Important Questions for Class 12 Maths Chapter 1 – Relations and functions are provided here for the students to score good marks in the class 12 board Maths examination. Here, all the important problems are covered as per the NCERT book. By practising and going through the important questions, students can get the confidence to solve the problems easily and efficiently. Get the class 12 Maths important questions for all the chapters at BYJU’S.
Here, we are going to discuss the important questions in class 12 chapter 1 – Relations and functions with solutions. Before that, let us learn the basic concepts covered in class 12 maths chapter 1. In relations and functions chapter 1, it covers the concept of relations, functions and binary operations.
What is the Relation?
Let us consider two sets A and B. Then a relation R from Set A into Set B is defined as the subset of A × B. The relation can be classified into three different types, such as:
 Reflexive Relation
 Symmetric Relation
 Transitive Relation
What is a Function?
A function is defined as the relation between two sets, in which every element in the set 1 associate to the elements in set 2. In other words, if “f” is a function from A to B, then every element in the set B is the image of an element in set A. The different types of functions are:
 One to One function
 One to Many functions
 Onto function
 One to one Correspondence
Also, check:
 Important 4 Marks Questions for CBSE Class 12 Maths
 Important 6 marks Questions for CBSE Class 12 Maths
Class 12 Chapter 1 – Relations and Functions Important Questions with Solutions
Go through the solved problems given below. Here, all the important questions for class 12 chapter 1 maths problems are given. The problems given here are frequently asked in the previous board examinations. Practice these problems well and solve the practice problems provided here.
Q.1: Show that the Signum Function f: R → R, given by
\(f(x)= \left\{\begin{matrix} 1 & for \ x >0 \\ 0 & for \ x =0 \ is \ neither\ oneone\ nor\ onto\\ 1 & for\ x<0 \end{matrix}\right.\)Solution:
Check for one to one function:
For example:
f(0) = 0
f(1) = 1
f(1) = 1
f(2) = 1
f(3) = 1
Since, for the different elements say f(1), f(2) and f(3), it shows the same image, then the function is not one to one function.
Check for Onto Function:
For the function,f: R →R
\(f(x)= \left\{\begin{matrix} 1 & for \ x >0 \\ 0 & for \ x =0 \\ 1 & for\ x<0 \end{matrix}\right.\)In this case, the value of f(x) is defined only if x is 1, 0, 1
For any other real numbers(for example y = 2, y = 100) there is no corresponding element x.
Thus, the function “f” is not onto function.
Hence, the given function “f” is neither oneone nor onto.
Q.2: If f: R → R is defined by f(x) = x^{2} − 3x + 2, find f(f(x)).
Solution:
Given function:
f(x) = x^{2} − 3x + 2.
To find f(f(x))
f(f(x)) = f(x)^{2} − 3f(x) + 2.
= (x^{2} – 3x + 2)^{2} – 3(x^{2} – 3x + 2) + 2
By using the formula (ab+c)^{2} = a^{2}+ b^{2}+ c^{2}2ab +2ac2ab, we get
= (x^{2})^{2} + (3x)^{2} + 2^{2}– 2x^{2} (3x) + 2x^{2}(2) – 2x^{2}(3x) – 3(x^{2} – 3x + 2) + 2
Now, substitute the values
= x^{4} + 9x^{2} + 4 – 6x^{3} – 12x + 4x^{2} – 3x^{2} + 9x – 6 + 2
= x^{4} – 6x^{3} + 9x^{2} + 4x^{2} – 3x^{2} – 12x + 9x – 6 + 2 + 4
Simplify the expression, we get,
f(f(x)) = x^{4} – 6x^{3} + 10x^{2} – 3x
Q.3: Let A = N × N and * be the binary operation on A defined by (a, b) * (c, d) = (a + c, b + d). Show that * is commutative and associative. Find the identity element for * on A, if any.
Solution:
Check the binary operation * is commutative :
We know that, * is commutative if (a, b) * (c, d) = (c, d) * (a, b) ∀ a, b, c, d ∈ R
L.H.S =(a, b) * (c, d)
=(a + c, b + d)
R. H. S = (c, d) * (a, b)
=(a + c, b + d)
Hence, L.H.S = R. H. S
Since (a, b) * (c, d) = (c, d) * (a, b) ∀ a, b, c, d ∈ R
* is commutative (a, b) * (c, d) = (a + c, b + d)
Check the binary operation * is associative :
We know that * is associative if (a, b) * ( (c, d) * (x, y) ) = ((a, b) * (c, d)) * (x, y) ∀ a, b, c, d, x, y ∈ R
L.H.S = (a, b) * ( (c, d) * (x, y) ) = (a+c+x, b+d+y)
R.H.S = ((a, b) * (c, d)) * (x, y) = (a+c+x, b+d+y)
Thus, L.H.S = R.H.S
Since (a, b) * ( (c, d) * (x, y) ) = ((a, b) * (c, d)) * (x, y) ∀ a, b, c, d, x, y ∈ R
Thus, the binary operation * is associative
Checking for Identity Element:
e is identity of * if (a, b) * e = e * (a, b) = (a, b)
where e = (x, y)
Thus, (a, b) * (x, y) = (x, y) * (a, b) = (a, b) (a + x, b + y)
= (x + a , b + y) = (a, b)
Now, (a + x, b + y) = (a, b)
Now comparing these, we get:
a+x = a
x = a a = 0
Next compare: b +y = b
y = bb = 0
Since A = N x N, where x and y are the natural numbers. But in this case, x and y is not a natural number. Thus, the identity element does not exist.
Therefore, the operation * does not have any identity element.
Q.4: Let f : N → Y be a function defined as f (x) = 4x + 3, where, Y = {y ∈ N: y = 4x + 3 for some x ∈ N}. Show that f is invertible. Find the inverse.
Solution:
Checking for Inverse:
f(x) = 4x + 3
Let f(x) = y
y = 4x + 3
y – 3 = 4x
4x = y – 3
x = (𝑦 − 3)/4
Let g(y) = (𝑦 − 3)/4
where g: Y → N
Now find gof:
gof= g(f(x))
= g(4x + 3) = [(4𝑥 + 3) − 3]/4
= [4𝑥 + 3 − 3]/4
=4x/4
= x = I_{N}
Now find fog:
fog= f(g(y))
= f [(𝑦 − 3)/4]
=4[(𝑦 − 3)/4] +3
= y – 3 + 3
= y + 0
= y = I_{y}
Thus, gof = I_{N}and fog = I_{y},
Hence, f is invertible
Also, the Inverse of f = g(y) = [𝒚 – 3]/ 4
Q. 5: Let A = R {3} and B = R – {1}. Consider the function f: A →B defined by f (x) = (x 2)/(x 3). Is f oneone and onto? Justify your answer.
Solution:
Given function:
f (x) = (x 2)/(x 3)
Checking for oneone function:
f (x_{1}) = (x_{1}– 2)/ (x_{1}– 3)
f (x_{2}) = (x_{2}2)/ (x_{2}3)
Putting f (x_{1}) = f (x_{2})
(x_{1}2)/(x_{1}3)= (x_{2}2 )/(x_{2} 3)
(x_{1}2) (x_{2}– 3) = (x_{1}– 3) (x_{2}2)
x_{1} (x_{2}– 3) 2 (x_{2} 3) = x_{1} (x_{2}– 2) – 3 (x_{2}– 2)
x_{1} x_{2} 3x_{1} 2x_{2} + 6 = x_{1} x_{2} – 2x_{1} 3x_{2} + 6
3x_{1}– 2x_{2} = 2x_{1} 3x_{2}
3x_{2} 2x_{2} = – 2x_{1} + 3x_{1}
x_{1}= x_{2}
Hence, if f (x_{1}) = f (x_{2}), then x_{1} = x_{2}
Thus, the function f is oneone function.
Checking for onto function:
f (x) = (x2)/(x3)
Let f(x) = y such that y B i.e. y ∈ R – {1}
So, y = (x 2)/(x 3)
y(x 3) = x 2
xy 3y = x 2
xy – x = 3y2
x (y 1) = 3y 2
x = (3y 2) /(y1)
For y = 1, x is not defined But it is given that. y ∈ R – {1}
Hence, x = (3y 2)/(y 1) ∈ R {3} Hence, f is onto.
Practice Problems for Class 12 Maths Chapter 1
 Show that the function f: R → R is given by f(x) = x^{3} is injective.

Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.

Let f: {1, 3, 4} → {1, 2, 5} and g: {1, 2, 5} → {1, 3} be given by f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}. Write down gof.
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