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

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 = IR.
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 = IR and fog = IR.
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 = Iw and fog = Iw
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.

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