While the algebraic manipulation leads to $x = 3$ and $x = -2/3$, we must check the domain. The question explicitly asks for 'positive solutions'. Always check the constraints given in the problem statement before finalizing your answer!

Q: In the vector problem, why is the angle between $\vec{a}_k$ and $\vec{a}_{k+1}$ equal to $2\pi/n$?

Imagine a regular polygon with $n$ sides inscribed in a circle. The center is the origin. The vectors $\vec{a}_k$ and $\vec{a}_{k+1}$ connect the center to two adjacent vertices. Since the polygon is regular, the central angle subtended by each side is $360^\circ / n$, or $2\pi/n$ radians. It is pure geometry!

Q: Why can't we just use energy conservation or some other physics principle for the normal to the ellipse?

This is a pure coordinate geometry problem. While physics often uses energy, here we rely on the slope property of the normal. The normal is perpendicular to the tangent, and the slope of the tangent to the ellipse at $(x_1, y_1)$ is $-b^2 x_1 / a^2 y_1$. Therefore, the normal slope is $a^2 y_1 / b^2 x_1$. Stick to the coordinate geometry toolkit!

Question

Match List I with List II:

List-I:

P. Let $y(x) = \cos(3 \cos^{-1} x), x \in [-1, 1], x 
eq \pm \frac{\sqrt{3}}{2}$. Then $\frac{1}{y(x)} \left\{ (x^2 - 1) \frac{d^2y(x)}{dx^2} + x \frac{dy(x)}{dx} ight\}$ equals

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 $\vec{a}_k$ be the position vector of the point $A_k, k = 1, 2, \dots, n$. If $|\sum_{k=1}^{n-1} (\vec{a}_k 	imes \vec{a}_{k+1})| = |\sum_{k=1}^{n-1} (\vec{a}_k \cdot \vec{a}_{k+1})|$, then the minimum value of $n$ is

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

S. Number of positive solutions satisfying the equation $	an^{-1} \left( \frac{1}{2x+1} ight) + 	an^{-1} \left( \frac{1}{4x+1} ight) = 	an^{-1} \left( \frac{2}{x^2} ight)$ is

List-II:

1. 1

2. 2

3. 8

4. 9

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future IITians! Today, we are not just solving a problem; we are embarking on a journey through the four pillars of JEE Advanced mathematics: Calculus, Vectors, Conic Sections, and Inverse Trigonometry.

This match-the-column question is a classic example of how the exam tests your ability to switch gears between different mathematical domains. Let us break it down, step by step, with the precision of a surgeon and the curiosity of a scientist.

### Part P: The Calculus Mirage

We begin with $y(x) = \cos(3 \cos^{-1} x)$. At first glance, this looks like a nightmare of chain rule differentiation.

But wait! Do not dive into the derivative yet. If we let $\cos^{-1} x = 	heta$, then $x = \cos 	heta$. The function transforms into $y = \cos 3	heta$.

We know the triple-angle identity: $\cos 3	heta = 4 \cos^3 	heta - 3 \cos 	heta$. Substituting $x$ back, we get $y = 4x^3 - 3x$.

Now, the differentiation becomes a breeze. The first derivative is:
$$\frac{dy}{dx} = 12x^2 - 3$$

The second derivative is:
$$\frac{d^2y}{dx^2} = 24x$$

When you plug these into the expression $\frac{1}{y} \{ (x^2 - 1) \frac{d^2y}{dx^2} + x \frac{dy}{dx} \}$, the algebra collapses beautifully. You are left with **$9$**, proving that even the most intimidating functions often hide a simple, elegant core.

### Part Q: The Symmetry of Vectors

Next, we step into the world of regular polygons. We have $n$ vertices, $\vec{a}_k$, centered at the origin.

The key here is visualization. The angle between consecutive position vectors $\vec{a}_k$ and $\vec{a}_{k+1}$ is $	heta = \frac{2\pi}{n}$.

The problem gives us a condition: $|\sum (\vec{a}_k 	imes \vec{a}_{k+1})| = |\sum (\vec{a}_k \cdot \vec{a}_{k+1})|$. The magnitude of the cross product is $R^2 \sin(\frac{2\pi}{n})$, and the dot product is $R^2 \cos(\frac{2\pi}{n})$.

Equating them gives $	an(\frac{2\pi}{n}) = 1$. This implies $\frac{2\pi}{n} = \frac{\pi}{4}$, leading us directly to **$n = 8$**. A regular octagon!

### Part R: The Geometry of the Normal

Now, let us tackle the ellipse $\frac{x^2}{6} + \frac{y^2}{3} = 1$. We need a normal perpendicular to $x + y = 8$.

The line has a slope of $-1$, so our normal must have a slope of $1$. The slope of the normal at $(x_1, y_1)$ is:
$$\frac{a^2 y_1}{b^2 x_1} = \frac{6y_1}{3x_1} = \frac{2y_1}{x_1}$$

Setting this to $1$, we find $x_1 = 2y_1$. Substituting this into the ellipse equation, we find the point of contact is $(2, 1)$.

The normal equation is $y - 1 = 1(x - 2)$, or $y = x - 1$. Since it passes through $P(h, 1)$, we substitute $x=h$ and $y=1$ to get $1 = h - 1$, which gives **$h = 2$**.

### Part S: The Inverse Trigonometric Trap

Finally, we arrive at the inverse trigonometry equation:
$$	an^{-1} \left( \frac{1}{2x+1} ight) + 	an^{-1} \left( \frac{1}{4x+1} ight) = 	an^{-1} \left( \frac{2}{x^2} ight)$$

We apply the identity $	an^{-1} A + 	an^{-1} B = 	an^{-1} \left( \frac{A+B}{1-AB} ight)$. After careful algebraic simplification, we arrive at:
$$\frac{6x+2}{8x^2+6x} = \frac{2}{x^2}$$

This simplifies to the quadratic $3x^2 - 7x - 6 = 0$. The roots are $x = 3$ and $x = -2/3$.

Since the problem asks for positive solutions, we reject the negative root. The answer is **$3$**. This problem teaches us that in JEE Advanced, the math is rarely just about calculation; it is about recognizing patterns, respecting geometric constraints, and staying calm under pressure.

Animated Solution for Mathematics - Vector Algebra: Match List I with List II:

List-I

List-II

Select Matching Pairs:

Visualized Solution