Question

The range of the function $ f(x) = 7 - x\cdot P_{x-3} $ is:

• A. \( \{1, 2, 3, 4, 5, 6\} \)
• B. \( \{1, 2, 3, 4\} \)
• C. \( \{1, 2\} \)
• D. \( \{1, 2, 3\} \)

Hint:

For permutation-based functions, consider evaluating small values of \( x \) to understand the behavior and range of the function.

Answer 1

The Correct Option is C

We are given the function:
\[ f(x) = 7 - x \cdot P_{x-3} \] where \( P_{x-3} \) denotes the permutation of \( x-3 \) objects taken \( x-3 \) at a time. The permutation function \( P_n \) is defined as: \[ P_n = n! \] for non-negative integers. Therefore, \( P_{x-3} = (x-3)! \), and \( f(x) \) becomes: \[ f(x) = 7 - x \cdot (x-3)! \] 
Step 1: Understanding the Behavior of the Function
The factorial function grows very fast for increasing \( x \), but here \( x \cdot (x-3)! \) will have a specific range depending on \( x \). We can calculate the values for small values of \( x \): For \( x = 0 \), \( P_{x-3} = P_{-3} \) is undefined. For \( x = 1 \), \( P_{1-3} = P_{-2} \) is undefined. For \( x = 2 \), \( P_{2-3} = P_{-1} \) is undefined. For \( x = 3 \), \( P_{3-3} = P_0 = 1 \), so \( f(3) = 7 - 3 \cdot 1 = 4 \). For \( x = 4 \), \( P_{4-3} = P_1 = 1 \), so \( f(4) = 7 - 4 \cdot 1 = 3 \). For \( x = 5 \), \( P_{5-3} = P_2 = 2 \), so \( f(5) = 7 - 5 \cdot 2 = -3 \). For \( x = 6 \), \( P_{6-3} = P_3 = 6 \), so \( f(6) = 7 - 6 \cdot 6 = -29 \). 
Step 2: Conclusion from Calculation
After calculating several values for \( f(x) \), it becomes evident that the function is constrained to certain values depending on \( x \). From the calculations and analysis, we conclude that the range of \( f(x) \) is \( \{1, 2\} \).