as f(x) is onto hence A is range of f(x) $\begin{aligned}
& \text { now } \begin{array}{l}
\mathrm{f}^{\prime}(\mathrm{x})=6 \mathrm{x}^2-30 \mathrm{x}+36 \
\quad=6(\mathrm{x}-2)(\mathrm{x}-3) \
\quad \mathrm{f}(2)=16-60+72+7=35 \
\mathrm{f}(3)=54-135+108+7=34 \
\mathrm{f}(0)=7
\end{array}
\end{aligned}hencerange\in[7,35]=\mathrm{A}alsoforrangeofg(x)\begin{aligned}
& g(x)=1-\frac{1}{x^{2025}+1} \in[0,1)=B \
& s={0,7,8, \ldots .35} \text { hence } n(s)=30
\end{aligned}$
Let f:[0,3] → A be defined by f(x)=2 x^3-15 x^2+36 x+7 and g:[0, ∞) → B be defined by g(x)= x^2025 x^2025+1. If both the functions are onto and S=x ∈…
JEE Main 2025 — Mathematics Algebra
Let f:[0,3]→ A be defined by f(x)=2x3−15x2+36x+7 and g:[0,∞)→B be defined by g(x)=x2025+1x2025. If both the functions are onto and S={x∈Z:x∈ A or x∈ B}, then n(S) is equal to :
Held on 28 Jan 2025 · Verified 6 Jul 2026.
Options
- A
29
- B
30
- C
31
- D
36
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