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

Rajasthan Board RBSE Solutions for Class 12 Maths Chapter 1 Relations and Functions Miscellaneous Exercise Textbook Exercise Questions and Answers.

RBSE Class 12 Maths Solutions 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.
Answer:
It is given that f: R → R is defined as f(x) = 10x + 7.
Clearly, g = f^-1 ...... (i)
For one-one,
Let f(x) = f(y), where x, y ∈ R.
⇒ 10x + 7 = 10y + 7
⇒ x = y
f is a one-one function.

For 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 is onto.
Since, f is one-one onto, so it is invertible, i.e. f^-1 exists.

Thus, gof = I_R and fog = I_R.
Hence, the required function g : 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.
Answer:
It is given that:

For 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,
For onto
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.

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, then find f(f(x))
Answer:
Here f(x) = x² - 3x + 2, f:R → R
Thus 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) = \(\frac{x}{1+|x|}\), x ∈ R is one-one and onto function.
Answer:
Here f: R → {x ∈ R:- 1 < x < 1} and f(x) = \(\frac{x}{1+|x|}\) (a) Let, x > 0 ⇒ |x| = x
∴ f(x) = \(\frac{x}{1+x}\)
x₁, x₂ ∈ R, then
f(x₁) = f(x₂) ⇒ \(\frac{x_{1}}{1+x_{1}}\) = \(\frac{x_{2}}{1+x_{2}}\)
⇒ x₁(1 + x₂) = x₂(1 + x₁)
⇒ x₁ + x₁x₂ = x₂ + x₁x₂ ⇒ x₁ = x₂
When x < 0 ⇒ | x | = - x then

⇒ x₁(1 - x₂) = x₂(1 - x₁)
⇒ x₁ - x₁x₂ = x₂ - x₂x₁
⇒ x₁ = x₂
∴ f is one-one.

(b) When x ≥ 0 ⇒ | x | = x (say)
y = f(x) ⇒ \(\frac{x}{1+x}\) = y
⇒ x = (1 + x)y ⇒ x = y + xy
⇒ x - xy = y ⇒ x(1 - y) = y
⇒ x = \(\frac{y}{1-y}\)
when x < 0 ⇒ |x| = - x
y = \(\frac{x}{1-x}\) ⇒ y(1 - x) = x
⇒ y - yx = x ⇒ y = x + xy
∴ y = x(1 + y) ⇒ x = \(\frac{y}{1+y}\)
Thus, each element of co-domain is image of any elements of domain in both cases.
∴ f is into.
Thus, f is one-one and into.
Hence Proved.

Question 5.
Show that the function f: R → R given by f(x) = x³ is injective.
Answer:
Here f: R → R,
f(x) = x³
x₁, x₂ ∈ R (domain) then
f(x₁ = f(x₂)
⇒ x₁³ = x₂³
⇒ x₁ = x₂
Thus, f is injective.
Hence Proved.

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 | ]
Answer:
Here f: N → Z and g = Z → Z
Now f(x) = - x, x ∈ N (domain)
and g(x) = |x|, x ∈ Z
Let x₁, x₂ ∈ Z (domain of g)
Then g(x₁) = g(x₂) ⇒ | x₁ | = | x₂ |
⇒ x₁ = x₂ or x₁ = - x₂ [ ∵ |- x₂| = x₂]
⇒ x₁ ≠ x₂
∴ g is not onto.

Example: g(1) = 1 = 1
and g(- 1) = |- 1 | = 1
i.e., g image of -1 and 1 is 1
∴ g is not one-one.
gof = N → Z
Let x₁, x₂ ∈ N (domain of gof)
∴ (gof) (x₁) = (gof) (x₂)
⇒ g[f(x₁)] = g[f(x₂)]
⇒ g(- x₁) = g(- x₂) (∵ f(x) = - x)
f(x) = - x
⇒ |- x₁| = |- x₂|
⇒ |x₁| = |x₂|
⇒ x₁ = x₂
∴ gof is one-one.
So, f(x) = - x and = g(x) = |x|

Question 7.
Give examples of two functions f: N → N and g: N → N such that gof is onto but f is not onto.

Answer:
Here f:N → N and g:N → N
Now f(x) = x + 1, x ∈ N (Domain)

f(x₁) = f(x₂)
⇒ x₁ + 1 = x₂ + 1
⇒ x₁ = x₂
∴ f is one-one.
For onto f,
Let y ∈ N (Co-domain) such as that
f(x) = y
Then y = x + 1 ⇒ x = y - 1
For y = 1
x = 0 ∉ N (Domain)
i.e., element 1 of co-domain N is not image.
∴ Range of f ⊂ N ( Co-domain)
[Here x = 1, then f(1) = 1 + 1 = 2 ∈ N]
∴ f is not onto.
Now for onto gof,

Let y ∈ N (Co-domain gof) is such as that
y = (gof) (x)
⇒ g[f(x)] = g(x + 1)
⇒ x + 1 - 1 = x ∈ N (domain)
When y = 1, then x = 1
Thus, 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 ⊂ B. Is R an equivalence relation on P(X)? Justify your answer.
Answer:
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.

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) is the power set of X. Show that X is the identity element for this operation and X is the only invertible Clement in P(X) with respect to the operation *.
Answer:
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 Be 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.
Hence Proved.

Question 10.
Find the number of all onto functions from the set {1, 2, 3, ....., n} to itself.
Answer:
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.

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 = {(a, 3), (b, 2), (c, 1)|
(ii) F = [(a, 2), (b, 1), (c, 1)}
Answer:
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.

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 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.
Answer:
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| and b * a = |b - a| = |- (a - b)| = | a - b |
∴ a* b = b* a
Hence, 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*b) = (|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 *.

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.
Answer:
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) = F ∪ A = A
∴ A * Φ = A = Φ * A for all A ∈ P(X)
Thus, F is the identity element for the given operation*.
Now, an element A ∈ P(X) will be invertible if there B ∈ P(X) such that A * B = Φ = B * A.
[As F is the identity element]
Now, we observed that
A * A = (A - A) ∪ (A - A) = Φ ∪ Φ = F for all A ∈ P(X).
Hence, all the elements A of P(X) are invertible with A^-1 = A.
Hence Proved.

Question 14.
Define a binary operation in set {0, 1, 2, 3, 4, 5} as:

Show that zero (0) is the identity of this operation and each element a ≠ 0 of the set is invertible with 6 - a; being the inverse of a.
Answer:
Here A = {0, 1, 2, 3, 4, 5} or

(i) Let identity element is e.
a*e = e*a = a
It e = 0, then
a*e = a + 0 = a
e*a = 0 + a = a
∴ a*e = e*a = a
∴ 0 is identity element.

(ii) let b is inverse of a, then
a*b = b*a = e
Now a*(6 - a) = a + (6 - a) - 6
= a + 6 - a - 6 = 0
and (6 - a)*a = (6 - a) + a - 6 = 0
∴ a*(6 - a) = (6 - a)*a = 0
Thus, 6 - a is inverse of each element a of A.
Hence Proved.

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 ? Given reason.
Justify your answer.
Answer:
Here A = {-1, 0, 1, 2}
B = {- 4, - 2, 0, 2}
f, g: A → B

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
Answer:
The given set is A = {1, 2, 3}.
The smallest relation containing (1, 2) and (1, 3) which is reflexive and symmetric, but not transtitive 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.
The correct answer 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
Answer:
It is given that A = {1, 2, 3}.
The smallest equivalence reaction 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 answer is B.

Question 18.
Let g: 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]?
Answer:
It is given that,

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

gof(x) = g(f(x)) = g(1) [as x > 0]
= [1] = 1
Thus, when x e (0,1), we have fog(x) = 0 and gof(x) = 1.
Hence, fog and gof do not coincide in (0, 1].

Question 19.
Number of binary operations on the st {a, b} are
(A) 10
(B) 16
(C) 20
(D) 8
Answer:
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.
The correct answer is B.