PSEB 12th Class Maths Solutions Chapter 1 Relations and Functions Miscellaneous Exercise

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 1 Relations and Functions Miscellaneous Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 1 Relations and Functions Miscellaneous Exercise

Question 1.
Let f: R → R be defined as f(x) = 10x + 7. Find the function g: R → R such that gof = fog = I_R.
Solution.
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.
⇒ x = \(\frac{y-7}{10}\) ∈ R
Therefore, for any y ∈ R, there exists x = \(\frac{y-7}{10}\) ∈ R such that
f(x) = f(\(\frac{y-7}{10}\))
= 10(\(\frac{y-7}{10}\)) + 7
= y – 7 + 7 = y
∴ f is onto.
Therefore f is one-one and onto.
Thus f is an invertible function.
Let us define g : R → R as g(y) = \(\frac{y-7}{10}\)
Now, we have
gof(x) = g(f(x)) = g(10x + 7)
= \(\frac{(10 x+7)-7}{10}=\frac{10 x}{10}\) = x
And, fog(y) = f(g(y))
= f(\(\frac{y-7}{10}\))
= 10(\(\frac{y-7}{10}\)) + 7
= y – 7 + 7 = y
∴ gof = I_R and fog = I_R
Hence, the required functiong:R → R is defined as g(y) = \(\frac{y-7}{10}\)

Question 2.
Let f: W → W be defined as f(n) = n – 1, if n 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.
Solution.
It is given that
f: W → W is defined as f(n) =

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 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.

Onto :
It is clear that any odd number 2r + 1 in co-domain W is the image of 2r in domain W and any even number 2r in co-domain W is the image of 2r + 1 in domain W.
∴ f is onto.
Hence, f is an invertible function.
Let us define g : W → W as

g(m) =

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)) = f(m – 1) = m – 1 + 1 = m
and when m is even
fog(m) = f(g(m)) = f(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 itself.

Question 3.
If f: R → R is defined by f(x) = x² – 3x + 2, find f(f(x)).
Solution.
It is given that f: R → R is defined as f(x) = x² – 3x + 2.
f(f(x)) = f(x² – 3x + 2)
= (x² – 3x + 2)² – 3(x² -3x + 2) + 2
= x⁴ + 9x² + 4 – 6x³ – 12x + 4x² – 3x² + 9x – 6 + 2
= x⁴ – 6x³ + 10x² – 3x.

Question 4.
Show that the function f: R → {x ∈ R: – 1 < x < 1} defined by f(x) = \(\frac{x}{1+|x|}\) ∈ R is one-one and onto function.
Solution.
It is given that f: R → {x ∈ R: – 1 < x < 1} is defined as f(x) = \(\frac{x}{1+|x|}\), x ∈ R.
Suppose f(x) = f(y), where x,y ∈ R ⇒ \(\frac{x}{1+|x|}=\frac{y}{1+|y|}\)
It can be observed that if x is positive and y is negative, then we have \(\frac{x}{1+x}=\frac{y}{1-y}\)
⇒ 2xy = x – y
Since x is positive and y is negative, then 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 x y
f(x) = f(y)
⇒ \(\frac{x}{1+x}=\frac{y}{1+y}\)
⇒ x + xy = y + xy
⇒ x = y
When x and y are both negative, we have
f(x) = f(y)
⇒ \(\frac{x}{1-x}=\frac{y}{1-y}\)
⇒ x – xy = y – yx
⇒ x = y
∴ f is one-one.
Now, let y ∈ R such that – 1 < y < 1.
If y is negative, then there exists x = \(\frac{y}{1+y}\) ∈ R such that
f(x) = f(\(\frac{y}{1+y}\))
= \(\frac{\left(\frac{y}{1+y}\right)}{1+\left|\frac{y}{1+y}\right|}=\frac{\frac{y}{1+y}}{1+\left(\frac{-y}{1+y}\right)}=\frac{y}{1+y-y}\) = y
If y is positive, then there exists x = \(\frac{y}{1-y}\) ∈ R such that
f(x) = \(f\left(\frac{y}{1-y}\right)=\frac{\left(\frac{y}{1-y}\right)}{1+\left(\frac{y}{1-y}\right)}\)

= \(\frac{\frac{y}{1-y}}{1+\left(\frac{y}{1-y}\right)}=\frac{y}{1-y+y}\) = y
∴ f is onto.
Hence, f is one-one and onto.

Question 5.
Show that the function f: R → R given by f(x) = x³ in injective.
Solution.
f: R → R is given as f(x) = x³.
Suppose f(x) = f(y), where x, y ∈ R.
⇒ x³ = y³ …………(i)
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 Eq. (i).
∴ x = y
Hence, f is injective.

c

Question 6.
Give examples of two functions f: N → Z and g: Z → Z such that gof is injective but g is not injective.
[Hint: consider f(x) x and g(x) = |x|].
Solution.
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.

Question 7.
Give examples of two functions f: N → N and g: N → N such that gof is onto but f is not onto.
[Hint: consider f(x) = x + 1 and g(x) = i]
Solution.
Define f: N → N by f(x) = x +1
and, g: N → N by g(x) =
We first show that f 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,
gof(x) = g(f(x)) = g(x + 1) = (x + 1) – 1 = x [∵ x ∈ N ⇒ (x + 1) > 1]
Then, it is clear that for y ∈ N, there exists x = y ∈ N such that gof(x) = y.
Hence, gof is onto.

Question 8.
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 c B. Is R an equivalence relation on P(X)? Justify your answer.
Solution.
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 c B and B c C.
⇒ A ⊂ C
⇒ ARC
R is transitive.
Hence, R is not an equivalence relation since it is not symmetric.

Question 9.
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) where 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*.
Solution.
It is given that * : P(X) × P(X) → P(X) is defined as A * B = A ∩ B ∀ A, B ∈ P(X).
We know that A * X = A ∩ X = A = X ∩ A ∀ A ∈ P(X).
⇒ A * X = A = X * A ∀ 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)
i.e., A ∩ B = X = B ∩ A
This case in 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.

Question 10.
Find the number of all onto functions from the set {1, 2, 3, n} to itself.
Solution.
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, 3,…, n} to itself is the same as the total number of permutations on symbols 1, 2,…, n, which is n!.

Question 11.
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 = {(o, 3), (6, 2), (c, 1)}
(ii) F = {(a, 2), (6, 1), (c, 1)}
Solution.
Given, S = {a, b, c}, and T = {1, 2, 3}
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.

Question 12.
Consider the binary operations * : R × 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 hut 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.
Solution.
It is given that *: R × 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
Therefore, the operation * is commutative..
It can be observed that
(1 * 2) * 3 = (|1 – 2|) * 3 = 1 * 3 = |1 – 3|= 2
1 * (2 * 3) = 1 * (|2 – 3|) = 1 * 1 =|1 – 1 |= 0
∴ (1 * 2) * 3 ≠ 1 * (2 * 3) (where 1, 2, 3 ∈ R)
Therefore, 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
Therefore, the operation o is not commutative.
Let a, b, c ∈ R. Then, we have
(a o b) o c = a o c = a
a o (b o c) = a o b = a
⇒ (a o b) o c = a o (b o c)
Therefore, 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 Therefore, the operation o does not distribute over *.

Question 13.
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]
Solution.
It is given that *: P(X) × P(X) → P(X) is defined as
A * B = (A – B) ∪ (B – A) ∀ A, B, ∈ P(X).
Let A ∈ P(X). Then, we have
A * (Φ) = (A – Φ) ∪ (Φ – A) = A ∪ Φ = A
Φ * A = (Φ – A) ∪ (A – Φ) = Φ ∪ A = A
A * Φ = A = Φ * A ∀ A ∈ P(X)
Thus, Φ is the identity element for the given operation *.
Now, an element A s 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) ∪ (A – A) = Φ ∪ Φ = Φ ∀ A ∈ P(X).
Hence, all the elements A of P(X) are invertible with A^-1 = A.

Question 14.
Define a binary operation * on the set {0, 1, 2, 3, 4, 5) as
a * b =
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.
Solution.
(i) e is the identity element if a * e = e * a = a
a * 0 = a + 0, 0 * a = 0 + a = a
⇒ a * 0 = 0 * a = a
∴ 0 is the identity of the operation.

(ii) b is the inverse of a if a * b = b * a = e
Now a * (6 – a) = a + (6 – a) – 6 = 0
(6 – a) * a = (6 – a) + a – 6 = 0
Hence, each element of a of the set is invertible with inverse.

Question 15.
Let A = {-1, 0, 1, 2}, B = {-4,-2, 0,2} and f, g: A → B be functions defined by f(x) = x² – x, x ∈ A and g(x) = 2|x – \(\frac{1}{2}\)| – 1, x ∈ A. Are f and g equal? Justify your answer.
[Hint: One may not be that two functions f: A → B and g: A → B
such that f(a) = g(a) ∀ a ∈ A, are called equal functions.]
Solution.
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 f(x) = x² – x, x ∈ A and
g(x) = 2 |x – \(\frac{1}{2}\)| – 1, x ∈ A
It is observed that
f(- 1) = (- 1)² – (- 1) = 1 + 1 = 2
and g(- 1) = 2|(- 1) – \(\frac{1}{2}\)| – 1
= 2(\(\frac{3}{2}\)) – 1 = 3 – 1 = 2
⇒ f(- 1) = g(- 1)

⇒ f(0) = (0)² – 0 = 0
and g(0) = 2|0 – \(\frac{1}{2}\)|
= 2(\(\frac{3}{2}\)) – 1 = 1 – 1 = 0

⇒ f(0) = g(0)
f(1) = (1)² – 1 = 1 – 1 = 0
and g(1)= 2|1 – \(\frac{1}{2}\)|
= 2(\(\frac{1}{2}\)) – 1 = 1 – 1 = o

⇒ f(1) = g(1)
f(2) = (2)² – 2 = 4 – 2 = 2
and g(2) = 2 |2 – \(\frac{1}{2}\)|
= 2(\(\frac{3}{2}\)) – 1 = 3 – 1 = 2
⇒ f(2) = g(2)
∴ f(a) = g(a) ∀ a ∈ A
Hence, the functions f and g are equal.

Question 16.
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
Solution.
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 relationR is not transitive as (3, 1), (1, 2) ∈ R but (3, 2) ∈ R.
Now, if we add any one of the two pairs (3, 2) and (2, 3) (or both) to relation R, then relation R will become transitive.
Hence, the total number of desired relation is one.
Thus, the correct option is (A).

Question 17.
Let A = {1, 2, 3}. Then, number of equivalence relations containing (1, 2) is
(A) 1
(B) 2
(C)3
(D) 4
Solution.
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 add 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. The correct option is (B).

Question 18.
Let f: R → R be the signum function defined as
f(x) =
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]?
Solution.
It is given that
f: R → R is defined as f(x) =
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
[x] = 1, if x = 1 and [x] = 0 if 0 < x < 1. ∴ fog(x) = f (g(x)) = f([x]) = gof(x) = g(f(x))= g(1) [∵ x > 0]
= [1] = 1 .
Thus, when x ∈ (0, 1), we have fog(x) = 0 and gof(x) = 1.
But fog (1) ≠ gof (1)
Hence, fog and gof do not coincide in (0,1].

Question 19.
Number of binary operations on the set {a, b} are (A) 10 (B) 16 (C) 20 (D) 8
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.
Thus, the correct option is (B).