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 $x \in B\}$, then $\mathrm{n}(\mathrm{S})$ is equal to : [JEE Main 2025, 28 Jan (Shift 2)]

Question

YESPYQ · Accepted Answer

Correct answer: c) 30. $\text{Given that }f(x)\text{ is onto }\ \text{therefore range of}f(x)=A\ {f}^{'}(x)=6{x}^{2}-30x+36\ =6(x-2)(x-3)\ f(x)=2{x}^{3}-15{x}^{2}+36x+7\ f(2)=16-60+72+7=35\ f(3)=54-135+108+7=34\ f(0)=7\ \text{hence range}\in [7,35]=\mathrm{A}\ \text{also for range of}g(x)\ g(x)=1-\frac{1}{{x}^{2025}+1}\in [0,1)=B\ S={0,7,8,\ldots ..35}\ \text{hence}n(s)=30$

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 …

Practice more JEE Maths PYQs