RD Sharma Class 12 Solutions - Functions | Mathematics (Maths) Class 12 - JEE PDF Download

 Page 1


2. Functions
Exercise 2.1
1 A. Question
Give an example of a function
Which is one – one but not onto.
Answer
TIP: – One – One Function: – A function  is said to be a one – one functions or an injection if different
elements of A have different images in B.
So,  is One – One function
? a?b
? f(a)?f(b) for all 
? f(a) = f(b)
? a = b for all 
Onto Function: – A function  is said to be a onto function or surjection if every element of A i.e, if f(A)
= B or range of f is the co – domain of f.
So,  is Surjection iff for each , there exists  such that f(a) = b
Now, Let,  given by f(x) = x
2
Check for Injectivity:
Let x,y be elements belongs to N i.e  such that
So, from definition
? f(x) = f(y)
? x
2
 = y
2
? x
2
 – y
2
 = 0
? (x – y)(x + y) = 0
As  therefore x + y>0
? x – y = 0
? x = y
Hence f is One – One function
Check for Surjectivity:
Let y be element belongs to N i.e  be arbitrary, then
? f(x) = y
? x
2
 = y
? 
? not belongs to N for non–perfect square value of y.
Therefore no non – perfect square value of y has a pre image in domain N.
Hence,  given by f(x) = x
2
 is One – One but not onto.
Page 2


2. Functions
Exercise 2.1
1 A. Question
Give an example of a function
Which is one – one but not onto.
Answer
TIP: – One – One Function: – A function  is said to be a one – one functions or an injection if different
elements of A have different images in B.
So,  is One – One function
? a?b
? f(a)?f(b) for all 
? f(a) = f(b)
? a = b for all 
Onto Function: – A function  is said to be a onto function or surjection if every element of A i.e, if f(A)
= B or range of f is the co – domain of f.
So,  is Surjection iff for each , there exists  such that f(a) = b
Now, Let,  given by f(x) = x
2
Check for Injectivity:
Let x,y be elements belongs to N i.e  such that
So, from definition
? f(x) = f(y)
? x
2
 = y
2
? x
2
 – y
2
 = 0
? (x – y)(x + y) = 0
As  therefore x + y>0
? x – y = 0
? x = y
Hence f is One – One function
Check for Surjectivity:
Let y be element belongs to N i.e  be arbitrary, then
? f(x) = y
? x
2
 = y
? 
? not belongs to N for non–perfect square value of y.
Therefore no non – perfect square value of y has a pre image in domain N.
Hence,  given by f(x) = x
2
 is One – One but not onto.
1 B. Question
Give an example of a function
Which is not one – one but onto.
Answer
TIP: – One – One Function: – A function  is said to be a one – one functions or an injection if different
elements of A have different images in B.
So,  is One – One function
? a?b
? f(a)?f(b) for all 
? f(a) = f(b)
? a = b for all 
Onto Function: – A function  is said to be a onto function or surjection if every element of A i.e, if f(A)
= B or range of f is the co – domain of f.
So,  is Surjection iff for each , there exists  such that f(a) = b
Now, Let,  given by f(x) = x
3
 – x
Check for Injectivity:
Let x,y be elements belongs to R i.e  such that
So, from definition
? f(x) = f(y)
? x
3
 – x = y
3
 – y
? x
3
 – y
3
 – (x – y) = 0
? (x – y)(x
2
 + xy + y
2
 – 1) = 0
As x
2
 + xy + y
2
 = 0
? therefore x
2
 + xy + y
2
 – 1= – 1
? x – y?0
? x ? y for some 
Hence f is not One – One function
Check for Surjectivity:
Let y be element belongs to R i.e  be arbitrary, then
? f(x) = y
? x
3
 – x = y
? x
3
 – x – y = 0
Now, we know that for 3 degree equation has a real root
So, let  be that root
? 
Page 3


2. Functions
Exercise 2.1
1 A. Question
Give an example of a function
Which is one – one but not onto.
Answer
TIP: – One – One Function: – A function  is said to be a one – one functions or an injection if different
elements of A have different images in B.
So,  is One – One function
? a?b
? f(a)?f(b) for all 
? f(a) = f(b)
? a = b for all 
Onto Function: – A function  is said to be a onto function or surjection if every element of A i.e, if f(A)
= B or range of f is the co – domain of f.
So,  is Surjection iff for each , there exists  such that f(a) = b
Now, Let,  given by f(x) = x
2
Check for Injectivity:
Let x,y be elements belongs to N i.e  such that
So, from definition
? f(x) = f(y)
? x
2
 = y
2
? x
2
 – y
2
 = 0
? (x – y)(x + y) = 0
As  therefore x + y>0
? x – y = 0
? x = y
Hence f is One – One function
Check for Surjectivity:
Let y be element belongs to N i.e  be arbitrary, then
? f(x) = y
? x
2
 = y
? 
? not belongs to N for non–perfect square value of y.
Therefore no non – perfect square value of y has a pre image in domain N.
Hence,  given by f(x) = x
2
 is One – One but not onto.
1 B. Question
Give an example of a function
Which is not one – one but onto.
Answer
TIP: – One – One Function: – A function  is said to be a one – one functions or an injection if different
elements of A have different images in B.
So,  is One – One function
? a?b
? f(a)?f(b) for all 
? f(a) = f(b)
? a = b for all 
Onto Function: – A function  is said to be a onto function or surjection if every element of A i.e, if f(A)
= B or range of f is the co – domain of f.
So,  is Surjection iff for each , there exists  such that f(a) = b
Now, Let,  given by f(x) = x
3
 – x
Check for Injectivity:
Let x,y be elements belongs to R i.e  such that
So, from definition
? f(x) = f(y)
? x
3
 – x = y
3
 – y
? x
3
 – y
3
 – (x – y) = 0
? (x – y)(x
2
 + xy + y
2
 – 1) = 0
As x
2
 + xy + y
2
 = 0
? therefore x
2
 + xy + y
2
 – 1= – 1
? x – y?0
? x ? y for some 
Hence f is not One – One function
Check for Surjectivity:
Let y be element belongs to R i.e  be arbitrary, then
? f(x) = y
? x
3
 – x = y
? x
3
 – x – y = 0
Now, we know that for 3 degree equation has a real root
So, let  be that root
? 
Thus for clearly , there exist  such that f(x) = y
Therefore f is onto
? Hence,  given by f(x) = x
3
 – x is not One – One but onto
1 C. Question
Give an example of a function
Which is neither one – one nor onto.
Answer
TIP: – One – One Function: – A function  is said to be a one – one functions or an injection if different
elements of A have different images in B.
So,  is One – One function
? a?b
? f(a)?f(b) for all 
? f(a) = f(b)
? a = b for all 
Onto Function: – A function  is said to be a onto function or surjection if every element of A i.e, if f(A)
= B or range of f is the co – domain of f.
So,  is Surjection iff for each , there exists  such that f(a) = b
Now, Let,  given by f(x) = 5
As we know
A constant function is neither one – one nor onto.
So, here f(x) = 5 is constant function
Therefore
 given by f(x) = 5 is neither one – one nor onto function.
2 A. Question
Which of the following functions from A to B are one – one and onto?
f
1
 = {(1, 3), (2, 5), (3, 7)}; A = {1, 2, 3}, B = {3, 5, 7}
Answer
TIP: – One – One Function: – A function  is said to be a one – one functions or an injection if different
elements of A have different images in B.
So,  is One – One function
? a?b
? f(a)?f(b) for all 
? f(a) = f(b)
? a = b for all 
Onto Function: – A function  is said to be a onto function or surjection if every element of A i.e, if f(A)
= B or range of f is the co – domain of f.
So,  is Surjection iff for each , there exists  such that f(a) = b
Page 4


2. Functions
Exercise 2.1
1 A. Question
Give an example of a function
Which is one – one but not onto.
Answer
TIP: – One – One Function: – A function  is said to be a one – one functions or an injection if different
elements of A have different images in B.
So,  is One – One function
? a?b
? f(a)?f(b) for all 
? f(a) = f(b)
? a = b for all 
Onto Function: – A function  is said to be a onto function or surjection if every element of A i.e, if f(A)
= B or range of f is the co – domain of f.
So,  is Surjection iff for each , there exists  such that f(a) = b
Now, Let,  given by f(x) = x
2
Check for Injectivity:
Let x,y be elements belongs to N i.e  such that
So, from definition
? f(x) = f(y)
? x
2
 = y
2
? x
2
 – y
2
 = 0
? (x – y)(x + y) = 0
As  therefore x + y>0
? x – y = 0
? x = y
Hence f is One – One function
Check for Surjectivity:
Let y be element belongs to N i.e  be arbitrary, then
? f(x) = y
? x
2
 = y
? 
? not belongs to N for non–perfect square value of y.
Therefore no non – perfect square value of y has a pre image in domain N.
Hence,  given by f(x) = x
2
 is One – One but not onto.
1 B. Question
Give an example of a function
Which is not one – one but onto.
Answer
TIP: – One – One Function: – A function  is said to be a one – one functions or an injection if different
elements of A have different images in B.
So,  is One – One function
? a?b
? f(a)?f(b) for all 
? f(a) = f(b)
? a = b for all 
Onto Function: – A function  is said to be a onto function or surjection if every element of A i.e, if f(A)
= B or range of f is the co – domain of f.
So,  is Surjection iff for each , there exists  such that f(a) = b
Now, Let,  given by f(x) = x
3
 – x
Check for Injectivity:
Let x,y be elements belongs to R i.e  such that
So, from definition
? f(x) = f(y)
? x
3
 – x = y
3
 – y
? x
3
 – y
3
 – (x – y) = 0
? (x – y)(x
2
 + xy + y
2
 – 1) = 0
As x
2
 + xy + y
2
 = 0
? therefore x
2
 + xy + y
2
 – 1= – 1
? x – y?0
? x ? y for some 
Hence f is not One – One function
Check for Surjectivity:
Let y be element belongs to R i.e  be arbitrary, then
? f(x) = y
? x
3
 – x = y
? x
3
 – x – y = 0
Now, we know that for 3 degree equation has a real root
So, let  be that root
? 
Thus for clearly , there exist  such that f(x) = y
Therefore f is onto
? Hence,  given by f(x) = x
3
 – x is not One – One but onto
1 C. Question
Give an example of a function
Which is neither one – one nor onto.
Answer
TIP: – One – One Function: – A function  is said to be a one – one functions or an injection if different
elements of A have different images in B.
So,  is One – One function
? a?b
? f(a)?f(b) for all 
? f(a) = f(b)
? a = b for all 
Onto Function: – A function  is said to be a onto function or surjection if every element of A i.e, if f(A)
= B or range of f is the co – domain of f.
So,  is Surjection iff for each , there exists  such that f(a) = b
Now, Let,  given by f(x) = 5
As we know
A constant function is neither one – one nor onto.
So, here f(x) = 5 is constant function
Therefore
 given by f(x) = 5 is neither one – one nor onto function.
2 A. Question
Which of the following functions from A to B are one – one and onto?
f
1
 = {(1, 3), (2, 5), (3, 7)}; A = {1, 2, 3}, B = {3, 5, 7}
Answer
TIP: – One – One Function: – A function  is said to be a one – one functions or an injection if different
elements of A have different images in B.
So,  is One – One function
? a?b
? f(a)?f(b) for all 
? f(a) = f(b)
? a = b for all 
Onto Function: – A function  is said to be a onto function or surjection if every element of A i.e, if f(A)
= B or range of f is the co – domain of f.
So,  is Surjection iff for each , there exists  such that f(a) = b
Now, As given,
f
1
 = {(1, 3), (2, 5), (3, 7)}
A = {1, 2, 3}, B = {3, 5, 7}
Thus we can see that,
Check for Injectivity:
Every element of A has a different image from B
Hence f is a One – One function
Check for Surjectivity:
Also, each element of B is an image of some element of A
Hence f is Onto.
2 B. Question
Which of the following functions from A to B are one – one and onto?
f
2
 = {(2, a), (3, b), (4, c)}; A = {2, 3, 4}, B = {a, b, c}
Answer
TIP: – One – One Function: – A function  is said to be a one – one functions or an injection if different
elements of A have different images in B.
So,  is One – One function
? a?b
? f(a)?f(b) for all 
? f(a) = f(b)
? a = b for all 
Onto Function: – A function  is said to be a onto function or surjection if every element of A i.e, if f(A)
= B or range of f is the co – domain of f.
So,  is Surjection iff for each , there exists  such that f(a) = b
Now, As given,
f
2
 = {(2, a), (3, b), (4, c)}
A = {2, 3, 4}, B = {a, b, c}
Thus we can see that
Check for Injectivity:
Every element of A has a different image from B
Hence f is a One – One function
Check for Surjectivity:
Also, each element of B is an image of some element of A
Hence f is Onto.
2 C. Question
Which of the following functions from A to B are one – one and onto?
f
3
 = {(a, x), (b, x), (c, z), (d, z)}; A = {a, b, c, d}, B = {x, y, z}
Page 5


2. Functions
Exercise 2.1
1 A. Question
Give an example of a function
Which is one – one but not onto.
Answer
TIP: – One – One Function: – A function  is said to be a one – one functions or an injection if different
elements of A have different images in B.
So,  is One – One function
? a?b
? f(a)?f(b) for all 
? f(a) = f(b)
? a = b for all 
Onto Function: – A function  is said to be a onto function or surjection if every element of A i.e, if f(A)
= B or range of f is the co – domain of f.
So,  is Surjection iff for each , there exists  such that f(a) = b
Now, Let,  given by f(x) = x
2
Check for Injectivity:
Let x,y be elements belongs to N i.e  such that
So, from definition
? f(x) = f(y)
? x
2
 = y
2
? x
2
 – y
2
 = 0
? (x – y)(x + y) = 0
As  therefore x + y>0
? x – y = 0
? x = y
Hence f is One – One function
Check for Surjectivity:
Let y be element belongs to N i.e  be arbitrary, then
? f(x) = y
? x
2
 = y
? 
? not belongs to N for non–perfect square value of y.
Therefore no non – perfect square value of y has a pre image in domain N.
Hence,  given by f(x) = x
2
 is One – One but not onto.
1 B. Question
Give an example of a function
Which is not one – one but onto.
Answer
TIP: – One – One Function: – A function  is said to be a one – one functions or an injection if different
elements of A have different images in B.
So,  is One – One function
? a?b
? f(a)?f(b) for all 
? f(a) = f(b)
? a = b for all 
Onto Function: – A function  is said to be a onto function or surjection if every element of A i.e, if f(A)
= B or range of f is the co – domain of f.
So,  is Surjection iff for each , there exists  such that f(a) = b
Now, Let,  given by f(x) = x
3
 – x
Check for Injectivity:
Let x,y be elements belongs to R i.e  such that
So, from definition
? f(x) = f(y)
? x
3
 – x = y
3
 – y
? x
3
 – y
3
 – (x – y) = 0
? (x – y)(x
2
 + xy + y
2
 – 1) = 0
As x
2
 + xy + y
2
 = 0
? therefore x
2
 + xy + y
2
 – 1= – 1
? x – y?0
? x ? y for some 
Hence f is not One – One function
Check for Surjectivity:
Let y be element belongs to R i.e  be arbitrary, then
? f(x) = y
? x
3
 – x = y
? x
3
 – x – y = 0
Now, we know that for 3 degree equation has a real root
So, let  be that root
? 
Thus for clearly , there exist  such that f(x) = y
Therefore f is onto
? Hence,  given by f(x) = x
3
 – x is not One – One but onto
1 C. Question
Give an example of a function
Which is neither one – one nor onto.
Answer
TIP: – One – One Function: – A function  is said to be a one – one functions or an injection if different
elements of A have different images in B.
So,  is One – One function
? a?b
? f(a)?f(b) for all 
? f(a) = f(b)
? a = b for all 
Onto Function: – A function  is said to be a onto function or surjection if every element of A i.e, if f(A)
= B or range of f is the co – domain of f.
So,  is Surjection iff for each , there exists  such that f(a) = b
Now, Let,  given by f(x) = 5
As we know
A constant function is neither one – one nor onto.
So, here f(x) = 5 is constant function
Therefore
 given by f(x) = 5 is neither one – one nor onto function.
2 A. Question
Which of the following functions from A to B are one – one and onto?
f
1
 = {(1, 3), (2, 5), (3, 7)}; A = {1, 2, 3}, B = {3, 5, 7}
Answer
TIP: – One – One Function: – A function  is said to be a one – one functions or an injection if different
elements of A have different images in B.
So,  is One – One function
? a?b
? f(a)?f(b) for all 
? f(a) = f(b)
? a = b for all 
Onto Function: – A function  is said to be a onto function or surjection if every element of A i.e, if f(A)
= B or range of f is the co – domain of f.
So,  is Surjection iff for each , there exists  such that f(a) = b
Now, As given,
f
1
 = {(1, 3), (2, 5), (3, 7)}
A = {1, 2, 3}, B = {3, 5, 7}
Thus we can see that,
Check for Injectivity:
Every element of A has a different image from B
Hence f is a One – One function
Check for Surjectivity:
Also, each element of B is an image of some element of A
Hence f is Onto.
2 B. Question
Which of the following functions from A to B are one – one and onto?
f
2
 = {(2, a), (3, b), (4, c)}; A = {2, 3, 4}, B = {a, b, c}
Answer
TIP: – One – One Function: – A function  is said to be a one – one functions or an injection if different
elements of A have different images in B.
So,  is One – One function
? a?b
? f(a)?f(b) for all 
? f(a) = f(b)
? a = b for all 
Onto Function: – A function  is said to be a onto function or surjection if every element of A i.e, if f(A)
= B or range of f is the co – domain of f.
So,  is Surjection iff for each , there exists  such that f(a) = b
Now, As given,
f
2
 = {(2, a), (3, b), (4, c)}
A = {2, 3, 4}, B = {a, b, c}
Thus we can see that
Check for Injectivity:
Every element of A has a different image from B
Hence f is a One – One function
Check for Surjectivity:
Also, each element of B is an image of some element of A
Hence f is Onto.
2 C. Question
Which of the following functions from A to B are one – one and onto?
f
3
 = {(a, x), (b, x), (c, z), (d, z)}; A = {a, b, c, d}, B = {x, y, z}
Answer
TIP: – One – One Function: – A function  is said to be a one – one functions or an injection if different
elements of A have different images in B.
So,  is One – One function
? a?b
? f(a)?f(b) for all 
? f(a) = f(b)
? a = b for all 
Onto Function: – A function  is said to be a onto function or surjection if every element of A i.e, if f(A)
= B or range of f is the co – domain of f.
So,  is Surjection iff for each , there exists  such that f(a) = b
Now, As given,
f
3
 = {(a, x), (b, x), (c, z), (d, z)}
A = {a, b, c, d}, B = {x, y, z}
Thus we can clearly see that
Check for Injectivity:
Every element of A does not have different image from B
Since,
f
3
(a) = x = f
3
(b) and f
3
(c) = z = f
3
(d)
Therefore f is not One – One function
Check for Surjectivity:
Also each element of B is not image of any element of A
Hence f is not Onto.
3. Question
Prove that the function f : N ? N, defined by f(x) = x
2
 + x + 1 is one – one but not onto.
Answer
TIP: – One – One Function: – A function  is said to be a one – one functions or an injection if different
elements of A have different images in B.
So,  is One – One function
? a?b
? f(a)?f(b) for all 
? f(a) = f(b)
? a = b for all 
Onto Function: – A function  is said to be a onto function or surjection if every element of A i.e, if f(A)
= B or range of f is the co – domain of f.
So,  is Surjection iff for each , there exists  such that f(a) = b
Now,  given by f(x) = x
2
 + x + 1