Question 1
Consider the function f(x)=∣x−3∣+2. If f(a)=f(b) where a=b, and both a and b are in the domain [−1,7], which of the following must be true?
- a+b=6 and one of a,b is less than 3 while the other is greater than 3 (correct answer)
- a+b=6 and both a and b are greater than or equal to 3
- a+b=3 and one of a,b is less than 3 while the other is greater than 3
- ∣a−3∣=∣b−3∣ but a+b could be any value in the given domain
Explanation: The function f(x)=∣x−3∣+2 has its vertex at x=3, where it achieves its minimum value of f(3)=2. For x<3, we have f(x)=−(x−3)+2=−x+5. For x≥3, we have f(x)=(x−3)+2=x−1. If f(a)=f(b) with a=b, then we need ∣a−3∣+2=∣b−3∣+2, which simplifies to ∣a−3∣=∣b−3∣. This occurs when either a−3=b−3 (giving a=b, which contradicts a=b) or a−3=−(b−3), which gives a−3=−b+3, so a+b=6. For this to happen with a=b, one value must be on the left side of the vertex (<3) and the other on the right side (>3). For example, if a=1, then f(1)=2+2=4, and we need f(b)=4. From b−1=4, we get b=5. Indeed, a+b=1+5=6. Choice D is incorrect because while ∣a−3∣=∣b−3∣ is necessary, it doesn't capture the full constraint that a+b=6.