### EX. NO. 1.4, RELATION AND FUNCTIONS, 10TH NEW SYLLABUS, Tamil Nadu, TN,

1. Determine whether the graph given below represent functions. Give reasons for your answers concerning each graph. Solution :
Since the graph intersects the vertical line (y-axis) at two points, it is not a function. Solution :
The given graph intersects the vertical line (y-axis) at one point. It is a function. Solution :
The graph intersects the y-axis at three points, hence it is not a function. Solution :
The graph intersects the vertical line at most one point. Hence it is a function.

2. Let f : A -> B be a function defined by
f (x) =  (x/2) - 1,
where A = {2, 4, 6, 10, 12}, B = {0, 1, 2, 4, 5, 9} .
Represent f by
(i) set of ordered pairs;
(ii) a table;
(iii) an arrow diagram;
(iv) a graph
Solution :
(i) set of ordered pairs
f (x) =  (x/2) - 1

 x = 2 f(2) = (2/2) - 1  = 1 - 1 f(2) = 0 x = 4 f(4) = (4/2) - 1  = 2 - 1 f(4) = 1 x = 6 f(6) = (6/2) - 1  = 3 - 1 f(6) = 2 x = 10 f(10) = (10/2) - 1  = 5 - 1 f(10) = 4 x = 12 f(12) = (12/2) - 1  = 6 - 1 f(12) = 5

Set of ordered pairs  = {(2, 0) (4, 1) (6, 2) (10, 4) (12, 5)}
(ii) a table (iii) an arrow diagram; (iv) a graph 3. Represent the function f = {(1, 2),(2, 2),(3, 2),(4, 3),(5, 4)} through (i) an arrow diagram (ii) a table form (iii) a graph
Solution :
(i) an arrow diagram (ii) a table form (iii) a graph 4. Show that the function f : N -> N defined by f (x) = 2x – 1 is one-one but not onto
Solution :
If for all a1, a2 ∈ A, f(a1) = f(a2) implies a1 = a2 then f is called one – one function.
Let x, y ∈ N, f(x)  = f(y)
f(x)  = 2x - 1  -----(1)
f(y)  = 2y - 1  -----(2)
(1)  = (2)
2x - 1  = 2y - 1
2x  = 2y
x  = y
Hence the function is one to one.
It is not onto :
If co-domain of the function = range of function, then the function is said to be onto.
Even numbers in the co-domain are not associated with the elements of domain. Hence it is not onto.

5. Show that the function f : N -> N defined by f (m) = m2 + m + 3 is one-one function.
Solution :
Let x, y ∈ N, f(x)  = f(y)
f (m) = m2 + m + 3
f(x)  = x2 + x + 3  -----(1)
f(y)  = y2 + y + 3  -----(2)
(1)  = (2)
x2 + x + 3  = y2 + y + 3
x2 + x  = y2 + y
x2 - y2 + x - y = 0
(x + y) (x - y) + (x - y)  = 0
(x - y) (x + y + 1)  = 0
x - y  = 0
x  = y
Hence it is one to one function.

6. Let A = {1, 2, 3, 4} and B = N . Let f : A -> B be defined by f (x) = x3 then, (i) find the range of f (ii) identify the type of function
Solution :
Given that :
f (x) = x3

 f (x) = x3 x = 1 f (1) = 13  = 1 f (x) = x3 x = 2 f (2) = 23  = 8 f (x) = x3 x = 3 f (3) = 33  = 27 f (x) = x3 x = 4 f (4) = 43  = 64

Range of f  = {1, 8, 27, 64}
Every element in A has associated with different elements of B. Hence it is one to one.

7. In each of the following cases state whether the function is bijective or not. Justify your answer.
(i) f : R -> R defined by f (x) = 2x +1
Solution :
Testing whether it is one to one :
If for all a1, a2 ∈ A, f(a1) = f(a2) implies a1 = a2 then f is called one – one function.
Let x, y ∈ R, f(x)  = f(y)
f(x)  = 2x + 1  ------(1)
f(y)  = 2y + 1  ------(2)
(1) = (2)
2x + 1  = 2y + 1
2x  = 2y
x  = y
So, it is one to one.
Testing whether it is onto :
Range of f = co-domain
If f : A -> B is an onto function then, the range of f = B . That is, f(A) = B.
Let x ∈ A, y ∈ B and x, y ∈ R. Then, x is pre-image and y is image.
Then, we have
y  = 2x + 1
Solve for x.
x  = (y - 1) /2
Here, y is a real number. When we subtract 1 from a real number and the result is divided by 2, again it is a real number.
For every real number of y, there is a real number x.
So, range of f(x) is equal to co-domain.
It is onto function.
Hence it is bijective function.
(ii) f : R -> R defined by f (x) = 3 – 4x2
Solution :
Testing whether it is one to one :
If for all a1, a2 ∈ A, f(a1) = f(a2) implies a1 = a2 then f is called one – one function.
Let x, y ∈ R, f(x)  = f(y)
f(x)  = 3 – 4x2  ------(1)
f(y)  = 3 – 4y2  ------(2)
(1) = (2)
3 – 4x2  =  3 – 4y2
-4x2  = –4y2
x2 - y2 =  0
(x - y)(x + y) =  0
x - y = 0  (or) x + y =  0
x = y (or) x = -y
It is not one to one.Hence it is not bijective function.

8. Let A = {−1, 1}and B = {0, 2} . If the function f : A -> B defined by f(x) = ax + b is an onto function? Find a and b.
Solution :
f(x) = ax + b
f(-1) = a(-1) + b  = 0
-a + b  = 0 ------(1)
f(1) = a(1) + b  = 2
a + b  = 2 ------(2)
(1) + (2)
2b  = 2
b  = 1
By applying the value of b in (1), we get
-a + 1  = 0
-a  = -1
a  = 1
Hence the values of a and b are 1 and 1 respectively.
9. If the function f is defined by find the values of
(i) f (3) (ii) f (0) (iii) f (−1.5) (iv) f (2)+ f (−2)
Solution : (i)  f(3)
Instead of x, we have 3. So we have to choose the function f(x)  = x + 2
f(3)  = 3 + 2
f(3)  = 5
(ii)  f(0)
0 lies between -1 and 1. So, the answer is 2.
(iii) f (−1.5)
f(x)  = x - 1
f(-1.5)  = -1.5 - 1
f(-1.5)  = -2.5
(iv) f (2)+ f (−2)
f (2) + f (−2)  = 4 + (-3) = 1

10. A function f : [−5,9] -> R is defined as follows: Find (i) f (−3) + f (2)
(ii) f (7) - f (1)
(iii) 2f (4) + f (8)
(iv)  [2f(-2) - f(6)] / [f(4) + f(-2)]
Solution :
(i) f (−3) + f(2)
f(x)  = 6x + 1 for f(-3) and f(x)  = 5x2 - 1 for f(2)
f (−3) + f(2)  = -17 + 19
f (−3) + f(2)  = 2
(ii) f (7) - f (1)
f(x)  = 3x - 4 for f(7) and f(x)  = 6x + 1 for f(1)
f (7) - f (1)  = 17 - 7 = 10
(iii) 2f (4) + f (8)
f(x)  = 5x2 - 1 for f(4)  and f(x) = 3x - 4 for f(8)
2f (4) + f (8)  = 2(79) + 20
= 158 + 20
2f (4) + f (8)  = 178
(iv)  [2f(-2) - f(6)] / [f(4) + f(-2)]
f(x)  = 6x + 1 for f(-2) and f(x) = 3x - 4 for f(6)
[2f(-2) - f(6)] / [f(4) + f(-2)]  = [2(-11) - 14] / [79 + (-11)]
= (-22 - 14) / (79 - 11)
= -36/68
= -9/17

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