Mathematics · Functions

JEE Main 2025 — 28 January, Evening Shift — Question 14

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 $g(x)=\frac{x^{2025}}{x^{2025}+1}$ . If both the functions are onto and $S=\{x \in \mathbf{Z}: x \in A$ or $x \in B\}$ , then $n(S)$ is equal to :

Option A:
30
Correct
Option B:
36
Option C:
29
Option D:
31

Answer: A

Step-by-step solution

as $f(x)$ is onto hence $A$ is range of $f(x)$

now $f^{\prime}(x)=6 x^{2}-30 x+36$

$\begin{aligned} & = 6 (x-2) (x-3) \\& f(2)=16-60+72+7=35 \end{aligned}$

$\mathrm{f}(3)=54-135+108+7=34$

$f(0)=7$

hence range $\in[7,35]=\mathrm{A}$

also for range of $g(x)$

$g(x)=1-\frac{1}{x^{2025}+1} \in[0,1)=B$

$s=\{0,1,7,8, \ldots . .35\}$ hence $n(s)=30$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 28 January, Evening Shift
Subject: Mathematics
Chapter: Functions
Topic: One-One, many-one, onto, into, bijective functions

Answer: A

Step-by-step solution

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