JEE Advanced 2025 — Paper 1 — MATH-IV solution

JEE Advanced2025Paper 1MATH-IV

Let $R$ denote the set of all real numbers. For a real number $x$ , let $[x]$ denote the greatest integer less than or equal to $x$ . 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 $f (x) = [\frac{10 x ^{3} - 45 x ^{2} + 60 x + 35}{n}]$ is continuous on the interval $[1, 2]$ , is	(1) $8$
(Q) The minimum value of $n$ for which $g (x) = (2 n^{2} - 13 n - 15) (x^{3} + 3 x)$ , $x \in R$ , is an increasing function on $R$ , is	(2) $9$
(R) The smallest natural number $n$ which is greater than $5$ , such that $x = 3$ is a point of local minima of $h (x) = (x^{2} - 9)^{n} (x^{2} + 2 x + 3)$ , is	(3) $5$
(S) Number of $x_{0} \in R$ such that $l (x) = k = 0 \sum 4 (sin ∣ x - k ∣ + cos x - k + \frac{1}{2})$ , $x \in R$ , is NOT differentiable at $x_{0}$ , is	(4) $6$
	(5) $10$

A
(P) $\to$ (1); (Q) $\to$ (3); (R) $\to$ (2); (S) $\to$ (5)
B
(P) $\to$ (2); (Q) $\to$ (1); (R) $\to$ (4); (S) $\to$ (3)
C
(P) $\to$ (5); (Q) $\to$ (1); (R) $\to$ (4); (S) $\to$ (3)
D
(P) $\to$ (2); (Q) $\to$ (3); (R) $\to$ (1); (S) $\to$ (5)

Solution

(P). Let $P (x) = 10 x^{3} - 45 x^{2} + 60 x + 35$ . Then $P^{'} (x) = 30 (x - 1) (x - 2)$ , which is negative on $(1, 2)$ . So $P$ is strictly decreasing on $[1, 2]$ from $P (1) = 60$ to $P (2) = 55$ . The range is $[55, 60]$ .

$f (x) = [P (x) / n]$ is continuous on $[1, 2]$ iff $P (x) / n$ never crosses an integer in the open interior of the range. Equivalently, $[55/ n, 60/ n]$ must contain at most one integer in its interior. For $n = 9$ : $[55/9, 60/9] = [6.11 \dots, 6.67 \dots]$ , no integer crossing, so $f$ is constant equal to $6$ and continuous. For $n = 8$ : $[55/8, 60/8] = [6.875, 7.5]$ , crosses the integer $7$ . For $n = 7$ : $[55/7, 60/7] = [7.86 \dots, 8.57 \dots]$ , crosses $8$ . Hence minimum $n = 9.$ (P) $\to$ (2).

(Q). $x^{3} + 3 x$ has derivative $3 x^{2} + 3 > 0$ , so it is strictly increasing. Hence $g$ is increasing iff $2 n^{2} - 13 n - 15 \geq 0$ . Roots of $2 n^{2} - 13 n - 15 = 0$ : $n = \frac{13 \pm 169 + 120}{4} = \frac{13 \pm 17}{4} = - 1$ or $\frac{15}{2} .$ So $2 n^{2} - 13 n - 15 \geq 0$ for $n \leq - 1$ or $n \geq 7.5$ . The minimum natural number is $n = 8.$ (Q) $\to$ (1).

(R). $h (x) = (x - 3)^{n} (x + 3)^{n} (x^{2} + 2 x + 3) .$ The factor $(x + 3)^{n}$ and $x^{2} + 2 x + 3$ are positive at $x = 3$ . Locally near $x = 3$ , the sign of $h (x) - h (3) = h (x)$ is governed by $(x - 3)^{n}$ . For $x = 3$ to be a local minimum, the function must transition from positive on the left to positive on the right of $3$ , which requires $n$ to be even. The smallest even $n > 5$ is $n = 6.$ (R) $\to$ (4).

(S). The function $cos ∣ y ∣$ is differentiable for every $y$ since $cos$ is even. So $cos ∣ x - k + 1/2∣$ is differentiable everywhere. The function $sin ∣ y ∣$ , however, fails to be differentiable at $y = 0$ (the left and right derivatives are $1$ and $- 1$ ). Hence $sin ∣ x - k ∣$ fails to be differentiable at $x = k$ .

Summing $sin ∣ x - k ∣$ for $k = 0, 1, 2, 3, 4$ , the non-differentiability points are $x = 0, 1, 2, 3, 4$ , giving $5$ points.

(S) $\to$ (3).

$∴$ The correct option is (B).

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