Part (P)

Let the given expression be $E$. We are given $y(x) = \cos(3\cos^{-1}x)$.

1. Let $\theta = \cos^{-1}x$. This implies $x = \cos\theta$, and since $x \in [-1, 1]$, $\theta \in [0, \pi]$.
2. The function becomes $y = \cos(3\theta)$. Using the triple angle identity, $\cos(3\theta) = 4\cos^3\theta - 3\cos\theta$.
3. Substituting $x = \cos\theta$, we get $y(x) = 4x^3 - 3x$. This is a polynomial function.
4. We can find the derivatives of $y(x)$: \frac{dy}{dx} = 12x^2 - 3 $$$$ \frac{d^2y}{dx^2} = 24x
5. Now, we substitute these into the expression in the numerator: $\left( {{x^2} - 1} \right){{\frac{d^2y}{d{x^2}}}} + x{{\frac{dy}{dx}}} = \left( {{x^2} - 1} \right)\left( {24x} \right) + x\left( {12{x^2} - 3} \right)$
6. Simplifying the expression: = 24x^3 - 24x + 12x^3 - 3x $$$$ = 36x^3 - 27x $$$$ = 9(4x^3 - 3x) = 9y(x)
7. The required expression is: ${1 \over {y\left( x \right)}}\left\{ {\left( {{x^2} - 1} \right){{{d^2}y\left( x \right)} \over {d{x^2}}} + x{{dy\left( x \right)} \over {dx}}} \right\} = {1 \over {y\left( x \right)}} \cdot 9y(x) = 9$ The condition $x \ne \pm \frac{\sqrt{3}}{2}$ ensures that $y(x) = 4x^3-3x \ne 0$, so the expression is well-defined.

Thus, (P) matches with (4).

Part (Q)

Let the vertices of the regular n-sided polygon be $A_1, A_2, ..., A_n$ on a circle of radius $R$ centered at the origin.

1. The position vectors are $\overrightarrow{a_k}$. We have $|\overrightarrow{a_k}| = R$ for all $k$.
2. The angle between any two consecutive position vectors $\overrightarrow{a_k}$ and $\overrightarrow{a_{k+1}}$ is constant, $\theta = \frac{2\pi}{n}$.
3. The dot product is $\overrightarrow{a_k} \cdot \overrightarrow{a_{k+1}} = |\overrightarrow{a_k}| |\overrightarrow{a_{k+1}}| \cos\theta = R^2 \cos(\frac{2\pi}{n})$.
4. The cross product is $\overrightarrow{a_k} \times \overrightarrow{a_{k+1}} = (|\overrightarrow{a_k}| |\overrightarrow{a_{k+1}}| \sin\theta) \hat{u} = R^2 \sin(\frac{2\pi}{n}) \hat{u}$, where $\hat{u}$ is a unit vector perpendicular to the plane of the polygon. The direction of this vector is the same for all $k=1, ..., n-1$.
5. Now we compute the sums: \sum_{k=1}^{n-1} (\overrightarrow{a_k} \cdot \overrightarrow{a_{k+1}}) = (n-1) R^2 \cos(\frac{2\pi}{n}) $$$$ \sum_{k=1}^{n-1} (\overrightarrow{a_k} \times \overrightarrow{a_{k+1}}) = (n-1) R^2 \sin(\frac{2\pi}{n}) \hat{u}
6. The given condition is $|\sum_{k=1}^{n-1} (\overrightarrow{a_k} \times \overrightarrow{a_{k+1}})| = |\sum_{k=1}^{n-1} (\overrightarrow{a_k} \cdot \overrightarrow{a_{k+1}})|$. |(n-1) R^2 \sin(\frac{2\pi}{n}) \hat{u}| = |(n-1) R^2 \cos(\frac{2\pi}{n})| $$$$ (n-1) R^2 |\sin(\frac{2\pi}{n})| = (n-1) R^2 |\cos(\frac{2\pi}{n})|
7. Since $n>2$, we can cancel $(n-1)R^2$, leaving $|\tan(\frac{2\pi}{n})| = 1$.
8. As $n>2$, we have $0 < \frac{2\pi}{n} < \pi$. In this interval, $\sin(\frac{2\pi}{n}) > 0$.
9. So, the equation becomes $\tan(\frac{2\pi}{n}) = \pm 1$.
10. Case 1: $\tan(\frac{2\pi}{n}) = 1$. This implies $\frac{2\pi}{n} = \frac{\pi}{4} + k\pi$. For $k=0$, $\frac{2\pi}{n} = \frac{\pi}{4} \implies n=8$. This is a valid integer solution with $n>2$.
11. Case 2: $\tan(\frac{2\pi}{n}) = -1$. This implies $\frac{2\pi}{n} = \frac{3\pi}{4} + k\pi$. For $k=0$, $\frac{2\pi}{n} = \frac{3\pi}{4} \implies n = 8/3$, which is not an integer.
12. The smallest integer value of $n>2$ that satisfies the condition is $n=8$.

Thus, (Q) matches with (3).

Part (R)

The equation of the ellipse is $\frac{x^2}{6} + \frac{y^2}{3} = 1$, so $a^2=6, b^2=3$.

1. The normal is perpendicular to the line $x+y=8$. The slope of this line is $m_L = -1$. For perpendicularity, the slope of the normal must be $m_N = -1/m_L = 1$.
2. Let the normal be the line $y=mx+c$. Here $m=1$. The line passes through $P(h,1)$, so its equation is $y-1 = 1(x-h)$, which gives $y = x + (1-h)$. So, $c=1-h$.
3. The condition for a line $y=mx+c$ to be a normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $c^2 = \frac{m^2(a^2-b^2)^2}{a^2+b^2m^2}$.
4. Substituting the values $m=1, c=1-h, a^2=6, b^2=3$: $(1-h)^2 = \frac{1^2(6-3)^2}{6+3(1^2)} = \frac{3^2}{9} = 1$
5. Solving for $h$: $(1-h)^2 = 1 \implies 1-h = \pm 1$. If $1-h = 1$, then $h=0$. If $1-h = -1$, then $h=2$.
6. The possible values for $h$ are 0 and 2. Looking at List-II, the values are {1, 2, 8, 9}. The only value from our solution that appears in the list is 2.

Thus, (R) matches with (2).

Part (S)

We need to find the number of positive solutions for the equation: ${\tan ^{ - 1}}\left( {{1 \over {2x + 1}}} \right) + {\tan ^{ - 1}}\left( {{1 \over {4x + 1}}} \right) = {\tan ^{ - 1}}\left( {{2 \over {{x^2}}}} \right)$

1. For positive solutions, $x>0$. In this case, $2x+1 > 0$ and $4x+1 > 0$. Let $A = \frac{1}{2x+1}$ and $B = \frac{1}{4x+1}$. Both $A$ and $B$ are positive.
2. The product $AB = \frac{1}{(2x+1)(4x+1)} = \frac{1}{8x^2+6x+1}$. Since $x>0$, $8x^2+6x+1 > 1$, so $0 < AB < 1$. We can use the formula $\tan^{-1}A + \tan^{-1}B = \tan^{-1}(\frac{A+B}{1-AB})$.
3. Applying the formula: $\tan^{-1}\left(\frac{\frac{1}{2x+1} + \frac{1}{4x+1}}{1 - \frac{1}{(2x+1)(4x+1)}}\right) = \tan^{-1}\left(\frac{2}{x^2}\right)$
4. Taking tangent on both sides: \frac{\frac{(4x+1) + (2x+1)}{(2x+1)(4x+1)}}{\frac{(2x+1)(4x+1)-1}{(2x+1)(4x+1)}} = \frac{2}{x^2} $$$$ \frac{6x+2}{8x^2+6x+1-1} = \frac{2}{x^2} $$$$ \frac{2(3x+1)}{2x(4x+3)} = \frac{2}{x^2} \implies \frac{3x+1}{x(4x+3)} = \frac{2}{x^2}
5. Since we are looking for $x>0$, we know $x \ne 0$. We can multiply both sides by $x^2(4x+3)$: x(3x+1) = 2(4x+3) $$$$ 3x^2 + x = 8x + 6 $$$$ 3x^2 - 7x - 6 = 0
6. Solving the quadratic equation: $x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(3)(-6)}}{2(3)} = \frac{7 \pm \sqrt{49+72}}{6} = \frac{7 \pm \sqrt{121}}{6} = \frac{7 \pm 11}{6}$
7. The roots are $x_1 = \frac{7+11}{6} = 3$ and $x_2 = \frac{7-11}{6} = -\frac{2}{3}$.
8. The question asks for the number of positive solutions. Only $x=3$ is positive.

Thus, there is 1 positive solution. (S) matches with (1).

Conclusion

The final matching is:

• (P) -> (4) (value is 9)
• (Q) -> (3) (value is 8)
• (R) -> (2) (value is 2)
• (S) -> (1) (value is 1)

This corresponds to the option $P = 4,Q = 3,R = 2,S = 1$, which is Option A.