NCERT Exemplar Class 12 Maths Practice Questions with Solutions [Chapter 1: Relations and Functions]: Prepare for Your Exams!

This article has the NCERT Exemplar Class 12 Maths Practice Questions with Solutions of Relations and Functions. Prepare with us!

Short Answer (S.A.)

1. Let A = {a, b, c} and the relation R be defined on A as follows: R = {(a, a), (b, c), (a, b)}. Then, write minimum number of ordered pairs to be added in R to make R reflexive and transitive. 

Solution: 

Given relation, R = {(a, a), (b, c), (a, b)}
To make R as reflexive we should add (b, b) and (c, c) to R. Also, to make R as transitive we should add (a, c) to R.
Hence, the minimum number of ordered pairs to be added are (b, b), (c, c) and (a, c) i.e. 3.

2. Let D be the domain of the real valued function f defined by f(x) = √(25 – x²). Then, write D. 

Solution: 

Given, f(x) = √(25 – x²)
The function is defined if 25 – x² ≥ 0
So, x² ≤ 25
-5 ≤ x ≤ 5
Therefore, the domain of the given function is [-5, 5].

3. Let f, g: R → R be defined by f(x) = 2x + 1 and g (x) = x² – 2, ∀ x ∈ R, respectively. Then, find g o f. 

Solution: 

Given,f(x) = 2x + 1 and g (x) = x² – 2, ∀ x ∈ R
Thus, g o f = g (f (x))
= g (2x + 1)
= (2x + 1)² – 2
= 4x² + 4x + 1 – 2
= 4x² + 4x – 1

4. Let f: R → R be the function defined by f (x) = 2x – 3, ∀ x ∈ R. write f ^–1. 

Solution: 

Given function,
f (x) = 2x – 3, ∀ x ∈ R
Let y = 2x – 3
x = (y + 3)/ 2
Thus,
f ^-1(x) = (x + 3)/ 2

5. If A = {a, b, c, d} and the function f = {(a, b), (b, d), (c, a), (d, c)}, write f ^–1. 

Solution: 

Given,
A = {a, b, c, d} and f = {(a, b), (b, d), (c, a), (d, c)}
So,
f ^-1 = {(b, a), (d, b), (a, c), (c, d)}

6. If f: R → R is defined by f (x) = x² – 3x + 2, write f (f (x)). 

Solution: 

Given, f (x) = x² – 3x + 2
Then,
f (f (x)) = f (x² – 3x + 2)
= (x² – 3x + 2)² – 3(x² – 3x + 2) + 2,
= x⁴ + 9x² + 4 – 6x³ + 4x² – 12x – 3x² + 9x – 6 + 2
= x⁴ – 6x³ + 10x² – 3x
Thus,
f (f (x)) = x⁴ – 6x³ + 10x² – 3x

7. Is g = {(1, 1), (2, 3), (3, 5), (4, 7)} a function? If g is described by g (x) = αx + β, then what value should be assigned to α and β. 

Solution: 

Given, g = {(1, 1), (2, 3), (3, 5), (4, 7)}
It’s seen that every element of domain has a unique image. So, g is function.

Now, also given that g (x) = αx + β

So, we have

g (1) = α(1) + β = 1

α + β = 1 …….. (i)

And, g (2) = α(2) + β = 3

2α + β = 3 …….. (ii)

Solving (i) and (ii), we have

α = 2 and β = -1

Therefore, g (x) = 2x – 1

8. Are the following set of ordered pairs functions? If so, examine whether the mapping is injective or surjective. 

(i) {(x, y): x is a person, y is the mother of x}. 

(ii){(a, b): a is a person, b is an ancestor of a}. 

Solution: 

(i) Given, {(x, y): x is a person, y is the mother of x}

It’s clearly seen that each person ‘x’ has only one biological mother.

Hence, the above set of ordered pairs make a function.

Now more than one person may have same mother. Thus, the function is many-many one and surjective.

(ii) Given, {(a, b): a is a person, b is an ancestor of a}

It’s clearly seen that any person ‘a’ has more than one ancestors.

Thus, it does not represent a function.

9. If the mappings f and g are given by f = {(1, 2), (3, 5), (4, 1)} and g = {(2, 3), (5, 1), (1, 3)}, write f o g. 

Solution: 

Given,

f = {(1, 2), (3, 5), (4, 1)} and g = {(2, 3), (5, 1), (1, 3)}

Now,

fog (2) = f(g(2)) = f(3) = 5

fog (5) = f(g(5)) = f(1) = 2

fog (1) = f(g(1)) = f(3) = 5

Thus,

fog = {(2, 5), (5, 2), (1, 5)}

10. Let C be the set of complex numbers. Prove that the mapping f: C → R given by f (z) = |z|, ∀ z ∈ C, is neither one-one nor onto. 

Solution: 

Given, f: C → R such that f (z) = |z|, ∀ z ∈ C

Now, let take z = 6 + 8i 

Then,

f (6 + 8i) = |6 + 8i| = √(6² + 8²) = √100 = 10

And, for z = 6 – 8i

f (6 – 8i) = |6 – 8i| = √(6² + 8²) = √100 = 10

Hence, f (z) is many-one.

Also, |z| ≥ 0, ∀ z ∈ C

But the co-domain given is ‘R’

Therefore, f(z) is not onto.

11. Let the function f: R → R be defined by f (x) = cos x, ∀ x ∈ R. Show that f is neither one-one nor onto. 

Solution: 

We have,

f: R → R, f(x) = cos x

Now,

f (x₁) = f (x₂)

cos x₁ = cos x₂

x₁ = 2nπ ± x₂, n ∈ Z

It’s seen that the above equation has infinite solutions for x₁ and x₂

Hence, f(x) is many one function.

Also the range of cos x is [-1, 1], which is subset of given co-domain R.

Therefore, the given function is not onto.

12. Let X = {1, 2, 3} and Y = {4, 5}. Find whether the following subsets of X × Y are functions from X to Y or not. 

(i) f = {(1, 4), (1, 5), (2, 4), (3, 5)} (ii) g = {(1, 4), (2, 4), (3, 4)} 

(iii) h = {(1,4), (2, 5), (3, 5)} (iv) k = {(1,4), (2, 5)}. 

Solution: 

Given, X = {1, 2, 3} and Y = {4, 5}

So, X × Y = {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}

(i) f = {(1, 4), (1, 5), (2, 4), (3, 5)}

f is not a function as f(1) = 4 and f(1) = 5

Hence, pre-image ‘1’ has not unique image.

(ii) g = {(1, 4), (2, 4), (3, 4)}

It’s seen clearly that g is a function in which each element of the domain has unique image.

(iii) h = {(1,4), (2, 5), (3, 5)}

It’s seen clearly that h is a function as each pre-image with a unique image.

And, function h is many-one as h(2) = h(3) = 5

(iv) k = {(1, 4),(2, 5)}

Function k is not a function as ‘3’ has not any image under the mapping.

13. If functions f: A → B and g: B → A satisfy g o f = I_A, then show that f is one-one and g is onto.

Solution: 

Given,

f: A → B and g: B → A satisfy g o f = I_A

It’s clearly seen that function ‘g’ is inverse of ‘f’.

So, ‘f’ has to be one-one and onto.

Hence, ‘g’ is also one-one and onto.

14. Let f: R → R be the function defined by f(x) = 1/(2 – cos x) ∀ x ∈ R. Then, find the range of f. 

Solution: 

Given,

f(x) = 1/(2 – cos x) ∀ x ∈ R

Let y = 1/(2 – cos x)

2y – ycos x = 1

cos x = (2y – 1)/ y

cos x = 2 – 1/y

Now, we know that -1 ≤ cos x ≤ 1

So,

-1 ≤ 2 – 1/y ≤ 1

-3 ≤ – 1/y ≤ -1

1 ≤ – 1/y ≤ 3

1/3 ≤ y ≤ 1

Thus, the range of the given function is [1/3, 1].

15. Let n be a fixed positive integer. Define a relation R in Z as follows: ∀ a, b ∈ Z, aRb if and only if a – b is divisible by n. Show that R is an equivalance relation.

Solution: 

Given ∀ a, b ∈ Z, aRb if and only if a – b is divisible by n.

Now, for

aRa ⇒ (a – a) is divisible by n, which is true for any integer a as ‘0’ is divisible by n.

Thus, R is reflective.

Now, aRb

So, (a – b) is divisible by n.

⇒ – (b – a) is divisible by n.

⇒ (b – a) is divisible by n

⇒ bRa

Thus, R is symmetric.

Let aRb and bRc

Then, (a – b) is divisible by n and (b – c) is divisible by n.

So, (a – b) + (b – c) is divisible by n.

⇒ (a – c) is divisible by n.

⇒ aRc

Thus, R is transitive.

So, R is an equivalence relation.

Long Answer (L.A.)

16. If A = {1, 2, 3, 4 }, define relations on A which have properties of being: 

(a) reflexive, transitive but not symmetric 

(b) symmetric but neither reflexive nor transitive 

(c) reflexive, symmetric and transitive. 

Solution: 

Given that, A = {1, 2, 3}.

(i) Let R₁ = {(1, 1), (1, 2), (1, 3), (2, 3), (2, 2), (1, 3), (3, 3)}

R₁ is reflexive as (1, 1), (2, 2) and (3, 3) lie is R₁.

R₁ is transitive as (1, 2) ∈ R₁, (2, 3) ∈ R₁ ⇒ (1, 3) ∈ R₁

Now, (1, 2) ∈ R₁ ⇒ (2, 1) ∉ R₁.

(ii) Let R₂ = {(1, 2), (2, 1)}

Now, (1, 2) ∈ R₂, (2, 1) ∈ R₂

So, it is symmetric,

And, clearly R₂ is not reflexive as (1, 1) ∉ R₂

Also, R₂ is not transitive as (1, 2) ∈ R₂, (2, 1) ∈ R₂ but (1, 1) ∉ R₂

(iii) Let R₃ = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}

R₃ is reflexive as (1, 1) (2, 2) and (3, 3) ∈ R₁

R₃ is symmetric as (1, 2), (1, 3), (2, 3) ∈ R₁ ⇒ (2, 1), (3, 1), (3, 2) ∈ R₁ 

Therefore, R₃ is reflexive, symmetric and transitive.

17. Let R be relation defined on the set of natural number N as follows: 

R = {(x, y): x ∈ N, y ∈ N, 2x + y = 41}. Find the domain and range of the relation R. Also verify whether R is reflexive, symmetric and transitive. 

Solution: 

Given function: R = {(x, y): x ∈ N, y ∈ N, 2x + y = 41}.

So, the domain = {1, 2, 3, ….., 20} [Since, y ∈ N ]

Finding the range, we have

R = {(1, 39), (2, 37), (3, 35), …., (19, 3), (20, 1)}

Thus, Range of the function = {1, 3, 5, ….., 39}

R is not reflexive as (2, 2) ∉ R as 2 x 2 + 2 ≠ 41

Also, R is not symmetric as (1, 39) ∈ R but (39, 1) ∉ R

Further R is not transitive as (11, 19) ∉ R, (19, 3) ∉ R; but (11, 3) ∉ R.

Thus, R is neither reflexive nor symmetric and nor transitive.

18. Given A = {2, 3, 4}, B = {2, 5, 6, 7}. Construct an example of each of the following: 

(a) an injective mapping from A to B 

(b) a mapping from A to B which is not injective 

(c) a mapping from B to A. 

Solution: 

Given, A = {2, 3, 4}, B = {2, 5, 6, 7}

(i) Let f: A → B denote a mapping

f = {(x, y): y = x + 3} or

f = {(2, 5), (3, 6), (4, 7)}, which is an injective mapping.

(ii) Let g: A → B denote a mapping such that g = {(2, 2), (3, 2), (4, 5)}, which is not an injective mapping.

(iii) Let h: B → A denote a mapping such that h = {(2, 2), (5, 3), (6, 4), (7, 4)}, which is one of the mapping from B to A.

19. Give an example of a map 

(i) which is one-one but not onto 

(ii) which is not one-one but onto 

(iii) which is neither one-one nor onto. 

Solution: 

(i) Let f: N → N, be a mapping defined by f (x) = x²

For f (x₁) = f (x₂)

Then, x₁² = x₂²

x₁ = x₂ (Since x₁ + x₂ = 0 is not possible)

Further ‘f’ is not onto, as for 1 ∈ N, there does not exist any x in N such that f (x) = 2x + 1.

(ii) Let f: R → [0, ∞), be a mapping defined by f(x) = |x|

Then, it’s clearly seen that f (x) is not one-one as f (2) = f (-2).

But |x| ≥ 0, so range is [0, ∞].

Therefore, f (x) is onto.

(iii) Let f: R → R, be a mapping defined by f (x) = x²

Then clearly f (x) is not one-one as f (1) = f (-1). Also range of f (x) is [0, ∞).

Therefore, f (x) is neither one-one nor onto.

20. Let A = R – {3}, B = R – {1}. Let f : A → B be defined by f (x) = x – 2/ x – 3 ∀ x ∈ A . Then show that f is bijective. 

Solution: 

Given,

A = R – {3}, B = R – {1}

And,

f : A → B be defined by f (x) = x – 2/ x – 3 ∀ x ∈ A

Hence, f (x) = (x – 3 + 1)/ (x – 3) = 1 + 1/ (x – 3)

Let f(x₁) = f (x₂)

So, f (x) is an injective function.

Now let y = (x – 2)/ (x -3)

x – 2 = xy – 3y

x(1 – y) = 2 – 3y

x = (3y – 2)/ (y – 1)

y ∈ R – {1} = B

Thus, f (x) is onto or subjective.

Therefore, f(x) is a bijective function.

21. Let A = [–1, 1]. Then, discuss whether the following functions defined on A are one-one, onto or bijective: 

(i) f(x) = x/2 (ii) g(x) = |x| 

(iii) h(x) = x|x| (iv) k(x) = x²

Solution: 

Given, A = [–1, 1]

(i) f: [-1, 1] → [-1, 1], f (x) = x/2

Let f (x₁) = f(x₂)

x₁/ 2 = x₂

So, f (x) is one-one.

Also x ∈ [-1, 1]

x/2 = f (x) = [-1/2, 1/2]

Hence, the range is a subset of co-domain ‘A’

So, f (x) is not onto.

Therefore, f (x) is not bijective.

(ii) g (x) = |x|

Let g (x₁) = g (x₂)

|x₁| = |x₂|

x₁ = ± x₂

So, g (x) is not one-one

Also g (x) = |x| ≥ 0, for all real x

Hence, the range is [0, 1], which is subset of co-domain ‘A’

So, f (x) is not onto.

Therefore, f (x) is not bijective.

(iii) h (x) = x|x|

Let h (x₁) = h (x₂)

x₁|x₁| = x₂|x₂|

If x₁, x₂ > 0

x₁² = x₂²

x₁² – x₂² = 0

(x₁ – x₂)(x₁ + x₂) = 0

x₁ = x₂ (as x₁ + x₂ ≠ 0)

Similarly for x₁, x₂ < 0, we have x₁ = x₂

It’s clearly seen that for x₁ and x₂ of opposite sign, x₁ ≠ x₂.

Hence, f (x) is one-one.

For x ∈ [0, 1], f (x) = x² ∈ [0, 1]

For x < 0, f (x) = -x² ∈ [-1, 0)

Hence, the range is [-1, 1].

So, h (x) is onto.

Therefore, h (x) is bijective.

(iv) k (x) = x²

Let k (x₁) = k (x₂)

x₁² = x₂²

x₁ = ± x₂

Therefore, k (x) is not one-one.

22. Each of the following defines a relation on N: 

(i) x is greater than y, x, y ∈ N 

(ii) x + y = 10, x, y ∈ N

(iii) x y is square of an integer x, y ∈ N 

(iv) x + 4y = 10 x, y ∈ N.

Determine which of the above relations are reflexive, symmetric and transitive.

Solution: 

(i) Given, x is greater than y; x, y ∈ N

If (x, x) ∈ R, then x > x, which is not true for any x ∈ N.

Thus, R is not reflexive.

Let (x, y) ∈ R

⇒ xRy

⇒ x > y

So, y > x is not true for any x, y ∈ N

Hence, R is not symmetric.

Let xRy and yRz

⇒ x > y and y > z

⇒ x > z

⇒ xRz

Hence, R is transitive.

(ii) x + y = 10; x, y ∈ N

Thus,

R = {(x, y); x + y = 10, x, y ∈ N}

R = {(1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)}

It’s clear (1, 1) ∉ R

So, R is not reflexive.

(x, y) ∈ R ⇒ (y, x) ∈ R

Therefore, R is symmetric.

Now (1, 9) ∈ R, (9, 1) ∈ R, but (1, 1) ∉ R

Therefore, R is not transitive.

(iii) Given, xy is square of an integer x, y ∈ N

R = {(x, y) : xy is a square of an integer x, y ∈ N}

It’s clearly (x, x) ∈ R, ∀ x ∈ N

As x² is square of an integer for any x ∈ N

Thus, R is reflexive.

If (x, y) ∈ R ⇒ (y, x) ∈ R

So, R is symmetric.

Now, if xy is square of an integer and yz is square of an integer.

Then, let xy = m² and yz = n² for some m, n ∈ Z

x = m²/y and z = x²/y

xz = m²n²/ y², which is square of an integer.

Thus, R is transitive.

(iv) x + 4y = 10; x, y ∈ N

R = {(x, y): x + 4y = 10; x, y ∈ N}

R = {(2, 2), (6, 1)}

It’s clearly seen (1, 1) ∉ R

Hence, R is not symmetric.

(x, y) ∈ R ⇒ x + 4y = 10

And (y, z) ∈ R ⇒ y + 4z = 10

⇒ x – 16z = -30

⇒ (x, z) ∉ R

Therefore, R is not transitive.

23. Let A = {1, 2, 3, … 9} and R be the relation in A ×A defined by (a, b) R (c, d) if a + d = b + c for (a, b), (c, d) in A ×A. Prove that R is an equivalence relation and also obtain the equivalent class [(2, 5)]. 

Solution: 

Given, A = {1, 2, 3, … 9} and (a, b) R (c, d) if a + d = b + c for (a, b), (c, d) ∈ A ×A.

Let (a, b) R(a, b)

So, a + b = b + a, ∀ a, b ∈ A which is true for any a, b ∈ A.

Thus, R is reflexive.

Let (a, b) R(c, d)

Then,

a + d = b + c

c + b = d + a

(c, d) R(a, b)

Thus, R is symmetric.

Let (a, b) R(c, d) and (c, d) R(e, f)

a + d = b + c and c + f = d + e

a + d = b + c and d + e = c + f

(a + d) – (d + e = (b + c) – (c + f)

a – e = b – f

a + f = b + e

(a, b) R(e, f)

So, R is transitive.

Therefore, R is an equivalence relation.

And, [(2,5)=(1,4)(2,5),(3,6),(4,7),(5,8),(6,9)] is the equivalent class under relation R.

24. Using the definition, prove that the function f : A→ B is invertible if and only if f is both one-one and onto. 

Solution: 

Let f: A → B be many-one function.

Let f(a) = p and f(b) = p

So, for inverse function we will have f^-1(p) = a and f^-1(p) = b

Thus, in this case inverse function is not defined as we have two images ‘a and b’ for one pre-image ‘p’.

But for f to be invertible it must be one-one.

Now, let f: A → B is not onto function.

Let B = {p, q, r} and range of f be {p, q}.

Here image ‘r’ has not any pre-image, which will have no image in set A.

And for f to be invertible it must be onto.

Thus, ‘f’ is invertible if and only if ‘f’ is both one-one and onto.

A function f = X → Y is invertible iff f is a bijective function.

25. Functions f , g : R → R are defined, respectively, by f (x) = x² + 3x + 1, g (x) = 2x – 3, find 

(i) f o g (ii) g o f (iii) f o f (iv) g o g 

Solution: 

Given, f(x) = x² + 3x + 1, g (x) = 2x – 3

(i) fog = f(g(x))

= f(2x – 3)

= (2x – 3)² + 3(2x – 3) + 1

= 4x² + 9 – 12x + 6x – 9 + 1

= 4x² – 6x + 1

(ii) gof = g(f(x))

= g(x² + 3x + 1)

= 2(x² + 3x + 1) – 3

= 2x² + 6x – 1

(iii) fof = f(f(x))

= f(x² + 3x + 1)

= (x² + 3x + 1)² + 3(x² + 3x + 1) + 1

= x⁴ + 9x² + 1 + 6x³ + 6x + 2x² + 3x² + 9x + 3 + 1

= x⁴ + 6x³ + 14x² + 15x + 5

(iv) gog = g(g(x))

= g(2x – 3)

= 2(2x – 3) – 3

= 4x – 6 – 3

= 4x – 9

26. Let * be the binary operation defined on Q. Find which of the following binary operations are commutative 

(i) a * b = a – b ∀ a, b ∈Q (ii) a * b = a² + b² ∀ a, b ∈ Q 

(iii) a * b = a + ab ∀ a, b ∈ Q (iv) a * b = (a – b)² ∀ a, b ∈ Q 

Solution: 

Given that * is a binary operation defined on Q.

(i) a * b = a – b, ∀ a, b ∈Q and b * a = b – a

So, a * b ≠ b * a

Thus, * is not commutative.

(ii) a * b = a² + b²

b * a = b² + a²

Thus, * is commutative.

(iii) a * b = a + ab

b * a = b + ab

So clearly, a + ab ≠ b + ab

Thus, * is not commutative.

(iv) a * b = (a – b)², ∀ a, b ∈Q

b * a = (b –a)²

Since, (a – b)² = (b – a)²

Thus, * is commutative.

27. Let * be binary operation defined on R by a * b = 1 + ab, ∀ a, b ∈ R. Then the operation * is 

(i) commutative but not associative 

(ii) associative but not commutative 

(iii) neither commutative nor associative 

(iv) both commutative and associative 

Solution: 

(i) Given that * is a binary operation defined on R by a * b = 1 + ab, ∀ a, b ∈ R

So, we have a * b = ab + 1 = b * a

So, * is a commutative binary operation.

Now, a * (b * c) = a * (1 + bc) = 1 + a (1 + bc) = 1 + a + abc

Also,

(a * b) * c = (1 + ab) * c = 1 + (1 + ab) c = 1 + c + abc

Thus, a * (b * c) ≠ (a * b) * c

Hence, * is not associative.

Therefore, * is commutative but not associative.

Objective Type Questions 

Choose the correct answer out of the given four options in each of the Exercises from 28 to 47 (M.C.Q.) 

28. Let T be the set of all triangles in the Euclidean plane, and let a relation R on T be defined as aRb if a is congruent to b ∀ a, b ∈ T. Then R is 

(A) reflexive but not transitive (B) transitive but not symmetric 

(C) equivalence (D) none of these 

Solution: 

(C) equivalence

Given aRb, if a is congruent to b, ∀ a, b ∈ T.

Then, we have aRa ⇒ a is congruent to a; which is always true.

So, R is reflexive.

Let aRb ⇒ a ~ b

b ~ a

bRa

So, R is symmetric.

Let aRb and bRc

a ~ b and b ~ c

a ~ c

aRc

So, R is transitive.

Therefore, R is equivalence relation.

29. Consider the non-empty set consisting of children in a family and a relation R defined as aRb if a is brother of b. Then R is 

(A) symmetric but not transitive (B) transitive but not symmetric 

(C) neither symmetric nor transitive (D) both symmetric and transitive

Solution: 

(B) transitive but not symmetric

aRb ⇒ a is brother of b.

This does not mean b is also a brother of a as b can be a sister of a.

Thus, R is not symmetric.

aRb ⇒ a is brother of b.

and bRc ⇒ b is brother of c.

So, a is brother of c.

Therefore, R is transitive.

30. The maximum number of equivalence relations on the set A = {1, 2, 3} are 

(A) 1 (B) 2 

(C) 3 (D) 5 

Solution: 

(D) 5
Given, set A = {1, 2, 3}
Now, the number of equivalence relations as follows
R₁ = {(1, 1), (2, 2), (3, 3)}
R₂ = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}
R₃ = {(1, 1), (2, 2), (3, 3), (1, 3), (3, 1)}

R₄ = {(1, 1), (2, 2), (3, 3), (2, 3), (3, 2)}

R₅ = {(1, 2, 3) ⇔ A x A = A²}

Thus, maximum number of equivalence relation is ‘5’.

31. If a relation R on the set {1, 2, 3} be defined by R = {(1, 2)}, then R is 

(A) reflexive (B) transitive 

(C) symmetric (D) none of these

Solution: 

(D) none of these
R on the set {1, 2, 3} be defined by R = {(1, 2)}
Hence, its clear that R is not reflexive, transitive and symmetric.