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← JEE Advanced 2025 · Paper 1
MathematicsList match+4 / -1Question 15 of 48

JEE Advanced 2025 · Paper 1, Mathematics question 15

Let R\mathbb{R} denote the set of all real numbers. For a real number xx, let [x][x] denote the greatest integer less than or equal to xx. Let nn denote a natural number.

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

List-I

  • P
    The minimum value of nn for which the function f(x)=[10x3−45x2+60x+35n]f(x) = \left[\dfrac{10x^3 - 45x^2 + 60x + 35}{n}\right] is continuous on the interval [1,2][1, 2], is
  • Q
    The minimum value of nn for which g(x)=(2n2−13n−15)(x3+3x)g(x) = (2n^2 - 13n - 15)(x^3 + 3x), x∈Rx \in \mathbb{R}, is an increasing function on R\mathbb{R}, is
  • R
    The smallest natural number nn which is greater than 55, such that x=3x = 3 is a point of local minima of h(x)=(x2−9)n(x2+2x+3)h(x) = (x^2 - 9)^n(x^2 + 2x + 3), is
  • S
    Number of x0∈Rx_0 \in \mathbb{R} such that l(x)=∑k=04(sin⁡∣x−k∣+cos⁡∣x−k+12∣)l(x) = \displaystyle\sum_{k=0}^{4}\left(\sin|x - k| + \cos\left|x - k + \frac12\right|\right), x∈Rx \in \mathbb{R}, is NOT differentiable at x0x_0, is

List-II

  • 1
    88
  • 2
    99
  • 3
    55
  • 4
    66
  • 5
    1010
Choose one option
Answer and worked solution

Official answer: B

(P) Let g(x)=10x3−45x2+60x+35g(x)=10x^3-45x^2+60x+35. Then g′(x)=30(x−1)(x−2)≤0g'(x)=30(x-1)(x-2)\le0 on [1,2][1,2], so gg falls from 60 to 55.

[g/n][g/n] is constant on [1,2][1,2] exactly when no multiple of nn lies in (55,60](55,60]:

  • n=1,…,8n=1,\dots,8 each have a multiple there (n=6n=6: 60; n=7n=7: 56; n=8n=8: 56).
  • n=9n=9 has none (its multiples nearby are 54 and 63).

So the minimum is n=9n=9 → (2).

(Q) x3+3xx^3+3x is increasing, so we need 2n2−13n−15=(2n−15)(n+1)>02n^2-13n-15=(2n-15)(n+1)>0, i.e. n≥8n\ge8 → (1).

(R) Near x=3x=3, h(x)=(x−3)n⋅(positive)h(x)=(x-3)^n\cdot(\text{positive}). This has a local minimum at 3 exactly when nn is even. The smallest even n>5n>5 is 6 → (4).

(S) cos⁡∣x−k+12∣=cos⁡(x−k+12)\cos|x-k+\frac12|=\cos(x-k+\frac12) is smooth. Each sin⁡∣x−k∣\sin|x-k| has a corner at x=kx=k, where the one-sided derivatives are ∓1\mp1. These corners are at distinct points, so they cannot cancel. That gives 5 points (x=0,…,4x=0,\dots,4) → (3).

The match is P→2, Q→1, R→4, S→3, which is option (B).