Example 1.6 Let X = { 1,2,3,4} and Y = { 2,4,6,8,10} and R = {(1,2),(2,4),(3,6),(4,8)}. Show that R is a function and find its domain, co-domain and range?
Solution
Since all elements in X have only one image in Y. Hence its a function.
Domain = {1,2,3,4} (All elements from X)
Co-domain = {2,4,6,8,10} (All elements from Y)
Range = {2,4,6,8} (only with respect to the relation given)
Example 1.7
A relation f:X →Y is defined by f (x) = X2-2 where, X = {-2,-1,0,3} and Y = R.(i) List the elements of f (ii) Is f a function?
Solution
They have given the formula f (x) = x2-2, so we need to put x values in that formula and can get answers.
(i) From above we have f = {(-2,2),(-1,-1),(0,-2),(3,7)}
(ii) Since each element in the domain of f has a unique image, f is a function.
Example 1.8 If X = {–5,1,3,4} and Y = {a,b,c}, then which of the following relations are functions from X to Y ?
(i) R1 = {(–5,a), (1,a), (3,b)}, (ii) R2 = {(–5,b), (1,b), (3,a),(4,c)}, (iii) R3 = {(–5,a), (1,a), (3,b),(4,c),(1,b)}
(i) R1 = {(–5,a), (1,a), (3,b)}
Solution
From arrow diagram we can understand that 4 (element of X) does not have image in Y. Hence R1 is not a function from X to Y.(ii) R2 = {(–5,b), (1,b), (3,a),(4,c)}
From arrow diagram we can understand that each elements in X have image on Y. Hence R2 is a function from X to Y.
From arrow diagram we can understand that 1 (element of X) have two images on Y (a and b). Hence R3 is not a function from X to Y.
Example 1.9 Given f(x) = 2x - x2 , find (i)f (1) (ii) f (x+1)
(iii) f (x) + f (1)
Solution
Exercise 1.3
1. Let f = {(x,y) | x,y ∈N and y = 2x} be a relation on N. Find the domain, co-domain and range. Is this relation a function?
Solution
Since x and y belongs to natural number(N), Domain = {1,2,3,4,5,6....} Co-domain = {1,2,3,4,5,6....} Also, we have y = 2x Simply we need to substitute values of x in y. When x = 1, When x = 2, When x = 3, When x = 4, y = 2(1) y = 2(2) y = 2(3) y = 2(4) y = 2. y = 4. y = 6 y = 8. From above, Range = {2,4,6,8,10......}2. Let X = {3, 4, 6, 8}. Determine whether the relation R = {( x, (f x)) | , x ∈X , f (x) = x2+1}is a function from X to N?
Solution
We have equation, f (x) = x2+1 —--------------(1)
X = {3, 4, 6, 8}
Just need to substitute the values of X in equation (1)
When x = 3 When x = 4 When x = 6 When x = 8
f(x) = 32+1 f(x) = 42+1 f(x) = 62+1 f(x) = 82+1
f(x) = 9+1 = 16+1 = 36+1 = 64+1
f(x) = 10 = 17 = 37 = 65
From above we get, R = {10,17,37,65}
Since R∈N, this is a function from X to N. (N is a natural number)
3. Given the function f : x→ x2-5x+6 , evaluate (i) f(-1) (ii) f(2a) (iii)f(2) (iv)f(x-1)
(i)f(-1)
Solution
We have, f(x) = x2-5x+6 —-----------(1)
Since they are asking f(-1), just substitute x = -1 in equation (1)
f(-1) = (-1)2-[(5)(-1)]+6
= 1-(-5)+6
= 1+5+6
= 6+6
f(-1) = 12
(ii)f(2a)
We have, f(x) = x2-5x+6 —-----------(1)
Since they are asking f(2a), just substitute x = 2a in equation (1)
f(2a) = (2a)2-[(5)(2a)]+6
= 4a2-(10a)+6
f(2a) = 4a2-10a+6
(iii)f(2)
We have, f(x) = x2-5x+6 —-----------(1)
Since they are asking f(2), just substitute x = 2 in equation (1)
f(2) = (2)2-[(5)(2)]+6
= 4-(10)+6
= 4-10+6
= -6+6
f(2) = 0
(iv)f(x-1)
We have, f(x) = x2-5x+6 —-----------(1)
Since they are asking f(x-1), just substitute x = x-1 in equation (1)
f(x-1) = (x-1)2-[(5)(x-1)]+6
= [x2+12-2(x)(1)] - (5x-5)+6
= [x2+1-2x]-5x+5+6
= x2+1-2x-5x+11
f(x-1) = x2-7x+12
4. A graph representing the function f (x) is given in Fig.1.16 it is clear that f(9) = 2.
(i) Find the following values of the function,
f(0) = 9 (From figure it’s very clear. In x-axis we need to take value 0 and see where the line passing through in y-axis)
f(7) = 6 (Here, x-axis we need to take 7 and look for the line passing through in y-axis)
f(2) = 6
f(10) = 0
(ii) For what value of x is f(x) = 1?
Its opposite to what we done in previous section. That is we need to look on y-axis 1 and where the line meets in x-axis. It’s clear from the graph,
x = 9.5
(iii) Describe the following (i) Domain (ii) Range.
From figure it’s very clear that,
Domain = {0,1,2,3,4,5,6,7,8,9,10}
Range = {0,1,2,3,4,5,6,7,8,9}
(iv) What is the image of 6 under f ?
From figure we can see that image of 6 is 5.
i.e., f(6) = 5
(ii) find x such that f(x) = 0.
Solution
We have, f(x) = 2x-3 —-------------(1)
Given that, f(x) = 0
Hence, f(x) = 2x-3 = 0
2x-3 = 0
2x = 3
x =3/2
(iii) find x such that f(x ) = x
Solution
We have, f(x) = 2x-3 —-------------(1)
Given that, f(x) = x
2x-3 = x
2x-x = 3
x = 3
(iv) find x such that f(x) = f(1-x)
Solution
We have, f(x) = 2x-3 —-------------(1)
Here, we need to find value of f(1-x)
To find value of f(1-x), we need to substitute x = 1-x in equation (1)
f(1-x) = [(2)(1-x)]-3
= (2-2x)-3
= 2-2x-3
= -2x-1
Now, they given that, f(x) = f(1-x) —--------------(2)
We know both f(x) and f(1-x) so substitute that values in equation (2)
2x-3 = -2x-1
2x+2x = -1+3
4x = 2
x = 2/4
x = 1/2
Solution
Since the box formed by cutting the edges is in the shape of cuboid.
From figure,
l = 24-2x
b = 24-2x
h = x
WKT, volume of cuboid = lbh
= (24-2x) (24-2x).x
= (24-2x)2 .x
= [242+(2x)2-[2(24)(2x)]].x
= [546+4x2-(48)(2x)].x
= (546+4x2-96x).x
= 546x+4x3-96x2
V = 4x3-96x2+546xLet, f(x2) = (f(x))2 —------------------(1)
WKT, f(x) = 3-2x
f(x2) = 3-2x2 (Just substitute x=x2)
(f(x))2 = (3-2x)2
= 32+(2x)2-[(2)(3)(2x)] Since, (a-b)2 = a2+b2-2ab
= 9+4x2-[(6)(2x)]
= 9+4x2-12x
(f(x))2 = 4x2-12x+9
Substitute f(x2) and (f(x))2 values in equation (1)
3-2x2 = 4x2-12x+9
0 = 4x2-12x+9-3+2x2
0 = 6(x2-2x+1)
0 x 6 = x2-2x+1
0 = x2-2x+1
x2-2x+1 = 0
On factorization we get,
(x-1)(x-1) = 0
(x-1)2 = 0
(x-1) = 0
x =1
9. A plane is flying at a speed of 500 km per hour. Express the distance ‘d ’ travelled by the plane as function of time t in hours.
Given
Velocity or speed = 500 km/h
Distance = d
Time taken = t
Solution
WKT, velocity = distance/time taken
500 = d/t
So, here as per question we need to express distance as a function of time. Hence interchanging the above equation we get,
d = 500 x t
d = 500t
10. The data in the adjacent table depicts the length of a person forehand and their corresponding height. Based on this data, a student finds a relationship between the height (y) and the forehand length(x) as y = ax+b , where a, b are constants.
(i) Check if this relation is a function.
(ii) Find a and b.
(iii) Find the height of a person whose forehand length is 40 cm.
(iv) Find the length of forehand of a person if the height is 53.3 inches.
y = ax+b
Solution
(i)Check if this relation is a function
Given, y = ax+b
From the table we can write length of forehand and height as a ordered pairs,
R = {(35,56),(45,65),(50,69.5),(55,74)}
From relation ‘R’ we can analyze that given relation is a function.
(ii)Find a and b.
Given, y = ax+b —-----------(1)
From relation ‘R’ we can take any ordered pairs to substitute in equation (1) and get values for a and b.
We can take first one (35,56). Here x = 35 and y = 56
Substitute x=35 and y=56 in equation (1)
56 = a(35) + b
56 = 35a+b
35a+b = 56 —-------------(2)
Since to solve equation with two variables we need two equation, so taking another ordered pairs from given table as (45,65)
Substitute x=45 and y=65 in equation (1)
65 = a(45) + b
65 = 45a+b
Solving (2) and (3) we get below equation, (refer image on right side)
-10a = -9
10a = 9
a = 9/10
a = 0.9
Substitute a=0.9 in equation (2)
35(0.9) +b = 56
31.5 + b = 56
b = 56 - 31.5
b = 24.5
Substitute a and b values in equation (1)
y = 0.9x + 24.5 —------------(4)
(iii) Find the height of a person whose forehand length is 40 cm.
We know that forehand length is ‘x’
Hence substitute x=40 in equation (4)
y = 0.9(40) + 24.5
= 36 + 24.5
y = 60.5 inches
Height of a person whose forehand length is 40 cm is 60.5 inches.
(iv) Find the length of forehand of a person if the height is 53.3 inches.
We know that height is ‘y’
Hence substitute y=53.3 in equation (4)
53.3 = 0.9x + 24.5
53.3 - 24.5 = 0.9x
28.8 = 0.9x
0.9x = 28.8
x = 28.8/0.9
x = 32cm
The length of forehand of a person with height 53.3 inches is 32cm
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