how can i solve these questions

lxl + lyl = 1

We need to get rid of the two absolute value signs from x and y:

Since we have two variables and each absolute value has 2 possiblities, we would end up with a total of 4 equations [abs(x) = x / -x & abs(y) = y / -y]:

x + y = 1, x - y = 1, -x + y = 1, & -x - y = 1

1) x + y = 1 --> y = -x + 1 --> Notice, we took positive x and y, so we are interested in values: x and y are >= to 0. If you graph this, you get a line that is only in the 1st quadrant.

2) x - y = 1 --> y = x - 1 --> We took positive x and negative y, so we are interested in values: x is >= 0 and y < 0. If you graph this, you get a line that is only in the 4th quadrant.

3) -x + y = 1 --> y = x + 1 --> Follow the pattern...negative x and positive y means the 2nd quadrant.

4) -x - y = 1 --> y = -x - 1 --> Follow the pattern...negative x and negative y means the 3rd quadrant.

Notice that each of the 4 lines has a slope of 1 or -1.

Now, if you graph each of these, you would find that it looks like the graph Phil posted of a square that has sides of sqrt(2).

xvxvxv, I can't see the first question. Can you tell me what the first question is so I can solve it?

"The area of the region enclosed by |x| + |y| = 1 is:"

The area is a square with a side of  √2....thus the area = ( √2)²  = 2

A graph might help........

but how you draw it .. and is the drawing the only way ! 

 AzizHusain 

amazing explain .. thank you I understand it 

but this is a simple of placement test so are there any way to solve it quickly 

because the test has a limit time =) ..؟ 

lxl + lyl = 1

if x = 0, then y = 1,

and if y = 0, then x = 1.

So (0,1) and (1,0) are definitely points.

Also, we can also do the same for:

if x = 0, then y= -1 as well since we have the absolute value sign,

and if y = 0, then x = -1 as well since we have the absolute sign.

So, we now have 4 main points (0,1), (1,0), (0, -1), & (-1,0). 

Now, we know x or y cannot be greater than 1 since lxl + lyl must be equal to 1 [if x = 2, then there would be no value for y that would satisfy the equation].

Furthermore, we notice that 0.5 + 0.5 also equals 1 so we get more points: (0.5, 0.5), (-0.5, 0.5), (-0.5, -0.5), & (0.5, -0.5).

We now have 8 points and if you connect them together, you get a square.

So, these 8 points create a square with sides of sqrt(2).

Very nice explanation, Aziz...!!!

Here is another way to look at this, analytically (if not particularly "mathematically")...Note that, when lyl =0,  lxl cannot be larger than 1. Thus, x cannot be more than +1 or less than -1. So, plot the two points on the graph (1,0) and (-1,0). Likewise, when lxl = 0, lyl cannot be more than 1 or less than -1. Then, also plot the two points (0,1) and (0, -1) on the graph. Then, let all four points be the vertices of a square. The sides of this square are the "bounds" of our graph. Note that every point lying on any of the four sides satisfies lxl + lyl = 1, and any point lying "outside" these bounds, doesn't. (They make the equation > 1).

This gives us a way to graph any equation of the form lxl + lyl = a....where a is some positive real number.

Does that help??

For the second question, relating the graph to f(x) = cosx with interval -2pi <= x <= 2pi:

Notice that for f(x) = cosx, if x = 0, then f(0) = 1. But, in this graph at cos(0), the value is 2 which suggests that the graph had to have been shifted up one unit. This graph is f(x) = cosx + 1 = cos(0) + 1 = 1 + 1 = 2.

thanx for all   CPhill and  AzizHusain  i was understanding the two questions ...