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Can anyone help me to solve the question?

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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Source: — Quantitative Reasoning |

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by [email protected] » Tue May 19, 2015 10:03 pm
Hi kellymichaelsohhh,

Are there any typos in this prompt? Is the "extra X" in the phrase "xf(1-x)" supposed to be there?

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by kellymichaelsohhh » Tue May 19, 2015 10:22 pm
[email protected] wrote:Hi kellymichaelsohhh,

Are there any typos in this prompt? Is the "extra X" in the phrase "xf(1-x)" supposed to be there?

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I'm not sure but the question was written exactly like what i post
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by Brent@GMATPrepNow » Wed May 20, 2015 8:51 am
kellymichaelsohhh wrote:Given that the equation f(x) + xf(1-x) = x holds for every real number x. The value of f(-1) + f(1) is ....
(A) -1
(B) 0
(C) 1/3
(D) 1
(E) 4/3
let's start with x = 0
The given info tells us that f(x) + xf(1-x) = x
So, f(0) + 0f(1-0) = 0
Simplify to get: f(0) + 0 = 0
This means that f(0) = 0

Now try x = 1
The given info tells us that f(x) + xf(1-x) = x
So, f(1) + 1f(1-1) = 1
Simplify to get: f(1) + 1f(0) = 1
We already determined that f(0) = 0, so when we plug in that value, we get:
f(1) + 1(0) = 1
Simplify to get: f(1) + 0 = 1
This means that f(1) = 1

Now try x = 2
So, f(2) + 2f(1-2) = 2
Simplify to get: f(2) + 2f(-1) = 2

Now try x = -1
So, f(-1) + (-1)f(1-(-1)) = -1
Simplify to get: f(-1) - f(2) = -1
Rearrange to get: -f(2) + f(-1) = -1

We now have two similar equations:
f(2) + 2f(-1) = 2
-f(2) + f(-1) = -1

If we ADD them, we get:
3f(-1) = 1
Divide both sides by 3 to get: f(-1) = 1/3

So, f(-1) + f(1) = 1/3 + 1
= [spoiler]4/3[/spoiler]

Answer: E

Cheers,
Brent
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by kellymichaelsohhh » Wed May 20, 2015 5:02 pm
Brent@GMATPrepNow wrote:
kellymichaelsohhh wrote:Given that the equation f(x) + xf(1-x) = x holds for every real number x. The value of f(-1) + f(1) is ....
(A) -1
(B) 0
(C) 1/3
(D) 1
(E) 4/3
let's start with x = 0
The given info tells us that f(x) + xf(1-x) = x
So, f(0) + 0f(1-0) = 0
Simplify to get: f(0) + 0 = 0
This means that f(0) = 0

Now try x = 1
The given info tells us that f(x) + xf(1-x) = x
So, f(1) + 1f(1-1) = 1
Simplify to get: f(1) + 1f(0) = 1
We already determined that f(0) = 0, so when we plug in that value, we get:
f(1) + 1(0) = 1
Simplify to get: f(1) + 0 = 1
This means that f(1) = 1

Now try x = 2
So, f(2) + 2f(1-2) = 2
Simplify to get: f(2) + 2f(-1) = 2

Now try x = -1
So, f(-1) + (-1)f(1-(-1)) = -1
Simplify to get: f(-1) - f(2) = -1
Rearrange to get: -f(2) + f(-1) = -1

We now have two similar equations:
f(2) + 2f(-1) = 2
-f(2) + f(-1) = -1

If we ADD them, we get:
3f(-1) = 1
Divide both sides by 3 to get: f(-1) = 1/3

So, f(-1) + f(1) = 1/3 + 1
= [spoiler]4/3[/spoiler]

Answer: E

Cheers,
Brent
Wowww its so wonderful! Thanks a lot haha. Can I ask you another question? Do you know how to calculate log without using calculator? For example: log(0.025)
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by kellymichaelsohhh » Wed May 20, 2015 5:02 pm
Brent@GMATPrepNow wrote:
kellymichaelsohhh wrote:Given that the equation f(x) + xf(1-x) = x holds for every real number x. The value of f(-1) + f(1) is ....
(A) -1
(B) 0
(C) 1/3
(D) 1
(E) 4/3
let's start with x = 0
The given info tells us that f(x) + xf(1-x) = x
So, f(0) + 0f(1-0) = 0
Simplify to get: f(0) + 0 = 0
This means that f(0) = 0

Now try x = 1
The given info tells us that f(x) + xf(1-x) = x
So, f(1) + 1f(1-1) = 1
Simplify to get: f(1) + 1f(0) = 1
We already determined that f(0) = 0, so when we plug in that value, we get:
f(1) + 1(0) = 1
Simplify to get: f(1) + 0 = 1
This means that f(1) = 1

Now try x = 2
So, f(2) + 2f(1-2) = 2
Simplify to get: f(2) + 2f(-1) = 2

Now try x = -1
So, f(-1) + (-1)f(1-(-1)) = -1
Simplify to get: f(-1) - f(2) = -1
Rearrange to get: -f(2) + f(-1) = -1

We now have two similar equations:
f(2) + 2f(-1) = 2
-f(2) + f(-1) = -1

If we ADD them, we get:
3f(-1) = 1
Divide both sides by 3 to get: f(-1) = 1/3

So, f(-1) + f(1) = 1/3 + 1
= [spoiler]4/3[/spoiler]

Answer: E

Cheers,
Brent
Wowww its so wonderful! Thanks a lot haha. Can I ask you another question? Do you know how to calculate log without using calculator? For example: log(0.025)
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by Brent@GMATPrepNow » Wed May 20, 2015 5:15 pm
kellymichaelsohhh wrote: Can I ask you another question? Do you know how to calculate log without using calculator? For example: log(0.025)
You won't be required to perform any log calculations on the GMAT. The closest to this would be a question that required us to find some approximate exponent.
For example:
If 10^x = 0.025, then what is the value of x?
A) -5.something
B) -4.something
C) -3.something
D) -2.something
E) -1.something
First recognize that 10^(-1) = 0.1
And 10^(-2) = 0.01

Since 0.025 is BETWEEN 0.1 and 0.01, x must be BETWEEN -1 and -2

Answer: E

Cheers,
Brent
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by kellymichaelsohhh » Wed May 20, 2015 5:52 pm
Brent@GMATPrepNow wrote:
kellymichaelsohhh wrote: Can I ask you another question? Do you know how to calculate log without using calculator? For example: log(0.025)
You won't be required to perform any log calculations on the GMAT. The closest to this would be a question that required us to find some approximate exponent.
For example:
If 10^x = 0.025, then what is the value of x?
A) -5.something
B) -4.something
C) -3.something
D) -2.something
E) -1.something
First recognize that 10^(-1) = 0.1
And 10^(-2) = 0.01

Since 0.025 is BETWEEN 0.1 and 0.01, x must be BETWEEN -1 and -2

Answer: E

Cheers,
Brent
Oh thank you!😃
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by nikhilgmat31 » Fri May 22, 2015 3:41 am
It is good question with cyclic functions.
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