JEE Advanced 2025 Paper 1 · Q15 · Maxima & Minima

Question

Let $[x]$ denote the greatest integer $\le x$ and let $n$ be a natural number. List-I: (P) Minimum $n$ such that $f(x) = \left[\frac{10x^3 - 45x^2 + 60x + 35}{n}\right]$ is continuous on $[1, 2]$. (Q) Minimum $n$ such that $g(x) = (2n^2 - 13n - 15)(x^3 + 3x)$ is increasing on $\mathbb{R}$. (R) Smallest natural $n > 5$ such that $x = 3$ is a local min of $h(x) = (x^2 - 9)^n (x^2 + 2x + 3)$. (S) Number of $x_0 \in \mathbb{R}$ at which $\ell(x) = \sum_{k=0}^{4} \left( \sin|x - k| + \cos\left|x - k + \frac{1}{2}\right| \right)$ is NOT differentiable. List-II: (1) $8$; (2) $9$; (3) $5$; (4) $6$; (5) $10$. Choose: (A) P->1,Q->3,R->2,S->5; (B) P->2,Q->1,R->4,S->3; (C) P->5,Q->1,R->4,S->3; (D) P->2,Q->3,R->1,S->5.

SolveKar AI · Accepted Answer

(P) Let $p(x) = 10x^3 - 45x^2 + 60x + 35$. $p'(x) = 30x^2 - 90x + 60 = 30(x-1)(x-2)$, so $p$ is increasing at the endpoints; compute $p(1) = 60$, $p(2) = 55$ (local max at $1$, local min at $2$). Range on $[1,2] = [55, 60]$. $f(x) = [p(x)/n]$ is continuous on $[1,2]$ iff $p(x)/n$ never crosses an integer, i.e., the interval $[55/n, 60/n]$ contains no integer in its interior. For $n=9$: $55/9 \approx 6.11$, $60/9 \approx 6.67$ — no integer crossed. So minimum $n = 9$. (P) $	o$ $9 = (2)$. (Q) $g'(x) = (2n^2 - 13n - 15)(3x^2 + 3) = 3(x^2+1)(2n^2 - 13n - 15)$. Always non-negative iff $2n^2 - 13n - 15 \ge 0$. Roots: $n = (13 \pm \sqrt{169 + 120})/4 = (13 \pm 17)/4$, i.e., $n = 7.5$ or $n = -1$. So need $n \ge 8$ (natural). Min $n = 8$. (Q) $	o$ $8 = (1)$. (R) $h(x) = (x-3)^n (x+3)^n (x^2 + 2x + 3)$. At $x = 3$, the factor $(x-3)^n$ controls behaviour locally. $(x+3)^n (x^2+2x+3)$ is positive at $x=3$ ($= 6^n \cdot 18 > 0$). For $x = 3$ to be a local minimum, the sign of $(x-3)^n$ must be non-negative on both sides — this requires $n$ even. Smallest natural $n > 5$ and even: $n = 6$. (R) $	o$ $6 = (4)$. (S) For each $k \in \{0,1,2,3,4\}$, $\sin|x-k|$ has a corner at $x = k…[truncated]

JEE Advanced 2025 Paper 1 · Q15 · Maxima & Minima

Solution