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Updated on: 26 Oct 2017, 06:46
Originally posted by papillon86 on 08 Nov 2009, 14:11.
Last edited by Bunuel on 26 Oct 2017, 06:46, edited 3 times in total.
Last edited by Bunuel on 26 Oct 2017, 06:46, edited 3 times in total.
Edited the question and added the OA
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
61% (01:57) correct
39%
(02:04)
wrong
based on 1596
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In the x-y plane, the area of the region bounded by the graph of |x + y| + |x - y| = 4 is
A. 8
B. 12
C. 16
D. 20
E. 24
A. 8
B. 12
C. 16
D. 20
E. 24
08 Nov 2009, 14:34
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papillon86
OK, there can be 4 cases:
|x+y| + |x-y| = 4
A. x+y+x-y = 4 --> x=2
B. x+y-x+y = 4 --> y=2
C. -x-y +x-y= 4 --> y=-2
D. -x-y-x+y=4 --> x=-2
The area bounded by 4 graphs x=2, x=-2, y=2, y=-2 will be square with the side of 4 so the area will be 4*4=16.
Attachment:
MSP17971c13h40gd024h6g10000466ge1e9df941i96.gif [ 1.86 KiB | Viewed 52928 times ]
Answer: C
08 Nov 2009, 17:41
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srini123
I think this one is different.
\(|\frac{x}{2}| + |\frac{y}{2}| = 5\)
After solving you'll get equation of four lines:
\(y=-10-x\)
\(y=10+x\)
\(y=10-x\)
\(y=x-10\)
These four lines will also make a square, BUT in this case the diagonal will be 20 so the \(Area=\frac{20*20}{2}=200\). Or the \(Side= \sqrt{200}\), area=200.
If you draw these four lines you'll see that the figure (square) which is bounded by them is turned by 90 degrees and has a center at the origin. So the side will not be 20.
Also you made a mistake in solving equation. The red part is not correct. You should have the equations written above.
In our original question when we were solving the equation |x+y| + |x-y| = 4 each time x or y were cancelling out so we get equations of a type x=some value twice and y=some value twice. And these equations give the lines which are parallel to the Y or X axis respectively so the figure bounded by them is a "horizontal" square (in your question it's "diagonal" square).
Hope it's clear.
