The function f is defined by \(\mathrm{f(x) = ax^2 + bx + c}\), where a, b, and c are constants....
GMAT Advanced Math : (Adv_Math) Questions
The function \(\mathrm{f}\) is defined by \(\mathrm{f(x) = ax^2 + bx + c}\), where \(\mathrm{a}\), \(\mathrm{b}\), and \(\mathrm{c}\) are constants. The graph of \(\mathrm{y = f(x)}\) in the xy-plane passes through the points \(\mathrm{(1, 3)}\) and \(\mathrm{(4, 9)}\). If \(\mathrm{a}\) is an integer greater than \(\mathrm{1}\), which of the following could be the value of \(\mathrm{a + b}\)?
\(\mathrm{-6}\)
\(\mathrm{-2}\)
\(\mathrm{6}\)
\(\mathrm{10}\)
1. TRANSLATE the problem information
- Given information:
- \(\mathrm{f(x) = ax^2 + bx + c}\) passes through \(\mathrm{(1, 3)}\) and \(\mathrm{(4, 9)}\)
- \(\mathrm{a}\) is an integer greater than \(\mathrm{1}\)
- Need to find possible value of \(\mathrm{a + b}\)
- What this tells us: Since the points lie on the graph, their coordinates must satisfy the function equation
2. TRANSLATE each point condition into equations
- For point \(\mathrm{(1, 3)}\): \(\mathrm{f(1) = 3}\)
- \(\mathrm{a(1)^2 + b(1) + c = 3}\)
- \(\mathrm{a + b + c = 3}\) ... (equation 1)
- For point \(\mathrm{(4, 9)}\): \(\mathrm{f(4) = 9}\)
- \(\mathrm{a(4)^2 + b(4) + c = 9}\)
- \(\mathrm{16a + 4b + c = 9}\) ... (equation 2)
3. INFER the strategy to eliminate variables
- We have two equations with three unknowns \(\mathrm{(a, b, c)}\)
- Since we only need \(\mathrm{a + b}\), we should eliminate \(\mathrm{c}\) first
- Subtracting equation 1 from equation 2 will eliminate \(\mathrm{c}\)
4. SIMPLIFY by eliminating c
- \(\mathrm{(16a + 4b + c) - (a + b + c) = 9 - 3}\)
- \(\mathrm{15a + 3b = 6}\)
- Divide both sides by 3: \(\mathrm{5a + b = 2}\)
5. INFER the relationship for a + b
- From \(\mathrm{5a + b = 2}\), we get: \(\mathrm{b = 2 - 5a}\)
- Therefore: \(\mathrm{a + b = a + (2 - 5a) = 2 - 4a}\)
6. APPLY CONSTRAINTS to find valid values
- Since \(\mathrm{a}\) must be an integer greater than \(\mathrm{1}\), test values:
- \(\mathrm{a = 2}\): \(\mathrm{a + b = 2 - 4(2) = -6}\) ← This matches choice (A)
- Check if other choices work:
- Choice (B): \(\mathrm{-2 = 2 - 4a}\) → \(\mathrm{a = 1}\), but \(\mathrm{a}\) must be \(\mathrm{\gt 1}\) ✗
- Choice (C): \(\mathrm{6 = 2 - 4a}\) → \(\mathrm{a = -1}\), but \(\mathrm{a}\) must be \(\mathrm{\gt 1}\) ✗
- Choice (D): \(\mathrm{10 = 2 - 4a}\) → \(\mathrm{a = -2}\), but \(\mathrm{a}\) must be \(\mathrm{\gt 1}\) ✗
Answer: A
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak INFER skill: Students substitute the points correctly but don't recognize they need to eliminate \(\mathrm{c}\) to find a relationship between \(\mathrm{a}\) and \(\mathrm{b}\).
Instead, they might try to solve for each variable individually or get overwhelmed by having three unknowns with only two equations. Without the key insight to subtract equations, they get stuck and resort to guessing.
This leads to confusion and random answer selection.
Second Most Common Error:
Poor APPLY CONSTRAINTS execution: Students correctly find \(\mathrm{a + b = 2 - 4a}\) but forget to check that \(\mathrm{a}\) must be an integer greater than \(\mathrm{1}\).
They might select the first answer choice that works mathematically without verifying the constraint. For example, they could work backwards from choice (B) \(\mathrm{-2}\), find \(\mathrm{a = 1}\), and not realize this violates the condition \(\mathrm{a \gt 1}\).
This may lead them to select Choice B (-2).
The Bottom Line:
This problem tests your ability to connect point-on-graph conditions with algebraic manipulation, requiring both systematic equation solving and careful attention to constraints.
\(\mathrm{-6}\)
\(\mathrm{-2}\)
\(\mathrm{6}\)
\(\mathrm{10}\)