MCQMediumJEE 2025Functions

JEE Mathematics 2025 Question with Solution

Let f:[0,3]Af: [0, 3] \to A be defined by f(x)=2x315x2+36x+7f(x) = 2x^3 - 15x^2 + 36x + 7 and g:[0,)Bg: [0, \infty) \to B be defined by g(x)=xx2025+1.g(x) = \frac{x}{x^{2025} + 1}. If both functions are onto and S={xZ:xA or xB}S = \{ x \in \mathbb{Z} : x \in A \text{ or } x \in B \}, then n(S)n(S) is equal to:](streamdown:incomplete-link)

  • A

    3030

  • B

    3636

  • C

    2929

  • D

    3131

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: f:[0,3]Af: [0,3] \to A with f(x)=2x315x2+36x+7f(x)=2x^3-15x^2+36x+7 and g:[0,)Bg:[0,\infty)\to B with g(x)=xx2025+1g(x)=\frac{x}{x^{2025}+1}. Both functions are onto.

Find: n(S)n(S) where S={xZ:xA or xB}S=\{x\in\mathbb{Z}: x\in A \text{ or } x\in B\}.

For f(x)f(x), find the range on [0,3][0,3].

f(x)=6x230x+36=6(x2)(x3)f'(x)=6x^2-30x+36=6(x-2)(x-3)

So the critical points in the interval are x=2x=2 and x=3x=3. Now evaluate:

f(0)=7f(0)=7 f(2)=2(8)15(4)+72+7=1660+72+7=35f(2)=2(8)-15(4)+72+7=16-60+72+7=35 f(3)=2(27)15(9)+108+7=54135+108+7=34f(3)=2(27)-15(9)+108+7=54-135+108+7=34

Hence the minimum value is 77 and the maximum value is 3535, so

A=[7,35]A=[7,35]

Therefore the integers in AA are 7,8,9,,357,8,9,\ldots,35, whose count is

357+1=2935-7+1=29

Now consider g(x)=xx2025+1g(x)=\frac{x}{x^{2025}+1} for x0x\ge 0.

g(0)=0g(0)=0

and as xx\to\infty,

g(x)0g(x)\to 0

Also, the function attains a positive maximum value less than 11, so the range includes 00 and all other values lie below 11. Thus the only integer belonging to BB is 00.

Therefore,

S={0,7,8,9,,35}S=\{0,7,8,9,\ldots,35\}

and hence

n(S)=29+1=30n(S)=29+1=30

Therefore, the correct option is A.](streamdown:incomplete-link)

Derivative-Based Range Analysis

Given: f:[0,3]Af: [0,3] \to A and g:[0,)Bg:[0,\infty)\to B are onto.

Find: The number of integers in ABA\cup B.

For ff, use derivative to locate extrema:

f(x)=6x230x+36=6(x2)(x3)f'(x)=6x^2-30x+36=6(x-2)(x-3)

Checking endpoint and critical-point values:

f(0)=7,f(2)=35,f(3)=34f(0)=7, \qquad f(2)=35, \qquad f(3)=34

Since ff is continuous on the closed interval [0,3][0,3], its range is the full interval from minimum to maximum:

A=[7,35]A=[7,35]

So the integers in AA are exactly 77 through 3535, giving

2929

integers.

For gg, differentiate:

g(x)=(x2025+1)x(2025x2024)(x2025+1)2g'(x)=\frac{(x^{2025}+1)-x(2025x^{2024})}{(x^{2025}+1)^2} =12024x2025(x2025+1)2=\frac{1-2024x^{2025}}{(x^{2025}+1)^2}

Thus the maximum occurs when

12024x2025=01-2024x^{2025}=0

that is,

x2025=12024x^{2025}=\frac{1}{2024}

At that point the maximum value is positive but less than 11. Also g(0)=0g(0)=0. Hence

B[0,1)B\subset [0,1)

and the only integer in BB is 00.

So,

S={xZ:xA or xB}={0,7,8,9,,35}S=\{x\in\mathbb{Z}:x\in A \text{ or } x\in B\}=\{0,7,8,9,\ldots,35\}

Therefore,

n(S)=29+1=30n(S)=29+1=30

So the correct option is A.](streamdown:incomplete-link)

Common mistakes

  • Taking the range of f(x)f(x) only from the endpoint values f(0)f(0) and f(3)f(3) is incorrect because a cubic can attain an interior maximum or minimum on the interval. Differentiate first and check the critical point x=2x=2 as well.

  • Assuming B=(0,1)B=(0,1) and forgetting that g(0)=0g(0)=0 is wrong. Since 00 is attained, 0B0\in B and must be counted in SS.

  • Counting integers in [7,35][7,35] as 357=2835-7=28 is an endpoint-counting error. For inclusive integer counts, use ba+1b-a+1, so the total is 357+1=2935-7+1=29.

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