Part (P)

The function is given by $f(x)=\left[\frac{10 x^3-45 x^2+60 x+35}{n}\right]$. We need to find the minimum natural number $n$ for which $f(x)$ is continuous on the interval $[1,2]$.

1. Let $u(x) = \frac{10 x^3-45 x^2+60 x+35}{n}$. The function $f(x) = [u(x)]$ is the greatest integer function. For $f(x)$ to be continuous on a closed interval, it must be a constant function. Let's assume $f(x) = k$ for some integer $k$.

2. For $f(x)$ to be constant, the range of $u(x)$ over $[1,2]$ must lie within an interval of the form $[k, k+1)$ for some integer $k$. That is, $k \le u(x) < k+1$ for all $x \in [1,2]$.

3. Let's find the range of the numerator, $v(x) = 10x^3-45x^2+60x+35$, on $[1,2]$. We first check for its monotonicity by finding its derivative: $v'(x) = 30x^2 - 90x + 60 = 30(x^2 - 3x + 2) = 30(x-1)(x-2)$ For $x \in (1,2)$, $v'(x) < 0$, which means $v(x)$ is a strictly decreasing function on $[1,2]$.

4. The extreme values of $v(x)$ will be at the endpoints of the interval: $v(1) = 10(1)^3 - 45(1)^2 + 60(1) + 35 = 10 - 45 + 60 + 35 = 60$. $v(2) = 10(2)^3 - 45(2)^2 + 60(2) + 35 = 80 - 180 + 120 + 35 = 55$. So, the range of $v(x)$ on $[1,2]$ is $[55, 60]$.

5. The range of $u(x) = v(x)/n$ on $[1,2]$ is $\left[\frac{55}{n}, \frac{60}{n}\right]$.

6. For $f(x)$ to be continuous, this range must be a subset of $[k, k+1)$ for some integer $k$. This gives us the conditions: $k \le \frac{55}{n} \quad \text{and} \quad \frac{60}{n} < k+1$

7. From these inequalities, we get: $n \le \frac{55}{k} \quad \text{and} \quad n > \frac{60}{k+1}$ So, we are looking for an integer $n$ such that $\frac{60}{k+1} < n \le \frac{55}{k}$. For such an integer $n$ to exist, we test integer values of $k$ (which must be positive since $n$ is positive).

   • If $k=5: \frac{60}{6} < n \le \frac{55}{5} \implies 10 < n \le 11$. This gives $n=11$.
   • If $k=6: \frac{60}{7} < n \le \frac{55}{6} \implies 8.57... < n \le 9.16...$. This gives $n=9$.
   • If $k=7: \frac{60}{8} < n \le \frac{55}{7} \implies 7.5 < n \le 7.85...$. No integer $n$.
   • If $k=8: \frac{60}{9} < n \le \frac{55}{8} \implies 6.66... < n \le 6.87...$. No integer $n$.

8. The possible values for $n$ are $9, 11, 13, \dots$. The minimum value of $n$ is 9.

   (P) matches with (2) which is 9.

Part (Q)

The function is $g(x)=\left(2 n^2-13 n-15\right)\left(x^3+3 x\right)$. We need the minimum value of $n \in \mathbb{N}$ for which $g(x)$ is an increasing function on $\mathbb{R}$.

1. For $g(x)$ to be an increasing function, its derivative $g'(x)$ must be non-negative for all $x \in \mathbb{R}$. $g'(x) = \frac{d}{dx} \left[ \left(2 n^2-13 n-15\right)\left(x^3+3 x\right) \right]$ Let $C = 2n^2-13n-15$. Then $g(x) = C(x^3+3x)$. $g'(x) = C \frac{d}{dx}(x^3+3x) = C(3x^2+3) = 3C(x^2+1)$

2. The term $x^2+1$ is always positive for all $x \in \mathbb{R}$. Therefore, the sign of $g'(x)$ is determined by the sign of the constant $C$.

3. For $g(x)$ to be increasing, we need $g'(x) \ge 0$, which implies $C \ge 0$. $2n^2 - 13n - 15 \ge 0$

4. We find the roots of the quadratic equation $2n^2 - 13n - 15 = 0$ using the quadratic formula: $n = \frac{-(-13) \pm \sqrt{(-13)^2 - 4(2)(-15)}}{2(2)} = \frac{13 \pm \sqrt{169 + 120}}{4} = \frac{13 \pm \sqrt{289}}{4} = \frac{13 \pm 17}{4}$ The roots are $n_1 = \frac{13+17}{4} = 7.5$ and $n_2 = \frac{13-17}{4} = -1$.

5. Since the quadratic has a positive leading coefficient, the inequality $2n^2 - 13n - 15 \ge 0$ holds for $n \le -1$ or $n \ge 7.5$.

6. Since $n$ must be a natural number ($n \in \mathbb{N}$), we must have $n \ge 7.5$. The smallest natural number satisfying this condition is $n=8$.

   (Q) matches with (1) which is 8.

Part (R)

The function is $h(x)=\left(x^2-9\right)^n\left(x^2+2 x+3\right)$. We are given that $n$ is a natural number, $n > 5$, and $x=3$ is a point of local minima.

1. For $x=3$ to be a point of local minima, we must have $h(x) \ge h(3)$ for all $x$ in a small neighborhood of 3.

2. Let's evaluate $h(3)$: $h(3) = (3^2-9)^n(3^2+2(3)+3) = 0^n(18) = 0$ (since $n>5$).

3. So, the condition for a local minimum is $h(x) \ge 0$ in a neighborhood of $x=3$.

4. Let's analyze the sign of $h(x)$ near $x=3$. The factor $\left(x^2+2 x+3\right)$ has discriminant $D=2^2-4(1)(3)=-8 < 0$ and a positive leading coefficient, so it is always positive.

5. The sign of $h(x)$ is therefore determined by the sign of $(x^2-9)^n$.

   • For $x$ near 3 but $x>3$, $x^2-9 > 0$, so $(x^2-9)^n > 0$. This means $h(x) > 0$.
   • For $x$ near 3 but $x<3$, $x^2-9 < 0$. The sign of $(x^2-9)^n$ depends on whether $n$ is even or odd.
     • If $n$ is odd, $(x^2-9)^n < 0$, which means $h(x) < 0$. In this case, $x=3$ is not a local minimum.
     • If $n$ is even, $(x^2-9)^n > 0$, which means $h(x) > 0$.

6. For $h(x) \ge 0$ to hold for all $x$ near 3, $n$ must be an even number.

7. We are given that $n$ is a natural number and $n>5$. The smallest even natural number greater than 5 is 6.

   (R) matches with (4) which is 6.

Part (S)

The function is $l(x)=\sum\limits_{k=0}^4\left(\sin |x-k|+\cos \left|x-k+\frac{1}{2}\right|\right)$. We need to find the number of points $x_0$ where $l(x)$ is not differentiable.

1. The differentiability of $l(x)$ depends on the differentiability of its individual terms.

2. Consider the function $f(y) = \sin|y|$. Let's check its differentiability at $y=0$.

   • Right derivative: $\lim_{h \to 0^+} \frac{\sin|h| - \sin|0|}{h} = \lim_{h \to 0^+} \frac{\sin h}{h} = 1$.
   • Left derivative: $\lim_{h \to 0^-} \frac{\sin|h| - \sin|0|}{h} = \lim_{h \to 0^-} \frac{\sin(-h)}{h} = \lim_{h \to 0^-} \frac{-\sin h}{h} = -1$. Since the left and right derivatives are not equal, $\sin|y|$ is not differentiable at $y=0$. Thus, $\sin|x-k|$ is not differentiable at $x=k$.

3. Consider the function $g(y) = \cos|y|$. Since $\cos y$ is an even function, $\cos|y| = \cos y$. The function $\cos y$ is differentiable for all $y \in \mathbb{R}$. Thus, $\cos\left|x-k+\frac{1}{2}\right|$ is differentiable for all $x \in \mathbb{R}$.

4. The function $l(x)$ is a sum of several functions. The terms $\cos\left|x-k+\frac{1}{2}\right|$ for $k=0,1,2,3,4$ are all differentiable everywhere.

5. The points of non-differentiability of $l(x)$ are determined by the terms $\sin|x-k|$. The term $\sin|x-k|$ is not differentiable only at $x=k$. For $k=0$, $\sin|x|$ is not differentiable at $x=0$. For $k=1$, $\sin|x-1|$ is not differentiable at $x=1$. For $k=2$, $\sin|x-2|$ is not differentiable at $x=2$. For $k=3$, $\sin|x-3|$ is not differentiable at $x=3$. For $k=4$, $\sin|x-4|$ is not differentiable at $x=4$.

6. At each point $x_0 \in \{0, 1, 2, 3, 4\}$, one of the terms in the sum is not differentiable while all other terms are differentiable. The sum of a non-differentiable function and differentiable functions is non-differentiable. Therefore, $l(x)$ is not differentiable at these five points.

7. The number of points $x_0$ where $l(x)$ is not differentiable is 5.

   (S) matches with (3) which is 5.

Conclusion

The final matching is:

• (P) → 9 → (2)
• (Q) → 8 → (1)
• (R) → 6 → (4)
• (S) → 5 → (3)

This corresponds to the option (P) → (2), (Q) → (1), (R) → (4), (S) → (3), which is option B.