Let f:[0,3] rightarrow A be defined by f(x)=2 | vedclass

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(A) Since $f(x)$ is onto,$A$ is the range of $f(x)$.$f'(x) = 6x^2 - 30x + 36 = 6(x-2)(x-3)$.Critical points are $x=2$ and $x=3$.Evaluating at boundari

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$30$

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(A) Since $f(x)$ is onto,$A$ is the range of $f(x)$.$f'(x) = 6x^2 - 30x + 36 = 6(x-2)(x-3)$.Critical points are $x=2$ and $x=3$.Evaluating at boundaries and critical points:$f(0) = 7$$f(2) = 2(8) - 15(4) + 36(2) + 7 = 16 - 60 + 72 + 7 = 35$$f(3) = 2(27) - 15(9) + 36(3) + 7 = 54 - 135 + 108 + 7 = 34$.Since $f(x)$ is continuous on $[0,3]$,the range $A = [7, 35]$.For $g(x) = \frac{x^{2025}}{x^{2025}+1}$,as $x \in [0, \infty)$,$x^{2025} \in [0, \infty)$.Thus,$g(x) = 1 - \frac{1}{x^{2025}+1}$.At $x=0$,$g(0) = 0$. As $x 	o \infty$,$g(x) 	o 1$.So,the range $B = [0, 1)$.$S = \{x \in \mathbb{Z} : x \in [7, 35] \cup [0, 1)\} = \{0, 7, 8, 9, \dots, 35\}$.The number of elements in $S$ is $1 + (35 - 7 + 1) = 1 + 29 = 30$.

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

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