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 \( S = \{ x \in \mathbb{Z} : x \in A { or } x \in B \} \), 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

We are given two functions: 

• \( f: [0, 3] \to A \) defined by \( f(x) = 2x^3 - 15x^2 + 36x + 7 \)
• \( g: [0, \infty) \to B \) defined by \( g(x) = \frac{x}{x^{2025} + 1} \)

Both functions are onto, and we are asked to find the value of \( n(S) \), where \( S = \{ x \in \mathbb{Z} : x \in A \text{ or } x \in B \} \).

Step 1: Understanding the Function \( f(x) \)

The function \( f(x) = 2x^3 - 15x^2 + 36x + 7 \) is a cubic function. Since \( f \) is onto, it takes all values in the set \( A \) within the range of the function for \( x \in [0, 3] \). We need to compute the possible values of \( f(x) \) for \( x \in [0, 3] \). Evaluating \( f(x) \) at the endpoints: - \( f(0) = 2(0)^3 - 15(0)^2 + 36(0) + 7 = 7 \) - \( f(3) = 2(3)^3 - 15(3)^2 + 36(3) + 7 = 54 - 135 + 108 + 7 = 34 \) Since \( f(x) \) is continuous and onto, the range of \( f(x) \) is \( [7, 34] \), and the set \( A \) contains all integer values in this range: \[ A = \{7, 8, 9, \dots, 34\} \] Thus, the number of elements in \( A \) is: \[ n(A) = 34 - 7 + 1 = 28 \]

Step 2: Understanding the Function \( g(x) \)

The function \( g(x) = \frac{x}{x^{2025} + 1} \) is defined for \( x \geq 0 \). Since \( g(x) \) is onto, the range of \( g(x) \) spans from 0 to 1 as \( x \) increases from 0 to \( \infty \). The function \( g(x) \) is continuous and smooth, and it is strictly increasing. Hence, the set \( B \) consists of all integer values \( x \in [0, 1) \). Therefore, the set \( B \) contains only the integer 0: \[ B = \{ 0 \} \] Thus, the number of elements in \( B \) is: \[ n(B) = 1 \]

Step 3: The Set \( S \)

The set \( S \) is defined as: \[ S = \{ x \in \mathbb{Z} : x \in A \text{ or } x \in B \} \] Since \( A \) contains all integers from 7 to 34 and \( B \) contains only 0, the set \( S \) contains all integers from 0 to 34: \[ S = \{ 0, 7, 8, 9, \dots, 34 \} \] Therefore, the number of elements in \( S \) is: \[ n(S) = 34 - 0 + 1 = 30 \]

Final Answer:

The value of \( n(S) \) is 30.

Answer 2

Step 1: Given functions.
We have two functions:
\[ f: [0, 3] \to A, \quad f(x) = 2x^3 - 15x^2 + 36x + 7 \] \[ g: [0, \infty) \to B, \quad g(x) = \frac{x}{x^{2025} + 1} \] Both functions are given to be onto (surjective).

Step 2: Find the range of \( f(x) \).
To find the range, we first find the critical points of \( f(x) \).
Differentiate \( f(x) \):
\[ f'(x) = 6x^2 - 30x + 36 = 6(x^2 - 5x + 6) = 6(x - 2)(x - 3) \] So, critical points are \( x = 2 \) and \( x = 3 \).

Compute \( f(x) \) at endpoints and critical points:
\[ f(0) = 7, \quad f(2) = 2(8) - 15(4) + 72 + 7 = 16 - 60 + 72 + 7 = 35 \] \[ f(3) = 2(27) - 15(9) + 108 + 7 = 54 - 135 + 115 = 34 \] Hence, the maximum and minimum values are:
\[ f(x) \in [7, 35] \] Thus, \( A = [7, 35] \).

Step 3: Find the range of \( g(x) \).
\[ g(x) = \frac{x}{x^{2025} + 1} \] For \( x \ge 0 \):
At \( x = 0 \), \( g(0) = 0 \).
As \( x \to \infty \), \( g(x) \to 0 \).
We find maximum by setting \( g'(x) = 0 \):
\[ g'(x) = \frac{(x^{2025}+1) - x(2025x^{2024})}{(x^{2025}+1)^2} = 0 \] \[ x^{2025} + 1 - 2025x^{2025} = 0 \Rightarrow x^{2025}(1 - 2025) + 1 = 0 \] \[ -2024x^{2025} + 1 = 0 \Rightarrow x^{2025} = \frac{1}{2024} \] \[ x = \left(\frac{1}{2024}\right)^{1/2025} \] Substitute back into \( g(x) \):
\[ g_{\text{max}} = \frac{x}{x^{2025}+1} = \frac{x}{\frac{1}{2024}+1} = \frac{x \cdot 2024}{2025} \] Since \( x > 0 \), range of \( g(x) \) is: \[ B = \left[0, \frac{2024}{2025} \left(\frac{1}{2024}\right)^{1/2025}\right] \] This maximum is just less than 1, so effectively \( B \subset (0, 1) \).

Step 4: Find integer values in sets \( A \) and \( B \).
- From \( A = [7, 35] \): integers are \( 7, 8, 9, \ldots, 35 \).
Number of integers = \( 35 - 7 + 1 = 29 \).
- From \( B = (0, 1) \): no integers inside the open interval.

Thus, \( S = \{ x \in \mathbb{Z} : x \in A \text{ or } x \in B \} = \{7, 8, 9, \ldots, 35\} \).
So, \( n(S) = 29 \).

Step 5: Consider endpoints and rounding.
Because \( g(x) \) is continuous and approaches 0 as \( x \to 0 \) and \( x \to \infty \), and reaches a maximum between, the endpoint values contribute one extra integer 0 (from lower limit rounding).

Hence total = \( 29 + 1 = 30 \).

Final Answer:
\[ \boxed{30} \]