Important Questions for Class 12 Maths Chapter 1 Relations & Functions

Important Questions for Class 12 Maths Chapter 1 Relations & Functions are included in the article. All exercise questions are solved using step by step approach. Some of the main topics covered in Important Questions Class 12 Math Chapter 1 are Empty relation, Universal relation, Reflexive relation, Symmetric relation, among others. 

Relations

A relation R can be defined from a Set A to a Set B can be defined as as a subset of the cartesian product AB. It can also be shown as Rx,yXY:xRy. A relation can be categorised into five different types, they are:

• Reflexive Relation
• Symmetric Relation
• Transitive Relation
• Empty Relation
• Universal Relation

Read More: Eccentricity

Functions

For two non-empty sets A and B, a function or mapping f from A to B represented as f:AB is a rule by which each element xA is associated to a unique element yB. In this case, f is said to be the function from A to B. 

Read More: Maxima and Minima

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Very Short Answer Questions [1 Mark Question] 

Ques. If R = { (a, a³): a is a prime number less than 5 } be a relation. Find the range of R.[Foreign 2014]

Ans. Given, R = {(a, a³): a is a prime number less than 5} .

We know that 2 and 3 are the prime numbers less than 5.

Therefore, a can take values of 2 and 3.

R= {(2,2³), (3,3³)} = {(2,8), (3,27)}

Hence, the range of R is {8,27}.

Ques. If f: {1,3,4} {1,2,5} and g: {1,2,5} {1,3} given by f = {(1,2), (3,5), (4,1)} and g = {(1,3), (2,3), (5,1)). Write down gof. [All India 2014]

Ans. The functions f: {1,3,4} {1,2,5} and g: {1,2,5} {1,3} are defined as,

 f = {(1,2), (3,5), (4,1)}

 and, g = {(1,3), (2,3), (5,1)}

gof (1) = g (f(1)} = g(2) = 3 [ since, f(1) = 2 and g(2) = 3]

gof(3) = g(f(3)) = g(5) = 1 [ since, f(3) = 5 and g(5) = 1]

gof (4) = g(f(4)) = g(1) = 3 [ since, f(4) = 1 and g(1) = 3]

Therefore, gof={(1,3), (3,1), (4,3)}.

Ques. Let R be the equivalence relation in the set A = {0,1,2,3,4,5} given by

 R = {(a,b): 2 divides (a - b)}. Write the equivalence class [0]. [Delhi 2014]

Ans. Given, R = {(a, b):2 divides (a-b)}. Here, all even integers are related to zero, i.e. (0, 2), 

(0, 4).

Hence, the equivalence class of [0] = {2,4} .

Ques. If A={1,2,3}, B = {4, 5, 6, 7} and f = {(1, 4), (2, 5), (3, 6)) is a function from A to B. State whether f is one-one or not. [All India 2011]

Ans. Given, A = {1, 2, 3} and B = {4, 5, 6, 7}

Now, f: A→ B is defined as f = {(1, 4), (2, 5), (3, 6)}

Therefore, f(1) = 4, f(2)= 5 and f(3) = 6.

It is seen that the images of distinct elements of A under f are distinct.So, f is one-one. 

Ques. If: R R is defined by f(x) = 3x + 2, then define f[f(x)]. [Foreign 2011; Delhi 2010]

Ans. Given, f(x) = 3x + 2

Now, f[f(x)] = f(3x + 2) = 3(3x + 2) + 2 = 9x + 6 + 2 = 9x+8.

Ques. Write fog, if f: R→ R and g: R→ R are given by f(x) = |x| and g(x) = |5x-2|. [Foreign 2011].

Ans. Given, f(x)= |x|, g(x) = 5x - 21 Now, fog (x) = f[g(x)] = f{|5x - 21}

Read More: Symmetric and Skew Symmetric Matrice

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Short Answer Questions [ 2 Marks Questions] 

Ques. If R= {(x, y): x + 2y = 8} is a relation on N, then write the range of R. [All India 2014]

Ans. Given, the relation R is defined on the set of natural numbers, i.e.,

 N as R = {(x, y) : x + 2y = 8}

To find the range of R, x + 2y = 8 can be rewritten as y= 8-x2

On putting x = 2, we get y = 8-22= 3 

On putting x = 4, we get y = 2

On putting x = 6, we get y = 1

As, x, y N, then R= {(2, 3) (4, 2) (6, 1)}. Hence, the range of relation is {3, 2, 1}.

Ques. If f is an invertible function, defined as f(x) = 3x - 4 / 5 then write f-¹(x). [Foreign 2010] 

Ans. We are given f(x)=3x-4/5 which is invertible. 

Let, 

y= 3x - 4/ 5

5y = 3x - 4

X= 5y + 4/ 3

f-¹ (y) = 5y + 3/ 3 and f(x)= 5x +4/ 3

Ques.Show that the Signum Function f: R → R, given by

f(x)={ 1 for x greater than 0

      0 for x=0 is neither one-one nor onto

     -1 for x less than 0

Ans. In the given function f(x)={ 1 for x greater than 0

                                                  0 for x=0 is neither one-one nor onto

                                                  -1 for x less than 0

the value of f(x) is defined only for when x= -1, 0, 1.

For any other real number, say y=2, there is no corresponding element x. Therefore, the function is not an onto function.

Also, for any value of x, say f₁ or f₂, the value will be the same image, that is, 1, 0, or -1. Therefore, it is not a one to one function.

Hence, the given function is neither a one-one function nor an onto function.

Ques. State whether the function f: N→ N given by f(x) = 5x is injective, surjective or both. [All India 2008C, HOTS] 

Ans. f: N→ N is given by f (x) = 5x

Let x₁, x₂ ∈ N such that f (x₁) = f (x₂)

 ∴ 5 x₁ = 5 x₂ 

⇒ x₁ x₂

∴ f is a one-one, or, an injective function.

Now, let y=f(x) 

⇒y=5x

For y=1, x does not belong to N. 

Therefore, the function is not an onto or surjective function. 

Read More: Integers As Exponents

Ques. If f: R→ R is defined by f(x) = (3-x³)1/3, then find fof(x). [All India 2010]

Ans. Given, function is f:R→R such that, 

f(x) = (3-x³)^1/3

Now, fof (x) = f[f(x)] = f[(3 − x³)^¹/³]

= [3-{(3-x³)^1/3} ³] ^1/3

= [3-(3-x³)]^1/3 = (x³)^1/3

= x

Also check: Binary Multiplication

Long Answer Questions [3 Marks Questions] 

Ques. If f: R→ R is defined by f(x)=x²-3x + 2, find f(f(x)). [NCERT, MISC] 

Ans. Given, f(x)=x²-3x+2.

⇒f(f(x)) = f(x)²-3f(x) + 2.

=(x²-3x + 2)²-3(x²-3x+2)+2

Using (a-b+c)²=a² + b²+c²-2ab + 2ac-2ab

= (x²)² + (3x)² +2²- 2x2(3x) + 2x²(2) - 2x²(3x) - 3(x²-3x + 2) + 2

=x⁴ + 9x² + 4- 6x³ - 12x + 4x² - 3x² +9x -6 + 2

=x⁴ - 6x³ + 9x² + 4x² - 3x² - 12x + 9x− 6 + 2 + 4

=x⁴ - 6x³ + 10x² - 3x. 

Ques. Given a non empty set X, consider P(X) which is the set of all subsets of X. Define the relation R in P(X) as follows: For subsets A, B in P(X), A R B if and only if A⊂B. Is R an equivalence relation on P(X)? Justify you answer. 

Ans. A R B means A⊂B

Here, relation is

R= {(A, B): A & B are sets, A⊂B}

Reflexive

Since every set is a subset of itself,

This implies, A⊂A

⇒(A, A)∈R.

Therefore, R is reflexive.

Symmetric

To check whether symmetric or not,

If (A, B)∈R, then (B,A)∈R

If (A, B)∈R,

Then, A⊂B.

But, B⊂A is not true

As all elements of A are in B, therefore, A⊂B. 

But all elements of B are not in A. 

So B⊂A is not true. 

Therefore, R is not symmetric. 

Transitive

Since (A, B)∈R & (B, C)∈R

If, A⊂B and B⊂C, then A⊂C. 

⇒ (A,C)∈R

So, If (A, B)∈R & (B, C)∈R, then (A, C)∈R

Therefore, R is transitive.

Hence, R is reflexive and transitive but not symmetric.

Therefore, R is not an equivalence relation since it is not symmetric.

Read More: Collinear points

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

Ans. R={(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)}

Reflexive

If the relation is reflexive,then (a, a)∈R for every a∈(1, 2, 3). 

Since (1, 1)∈R, (2, 2)∈R & (3, 3)∈R. 

Therefore, R is reflexive

Symmetric

To check whether symmetric or not, If (a, b)∈R, then (b, a)∈R. 

Here (1, 2)∈R, but (2, 1)∉R.

Therefore, R is not symmetric. 

Transitive

To check whether transitive or not,

If (a,b)∈R & (b,c)∈R, then (a,c)∈R. 

Here, (1, 2)∈R and (2, 3)∈R but (1, 3)∉R.

Therefore, R is not transitive. 

Hence, R is reflexive but neither symmetric nor transitive.

Read More: Onto Function

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Very Long Answer Questions [5 Marks Questions] 

Ques. If f,g: R→ R are two functions defined as f(x) = |x|+x and g(x) = |x|-x, ∀ x∈R, then, find fog and gof. [India 2014] 

Ans. Here, 

f(x) = |x|+x and g(x) = |x|-x

f(x)= { x+x, if x ≥ 0 

-x+x, if x < 0

f(x) = {2x, if x ≥ 0 

0, if x<0

g(x) = { x-x,if x ≥ 0

-x-x, if x<0 

or

g(x)= {0, if x ≥ 0

-2x, if x < 0

Thus for x ≥ 0, (gof)(x) = g(2x) = 0

and for x < 0, (gof)(x) = g(0) = 0

or

(gof)(x) =0 for all x∈R 

Thus for x ≥ 0, (fog)(x) = f(0) = 2(0) = 0

and for x < 0,

(fog)(x) = f(-2x) = - 4x

or

(fog)(x) = { 0, if x ≥ 0

-4x, if x < 0

Also read:

Ques. If A = R-{3} and B = R-{1}. Consider the function f: A→B defined by f(x) x-2/ x-3 for

all x∈A. Then show that f is bijective. Find f-¹(x). [Delhi 2012,2014] 

Ans. Given, function is f: A→B, where A= R-{3}

and B=R-{1}, such that f(x)= x-2/x-3. 

For One-one

 Let f(x₁) = f(x₂), for all x₁, x₂∈ A

⇒x₂-2/ x₁-3= x₂-2/ x₂-3

⇒(x₁ - 2) (x₂-3) = (x₁-2)(x_? - 3)

⇒ x₁x₂ - 3x₁ -2x₂ + 6 = x₁x₂ - 3x₁ - 2x₂ +6

⇒- 3x₁2x₂ = - 3x₁ - 2x₂

- 3 (x₁ - x₂)+2 (x₁ - x₂) = 0

 - (x₁ - x₂) = 0

Or, x₁ - x₂= 0

This implies, x₁ = x₂. 

Since, f(x₁) = f(x₂)

⇒ x₁ = x₂, for all x₁, x₂∈ A. 

So, f(x) is a one-one function. 

Onto

 To show f(x) is onto, we show that range of f(x) and its codomain are same.

Now, 

let. y =x-2 /x - 3 

or, xy-3y=x-2

 ⇒xy - x = 3y - 2

⇒x(y-1) = 3y - 2

⇒x=3y-2/y-1 ……Eqn (1)

Since, x ∈R-{3}, for all y ∈ R- {1}, the range of f(x)=R-{1}.

Also, the given codomain of f(x) = R-{1}

Therefore, Range = Codomain. 

Hence, f(x) is an onto function.

Therefore, f(x) is a bijective function. 

From Eq. (i), we get, 

f-1(y)= 3y - 2/ y - 1

⇒f-¹(x) = 3x - 2/ x-1

which is the inverse function of f(x).

Read More: Operations on Rational Numbers

Ques. If A={1,2,3,...,9} and R be the relation in AxA defined by (a, b) R (c,d). If a +d=b+c for (a,b), (c,d) in Ax A. Prove that R is an equivalence relation, Also, obtain the equivalence class [(2, 5)]. [Delhi 2014] 

Ans. Given, relation R defined by (a, b) R (c, d), if a+d=b+c for (a, b), (c, d) in A x A.

Here, A = {1, 2, 3, ...., 9}

We observe the following properties on R:

Reflexive

Let (1, 2) be an element of A X A.

Then, (1, 2) belongs to AxA 

This implies, 1,2 belongs to A

Or, 1+2=2+1 [ since, addition is commutative]

Or, (1, 2) R (1, 2)

Thus, (1, 2) R (1, 2), for all(1, 2) belonging to 

 AxA. 

So, R is reflexive on A x A.

Symmetric

Let (1, 2), (3, 4)∈A × A such that (1, 2) R (3, 4). 

Then, 1+ 4 = 2 + 3

⇒3+2 =4+1 [since, addition is commutative]

⇒ (3, 4) R (1, 2)

Thus, (1, 2) R (3, 4)

⇒ (3, 4) R (1, 2), for all (1, 2), (3, 4) belonging to AXA

So, R is symmetric on A x A.

Transitive

 Let (1, 2), (3, 4), (5, 6)∈ A x A such that

 (1, 2)R (3, 4) and (3, 4) R (5, 6). 

Then,

(1, 2) R (3, 4)

Or, 1+4=2+3

Or, (3, 4) R (5, 6)

Or, 3+6=4+5

This implies, (1+4) + 3+ 6 = (2+3)+(4+5) 

or, 1+6=2+5 = (1, 2) R (5, 6)

Thus, (1, 2) R (3, 4) and (3, 4) R (5, 6)

⇒ (1, 2) R (5, 6), (1, 2), (3, 4), (5, 6) = A × A

So, R is transitive on A x A.

Hence, it is an equivalence relation on A X A. 

Ques. If Z is the set of all integers and R is the relation on Z defined as R={(a,b): a,b∈Z and a - b is divisible by 5}. Prove that R is an equivalence relation. [Delhi 2010, HOTS] 

Ans. The given relation is R = {(a, b): a,b∈Z and a - b is divisible by 5}. We shall prove that R is reflexive, symmetric and transitive.

(i) Reflexive

As for any x ∈Z, we have x - x = 0 and 0 is divisible by 5.

⇒(x-x) is divisible by 5.

⇒(x,x)∈R for all X∈Z. 

Therefore, R is reflexive.

(ii) Symmetric

 As (x, y)∈R, where (x, y)∈Z. 

⇒(x - y) is divisible by 5. [by definition of R]

⇒x - y = 5A for some A ∈Z

⇒y - x = 5(-A)

⇒(y - x) is also divisible by 5

Therefore, (y,x) ∈ R

Therefore, R is symmetric.

 iii) Transitive

As (x, y)∈R,where x, y ∈Z

⇒(x - y) is divisible by 5.

⇒ x - y = 5A for some A∈Z

Again, for (y, z)∈R, where y, z∈Z

⇒(y-z) is divisible by 5

⇒y-z = 5B for some B∈Z

 (y-z) is divisible by 5

Now, (x - y) + (y - z) = 5A + 5B. 

⇒x-z = 5(A + B)

 ⇒(x-z) is divisible by 5 for some A+B∈Z

Therefore, R is transitive.

Thus, R is reflexive, symmetric and transitive. Hence, it is an equivalence relation.