Question

If \( 0 \leq x \leq 5 \), then the greatest value of \( \alpha \) and the least value of \( \beta \) satisfying the inequalities \( \alpha \leq 3x + 5 \leq \beta \) are, respectively,

• A. \(0,5\)
• B. \(10,15\)
• C. \(5,10\)
• D. \(5,15\)
• E. \(5,20\)

Hint:

When solving inequalities involving a linear function within a specific interval, always evaluate the function at the boundaries of the interval. This will give you the minimum and maximum values the function can take. These values directly determine the limits for any variables compared against this function, helping in identifying the range for parameters like \( \alpha \) and \( \beta \) in inequalities.

Answer 1

Answer: (E)

To determine the values of \( \alpha \) and \( \beta \), we analyze the behavior of the function \( f(x) = 3x + 5 \) within the given interval \( 0 \leq x \leq 5 \).
Step 1: Calculate the minimum and maximum values of \( f(x) \) over the interval. \[ {Minimum at } x = 0: \quad f(0) = 3 \cdot 0 + 5 = 5. \] \[ {Maximum at } x = 5: \quad f(5) = 3 \cdot 5 + 5 = 20. \] Thus, the function \( f(x) \) ranges from 5 to 20 over the interval.
Step 2: Find \( \alpha \) and \( \beta \) such that \( \alpha \leq 5 \) and \( 20 \leq \beta \).
The greatest possible value of \( \alpha \) that satisfies \( \alpha \leq 5 \) is 5.
The least possible value of \( \beta \) that satisfies \( 20 \leq \beta \) is 20.
Conclusion: The greatest value of \( \alpha \) is 5 and the least value of \( \beta \) is 20, matching option (E).