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Let S be the set of all points (x, y) in the xy plane such that |x| +

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Let S be the set of all points (x, y) in the xy plane such that |x| + [#permalink] Let S be the set of all points (x, y) in the xy plane such that |x| +   27 Mar 2020, 05:45
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Let S be the set of all points (x, y) in the xy plane such that |x| + |y| ≤ 2 and |x| ≥ 1. Then what is the area of the region represented by S?

A. 1/2
B. 1
C. 2
D. 3
E. 4
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C
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Re: Let S be the set of all points (x, y) in the xy plane such that |x| + [#permalink] Let S be the set of all points (x, y) in the xy plane such that |x| +   27 Mar 2020, 07:13
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for the condition this to be true lxl +lyl <=2, the area will be a square with vertices (-2,0)(2,0)(0,-2)(0,2)

Now we need to subtract the the area which does not satisfy the condition lxl >=1

Area of Set will be the area of Non Overlapping Part

2 x (1/2) x base x height

2 x (1/2) x 1 x 2 = 2
Attachments

Area.PNG
Area.PNG [ 11.53 KiB | Viewed 7482 times ]

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Re: Let S be the set of all points (x, y) in the xy plane such that |x| + [#permalink] Let S be the set of all points (x, y) in the xy plane such that |x| +   29 Mar 2020, 11:43
Possible values are (1,1);(1,-1);(-1,-1);(-1,1);(-2,0)and(2,0)

So there are two triangle each with an area of 1 so 1+1 = 2
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Let S be the set of all points (x, y) in the xy plane such that |x| + [#permalink] Let S be the set of all points (x, y) in the xy plane such that |x| +   30 Jul 2023, 04:23
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↧↧↧ Detailed Video Solution to the Problem ↧↧↧




Given that S be the set of all points (x, y) in the xy plane such that |x| + |y| ≤ 2 and |x| ≥ 1 and we need to find what is the area of the region represented by S

Plotting |x| + |y| ≤ 2

We will have four case for values of x and y each being ≥ 0 and ≤ 0 respectively (in four quadrants)

    1. x≥0, y≥0 => |x| = x and |y| = y => x + y ≤ 2 (Quadrant 1). We will plot the line x + y = 2 and will consider the area below the line as we have ≤ sign
[Note: we need to keep the sign of y positive while considering the sign of the inequality]
    2. x≤0, y≥0 => |x| = -x and |y| = y => -x + y ≤ 2 (Quadrant 2). We will plot the line -x + y = 2 and will consider the area below the line as we have ≤ sign
    3. x≤0, y≤0 => |x| = -x and |y| = -y => -x - y ≤ 2 (Quadrant 3) => x + y ≥ -2. We will plot the line x + y = -2 and will consider the area above the line as we have ≥ sign
    4. x≥0, y≤0 => |x| = x and |y| = -y => x - y ≤ 2 (Quadrant 4) => y ≥ x - 2. We will plot the line x - y = 2 and will consider the area above the line as we have ≥ sign
=> We get the square as the enclosed figure shown in the image below

Attachment:
Mod x + Mod y LTEQ 2.jpg
Mod x + Mod y LTEQ 2.jpg [ 32.92 KiB | Viewed 2970 times ]


Plotting |x| ≥ 1

We will have two cases x≥0 and x<0

    1. x≥0 => |x| = x => x ≥ 1 and Intersection of x≥0 and x ≥ 1 equals x ≥ 1 => Area will be to the right of x = 1
    2. x<0 => |x| = -x => -x ≥ 1 => x ≤ - 1 and Intersection of x<0 and x ≤ - 1 equals x ≤ - 1 => Area will be to the left of x = -
1

Attachment:
Mod x + Mod y LTEQ 2 intersection with mod x GTEQ 1.jpg
Mod x + Mod y LTEQ 2 intersection with mod x GTEQ 1.jpg [ 21.96 KiB | Viewed 2895 times ]


Point of intersection of both the graphs

There will be four points of intersections

    1. x=1 and x+y=2 will intersect at (1,1)
    2. x=-1 and -x+y=2 will intersect at (-1,1)
    3. x=-1 and -x-y=2 will intersect at (-1,-1)
    4. x=1 and x-y=2 will intersect at (1,-1)

So, common area between the two graphs will be given by the two shaded triangles shown below:

Attachment:
Common area.jpg
Common area.jpg [ 23.42 KiB | Viewed 2833 times ]


Area of each triangle = \(\frac{1}{2}\) * Base * Height = \(\frac{1}{2}\) * 2 * 1 = 1

=> Total area of two triangles= 2*1 = 2 units

So, Answer will be C
Hope it helps!

Watch the following video to learn Basics of Absolute Values

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Let S be the set of all points (x, y) in the xy plane such that |x| + [#permalink]
30 Jul 2023, 04:23
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Bunuel
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