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A function f(x) satisfies 2f(x)+3f(1-x)=x^2. What is the expression of

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MathRevolution
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A function f(x) satisfies 2f(x)+3f(1-x)=x^2. What is the expression of [#permalink] New post   08 Oct 2019, 00:45
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[GMAT math practice question]

A function \(f(x)\) satisfies \(2f(x)+3f(1-x)=x^2.\) What is the expression of \(f(x)\)?

\(A. \frac{(x^2 +6x +3)}{7}\)

\(B. \frac{(x^2 -6x +3)}{7}\)

\(C. \frac{(x^2 -6x +1)}{3}\)

\(D. \frac{(x^2 -6x +3)}{5}\)

\(E. \frac{(x^2 +6x +3)}{5}\)
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D
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nick1816
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A function f(x) satisfies 2f(x)+3f(1-x)=x^2. What is the expression of [#permalink] New post   08 Oct 2019, 01:10
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\(2f(x)+3f(1-x)=x^2.\) .......(1)


put x=1-x, we get

\(2f(1-x)+3f(x)=(1-x)^2.\) ........(2)

Multiply equation (1) by 2 and equation (2) by 3 and subtract both equations

we'll get

f(x)= \(\frac{(x^2 -6x +3)}{5}\)


MathRevolution wrote:
[GMAT math practice question]

A function \(f(x)\) satisfies \(2f(x)+3f(1-x)=x^2.\) What is the expression of \(f(x)\)?

\(A. \frac{(x^2 +6x +3)}{7}\)

\(B. \frac{(x^2 -6x +3)}{7}\)

\(C. \frac{(x^2 -6x +1)}{3}\)

\(D. \frac{(x^2 -6x +3)}{5}\)

\(E. \frac{(x^2 +6x +3)}{5}\)
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Re: A function f(x) satisfies 2f(x)+3f(1-x)=x^2. What is the expression of [#permalink] New post   08 Oct 2019, 01:53
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MathRevolution wrote:
[GMAT math practice question]

A function \(f(x)\) satisfies \(2f(x)+3f(1-x)=x^2.\) What is the expression of \(f(x)\)?

\(A. \frac{(x^2 +6x +3)}{7}\)

\(B. \frac{(x^2 -6x +3)}{7}\)

\(C. \frac{(x^2 -6x +1)}{3}\)

\(D. \frac{(x^2 -6x +3)}{5}\)

\(E. \frac{(x^2 +6x +3)}{5}\)


Given: A function \(f(x)\) satisfies \(2f(x)+3f(1-x)=x^2.\)

Asked: What is the expression of \(f(x)\)?

For x = 0
2f(0) + 3f(1) = 0

For x = 1
2f(1) + 3f(0) = 1

f(0) - f(1) = 1
f(0) + f(1) = .2

f(0) = .6
f(1) = -.4

Only D \(D. \frac{(x^2 -6x +3)}{5}\) satisfies the conditions

IMO D
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Re: A function f(x) satisfies 2f(x)+3f(1-x)=x^2. What is the expression of [#permalink] New post   10 Oct 2019, 01:34
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=>

We have \(4f(x)+6f(1-x)=2x^2\) when we multiply the equation in the original condition by \(2\).

If we substitute \(x\) by \(1-x,\) then we have \(2f(1-x)+3f(x) = (1-x)^2.\)

If we multiply the equation by \(3\), we have \(6f(1-x)+9f(x) = 3(1-x)^2.\)

If we subtract equations \([4f(x)+6f(1-x)=2x^2] - [6f(1-x)+9f(x) = 3(1-x)^2]\), then we have \(-5f(x) = 2x^2 – 3(1-x)^2, -5f(x) = 2x^2 - 3(1-2x+x^2), -5f(x) = 2x^2 - 3 + 6x - 3x^2, -5f(x) = -x^2 + 6x – 3\) or \(f(x) = (\frac{1}{5})(x^2-6x+3).\)

Therefore, the answer is D.
Answer: D
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Re: A function f(x) satisfies 2f(x)+3f(1-x)=x^2. What is the expression of [#permalink] New post   27 Nov 2019, 07:23
MathRevolution, could you please explain why you chose those operations: substituting, multiplying by 3, etc.?
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Re: A function f(x) satisfies 2f(x)+3f(1-x)=x^2. What is the expression of [#permalink] New post   15 Jul 2021, 06:02
MathRevolution wrote:
=>

We have \(4f(x)+6f(1-x)=2x^2\) when we multiply the equation in the original condition by \(2\).

If we substitute \(x\) by \(1-x,\) then we have \(2f(1-x)+3f(x) = (1-x)^2.\)

If we multiply the equation by \(3\), we have \(6f(1-x)+9f(x) = 3(1-x)^2.\)

If we subtract equations \([4f(x)+6f(1-x)=2x^2] - [6f(1-x)+9f(x) = 3(1-x)^2]\), then we have \(-5f(x) = 2x^2 – 3(1-x)^2, -5f(x) = 2x^2 - 3(1-2x+x^2), -5f(x) = 2x^2 - 3 + 6x - 3x^2, -5f(x) = -x^2 + 6x – 3\) or \(f(x) = (\frac{1}{5})(x^2-6x+3).\)

Therefore, the answer is D.
Answer: D

In this part If we substitute \(x\) by \(1-x,\) then we have \(2f(1-x)+3f(x) = (1-x)^2.\)
What made you want to substitute (1-x) in for x?
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Re: A function f(x) satisfies 2f(x)+3f(1-x)=x^2. What is the expression of [#permalink] New post   02 Aug 2022, 10:57
nick1816 wrote:
\(2f(x)+3f(1-x)=x^2.\) .......(1)


put x=1-x, we get

\(2f(1-x)+3f(x)=(1-x)^2.\) ........(2)

Multiply equation (1) by 2 and equation (2) by 3 and subtract both equations

we'll get

f(x)= \(\frac{(x^2 -6x +3)}{5}\)


MathRevolution wrote:
[GMAT math practice question]

A function \(f(x)\) satisfies \(2f(x)+3f(1-x)=x^2.\) What is the expression of \(f(x)\)?

\(A. \frac{(x^2 +6x +3)}{7}\)

\(B. \frac{(x^2 -6x +3)}{7}\)

\(C. \frac{(x^2 -6x +1)}{3}\)

\(D. \frac{(x^2 -6x +3)}{5}\)

\(E. \frac{(x^2 +6x +3)}{5}\)


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Re: A function f(x) satisfies 2f(x)+3f(1-x)=x^2. What is the expression of [#permalink] New post   02 Aug 2022, 10:59
Sid116 wrote:
nick1816 wrote:
\(2f(x)+3f(1-x)=x^2.\) .......(1)


put x=1-x, we get

\(2f(1-x)+3f(x)=(1-x)^2.\) ........(2)

Multiply equation (1) by 2 and equation (2) by 3 and subtract both equations

we'll get

f(x)= \(\frac{(x^2 -6x +3)}{5}\)


MathRevolution wrote:
[GMAT math practice question]

A function \(f(x)\) satisfies \(2f(x)+3f(1-x)=x^2.\) What is the expression of \(f(x)\)?

\(A. \frac{(x^2 +6x +3)}{7}\)

\(B. \frac{(x^2 -6x +3)}{7}\)

\(C. \frac{(x^2 -6x +1)}{3}\)

\(D. \frac{(x^2 -6x +3)}{5}\)

\(E. \frac{(x^2 +6x +3)}{5}\)


Posted from my mobile device
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A function f(x) satisfies 2f(x)+3f(1-x)=x^2. What is the expression of [#permalink] New post   20 Jul 2023, 07:17
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A simple and elegant way to solve function questions
A function f(x) satisfies 2f(x)+3f(1−x)=x^2. What is the expression of f(x)?
Lets assume that x = 1/2 (why 1/2 because in that case x = 1-x = 1/2)
this way we get f(x) = f(1-x)
so substituting x = 1/2 in the given condition
2f(1/2) + 3f(1/2) = 1/4
5f(1/2) = 1/4
or f(1/2) = 1/20
Now put x = 1/2 to check which option gives f(1/2) = 1/20
only expression given in option D produces f(1/2) = 1/20
Answer D
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A function f(x) satisfies 2f(x)+3f(1-x)=x^2. What is the expression of [#permalink]
20 Jul 2023, 07:17
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