Question

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 \[ S = \{ x \in \mathbb{Z} : x \in A \text{ or } x \in B \}, \] then \( n(S) \) is equal to:

• A. 30
• B. 36
• C. 29
• D. 31

Hint:

When dealing with ranges and functions, always consider the behavior of the function and its derivative to understand its range.

Answer 1

The Correct Option is A

Since \( f(x) \) is onto, the range of \( f(x) \) is \( A \). 
Now, the derivative \( f'(x) = 6x^2 - 30x + 36 \), which factors as: \[ f'(x) = 6(x-2)(x-3) \] 
Evaluating \( f(x) \) at various points: 
\[ f(2) = 16 - 60 + 72 + 7 = 35, \quad f(3) = 54 - 135 + 108 + 7 = 34, \quad f(0) = 7 \] 
Hence, the range of \( f(x) \) is \( [7, 35] \). 
Now, for \( g(x) \), we have: \[ g(x) = \frac{1}{x^{2025} + 1}, \quad g(x) \in [0, 1] \] 
Thus, the range of \( g(x) \) is \( [0, 1] \), and the set \( S = \{ 0, 7, 8, \ldots, 35 \} \). Therefore, the number of elements in \( S \) is 30.