If f(x) = 4x^2 + px, what is the minimum value of f(x)?

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Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 1 variable (p) and 0 equations, D is most likely to be the answer. So, we should consider each of the conditions on their own first.

Condition 1)
Since \(f(1) = -4,\) we have \(4 + p = -4\) and p = \(-8.\)
\(f(x) = 4x^2 -8x = 4( x^2 - 2x + 1 - 1 ) = 4(x-1)^2 – 4.\)
The minimum of \(f(x)\) is \(-4\).
Condition 1) is sufficient.

Condition 2)
Since \(f(2) = 0\), we have \(16 + 2p = 0\) and \(p = -8.\)
\(f(x) = 4x^2 -8x = 4( x^2 - 2x + 1 - 1 ) = 4(x-1)^2 – 4.\)
The minimum of \(f(x)\) is \(-4.\)
Condition 2) is sufficient.

Therefore, D is the answer.

Answer: D

Note: Tip 1) of the VA method states that D is most likely to be the answer if condition 1) gives the same information as condition 2).


If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.