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by Aman verma » Sun Nov 24, 2013 2:15 am
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Q: The graph of the function y = f(x) is symmetrical about the line x = 2, then which is/are correct:

(i) f(x + 2) = f(x - 2)
(ii) f(2 + x) = f(4 - x)
(iii)f(x) = f(-x)

(A) (i) only

(B) (ii)only

(C) (iii)only

(D) (i),(iii) Both

(E) (i),(ii) and (iii)

NB: I have made up the graph, but I am not sure it's absolutely correct. Also I request to provide the graphs of the correct solution besides that of the question.
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by Aman verma » Sun Nov 24, 2013 9:36 am
Can anyone suggest a solution to this one ?OA[spoiler](B)[/spoiler]
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by GMATSUCKER » Mon Nov 25, 2013 12:43 am
Aman verma wrote:Can anyone suggest a solution to this one ?OA[spoiler](B)[/spoiler]
Is this a GMAT question ? If this is not a GMAT question then what's the source ?
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by Aman verma » Mon Nov 25, 2013 10:31 am
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Okay, I came up with this graph which satisfies option (B) but unfortunately it satisfies all the three equations. The solution to this problem lies in creating the correct graph and then we can check the equation. Alternatively we can solve this algebraically, but I don't know how to do so. So what could be the possible solution. How to solve this. What is the correct graph ?
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by GMATGuruNY » Mon Nov 25, 2013 12:10 pm
I believe that the problem should read as follows:
Aman verma wrote: Q: If the graph of the function y = f(x) is symmetrical about the line x = 2, then which of the following MUST be true?

(i) f(x + 2) = f(x - 2)
(ii) f(2 + x) = f(2 - x)
(iii)f(x) = f(-x)

(A) (i) only

(B) (ii)only

(C) (iii)only

(D) (i),(iii) Both

(E) (i),(ii) and (iii)
Symmetry about x=2 implies the following:
If the value of x is k places TO THE LEFT OR RIGHT of x=2, the value of f(x) must be THE SAME in each case.
For example:
f(1) = f(3).
Here, the x-values (1 and 3) are each ONE PLACE away from x=2, so the value of f(x) must be the same in each case.
f(0) = f(4).
Here, the x-values (0 and 4) are each TWO PLACES away from x=2, so the value of f(x) must be the same in each case.
f(-1) = f(5).
Here, the x-values (-1 and 5) are each THREE PLACES away from x=2, so the value of f(x) must be the same in each case.

Implication:
x-values that are the SAME DISTANCE from x=2 MUST yield the same value for f(x).
x-values that are DIFFERENT DISTANCES from x=2 do NOT have to yield the same value for f(x).

(i) f(x + 2) = f(x - 2)
If x=1, we get:
f(1+2) = f(1-2).
f(3) = f(-1).
Doesn't work:
3 and -1 are DIFFERENT distances from x=2, so f(3) does not have to be equal to f(-1).
The correct answer choice cannot include (i).
Eliminate A, D and E.

(ii) f(2 + x) = f(2 - x)
If x=1, we get:
f(2+1) = f(2-1).
f(3) = f(1).
This works.
3 and 1 are each the SAME distance from x=2, so f(3) must be equal to f(1).

If x=10, we get:
f(2+10) = f(2-10).
f(12) = f(-8).
This works.
12 and -8 are each the SAME distance from x=2, so f(12) must be equal to f(-8).

The cases above illustrate that the correct answer choice must include (ii).
Eliminate C.

The correct answer is B.
Last edited by GMATGuruNY on Mon Nov 25, 2013 2:02 pm, edited 1 time in total.
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by Aman verma » Mon Nov 25, 2013 1:57 pm
GMATGuruNY wrote:I believe that the problem should read as follows:
Aman verma wrote: Q: The graph of the function y = f(x) is symmetrical about the line x = 2, then which of the following MUST be true?

(i) f(x + 2) = f(x - 2)
(ii) f(2 + x) = f(2 - x)
(iii)f(x) = f(-x)

(A) (i) only

(B) (ii)only

(C) (iii)only

(D) (i),(iii) Both

(E) (i),(ii) and (iii)
Symmetry about x=2 implies the following:
If the value of x is k places TO THE LEFT OR RIGHT of x=2, the value of f(x) must be THE SAME in each case.
For example:
f(1) = f(3).
Here, the x-values (1 and 3) are each ONE PLACE away from x=2, so the value of f(x) must be the same in each case.
f(0) = f(4).
Here, the x-values (0 and 4) are each TWO PLACES away from x=2, so the value of f(x) must be the same in each case.
f(-1) = f(5).
Here, the x-values (-1 and 5) are each THREE PLACES away from x=2, so the value of f(x) must be the same in each case.

Implication:
x-values that are the SAME DISTANCE from x=2 MUST yield the same value for f(x).
x-values that are DIFFERENT DISTANCES from x=2 do NOT have to yield the same value for f(x).

(i) f(x + 2) = f(x - 2)
If x=1, we get:
f(1+2) = f(1-2).
f(3) = f(-1).
Doesn't work:
3 and -1 are DIFFERENT distances from x=2, so f(3) does not have to be equal to f(-1).
The correct answer choice cannot include (i).
Eliminate A, D and E.

(ii) f(2 + x) = f(2 - x)
If x=1, we get:
f(2+1) = f(2-1).
f(3) = f(1).
This works.
3 and 1 are each the SAME distance from x=2, so f(3) must be equal to f(1).

If x=10, we get:
f(2+10) = f(2-10).
f(12) = f(-8).
This works.
12 and -8 are each the SAME distance from x=2, so f(12) must be equal to f(-8).

The cases above illustrate that the correct answer choice must include (ii).
Eliminate C.

The correct answer is B.
Awesome!

And I guess there must be some misprint in my book. Cause even I was not getting any logical answer for option (ii). Thanks for clarifying the underlying concept and pointing out the mistake for option(ii). That explanation of K places was awesome. I am really grateful to you. I was having difficulty interpreting graphs of this type. Now please tell me which graph is correct:-the 1st one or the 2nd one ? I now personally think the first one is correct which is similar to that of Modulus X.
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