This article has the NCERT Exemplar Class 12 Maths Practice Questions with Solutions of Relations and Functions. Prepare with us!
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 = IA, then show that f is one-one and g is onto.
Solution:
Given,
f: A → B and g: B → A satisfy g o f = IA
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.
