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

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Step 1: Identify the Range of $ f(x) $ We are given that $ f(x) $ is onto, meaning its range is $ A $. The derivative of $ f(x) $ is: $$ f'(x) = 6x^2 - 30x + 36 $$ Factoring the expression: $$ f'(x) = 6(x-2)(x-3) $$ Evaluating $ f(x) $ at key points: $$ f(2) = 16 - 60 + 72 + 7 = 35 $$ $$ f(3) = 54 - 135 + 108 + 7 = 34 $$ $$ f(0) = 7 $$ Therefore, the range of $ f(x) $ is: $$ [7, 35] $$ Step 2: Identify the Range of $ g(x) $ The given function for $ g(x) $ is: $$ g(x) = \frac{1}{x^{2025} + 1} $$ Since the denominator is always greater than or equal to 1, the range of $ g(x) $ is: $$ [0, 1] $$ Step 3: Compute the Number of Elements in Set $ S $ The set $ S $ is defined as: $$ S = \{ 0, 7, 8, \ldots, 35 \} $$ The sequence starts at 7 and ends at 35, inclusive. The total number of terms is: $$ 35 - 7 + 1 = 29 $$ Including the element 0 in the set, $$ |S| = 30 $$ Final Answer: 30

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

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