Match List Matrix Inverse Trigonometric Derivatives Polygon Vectors Normal Ellipse

To determine the correct match between List-I and List-II, we solve each part step-by-step:

---

Part P:

Given $y(x) = \cos(3 \cos^{-1} x)$ for $x \in [-1,1]$.

Let $\theta = \cos^{-1} x$, so $x = \cos \theta$. Then, $y(x) = \cos(3\theta) = 4\cos^3\theta - 3\cos\theta = 4x^3 - 3x$

Differentiating $y(x)$ with respect to $x$: $\frac{dy(x)}{dx} = 12x^2 - 3$ $\frac{d^2 y(x)}{dx^2} = 24x$

Now, evaluate the expression: $(x^2 - 1) \frac{d^2 y(x)}{dx^2} + x \frac{dy(x)}{dx} = (x^2 - 1)(24x) + x(12x^2 - 3)$ $= 24x^3 - 24x + 12x^3 - 3x$ $= 36x^3 - 27x = 9(4x^3 - 3x) = 9 y(x)$

Thus, $\frac{1}{y(x)} \left\{ (x^2 - 1) \frac{d^2 y(x)}{dx^2} + x \frac{dy(x)}{dx} \right\} = \frac{9 y(x)}{y(x)} = 9$

Therefore, P $\rightarrow$ 4.

---

Part Q:

Let the vertices of a regular $n$-gon centered at the origin be represented by position vectors $\vec{a}_k$ of length $R$.

The angle between adjacent vertices $\vec{a}_k$ and $\vec{a}_{k+1}$ is $\theta = \frac{2\pi}{n}$.

1. Cross Product Sum: Since all $\vec{a}_k \times \vec{a}_{k+1}$ point in the same normal direction to the plane (say $\hat{k}$): $\vec{a}_k \times \vec{a}_{k+1} = R^2 \sin\left(\frac{2\pi}{n}\right) \hat{k}$ $\left| \sum_{k=1}^{n-1} (\vec{a}_k \times \vec{a}_{k+1}) \right| = (n-1) R^2 \sin\left(\frac{2\pi}{n}\right)$

2. Dot Product Sum: $\vec{a}_k \cdot \vec{a}_{k+1} = R^2 \cos\left(\frac{2\pi}{n}\right)$ $\left| \sum_{k=1}^{n-1} (\vec{a}_k \cdot \vec{a}_{k+1}) \right| = (n-1) R^2 \left| \cos\left(\frac{2\pi}{n}\right) \right|$

Equating the two magnitudes: $(n-1) R^2 \sin\left(\frac{2\pi}{n}\right) = (n-1) R^2 \left| \cos\left(\frac{2\pi}{n}\right) \right|$

Since $n > 2$, $n-1 \neq 0$, which gives: $\tan\left(\frac{2\pi}{n}\right) = \pm 1$

For $n > 2$, $\frac{2\pi}{n} \in (0, \pi)$: $\frac{2\pi}{n} = \frac{\pi}{4} \implies n = 8$

Thus, the minimum value of $n$ is $8$.

Therefore, Q $\rightarrow$ 3.

---

Part R:

The point $P(h, 1)$ lies on the ellipse $\frac{x^2}{6} + \frac{y^2}{3} = 1$: $\frac{h^2}{6} + \frac{1^2}{3} = 1 \implies \frac{h^2}{6} = \frac{2}{3} \implies h^2 = 4 \implies h = \pm 2$

The equation of the normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at $(x_1, y_1) = (h, 1)$ is: $\frac{a^2 x}{x_1} - \frac{b^2 y}{y_1} = a^2 - b^2$ $\frac{6x}{h} - \frac{3y}{1} = 6 - 3 = 3 \implies y = \frac{2}{h}x - 1$

The slope of the normal is $m_N = \frac{2}{h}$.

Since the normal is perpendicular to $x + y = 8$ (which has slope $m_L = -1$): $m_N \cdot m_L = -1 \implies \left(\frac{2}{h}\right)(-1) = -1 \implies h = 2$

Therefore, R $\rightarrow$ 2.

---

Part S:

We solve for positive solutions ($x > 0$) of the equation: $\tan^{-1}\left(\frac{1}{2x+1}\right) + \tan^{-1}\left(\frac{1}{4x+1}\right) = \tan^{-1}\left(\frac{2}{x^2}\right)$

Using the identity $\tan^{-1} A + \tan^{-1} B = \tan^{-1}\left(\frac{A+B}{1-AB}\right)$ for $AB < 1$: $\tan^{-1}\left(\frac{\frac{1}{2x+1} + \frac{1}{4x+1}}{1 - \frac{1}{(2x+1)(4x+1)}}\right) = \tan^{-1}\left(\frac{(4x+1) + (2x+1)}{(2x+1)(4x+1) - 1}\right) = \tan^{-1}\left(\frac{6x+2}{8x^2+6x}\right) = \tan^{-1}\left(\frac{3x+1}{x(4x+3)}\right)$

Equating to the right-hand side: $\frac{3x+1}{x(4x+3)} = \frac{2}{x^2}$

Since $x > 0$, multiply both sides by $x^2$: $x(3x+1) = 2(4x+3)$ $3x^2 + x = 8x + 6$ $3x^2 - 7x - 6 = 0$ $(3x+2)(x-3) = 0$

Since $x > 0$, we have $x = 3$.

Thus, there is only $1$ positive solution.

Therefore, S $\rightarrow$ 1.

---

Conclusion:

• P $\rightarrow$ 4
• Q $\rightarrow$ 3
• R $\rightarrow$ 2
• S $\rightarrow$ 1

This corresponds to Option A.