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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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Re: A function f(x) satisfies 2f(x)+3f(1-x)=x^2. What is the expression of [#permalink]
=>

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 $$f(x) = (\frac{1}{5})(x^2-6x+3).$$

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Re: A function f(x) satisfies 2f(x)+3f(1-x)=x^2. What is the expression of [#permalink]
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]
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 $$f(x) = (\frac{1}{5})(x^2-6x+3).$$

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]
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]
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 $$f(x)$$?

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

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

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

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

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

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