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

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If f(x) = x^2 - x + 1, is f(p)>f(q)? [#permalink]

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New post Updated on: 03 Jun 2018, 23:47
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C
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Question Stats:

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[GMAT math practice question]

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

\(1) p<q\)
\(2) p^2>q^2\)

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Originally posted by MathRevolution on 31 May 2018, 04:37.
Last edited by MathRevolution on 03 Jun 2018, 23:47, edited 1 time in total.
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Re: If f(x) = x^2 - x + 1, is f(p)>f(q)? [#permalink]

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New post 31 May 2018, 05:55
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 Example
Example: 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 Example
Example: \(p^2\) = 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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Re: If f(x) = x^2 - x + 1, is f(p)>f(q)? [#permalink]

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New post 31 May 2018, 08:01
I don't get how answer is D.

Statement 1. Here with two negative numbers satisfying, p<q, answer is yes, and with positive values, answer is No. Insufficient.

Statement 2. Put any value, the answer would be a yes. Hence, sufficient.

Hence, answer should be B.

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Re: If f(x) = x^2 - x + 1, is f(p)>f(q)? [#permalink]

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New post 31 May 2018, 11:43
MathRevolution wrote:
[GMAT math practice question]

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

\(1) p<q\)
\(2) p^2>q^2\)



Please check answer should be C
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Re: If f(x) = x^2 - x + 1, is f(p)>f(q)? [#permalink]

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New post 01 Jun 2018, 12:39
Statement 1...let p=-4 & q=5
F(p)=21 & f(q)=21
Statement 2...let p=5 & q=-4
F(p)=21 & f(q)=21
If we will assume any two positive values, these statements will give different answers for different values.
Hence ans E

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Re: If f(x) = x^2 - x + 1, is f(p)>f(q)? [#permalink]

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New post 02 Jun 2018, 03:17
As \(f(x) = x^2 - x + 1, f(p) = p^2 - p + 1\) and \(f(q) = q^2 - q + 1\)
f(p) > f(q) can be written as f(p) - f(q) > 0.
This can be further written as \(p^2 - p + 1 - q^2 + q - 1 > 0.\)
On further simplification, we get (p - q)(p + q - 1) > 0.

From statement 1, we get p - q < 0.
This is not sufficient as we do not the value of (p + q - 1). It can > 0 or < 0.
So, Statement 1 is insufficient.

From statement 2, we get (p + q )(p - q) > 0
(p + q) > 0 and (p - q) > 0 OR (p + q) < 0 and (p - q) < 0
Case 1: (p + q) > 0. Add -1 on both sides.
(p + q - 1) > -1 this means it can be 0 too. So, may or may not be
Case 2: (p + q) < 0. Add -1 on both sides.
(p + q - 1) < -1 and we have (p - q) < 0. So, > 0.
But as in one of the case we are not sure. Statement 2 is insufficient.

From Statement 1 and Statement 2,
As (p - q) < 0 from Statement 1. This implies (p - q) also < 0. ( As \(p^2 - q^2 > 0\)).
p + q - 1 < -1. This assures us that (p + q - 1) is definitely < 0.
Therefore, we can conclude that (p - q)(p + q - 1) > 0.

So, C is the correct answer.
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Re: If f(x) = x^2 - x + 1, is f(p)>f(q)? [#permalink]

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New post 02 Jun 2018, 05:53
I am also getting C option as an answer, but given OA is D. ?

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If f(x) = x^2 - x + 1, is f(p)>f(q)? [#permalink]

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New post 04 Jun 2018, 12:46
=>

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.
The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.
Modifying the question:
f(p) > f(q)
=> f(p) – f(q) > 0
=> ( p^2 – p + 1 ) – ( q^2 – q + 1 ) > 0
=> ( p^2 – q^2 ) – ( p – q ) > 0
=> ( p + q )( p – q ) – ( p – q ) > 0
=> ( p + q - 1 )( p – q ) > 0
Since we have 2 variables (p and q) and 0 equations, C is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.
Conditions 1) & 2):
Condition 2) tells us that p^2 – q^2 = (p + q)(p – q) > 0. Since p – q < 0 by condition 1), we must have p + q < 0. It follows that p + q – 1 < 0, and
f(p) – f(q) = (p + q – 1)(p – q) > 0.
So, f(p) > f(q), and both conditions are sufficient, when taken together.
Therefore, C is the answer.
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If f(x) = x^2 - x + 1, is f(p)>f(q)?   [#permalink] 04 Jun 2018, 12:46
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