Is this one merely a plug and chug question ? whats the shortcut for this one? (answer is 'D' btw)
Help me solve this GMAT Quant Question (GMATPrep) #6
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- EricImasogie
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Let's use the INPUT-OUTPUT approach (aka plug and chug).For which of the following functions f is f(x)=f(1-x) for all x?
A) f(x) = 1-x
B) f(x) = 1-x²
C) f(x) = x² - (1-x)²
D) f(x) = x²(1-x)²
E) f(x) = x/(1-x)
So, let's use a "nice" value for x.
How about x = 0?
So, we can reword the question as, For which of the following functions is f(0)=f(1-0)
In other words, we're looking for a function such that f(0) = f(1)
A) f(x)=1-x
f(0)=1-0 = 1
f(1)=1-1 = 0
Since f(0) doesn't equal f(1), eliminate A
B) f(x) = 1 - x²
f(0) = 1 - 0² = 1
f(1) = 1 - 1² = 0
Since f(0) doesn't equal f(1), eliminate B
C) f(x) = x² - (1-x)²
f(0) = 0² - (1-0)² = -1
f(1) = 1² - (1-1)² = 1
Since f(0) doesn't equal f(1), eliminate C
D) f(x) = x²(1-x)²
f(0) = 0^2(1-0)^2 = 0
f(1) = 1^2(1-1)^2 = 0
Since f(0) equals f(1), keep D for now
E) f(x) = x/(1-x)
f(0) = 0/(1-0) = 0
f(1) = 1/(1-1) = undefined
Since f(0) doesn't equal f(1), eliminate E
Since only D satisfies the condition that f(x)=f(1-x) when x=0, the correct answer is D
Here's a similar question: https://www.beatthegmat.com/number-systems-t270738.html
Cheers,
Brent
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Hi EricImasogie,
This question comes up every few months in the Forums. Here's a recent discussion about how to approach it and the inherent math behind the logic:
https://www.beatthegmat.com/function-pro ... 81275.html
GMAT assassins aren't born, they're made,
Rich
This question comes up every few months in the Forums. Here's a recent discussion about how to approach it and the inherent math behind the logic:
https://www.beatthegmat.com/function-pro ... 81275.html
GMAT assassins aren't born, they're made,
Rich
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Using D) f(x) = x²(1-x)²Brent@GMATPrepNow wrote:Let's use the INPUT-OUTPUT approach (aka plug and chug).For which of the following functions f is f(x)=f(1-x) for all x?
A) f(x) = 1-x
B) f(x) = 1-x²
C) f(x) = x² - (1-x)²
D) f(x) = x²(1-x)²
E) f(x) = x/(1-x)
So, let's use a "nice" value for x.
How about x = 0?
So, we can reword the question as, For which of the following functions is f(0)=f(1-0)
In other words, we're looking for a function such that f(0) = f(1)
A) f(x)=1-x
f(0)=1-0 = 1
f(1)=1-1 = 0
Since f(0) doesn't equal f(1), eliminate A
B) f(x) = 1 - x²
f(0) = 1 - 0² = 1
f(1) = 1 - 1² = 0
Since f(0) doesn't equal f(1), eliminate B
C) f(x) = x² - (1-x)²
f(0) = 0² - (1-0)² = -1
f(1) = 1² - (1-1)² = 1
Since f(0) doesn't equal f(1), eliminate C
D) f(x) = x²(1-x)²
f(0) = 0^2(1-0)^2 = 0
f(1) = 1^2(1-1)^2 = 0
Since f(0) equals f(1), keep D for now
E) f(x) = x/(1-x)
f(0) = 0/(1-0) = 0
f(1) = 1/(1-1) = undefined
Since f(0) doesn't equal f(1), eliminate E
Since only D satisfies the condition that f(x)=f(1-x) when x=0, the correct answer is D
Here's a similar question: https://www.beatthegmat.com/number-systems-t270738.html
Cheers,
Brent
If x = -1:
f(-1) = 1²(2)² = 4
Using f(x)=f(1-x)
If x = -1
f(1-x) = f(2) = 16
Therefore surely D cannot be correct for all values of x?
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f(1-x) = f(2) = (2)²(-1)² = 4Mathsbuddy wrote: Using D) f(x) = x²(1-x)²
If x = -1:
f(-1) = 1²(2)² = 4
Using f(x)=f(1-x)
If x = -1
f(1-x) = f(2) = 16
Therefore surely D cannot be correct for all values of x?
Cheers,
Brent
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Here's another way to show that D is the correct answer:
Given: f(x) = x²(1-x)²
Now compare f(x) and f(1-x)
f(x) = x²(1-x)² = x²(1-x)²
f(1-x) = (1-x)²[1-(1-x)]²
= (1-x)² x²
Since x²(1-x)² = (1-x)² x², we can be certain that f(x) = f(1-x) for all values of x
Cheers,
Brent
Given: f(x) = x²(1-x)²
Now compare f(x) and f(1-x)
f(x) = x²(1-x)² = x²(1-x)²
f(1-x) = (1-x)²[1-(1-x)]²
= (1-x)² x²
Since x²(1-x)² = (1-x)² x², we can be certain that f(x) = f(1-x) for all values of x
Cheers,
Brent
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This must be the most frequently asked OG/GMATPrep problem!
Plugging and chugging is one way. Another is noticing that 1 - (1 - x) = x, so D works perfectly. If we plug in x, we get
f(x) = x²(1-x)²
If we plug in (1 - x), we get
f(1-x) = (1-x)²(1-(1-x))² = (1-x)²x²
So f(x) = f(1-x), as they're just the same thing presented in a different order.
Plugging and chugging is one way. Another is noticing that 1 - (1 - x) = x, so D works perfectly. If we plug in x, we get
f(x) = x²(1-x)²
If we plug in (1 - x), we get
f(1-x) = (1-x)²(1-(1-x))² = (1-x)²x²
So f(x) = f(1-x), as they're just the same thing presented in a different order.