MathRevolution wrote:

[GMAT math practice question]

If \(f(x) = x^2 - x + 1\), is \(f(p)>f(q)\)?

\(1) p<q\)

\(2) p^2>q^2\)

The question is an "IS" question. If we can answer with a conclusive yes or a conclusive no, the data is sufficient.

Statement 1: p < q

Approach:

Counter ExampleExample: p = 0, q = 2; p < q;

f(p) = 1 and f(q) = 3. f(p) < f(q). Answer No.

Counter example: p = -3, q = 2; p < q;

f(p) = 13 and f(q) = 3; f(p) > f(q). Answer Yes.

Statement 1 alone is NOT sufficient.

Statement 2: \(p^2>q^2\)

Approach:

Counter ExampleExample: \(p^2\) = 9, \(q^2\) = 4; \(p^2>q^2\);

Possibility 1: p = -3 and q = 2; f(p) = 13 and f(q) = 3; f(p) > f(q). Answer Yes.

Possibility 2: p = 3 and q = -2; f(p) = 7 and f(q) = 7; f(p) = f(q). Answer No.

Statement 2 alone is NOT sufficient.

Combining the 2 statements: p < q and \(p^2>q^2\)

From statement 2, we can infer that |p| > |q|

If the magnitude of p is greater than q and p < q, we can infer that p is negative and q could be positive, negative or zero.

Let us take the first two terms of the function to compare f(p) and f(q).

The first term is x^2. If p^2 > q^2, the first term of f(p) > f(q)

The second term is -x. If p is a larger magnitude negative number, effectively we will be adding a larger number. If q is a negative number, its magnitude will lesser than p. So, we will be adding a lesser value. If q is 0, we will add lesser than what p would have contributed to f(p). If q is positive, we will be taking away value from f(q). In all 3 scenarios it pans out that the value added by the second term of f(p) > f(q).

Therefore, we can conclude that f(p) > f(q)

Choice C is the answer.

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An IIM C Alumnus - Class of '94

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