NCERT Solutions Class 11 Maths Chapter 2 - Relations and Functions

Q1: Show that the function f: R → R given by f(x) = x³ is injective.
Ans:  f : R → R is given as f(x) = x³.
For one – one
Suppose f(x) = f(y), where x, y ∈ R.
⇒ x³ = y³   ...........(1)
Now, we need to show that x = y.
Suppose x ≠ y, their cubes will also not be equal.
⇒ x³ ≠ y³
However, this will be a contradiction to (1).
∴ x = y
Hence, f is injective.

Q2: 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), ARB if and only if A ⊂ B. Is R an equivalence relation on P(X)? Justify you answer.
Ans: Since every set is a subset of itself, ARA for all A ∈ P(X).
∴ R is reflexive.
Let ARB ⇒ A ⊂ B.
This cannot be implied to B ⊂ A.
For instance, if A = {1, 2} and B = {1, 2, 3}, then it cannot be implied that B is related to A.
∴ R is not symmetric.
Further, if ARB and BRC, then A ⊂ B and B ⊂ C.
⇒ A ⊂ C
⇒ ARC
∴ R is transitive.
Hence, R is not an equivalence relation as it is not symmetric.

Q3: Find the number of all onto functions from the set {1, 2, 3, … , n) to itself.  
Ans: Onto functions from the set {1, 2, 3, … , n} to itself is simply a permutation on n symbols 1, 2, …, n. Thus, the total number of onto maps from {1, 2, … , n} to itself is the same as the total number of permutations on n symbols 1, 2, …, n, which is n. 9

Q4: Let A = {−1, 0, 1, 2}, B = {−4, −2, 0, 2} and f, ??: A → B be functions defined by f(x) = x² − x, x ∈ A 

Are f and g equal? Justify your answer. (Hint: One may note that two function f: A → B and ??: A → B such that f(a) = ??(a) ∀ a ∈ A, are called equal functions). 
Ans: It is given that A = {-1, 0, 1, 2}, B = {-4, -2, 0, 2}
Also, it is given that f, g : A → B are defined by

It is observed that




Hence, the functions f and g are equal.

Q5: Let A = {1, 2, 3}. Then number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is
(a) 1
(b) 2
(c) 3
(d) 4
Ans: (a)
Solution. The given set is A = {1, 2, 3}.
The smallest relation containing (1, 2) and (1, 3) which is reflexive and symmetric, but not transitive is given by:
R = {(1, 1), (2, 2), (3, 3), (1, 2), (1, 3), (2, 1), (3, 1)}
This is because relation R is reflexive as (1, 1), (2, 2), (3, 3) ∈ R.
Relation R is symmetric since (1, 2), (2, 1) ∈ R and (1, 3), (3, 1) ∈ R.
But relation R is not transitive as (3, 1), (1, 2) ∈ R, but (3, 2) ∉ R.
Now, if we add any two pairs (3, 2) and (2, 3) (or both) to relation R, then relation R will become transitive.
Hence, the total number of desired relations is one.

Q6: Let A = {1, 2, 3}. Then number of equivalence relations containing (1, 2) is
(a) 1
(b) 2
(c) 3
(d) 4
Ans: (b)
It is given that A = {1, 2, 3}.
The smallest equivalence relation containing (1, 2) is given by,
R₁ = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}
Now, we are left with only four pairs i.e., (2, 3), (3, 2), (1, 3), and (3, 1).
If we odd any one pair [say (2, 3)] to R₁, then for symmetry we must add (3, 2).
Also, for transitivity we are required to add (1, 3) and (3, 1).
Hence, the only equivalence relation (bigger than R₁) is the universal relation.
This shows that the total number of equivalence relations containing (1, 2) is two.

Q7: Show that the function f : R → {x ∈ R : – 1 < x < 1} defined by f(x) = x/(x+|x|), x ∈ R is one one and onto function.
Ans: It is given that
f(x)∈(−1,1)
 
we know that :


Checking - one-one
case 1: for x≥0

on putting f(x1)=f(x2), we have: 

Similarly,
In case 2: for x<0, we get:
x₁=x₂
Hence, if f(x₁)=f(x₂), then x₁=x₂
∴f is one-one.
Checking onto
case 1: for x≥0

Similarly, in case 2: for x<0 , we have:

∵y∈(−1,1)
∴ ∀ y∈(−1,1), x is defined.
∴f is onto.
Hence, f(x) is one-one and onto (proved).

Old NCERT Questions:

Q1: Let f: R → R be defined as f(x) = 10x + 7. Find the function g: R → R such that gof = fog = I_R.
Ans: It is given that f: R → R is defined as f(x) = 10x + 7.
One-one:
Let f(x) = f(y), where x, y ∈ R.
⇒ 10x + 7 = 10y + 7
⇒ x = y
∴ f is a one-one function.
Onto:
For y ∈ R, let y = 10x + 7.

Therefore, for any y ∈ R, there exists

∴ f is onto.
Therefore, f is one-one and onto.
Thus, f is an invertible function.
Let us define g: R → R as f (y) = y-7/10.
Now, we have

gof = I_R, fog =  I_R
Hence, the required function g: R → R is defined as g(y) = y - 7/10

Q2: Let f: W → W be defined as f(n) = n − 1, if is odd and f(n) = n + 1, if n is even. Show that f is invertible. Find the inverse of f. Here, W is the set of all whole numbers.
Ans: It is given that:

One-one:
Let f(n) = f(m).
It can be observed that if n is odd and m is even, then we will have n − 1 = m + 1.
⇒ n − m = 2
However, this is impossible.
Similarly, the possibility of n being even and m being odd can also be ignored under a similar argument.
∴ Both n and m must be either odd or even.
Now, if both n and m are odd, then we have:
f(n) = f(m) ⇒ n − 1 = m − 1 ⇒ n = m
Again, if both n and m are even, then we have:
f(n) = f(m) ⇒ n + 1 = m + 1 ⇒ n = m
∴ f is one-one.
It is clear that any odd number 2r + 1 in co-domain N is the image of 2r in domain N and any even number 2r in co-domain N is the image of 2r + 1 in domain N.
∴ f is onto.
Hence, f is an invertible function.
Let us define g: W → W as:

Now, when n is odd:
gof(n) = g(f(n)) = g(n - 1) = n - 1 + 1 = n
And, when n is even:
gof(n) = g(f(n)) = g(n + 1) = n + 1 - 1 = n
Similarly, when m is odd:
fog(m) = f(g(m)) = g(m - 1) = m - 1 + 1 = m
When m is even:
fog(m) = f(g(m)) = g(m + 1) = m + 1 - 1 = m
∴ gof = I_w and fog = I_w
Thus, f is invertible and the inverse of f is given by f^—1 = g, which is the same as f.
Hence, the inverse of f is f itself.

Q3: If f: R → R is defined by f(x) = x² − 3x + 2, find f(f(x)).
Ans: f (f(x)) = f(x² - 3x + 2)
= (x² - 3x + 2)² - 3(x² - 3x + 2) + 2
= (x⁴ + 9 x 2 + 4 - 6 x³ - 12x + 4x²) + (- 3x² + 9x - 6) + 2
= x⁴ - 6x³ + 10x² - 3x

Q4: Show that function f: R → {x ∈ R: −1 < x < 1} defined by f(x) = x/1+|x|, x ∈ R is one – one and onto function.
Ans: It is given that f: R → {x ∈ R: −1 < x < 1} is defined as

For one – one
Suppose f(x) = f(y), where x, y ∈ R.

It can be observed that if x is positive and y is negative,
Then, we have

Since x is positive and y is negative:
x > y ⇒ x − y > 0
But, 2xy is negative.
Then, 2xy ≠ x - y.
Thus, the case of x being positive and y being negative can be ruled out.
Under a similar argument, x being negative and y being positive can also be ruled out.
∴ x and y have to be either positive or negative.
When x and y are both positive, we have

When x and y are both negative, we have

∴ f is one – one.
For onto
Now, let y ∈ R such that −1 < y < 1.

∴ f is onto.
Hence, f is one – one and onto.

Q5: Give examples of two functions f: N → Z and g: Z → Z such that g o f is injective but g is not injective. (Hint: Consider f(x) = x and g(x) = |x|)
Ans: Define f: N → Z as f(x) = x and g: Z → Z as g(x) = |x|.
We first show that g is not injective.
It can be observed that:
g(-1) = |-1| = 1
g(1) = |1| = 1
∴ g(−1) = g(1), but −1 ≠ 1.
∴ g is not injective.
Now, gof: N → Z is defined as gof (x) = g(f(x)) = g(x) = |x|.
Let x, y ∈ N such that gof(x) = gof(y).
⇒ |x| = |y|
Since x and y ∈ N, both are positive.
∴ |x| = |y| ⇒x = y
Hence, gof is injective

Q6: Given examples of two functions f: N → N and ??: N → N such that ?? of is onto but f is not onto.

Ans: Define f: N → N by f(x) = x + 1

We first show that g is not onto.
For this, consider element 1 in co-domain N. It is clear that this element is not an image of any of the elements in domain N.
∴ f is not onto.
Now, gof: N → N is defined by,

Then, it is clear that for y ∈ N, there exists x = y ∈ N such that gof(x) = y.
Hence, gof is onto.

Q7: Given a non-empty set X, consider the binary operation *: P(X) × P(X) → P(X) given by A * B = A ∩ B ∀ A, B in P(X) is the power set of X. Show that X is the identity element for this operation and X is the only invertible element in P(X) with respect to the operation*.
Ans: It is given the binary operation *:
P(X) × P(X) → P(X) given by A * B = A ∩ B ∀ A, B in P(X)
We know that A ∩ X = A = X ∩ A for all A ∈ P(X)
⇒ A * X = A = X * A for all A ∈ P(X)
Thus, X is the identity element for the given binary operation *.
Now, an element A ∈ P(X) is invertible if there exists B ∈ P(X) such that
A * B = X = B * A [As X is the identity element]
or A ∩ B = X = B ∩ A
This case is possible only when A = X = B.
Thus, X is the only invertible element in P(X) with respect to the given operation*.
Hence, the given result is proved.

Q8: Let S = {a, b, c} and T = {1, 2, 3}. Find F⁻¹ of the following functions F from S to T, if it exists.
(i) F = {(a, 3), (b, 2), (c, 1)}
(ii) F = {(a, 2), (b, 1), (c, 1)}
Ans: S = {a, b, c}, T = {1, 2, 3}
(i) F : S → T is defined as F = {(a, 3), (b, 2), (c, 1)}
⇒ F (a) = 3, F (b) = 2, F(c) = 1
Therefore, F⁻¹: T → S is given by F⁻¹ = {(3, a), (2, b), (1, c)}.
(ii) F : S → T is defined as F = {(a, 2), (b, 1), (c, 1)}
Since F (b) = F (c) = 1, F is not one – one.
Hence, F is not invertible i.e., F⁻¹ does not exist.

Q9: Consider the binary operations*: R ×R → and o: R × R → R defined as a*b = |a - b| and a o b = a, "a,b ∈ R. Show that * is commutative but not associative, o is associative but not commutative. Further, show that "a, b, c ∈ R, a*(b o c) = (a * b) o (a * c). [If it is so, we say that the operation * distributes over the operation o]. Does o distribute over *? Justify your answer.
Ans: It is given that *: R × R → and o: R × R → R is defined as
a * b = |a - b| and a o b = a, "a, b ∈ R.
For a, b ∈ R, we have:
a * b = |a - b|
b * a = |b - a| = |-(a - b)| = |a - b|
∴ a * b = b * a
∴ The operation * is commutative.
It can be observed that,
(1 * 2) * 3 = (|1 - 2|) * 3 = 1 * 3 = |1 - 3| = 2
and 1 * (2 * 3) = 1 * (|2 - 3|) = 1 * 1 = |1 - 1| = 0
∴ (1 * 2) * 3 * 1 * (2 * 3) where 1, 2, 3 ∈ R.
Hence, the operation * is not associative.
Now. consider the operation o
It can be observed that 1 o 2 = 1 and 2 o 1 = 2.
∴ 1 o 2 ≠ 2 o 1 where 1, 2 ∈ R.
Hence, the operation o is not commutative.
Let a.b.c ∈ R. Then, we have
(a o b)o c = a o c = a
and a o (b o c) = a o b = a
∴ a o b) o c = a o (b o c). where a, b, c ∈ R
Hence, the operation o is associative.
Now. let a, b, c ∈ R. then we have
a * (b o c) = a * b =|a — b|
(a * b) o (a * c) = (|a — b|) o (|a — c|) = |a — b|
Hence, a * (b o c) = (a * b)o (a * c).
Now, 1 o (2 * 3) = 1 o (|2 - 3|) = 1 o 1 = 1
(1 o 2) * (1 o 3) = 1 * 1 = |1 - 1| = 0
∴ 1 o (2 * 3) # (1 o 2) * (1 o 3) where 1, 2, 3 ∈ R
Hence, the operation o does not distribute over *.

Q10: Given a non -empty set X, let *: P(X) × P(X) → P(X) be defined as A * B = (A − B) ∪ (B − A), ∀ A, B ∈ P(X). Show that the empty set Φ is the identity for the operation * and all the elements A of P(X) are invertible with A−1 = A.
(Hint: (A − Φ) ∪ (Φ − A) = A and (A − A) ∪ (A − A) = A * A = Φ).
Ans: It is given that *: P(X) × P(X) → P(X) is defined as A *B = (A -B) U (B - A) ∀ A, B ∈ P (X).
Let A ∈ P(X). Then, we have
A * Φ = (A - Φ) U (Φ - A)= A U Φ = A
Φ * A = (Φ - A) U (A - Φ) = Φ U A = A
∴ A * Φ = A = Φ * A for all A ∈ P(X)
Thus, Φ is the identity element for the given operation *.
Now, an elements A ∈ P(X) will be invertible if there exists B ∈ P(X) such that
A * B = Φ = B * A. [As Φ is the identity element]
Now. we observed that
A * A = (A - A) U (A - A) = Φ U Φ = 0 for all A ∈ P(X).
Hence, all the elements A of P(X) are invertible with A^-1 = A.

Q11: Define a binary operation *on the set {0, 1, 2, 3, 4, 5} as 

Show that zero is the identity for this operation and each element a ≠ 0 of the set is invertible with 6 − a being the inverse of a. 
Ans: Let X = {0, 1, 2, 3, 4, 5}.
The operation * on X is defined as

An element e ∈ X is the identity element for the operation *. if
a * e = a = e * a for all a ∈ X
For a ∈ X, we have
a * 0 = a + 0 = a [a ∈ X ⇒ a + 0 < 6 ]
0 * a = 0 + a = a [a ∈ X ⇒ 0 + a < 6 ]
∴ a * 0 = a = 0 * a for all a ∈ X
Thus, 0 is the identity element for the given operation *.
An element a ∈ X is invertible if there exists b ∈ X such that a * b = 0 = b * a.

⇒ a = - b or b = 6 - a
But, X = {0, 1, 2, 3, 4, 5} and a, b ∈ X. Then, a ≠ -b.
∴ b = 6 - a is the inverse o f a for all a E X .
Hence, the inverse o f an element a ∈ X. a ≠ 0 is 6 - a i.e., a^-1 = 6 - a.

Q12: Let f: R → R be the Signum Function defined as  

 and g: R → R be the Greatest Integer Function given by g(x) = [x], where [x] is greatest integer less than or equal to x. Then does fog and gof coincide in (0, 1]?
Ans: It is given that,
f: R → R is defined as

Also, g: R → R is defined as g(x) = [x], where [x] is the greatest integer less than or equal to x.
Now, let x ∈ (0, 1].
Then, we have:

Thus, when x ∈ (0, 1), we have fog(x) = 0and gof (x) = 1.
Hence, fog and gof do not coincide in (0, 1].

Q13: Number of binary operations on the set {a, b} are
(a) 10
(b) 16
(c) 20
(d) 8
Ans: (b)
Solution. A binary operation * on {a, b} is a function from {a, b} × {a, b} → {a, b} i.e., * is a function from {(a, a), (a, b), (b, a), (b, b)} → {a, b}.
Hence, the total number of binary operations on the set {a, b} is 24 i.e., 16.