Question 15

(P) → (2): Let $y(x) = 10x^3 - 45x^2 + 60x + 35$. $y'(x) = 30(x-1)(x-2)$. On $[1, 2]$, $y(x)$ is decreasing. $y(1) = 60$ and $y(2) = 55$. For $f(x) = [y(x)/n]$ to be continuous, the range $[55/n, 60/n]$ must not contain an integer in its interior. For $n=9$, the range is $[6.11, 6.66]$, which contains no integer. For $n=8$, the range is $[6.875, 7.5]$, which contains the integer 7, causing a discontinuity. Thus, minimum $n = 9$.

(Q) → (1): $g'(x) = (2n^2 - 13n - 15) \cdot 3(x^2 + 1)$. For $g(x)$ to be increasing, $2n^2 - 13n - 15 \geq 0 \implies (2n - 15)(n + 1) \geq 0$. For $n \in \mathbb{N}$, $n \geq 15/2$, so minimum $n = 8$.

(R) → (4): $h(x) = (x-3)^n (x+3)^n (x^2+2x+3)$. Near $x=3$, $h(x) \approx (x-3)^n \cdot 6^n \cdot 18$. For $x=3$ to be a local minimum, $h(x)$ must not be negative in the neighborhood, implying $n$ must be even. Given $n > 5$, the smallest even natural number is $n = 6$.

(S) → (3): $\cos |u| = \cos u$, which is differentiable everywhere. Thus, non-differentiability of $l(x)$ arises only from $\sin |x-k|$ at $x=k$. For $k = 0, 1, 2, 3, 4$, there are 5 such points.

Matching: P-2, Q-1, R-4, S-3. Correct Option: B