The curves in Fig.1.35(a) and Fig.1.35(c) represent a one – one function as the horizontal lines meet the curves in only one point P.
The curve in Fig.1.35(b) does not represent a one–one function, since, the horizontal line meet the curve in two points P and Q.

A = {1, 2, 3}, B = {4, 5, 6, 7}; f = {(1, 4),(2, 5),(3, 6)}
Then f is a function from A to B and for different elements in A, there are different images in B. Hence f is one–one function. Note that the element 7 in the co-domain does not have any pre-image in the domain. Hence f is not on to.
Therefore f is one–one but not an onto function.

f(x) = 3x - 5 can be written as f{(x, 3x - 5) |x \(\in \) R }
(a, 4) means the image of a is 4. That is, f(a) = 4
3a - 5 = 4 ⇒ a = 3
(1,b) means the image of 1 is b. That is, f(1) = b ⇒ b =-2
3(1) - 5 = b ⇒ b = -2

It is not a function. Since, a vertical line intersects the curve at more than one points.

A = {1, 2, 3, 4} B = N
(i) f: A⟶ B,
f(x) = x³
f(1) = 1³ =1
f(2) = 2³ = 8
f(3) = 3³ = 27
f(4) = 4³ = 64
The range of f = {1, 8,27, 64}
(ii) 'f is a function from 'A' to 'B' and all the elements in A having different images in 'B' 
f : A⟶ B, is a function. It is one-to-one and into function and also called a cubic function.

A = {-1, 1}, B = {0, 2}
f: A ⟶ B, f(x) = ax + b
when x = -1, f(-1) = 9
when x = 1, f(1) = 2
Since, there is a constant difference between x and f (x), it is an onto function
Substituting the values, we get
-a + b = 0
a + b = 2
Solving these equations, we get a = 1, b = 1

Distance travelled by an object is given to be S(t) =\(\frac{1}{2}\) gt² + at + b
'g'- acceleration due to gravity is a constant.
'g' is the acceleration due to gravity 'a' and 'b' are constants
't' is a variable. (t - time)
At different values of 't' , S(t) is having different values. (images in codomain) clearly t - S (t) is one-to-one function
Given S(t) =\(\frac{1}{2}\) gt² + at + b (a, b are constants)
t = time in seconds
Let us take t = 1, 2,3 ..... Seconds
When t = 1 \(S(1)=\frac{g}{2}+a+b \)
When t = 2 \(S(2)=2 g+2 a+b \)
When t = 3  \(S(3)=\frac{9}{2} g+3 a+b \)  and so on.

AII the elements of t, having different images in S(t). Hence it is an One-to-one function.

Since A x A has 9 elements,
A would have 3 elements 
A x A contains (- 1, 0) and (0, 1)
-1, 0 \(\in \) A
Similarly (0, 1) is in A x A
So, 0,1 \(\in \) A
From (1) and (2). = -1, 0, 1 \(\in \) A
A = {-1, 0 , 1}
A x A = {-1, 0, 1} x {1, 0, -1}
= {(-1, 1), (-1, 0), (-1, -1), (0, -1),(0,0), (0, 1), (1, -1), (1, 0),(1, 1)}
The remaining elements of A x A are 
{(- 1, - 1), (- 1, 1), (0, - 1), (0, 0), (1, - 1), (1,0),(1, 1)}

 f(x) = \(\frac { 2x+1 }{ x-9 } \)
If x - 9 = 0,then
x = 9

The domain is all values of x that make the expression defined.
\(i.e., (-\infty, 9) \cup(9, \infty) \)
\(i.e., (x / x \neq 9) \Rightarrow R-\{9\}\)

i) f: R ⟶ R and f(x) = 2x + 1 
when x = -1,f(- 1) = 2(- 1) + 1 =- 1 \( \in\) R
when x =  0, f(0) = 2 (0) + 1 = 1 \( \in\) R
when x = 1,f (1) = 2(1) + 1 = 3 \( \in\) R and soon
For every value of x \( \in\) R, f (x) also \( \in\) R.
The function is well defined and it is one-to-one function (Injective)
For f (x) : R ⟶ R , the domain and range are also well defined. So it is an onto function (Surjective)
Thus, the function is one - to one onto i.e. Bijective function.
(ii) f : R ⟶ R , f(x) = 3 - 4x²
when x = 0, f (0) = 3 - 4(0) = 3 \( \in\) R
when x = 1, f (1) = 3 - 4(1)² = -1  \( \in\) R
when x = 2, f (2) = 3 - 4(2)² = 13  \( \in\) R
when x = -1, f (-1) = 3 - 4(-1)² = -1  \( \in\) R
when x = -2, f (-2) = 3 - 4(-2)² = -13  \( \in\) R
From this, it is clear that two or more elements having same image in the co-domain So, it is not one-to-one and it is many-to-one function. Hence it is not Bijective.