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JEE Main 202528 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
28 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Functions

JEE Main 2025 · 28 January · Evening Shift · Mathematics · Functions

Let f:[0,3] arrow A be defined by f(x)=2 x^3-15 x^2+36 x+7 and...

Let f:[0,3]→f:[0,3] \rightarrow A be defined by f(x)=2x3−15x2+36x+7f(x)=2 x^3-15 x^2+36 x+7 and g:[0,∞)→Bg:[0, \infty) \rightarrow B be defined by g(x)=x2025x2025+1\mathrm{g}(x)=\frac{x^{2025}}{x^{2025}+1}. If both the functions are onto and S={x∈Z:x∈ A or x∈ B}\mathrm{S}=\{x \in \mathbf{Z}: x \in \mathrm{~A}\text{ or } x \in \mathrm{~B}\}, then n(S)\mathrm{n}(\mathrm{S}) is equal to :

Options

  1. A2929
  2. B3030
  3. C3131
  4. D3636

Correct answer

B. 3030

Step-by-step solution

as f(x)f(x) is onto hence Aisrangeofis range off(x)  now f′(x)=6x2−30x+36=6(x−2)(x−3)f(2)=16−60+72+7=35f(3)=54−135+108+7=34f(0)=7\begin{aligned} & \text{ now } \begin{array}{l} \mathrm{f}^{\prime}(\mathrm{x})=6 \mathrm{x}^2-30 \mathrm{x}+36 \\ \quad=6(\mathrm{x}-2)(\mathrm{x}-3) \\ \quad \mathrm{f}(2)=16-60+72+7=35 \\ \mathrm{f}(3)=54-135+108+7=34 \\ \mathrm{f}(0)=7 \end{array} \end{aligned} hence range ∈[7,35]=\in[7,35]=\mathrm{A}alsoforrangeofalso for range ofg(x) g(x)=1−1x2025+1∈[0,1)=Bs={0,7,8,….35} hence n(s)=30\begin{aligned} & g(x)=1-\frac{1}{x^{2025}+1} \in[0,1)=B \\ & s=\{0,7,8, \ldots .35\} \text{ hence } n(s)=30 \end{aligned}$
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