Q34 / 75 · JEE Main 2025 (28 Jan Shift 2) · Mathematics

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

  • A
    $29$
  • Correct answer
    B
    $30$
  • C
    $31$
  • D
    $36$

Solution

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

Hence, the correct answer is option: 'B'.

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