Let ℝ denote the set of all real numbers. For a real number x, let [x] denote the greatest integer ≤ x. Let n denote a natural number.

Match each entry in List-I to the correct entry in List-II and choose the correct option.

List-I	List-II
(P) The minimum value of n for which the function f(x) = [ (10x³ − 45x² + 60x + 35) / n ] is continuous on the interval [1, 2], is	(1) 8
(Q) The minimum value of n for which g(x) = (2n² − 13n − 15)(x³ + 3x), x ∈ ℝ, is an increasing function on ℝ, is	(2) 9
(R) The smallest natural number n which greater than 5, such that x = 3 is a point of local minima of h(x) = (x² − 9)ⁿ (x² + 2x + 3), is	(3) 5
(S) Number of x₀ ∈ ℝ such that f(x) = Σ (k = 0 to 4) [ sin|x − k| + cos|x − k + 1/2| ], is NOT differentiable at x₀, is	(4) 6
	(5) 10

We are given the four entries in List-I and need to match them with the correct values in List-II.

Analysis of Entry (P):
Let $g\left(x\right)=10x^3-45x^2+60x+35$.
To understand the behavior of $g\left(x\right)$ on the interval $\left[1,2\right]$, we find its derivative:
$g^{\prime }\left(x\right)=30x^2-90x+60=30\left(x-1\right)\left(x-2\right)$.
For $x\in \left[1,2\right]$, we have $g^{\prime }\left(x\right)\le 0$, meaning $g\left(x\right)$ is a decreasing function on this interval.
The boundary values of $g\left(x\right)$ are:
$g\left(1\right)=10-45+60+35=60$
$g\left(2\right)=10\left(8\right)-45\left(4\right)+60\left(2\right)+35=80-180+120+35=55$.
Thus, the range of $\frac{g\left(x\right)}{n}$ on $\left[1,2\right]$ is $\left[\frac{55}{n},\frac{60}{n}\right]$.

For the greatest integer function $f\left(x\right)=\left[\frac{g\left(x\right)}{n}\right]$ to be continuous on $\left[1,2\right]$, it must be constant. This requires that the interval $\left[\frac{55}{n},\frac{60}{n}\right]$ does not contain any integers in its interior, and if it contains an integer as a boundary, it must not cause a discontinuity.
Let's check the natural number values for $n$:
If $n=8$, the interval is $\left[6.875,7.5\right]$, which contains the integer $7$, causing a discontinuity.
If $n=9$, the interval is $\left[6.11,6.67\right]$. This interval does not contain any integers since $6<6.11$ and $6.67<7$.
Thus, the minimum value of $n$ is $9$.
Therefore, (P) → (2).

Analysis of Entry (Q):
We are given $g\left(x\right)=\left(2n^2-13n-15\right)\left(x^3+3x\right)$.
Let $k\left(x\right)=x^3+3x$. Its derivative is $k^{\prime }\left(x\right)=3x^2+3>0$, so $x^3+3x$ is strictly increasing on $ℝ$.
For $g\left(x\right)$ to be an increasing function, the coefficient must be non-negative:
$2n^2-13n-15\ge 0$
Factoring the expression:
$\left(2n-15\right)\left(n+1\right)\ge 0$.
Since $n$ is a natural number, $n+1>0$. Hence, we must have:
$2n-15\ge 0\Rightarrow n\ge 7.5$.
The minimum natural number satisfying this inequality is $n=8$.
Therefore, (Q) → (1).

Analysis of Entry (R):
We are given $h\left(x\right)={\left(x^2-9\right)}^n\left(x^2+2x+3\right)={\left(x-3\right)}^n{\left(x+3\right)}^n\left(x^2+2x+3\right)$.
For a neighborhood around $x=3$, let $P\left(x\right)={\left(x+3\right)}^n\left(x^2+2x+3\right)$.
Since $P\left(3\right)=6^n\left(18\right)>0$, $P\left(x\right)$ remains strictly positive in a small interval around $x=3$.
Thus, the sign and extremum properties of $h\left(x\right)$ near $x=3$ depend entirely on ${\left(x-3\right)}^n$.
For $x=3$ to be a local minimum, we must have $h\left(x\right)\ge h\left(3\right)=0$ in a neighborhood of $x=3$.
This requires ${\left(x-3\right)}^n\ge 0$ for all $x$ near $3$, which is true if and only if $n$ is an even integer.
We are looking for the smallest natural number $n>5$. Since $n$ must be even, the smallest such number is $n=6$.
Therefore, (R) → (4).

Analysis of Entry (S):
We have the function:
$f\left(x\right)=\sum _{k=0}^4\left[\sin \left|x-k\right|+\cos \left|x-k+\frac{1}{2}\right|\right]$.
Let us evaluate each term of the sum:
1. $\cos \left|\theta \right|=\cos \left(\theta \right)$ since cosine is an even function. Therefore, $\cos \left|x-k+\frac{1}{2}\right|=\cos \left(x-k+\frac{1}{2}\right)$ is differentiable everywhere on $ℝ$.
2. For $\sin \left|x-k\right|$, at $x=k$:
- The right-hand derivative is $\underset{h\to 0^{+}}{\lim }\frac{\sin \left(h\right)-0}{h}=1$.
- The left-hand derivative is $\underset{h\to 0^{+}}{\lim }\frac{\sin \left(h\right)-0}{-h}=-1$.
Thus, $\sin \left|x-k\right|$ is not differentiable at $x=k$.
Since the summation runs from $k=0$ to $4$, the points of non-differentiability are exactly $x\in \left\{0,1,2,3,4\right\}$.
The number of such points $x_0$ is $5$.
Therefore, (S) → (3).

Combining all the results, we get the matching:
(P)→(2), (Q)→(1), (R)→(4), (S)→(3).