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Area of a Graphs

Before we proceed ahead with calculating the area of the combination of graphs, we should be clear with the quadrants and the signs of X and Y in the same.
 
Description: 11-15.tif

To find the area of graphs, we first need to sketch the graphs of the equations and then by using geometry/ coordinate geometry we can find the area of the enclosed figure.

 
Example-1
Find out the area of the region enclosed by y = |x| and y = 2.
Solution
Following is the area enclosed by the equations given above:
Description: 11-16.tif
 
The area of the enclosed figures will be, Area = ½ × 4 × 2 = 4 sq. units
 
 
Example-2
In the X – Y plane, the area of the region bounded by the graph |x + y| + |x − y| = 4 is
  1. 8
  2. 12
  3. 16
  4. 20
Solution
Let x ≥ 0, y ≥ 0 and x ≥ y
Then |x + y| + |x − y| = 4
⇒ x + y + x – y = 4 ⇒ x = 2
Similarly, x ≥ 0, y ≥ 0, x ≤ y
x + y + y – x = 4 ⇒ y = 2
The area in the first quadrant is 4.
 
By using symmetry, the total area in all the four quadrants = 4 × 4 = 16 sq. units
 
 
Example-3
If p, q, and r are any real numbers, then
  1. max (p, q) < max (p, q, r)
  2. min (p, q) = Description: 3336.png(p + q −|p − q|)
  3. min (p, q) < min (p, q, r)
  4. None of these
Solution
If we take r p, q, then (1) and (3) cannot hold. For (2), if p ≥ q, then |p − q| = p – q.
 
Description: 3340.png(p + q |p  q|) = Description: 3345.png(p + q  p + q|) = q = min (p, q)
Similarly, If p < q then |p – q| = q – p.
 
(p + q − |p − q|) = Description: 3349.png(p + q  q + p) = p = min (p, q)
So, option (b) is the answer.
 
 
Example-4
If a b c d = 1, a > 0, b > 0, c > 0, d > 0, then what is the minimum value of (a + 1) (b + 1) (c + 1) (d + 1)?
  1. 1
  2. 8
  3. 16
  4. None of these
Solution
The minimum value will occur when a = b = c = d = 1
 
So, the minimum value of (a + 1) (b + 1) (c + 1) (d + 1) = 16
 
 
Example-5
Consider a triangle drawn on the X-Y plane with its three vertices (41, 0), (0, 41) and (0, 0), each vertex being represented by its (X,Y) coordinates. What is the number of points with integer coordinates inside the triangle (excluding all the points on the boundary)?
  1. 780
  2. 800
  3. 820
  4. 741
Solution
The equation formed from the data is x + y < 41
 
The values which will satisfy this equation are
(1, 39), (1, 38) …(1,1)
(2, 38), (2, 37), …(2,1)
(39, 1)
 
So the total number of cases are 39 + 38 + 37 … + 1 Description: 3353.png
 
 
Example-6
A telecom service provider engages male and female operators for answering 1000 calls per day. A male operator can handle 40 calls per day whereas a female operator can handle 50 calls per day. The male and the female operators get a fixed wage of Rs 250 and Rs 300 per day respectively. In addition, a male operator gets Rs 15 per call he answers and female operator gets Rs 10 per call she answers. To minimize the total cost, how many male operators should the service provider employ assuming he has to employ more than 7 of the 12 female operators available for the job?
  1. 15
  2. 14
  3. 12
  4. 10
Solution
First let us form both the equations:
40 m + 50 f = 1000
250 m + 300 f + 40 × 15 m + 50 × 10 × f = A
850 m + 8000 f = A
 
Where m and f are the number of males and females and A is the amount paid by the service provider.
 
Then the possible values for f are 8, 9, 10, 11, 12
 
If f = 8, then m = 15
 
If f = 9, 10 and 11 then m will not be an integer while
f = 12 then m will be 10.
 
By putting f = 8 m = 15 and A = 18800. When f = 12 and m = 10 then A = 18100.
 
Hence, the number of males will be 10.
 
 
Example-7
If ab and c are the sides of a triangle, then what is the maximum value of the expressionDescription: 3362.png?
  1. 1
  2. 3/2
  3. 2
  4. 5/2
Solution
Assume 2s = a + b + c. We know that b + c > a, so we get 2(b + c) > a b + c = 2s
 
Hence, b + c > s
 
Similarly, c + a > s, a + b > s
 
Hence, Description: 3371.png Description: 3375.png
 
 
Example-8
At how many distinct points the graphs of y = x-1 and y = logex intersect? (CAT 2003)
Solution
We can see that the graphs of y = x-1 and y = logex intersect just once.
 
 
Example-9
How many integral solution is/are possible for the equation |y−18| + |y – 9| + |y +9| + | y + 18| = 54?
Solution
Here, |yN| should be seen as nothing but the distance of the point y from the point N on the number line, a person standing at a point N.
 
Hence |y − 18| + |y – 9| + |y + 9| + |y + 18| is the sum of the distances of the point y from 18, 9, −9 and –18.
 
Now, for any point y, where p ≤ y ≤ q, the sum of the distances from p and q is q – p.
 
So, for point y, where −9 ≤ y ≤ 9, the sum of the distances of y from –18, −9, 9 and 18 is [18−(−18)] + [9− (−9)] = 54
 
For points outside this limitation, this expression will have different values.
 
Hence, the required numbers are –9, −8,…, 8, 9. So, there will be a total of 19 values.
 




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