For which of the following functions f is f(x) = f(1-x) for all x? : Problem Solving (PS)

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Okay, so you need to check if f(x) = f(1-x)

Let's do this one by one.

Case A: - Wrong
\(f(x) = 1-x\)
\(f(1-x) = 1-(1-x) = x\)

So we clearly see that f(x) is not f(1-x)

Case B: Wrong

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

Case C: Wrong

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

Case D: Right

\(f(x) = x^2 (1-x)^2\)
\(f(1-x) =(1-x)^2 (1-(1-x))^2= (1-x)^2(x)^2=f(x)\)

Let's check E just to be sure.

Case E: Wrong

\(f(x) = \frac{x}{1-x}\)
\(f(1-x) = \frac{1-x}{1-(1-x)}=\frac{1-x}{x}\)

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Bunuel:

sometimes, I almost die from the awesomeness of your posts.

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This one is quite simple. We have to substitute (1 - x) for x in each of the options and see which option yields the same result for both f(x) and f(1 - x).

However, careful observation of the options will easily tell you that only D will fulfill this condition because in D, the x now becomes (1 - x) and (1 - x) now becomes x. The overall outcome will remain the same. You don't even need to expand or do any calculations whatsoever.

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You need to check for every option.
Either you can replace x in every option with 1-x or you can take x = 1 and then check for which option f(1) = f(1-1) = f(0).
In A f(1) = 1-1 = 0. f(0) = 1-0 = 1.
In B f(1) = 1 - 1^2 = 0. f(0) = 1 - 0 = 1
In C f(1) = 1^2 - (1 - 1 )^2 = 1. f(0) = 0 - (1-0)^2 = 0 - 1 = -1
in D f(1) = 1(1-1)^2 = 0. f(0) = 0.(1-0)^2 = 0. ( Right answer )
And for option E you can take x = 2 because if you take x = 1 in denominator the denominator becomes zero, which makes the f(x) undefined.
f(2) = 2/(1 - 2) = -2. f(1-2) = f(-1) = -1/(1-(-1)) = -1/2.

D is the right answer.

Please give a kudo if you like my explanation.

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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 - x^2
C. f(x) = x^2 - (1 - x)^2
D. f(x) = x^2*(1 - x)^2
E. f(x) = x/(1 - x)

Since we are not given any restrictions on the value of x, let’s let x = 1. Thus, we are determining for which of the following functions is f(1) = f(1-1), i.e., f(1) = f(0). Next, we can test each answer choice using our value x = 1.

A. f(x) = 1 - x

f(1) = 1 - 1 = 0

f(0) = 1 - 0 = 1

Since 0 does not equal 1, A is not correct.

B. f(x) = 1 - x^2

f(1) = 1 - 1^2 = 1 - 1 = 0

f(0) = 1 - 0^2 = 1 - 0 = 1

Since 0 does not equal 1, B is not correct.

C. f(x) = x^2 - (1 - x)^2

f(1) = 1^2 - (1 - 1)^2 = 1 - 0 = 1

f(0) = 0^2 - (1 - 0)^2 = 0 - 1 = -1

Since 1 does not equal -1, C is not correct.

D. f(x) = x^2*(1 - x)^2

f(1) = 1^2*(1 - 1)^2 = 1(0)= 0

f(0) = 0^2*(1 - 0)^2 = 0(2) = 0

Since 0 equals 0, D is correct.

Alternate Solution:

Let’s test each answer choice using x and 1 - x.

A. f(x) = 1 - x

f(x) = 1 - x

f(1 - x) = 1 - (1 - x) = x

Since 1 - x does not equal x, A is not correct.

B. f(x) = 1 - x^2

f(x) = 1 - x^2

f(1 - x) = 1 - (1 - x)^2 = 1 - (1 + x^2 -2x) = 2x - x^2

Since 1 - x^2 does not equal 2x - x^2, B is not correct.

C. f(x) = x^2 - (1 - x)^2

f(x) = x^2 - (1 - x)^2 = x^2 - (1 + x^2 - 2x) = 2x - 1

f(1 - x) = (1 - x)^2 - (1 - (1 - x))^2 = 1 + x^2 - 2x - x^2 = 1 - 2x

Since 2x - 1 does not equal 1 - 2x, C is not correct.

D. f(x) = x^2*(1 - x)^2

f(x) = x^2*(1 - x)^2

f(1 - x) = (1 - x)^2*(1 - (1 - x))^2 = (1 - x)^2*x^2

Since x^2*(1 - x)^2 equals (1 - x)^2*x^2, D is correct.

Answer: D

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For which of the following functions f is f(x) = f(1-x) for all x?

A. \(f(x) = 1 - x\)

B. \(f(x) = 1 - x^2\)

C. \(f(x) = x^2 - (1 - x)^2\)

D. \(f(x) = x^2*(1 - x)^2\)

E. \(f(x) = \frac{x}{(1 - x)}\)

Let's try plugging in an easy 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 - x^2
f(0) = 1 - 0^2 = 1
f(1) = 1 - 1^2 = 0
Since f(0) doesn't equal f(1), eliminate B

C) f(x) = x^2 - (1-x)^2
f(0) = 0^2 - (1-0)^2 = -1
f(1) = 1^2 - (1-1)^2 = 1
Since f(0) doesn't equal f(1), eliminate C

D) f(x) = x^2(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

Cheers,
Brent

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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 - x^2
C. f (x) = x^2 - (1 - x)^2
D. f (x) = x^2 (1 - x)^2
E. f (x) = x/(1 - x)

I've already seen quite a few GMAT questions of this type. They are quite easy to solve once you understand the concept.

\(f(x)="some \ expression \ with \ variable \ x"\) means that the value of \(f(x)\) can be found by calculating the expression for a given \(x\).

For example, if \(f(x)=3x+2\), what is the value of \(f(3)\)? Just plug \(3\) in for \(x\): \(f(3)=3*3+2=11\), so if the function is \(f(x)=3x+2\), then \(f(3)=11\).

There are some functions for which \(f(x)=f(-x)\). For example, if we define \(f(x)\) as \(f(x)=3x^2+2\), the value of \(f(x)\) will always be positive and will give the following values:

For \(x=-5\), \(f(x)=3(-5)^2+2=77\);
For \(x=0\), \(f(0)=3*0^2+2=2\).

Notice that in this case, \(f(x)\) is equal to \(f(-x)\), meaning that for positive values of \(x\), you'll get the same values of \(f(x)\) as for negative values of \(x\).

Now, in the original question, we are told to define the expression for which \(f(x)=f(1-x)\), meaning that plugging in \(x\) and \(1-x\) must yield the same result.

Let's analyze the options:

(A) \(f(x)=1-x\) → \(f(1-x)=1-(1-x)=x\). Since \(1-x\) and \(x\) are different, \(f(x)\) does not equal \(f(1-x)\).

(B) \(f(x)=1-x^2\) → \(f(1-x)=1-(1-x)^2=1-1+2x-x^2=2x-x^2\). Since \(1-x^2\) and \(2x-x^2\) are different, no match.

(C) \(f(x)=x^2-(1-x)^2=x^2-1+2x+x^2=2x-1\) → \(f(1-x)=2(1-x)-1=1-2x\). Since \(2x-1\) and \(1-2x\) are different, no match.

(D) \(f(x)=x^2*(1-x)^2\) → \(f(1-x)=(1-x)^2*(1-1+x)^2=(1-x)^2x^2\). Since \(x^2(1-x)^2\) and \( (1-x)^2x^2\) are the same, this is correct.

Still let's check (E)

(E) \(f(x)=\frac{x}{1-x}\) → \(f(1-x)=\frac{1-x}{1-1+x}=\frac{1-x}{x}\). Since \(\frac{x}{1-x}\) and \(\frac{1-x}{x}\) are different, no match.

This problem can also be solved by simple number picking.

Since the question states that "f(x) = f(1-x) for all x," it must hold for any choice of \(x\).

Let \(x\) be 2. Then, \(1-x=1-2=-1\), so we should check whether \(f(2)=f(-1)\).

(A) \(f(2)=1-x=1-2=-1\) and \(f(-1)=1-(-1)=2\) → \(-1\neq{2}\)

(B) \(f(2)=1-x^2=1-4=-3\) and \(f(-1)=1-1=0\) → \(-3\neq{0}\)

(C) \(f(2)=x^2-(1-x)^2=x^2-1+2x+x^2=2x-1=22-1=3\) and \(f(-1)=2(-1)-1=-3\) → \(3\neq{-3}\)

(D) \(f(2)=x^2*(1-x)^2=(-2)^2*(-1)^2=4\) and \(f(-1)=(-1)^2*2^2=4\) → \(4=4\), correct.

(E) \(f(2)=\frac{x}{1-x}=\frac{2}{1-2}=-2\) and \(f(-1)=\frac{-1}{1-(-1)}=-\frac{1}{2}\) → \(-2\neq{-\frac{1}{2}}\).

It might happen that for some choices of \(x\), other options appear "correct" as well. If this happens, simply pick another value for \(x\) and recheck only these "correct" options.

Similar questions:
https://gmatclub.com/forum/for-which-of ... 93184.html
https://gmatclub.com/forum/for-which-of ... 24491.html
https://gmatclub.com/forum/let-the-func ... 43311.html

Hope it helps.

Bunuel for option D

\(f(x)=x^2* (1-x)^2 = x^2 * ( 1-2x+x^2)\) = now plugged in (1-x) for x --> \((1-x)^2 * (1-2(1-x)+(1-x)^2) \)

got \((1-x)^2 * (1-2+2x+1-2x+x^2) \) obviously smth is not right abt it

what am doing wrong ?

And why in Option C you didn`t plug in 1-x to get 2x-1 whereas in D you plugged in x-1 straight away without expanding (1-x)^2 ?

chetan2u VeritasKarishma

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dave13

Bunuel for option D

\(f(x)=x^2* (1-x)^2 = x^2 * ( 1-2x+x^2)\) = now plugged in (1-x) for x --> \((1-x)^2 * (1-2(1-x)+(1-x)^2) \)

got \((1-x)^2 * (1-2+2x+1-2x+x^2) \) obviously smth is not right abt it

what am doing wrong ?

And why in Option C you didn`t plug in 1-x to get 2x-1 whereas in D you plugged in x-1 straight away without expanding (1-x)^2 ?

chetan2u VeritasKarishma

In C, even if you substitute x =1-x, you will get the same answer.

In D, 1-2+2x+1-2x+x^2 and you will still get your answer as everything will get canceled except x^2
\((1-x)^2=1+x^2-2x....1+(1-x)^2-2(1-x)=1+1+x^2-2x-2+2x=x^2\)

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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 - x^2
C. f (x) = x^2 - (1 - x)^2
D. f (x) = x^2 (1 - x)^2
E. f (x) = x/(1 - x)

I've already seen quite a few GMAT questions of this type. They are quite easy to solve once you understand the concept.

\(f(x)="some \ expression \ with \ variable \ x"\) means that the value of \(f(x)\) can be found by calculating the expression for a given \(x\).

For example, if \(f(x)=3x+2\), what is the value of \(f(3)\)? Just plug \(3\) in for \(x\): \(f(3)=3*3+2=11\), so if the function is \(f(x)=3x+2\), then \(f(3)=11\).

There are some functions for which \(f(x)=f(-x)\). For example, if we define \(f(x)\) as \(f(x)=3x^2+2\), the value of \(f(x)\) will always be positive and will give the following values:

For \(x=-5\), \(f(x)=3(-5)^2+2=77\);
For \(x=0\), \(f(0)=3*0^2+2=2\).

Notice that in this case, \(f(x)\) is equal to \(f(-x)\), meaning that for positive values of \(x\), you'll get the same values of \(f(x)\) as for negative values of \(x\).

Now, in the original question, we are told to define the expression for which \(f(x)=f(1-x)\), meaning that plugging in \(x\) and \(1-x\) must yield the same result.

Let's analyze the options:

(A) \(f(x)=1-x\) → \(f(1-x)=1-(1-x)=x\). Since \(1-x\) and \(x\) are different, \(f(x)\) does not equal \(f(1-x)\).

(B) \(f(x)=1-x^2\) → \(f(1-x)=1-(1-x)^2=1-1+2x-x^2=2x-x^2\). Since \(1-x^2\) and \(2x-x^2\) are different, no match.

(C) \(f(x)=x^2-(1-x)^2=x^2-1+2x+x^2=2x-1\) → \(f(1-x)=2(1-x)-1=1-2x\). Since \(2x-1\) and \(1-2x\) are different, no match.

(D) \(f(x)=x^2*(1-x)^2\) → \(f(1-x)=(1-x)^2*(1-1+x)^2=(1-x)^2x^2\). Since \(x^2(1-x)^2\) and \( (1-x)^2x^2\) are the same, this is correct.

Still let's check (E)

(E) \(f(x)=\frac{x}{1-x}\) → \(f(1-x)=\frac{1-x}{1-1+x}=\frac{1-x}{x}\). Since \(\frac{x}{1-x}\) and \(\frac{1-x}{x}\) are different, no match.

This problem can also be solved by simple number picking.

Since the question states that "f(x) = f(1-x) for all x," it must hold for any choice of \(x\).

Let \(x\) be 2. Then, \(1-x=1-2=-1\), so we should check whether \(f(2)=f(-1)\).

(A) \(f(2)=1-x=1-2=-1\) and \(f(-1)=1-(-1)=2\) → \(-1\neq{2}\)

(B) \(f(2)=1-x^2=1-4=-3\) and \(f(-1)=1-1=0\) → \(-3\neq{0}\)

(C) \(f(2)=x^2-(1-x)^2=x^2-1+2x+x^2=2x-1=22-1=3\) and \(f(-1)=2(-1)-1=-3\) → \(3\neq{-3}\)

(D) \(f(2)=x^2*(1-x)^2=(-2)^2*(-1)^2=4\) and \(f(-1)=(-1)^2*2^2=4\) → \(4=4\), correct.

(E) \(f(2)=\frac{x}{1-x}=\frac{2}{1-2}=-2\) and \(f(-1)=\frac{-1}{1-(-1)}=-\frac{1}{2}\) → \(-2\neq{-\frac{1}{2}}\).

It might happen that for some choices of \(x\), other options appear "correct" as well. If this happens, simply pick another value for \(x\) and recheck only these "correct" options.

Similar questions:
https://gmatclub.com/forum/for-which-of ... 93184.html
https://gmatclub.com/forum/for-which-of ... 24491.html
https://gmatclub.com/forum/let-the-func ... 43311.html

Hope it helps.

Bunuel for option D

\(f(x)=x^2* (1-x)^2 = x^2 * ( 1-2x+x^2)\) = now plugged in (1-x) for x --> \((1-x)^2 * (1-2(1-x)+(1-x)^2) \)

got \((1-x)^2 * (1-2+2x+1-2x+x^2) \) obviously smth is not right abt it

what am doing wrong ?

And why in Option C you didn`t plug in 1-x to get 2x-1 whereas in D you plugged in x-1 straight away without expanding (1-x)^2 ?

chetan2u VeritasKarishma

When you replace x by (1 - x) in (1 - x), you get x back!

(1 - x) becomes (1 - (1-x)) which is (1 - 1 + x) which is x.

Hence

C. f (x) = x^2 - (1 - x)^2

becomes (1-x)^2 - x^2

D. f (x) = x^2 * (1 - x)^2

becomes (1-x)^2 * x^2

In (C) the expression changes but (D) doesn't because multiplication is commutative (xy = yx)

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Video solution from Quant Reasoning starts at 19:38
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For which of the following functions f is f(x) = f (1-x) for all x?

Lets try plugging in values to solve this question. Since there is no condition given for x , we can assume any values for x.
Assume x= 2, then f(2) = f(1-2) = f(-1). So we need to check in which functions given below, f(2)= f(-1)?

A. f (x) = 1 - x
f(2)=1-2 =-1
f(-1) = 1+1 =2 ==> f(2) ≠ f(-1)

B. f (x) = 1 - x^2
f(2) = 1-4 = -3
f(-1) = 1-1 = 0 ==> f(2) ≠ f(-1)

C. f (x) = x^2 - (1 - x)^2
f(2)= 4-(-1)^2 = 4-1 = 3
f(-1) = 1-4 = -3 ==> f(2) ≠ f(-1)

D. f (x) = x^2 (1 - x)^2
f(2) = 4*1=4
f(-1) = 1*4 =4 ==> f(2)= f(-1)

E. f (x) = x/(1 - x)
f(2)= 2/-1 = -2
f(-1) = -1/2 ==> f(2) ≠ f(-1)

Option D is the correct answer.

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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 - x^2
C. f(x) = x^2 - (1 - x)^2
D. f(x) = x^2*(1 - x)^2
E. f(x) = x/(1 - x)

Since we are not given any restrictions on the value of x, let’s let x = 1. Thus, we are determining for which of the following functions is f(1) = f(1-1), i.e., f(1) = f(0). Next, we can test each answer choice using our value x = 1.

A. f(x) = 1 - x

f(1) = 1 - 1 = 0

f(0) = 1 - 0 = 1

Since 0 does not equal 1, A is not correct.

B. f(x) = 1 - x^2

f(1) = 1 - 1^2 = 1 - 1 = 0

f(0) = 1 - 0^2 = 1 - 0 = 1

Since 0 does not equal 1, B is not correct.

C. f(x) = x^2 - (1 - x)^2

f(1) = 1^2 - (1 - 1)^2 = 1 - 0 = 1

f(0) = 0^2 - (1 - 0)^2 = 0 - 1 = -1

Since 1 does not equal -1, C is not correct.

D. f(x) = x^2*(1 - x)^2

f(1) = 1^2*(1 - 1)^2 = 1(0)= 0

f(0) = 0^2*(1 - 0)^2 = 0(2) = 0

Since 0 equals 0, D is correct.

Alternate Solution:

Let’s test each answer choice using x and 1 - x.

A. f(x) = 1 - x

f(x) = 1 - x

f(1 - x) = 1 - (1 - x) = x

Since 1 - x does not equal x, A is not correct.

B. f(x) = 1 - x^2

f(x) = 1 - x^2

f(1 - x) = 1 - (1 - x)^2 = 1 - (1 + x^2 -2x) = 2x - x^2

Since 1 - x^2 does not equal 2x - x^2, B is not correct.

C. f(x) = x^2 - (1 - x)^2

f(x) = x^2 - (1 - x)^2 = x^2 - (1 + x^2 - 2x) = 2x - 1

f(1 - x) = (1 - x)^2 - (1 - (1 - x))^2 = 1 + x^2 - 2x - x^2 = 1 - 2x

Since 2x - 1 does not equal 1 - 2x, C is not correct.

D. f(x) = x^2*(1 - x)^2

f(x) = x^2*(1 - x)^2

f(1 - x) = (1 - x)^2*(1 - (1 - x))^2 = (1 - x)^2*x^2

Since x^2*(1 - x)^2 equals (1 - x)^2*x^2, D is correct.

Answer: D

Hi Bunuel and ScottTargetTestPrep,
Can we infer that if the function question stem doesn't specify any restrictions on the value of a given variable, we can pluging any number to test the function?

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KarishmaB

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Is there a faster way to solve this question rather than replacing each "x" by (1-x)? Thanks!

For which of the following functions \(f\) is \(f(x) = f(1-x)\) for all x?

A. \(f(x) = 1-x\)
B. \(f(x) = 1-x^2\)
C. \(f(x) = x^2 - (1-x)^2\)
D. \(f(x) = (x^2)(1-x)^2\)
E. \(f(x) = x / (1-x)\)

Tip: Try to first focus on the options where terms are added/multiplied rather than subtracted/divided. They are more symmetrical and a substitution may not change the expression. I will intuitively check D first since it involves multiplication of the terms.

Thank you KarishmaB for your amazing tip!
Is this tip is suggested for all exercises involving algebra and number plugging?

KarishmaB

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Quote:

KarishmaB metallicafanIs there a faster way to solve this question rather than replacing each "x" by (1-x)? Thanks!

For which of the following functions \(f\) is \(f(x) = f(1-x)\) for all x?

A. \(f(x) = 1-x\)
B. \(f(x) = 1-x^2\)
C. \(f(x) = x^2 - (1-x)^2\)
D. \(f(x) = (x^2)(1-x)^2\)
E. \(f(x) = x / (1-x)\)Tip: Try to first focus on the options where terms are added/multiplied rather than subtracted/divided. They are more symmetrical and a substitution may not change the expression. I will intuitively check D first since it involves multiplication of the terms.Thank you KarishmaB for your amazing tip!
Is this tip is suggested for all exercises involving algebra and number plugging?

The tip is useful when we are looking for symmetry. e.g. \(f(x) = f(1-x)\) for all x?

I know that a + b = b + a but (a - b) is not equal to (b - a).
Similarly, a * b = b * a but a/b is not equal to b/a.
Hence addition and multiplications are more likely to give me same result when the terms are exchanged.

It could be applicable in any topic depending on context.

DanTheGMATMan

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Usually start with the function that involves multiplication- it's easier for the sides to be equal because you can multiply in an order, etc. Doesn't always work but that's a tip:

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For which of the following functions f is f(x) = f(1-x) for all x?

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