# Miscellaneous Exercise on Chapter 1 Solutions

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Miscellaneous Exercise on Chapter 1
1. Let f : R → R be defined as {f (x) = 10x + 7}. Find the function g : R → R such that {gof = fog} = IR.
To solve this problem, we’ll first define g and then prove that {gof = fog} = IR
Step 1: Define g
Consider an arbitrary element in the co-domain of f, yR
By definition of f, we have
{y = 10x + 7}, for some x in the domain R
{⇒ 10x + 7 = y}
{⇒ 10x = y - 7}
{⇒ x = \dfrac{y - 7}{10}}
Now let’s define g : R → R as
{g(y) = \dfrac{y - 7}{10}}
Step 2: To prove that {gof = fog} = IR
We have
gof(x)
{= g(f(x))}
{= f(10x + 7)}
{= \dfrac{(10x + 7) - 7}{10}}
{= \dfrac{10x + 7 - 7}{10}}
{= \dfrac{10x}{10}}
{= x}
= IR(x)
Also, we have
fog(y)
{= f(g(y))}
{= f\left(\dfrac{y - 7}{10}\right)}
{= 10 × \left(\dfrac{y - 7}{10}\right) + 7} {(∵ f(x) = 10x + 7)}
{= y - 7 + 7}
{= y}
= IR(y)
{gof = fog = } = IR
f is invertible and g : R → R is the inverse of f and is defined as
{g(y) = \dfrac{y - 7}{10}}

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.
f can be proved to be invertible in two ways.
Method 1:
Define an inverse function g for f and then prove that {gof = fog} = IW
Prove that f is both one-one and onto. This implies that f is invertible.
Method 1
Step 1: Define g
Consider an arbitrary element in the co-domain of f, yW.
The function f : W → W is defined as
{f(n) = \begin{cases}n - 1, & \text{if }n\text{ is odd}\\n + 1, & \text{if }n\text{ is even}\end{cases}}
Thus, by definition of f, we have two cases
When n is odd
When n is even
{f(n) = n - 1}
f(n) = n + 1
{⇒ y = n - 1} (this also implies that y is even)
{⇒ y = n + 1} (this also implies that y is odd)
{⇒ n - 1 = y}
{⇒ n + 1 = y}
{⇒ n = y + 1}
{⇒ n = y - 1}
Let’s now consider the function g : W → W is defined as
{g(y) = \begin{cases}y - 1, & \text{if }y\text{ is odd}\\n + 1, & \text{if }y\text{ is even}\end{cases}}
Step 2: To prove that {gof = fog} = IW
Note: We have to prove these cases both when n is odd as well as when n is even.
When n is odd, we have
gof(n)
{= g(f(n))}
{= g(n - 1)} (∵ n is odd)
{= (n - 1) + 1} (∵ {n - 1} is even)
= n
= IW(n)
Also, when n is even, we have
gof(n)
{= g(f(n))}
{= g(n + 1)} (∵ n is even)
{= (n + 1) - 1} (∵ {n + 1} is odd)
= n
= IW(n)
Now, when y is odd, we have
fog(y)
{= f(g(y))}
{= f(y - 1)} (∵ y is odd)
{= (y - 1) + 1} (∵ {y - 1} is even)
= y
= IW(y)
Also, when y is even, we have
fog(y)
{= f(g(y))}
{= f(y + 1)} (∵ y is even)
{= (y + 1) - 1} (∵ {y + 1} is odd)
= y
= IW(y)
Thus we see that gof = fog = IW
f is invertible
Also, from the definition of both f and g, we see that f = g.
Method 2:
As discussed earlier, in this method, we prove that f is one-one and onto. This implies that f is invertible.
To show that f is one-one
Consider two whole numbers n_1, n_2W such that {f(n_1) = f(n_2)}
Now this need to be verified under the following scenario.
i.
When n_1 is odd and n_2 is even.
ii.
When n_1 is even and n_2 is odd.
iii.
When both n_1 and n_2 are odd.
iv.
When both n_1 and n_2 are even.
i. When n_1 is odd and n_2 is even.
{f(n_1) = f(n_2)}
{⇒ n_1 + 1 = n_2 - 1}
{⇒ 1 + 1 = n_2 - n_1}
{⇒ 2 = n_2 - n_1}
⇒ Difference of an even and odd whole numbers is even.
So, this case is false
ii. When n_1 is even and n_2 is odd.
{f(n_1) = f(n_2)}
{⇒ n_1 - 1 = n_2 + 1}
{⇒ n_1 - n_2 = 1 + 1}
{⇒ n_1 - n_2 = 2}
⇒ Difference of an odd and even whole numbers is even.
So, this case is false
iii. When both n_1 and n_2 are odd.
{f(n_1) = f(n_2)}
{⇒ n_1 + 1 = n_2 + 1}
{⇒ n_1 = n_2}
iv. When both n_1 and n_2 are even.
{f(n_1) = f(n_2)}
{⇒ n_1 - 1 = n_2 - 1}
{⇒ n_1 = n_2}
So, we have proved that when
{f(n_1) = f(n_2)}
{⇒ n_1 = n_2}
f is one-one
To show that f is onto
As f : W → W is defined as
{f(n)= \begin{cases}n - 1, & \text{if }n\text{ is odd}\\n + 1, & \text{if }n\text{ is even}\end{cases}}
For any element y ∈ W in the co-domain, there should exist a corresponding element n in the domain W defined under the following scenarios.
When n is odd
When n is even
{f(n) = n - 1}
f(n) = n + 1
{⇒ y = n - 1} (this also implies that y is even)
{⇒ y = n + 1} (this also implies that y is odd)
{⇒ n - 1 = y}
{⇒ n + 1 = y}
{⇒ n = y + 1}
{⇒ n = y - 1}
This can be put together as
{n = \begin{cases}y - 1, & \text{if }y\text{ is odd}\\y + 1, & \text{if }y\text{ is even}\end{cases}}
Thus, every whole number yW in the co-domain will be an image of a whole number nW in the domain.
f is onto.
As f is both one-one and onto, f is invertible and the inverse of f is defined as
{f^{-1}(y) = \begin{cases}y - 1, & \text{if }y\text{ is odd}\\y + 1, & \text{if }y\text{ is even}\end{cases}}

3. If f : R → R is defined by f(x) = x^2 – 3x + 2, find f(f(x)).
The function f is defined as f(x) = x^2 – 3x + 2
{∴ f(f(x))}
{= f(x^2 - 3x + 2)}
{= (x^2 - 3x + 2)^2} {- 3(x^2 - 3x + 2)} {+ 2}
{= x^4 + 9x^2 + 4} {- 6x^3 - 12x + 4x^2} {- 3x^2 + 9x - 6} + 2 {(∵ (a - b + c)^2} {= a^2 + b^2 + c^2} {- 2ab - 2bc + 2ca)}
{= x^4 - 6x^3} {+ 9x^2 + 4x^2 - 3x^2} {- 12x + 9x} + 4 – 6 + 2
{= x^4 - 6x^3 + 10x^2 - 3x}

4. Show that the function f : R → {x ∈ R : -1 < x < 1} defined by {f(x) = \dfrac{x}{1 + |x|}}, x ∈ R is one one and onto function.
It is given that the function f : R → {x ∈ R : -1 < x < 1} is defined as
{f(x) = \dfrac{x}{1 + |x|}}
This can be re-defined as follows:
{f(x) = \begin{cases}\dfrac{x}{1 + x}, & if x \geq 0 \\ \dfrac{x}{1 - x}, & if x \lt 0\end{cases}}
To show that f is one-one
Consider two real numbers x_1, x_2R such that
{f(x_1) = f(x_2)}
Now, there are four possible scenarios.
Case a: When x_1 > 0 and x_2 > 0
{f(x_1) = f(x_2)}
{⇒ \dfrac{x_1}{1 + x_1} = \dfrac{x_2}{1 + x_2}}
{⇒ x_1(1 + x_2) = x_2(1 + x_1)}
{⇒ x_1 + x_1x_2 = x_2 + x_1x_2}
{⇒ x_1 = x_2}
Case b: When x_1 < 0 and x_2 < 0
{f(x_1) = f(x_2)}
{⇒ \dfrac{x_1}{1 - x_1} = \dfrac{x_2}{1 - x_2}}
{⇒ x_1(1 - x_2) = x_2(1 - x_1)}
{⇒ x_1 - x_1x_2 = x_2 - x_1x_2}
{⇒ x_1 = x_2}
Case c: When x_1 > 0 and x_2 < 0
Here x_1 is positive and x_2 is negative.
{⇒ x_1}x_2
{⇒ \dfrac{x_1}{1 + x_1}}\dfrac{x_2}{1 - x_2}
{⇒ f(x_1)}f(x_2)
⇒ Distinct elements have distinct images.
Case d: When x_1 < 0 and x_2 > 0
Here x_1 is negative and x_2 is positive.
{⇒ x_1}x_2
{⇒ \dfrac{x_1}{1 - x_1}}\dfrac{x_2}{1 + x_2}
{⇒ f(x_1)}f(x_2)
⇒ Distinct elements have distinct images.
∴ From all the above 4 cases, f is one-one
To show that f is onto
The range of f is given as {xR; -1 < x < 1}
y ∈ (-1, 1)
Now y
{= f(x)}
{= \dfrac{x}{1 + |x|}}
{= \dfrac{x}{1 \pm x}}
{⇒ x}
{= \dfrac{y}{1 \pm y}}
As y ≠ -1, y ≠ 1, {1 \pm y} ≠ 0
{x = \dfrac{y}{1 \pm y}}R
Thus, every element y in the co-domain of f will be an image of an element x in the domain R
{⇒ f} is onto

5. Show that the function f : R → R given by {f(x) = x^3} is injective.
The function f : R → R is defined as
{f(x) = x^3}
Consider two real numbers x_1, x_2R such that
{f(x_1) = f(x_2)}
{⇒ x_1^3 = x_2^3}
{⇒ x_1 = x_2}
f is injective (one-one)
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|}).
Let the function f : N → Z be defined as
{f(x) = x}
and the function g : Z → Z be defined as
{g(x) = |x|}
Now, consider two integer x_1, x_2Z such that
{g(x_1) = g(x_2)}
{⇒ |x_1| = |x_2|}
{⇒ x_1 = {\pm}x_2}
{⇒ x_1 = x_2} or {x_1}x_2
{⇒ x_1} and x_2 are not necessarily unique.
{⇒ g} is not injective.
Now, as f : N → Z and g : Z → Z, gof : N → Z
Now, consider two elements x_1, x_2N such that
{gof(x_1) = gof(x_2)}
{⇒ g(f(x_1)) = g(f(x_2))}
{⇒ g(x_1) = g(x_2)} {(∵ f(x) = x)}
{⇒ |x_1| = |x_2|}
{⇒ x_1 = x_2} (∵ as both x_1 and x_2N, {|x_1| = x_1}, {|x_2| = x_2}, as N contains only positive numbers)
{∴ gof} is injective.

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) = \begin{cases}x-1 & \text{if }x \gt 1\\1 & \text{if }x =1\end{cases}}
Let the function f : N → N be defined as
{f(x) = x + 1}
and the function g : Z → Z be defined as
{g(x) = \begin{cases}x-1 & \text{if }x \gt 1\\1 & \text{if }x =1\end{cases}}
To show that f is not onto.
As f : N → N is defined as
{f(x) = x + 1}
Consider an element yN in the co-domain N
{⇒ y = f(x) = x + 1}
{⇒ x = y - 1}
Now consider the element y in the co-domain y = 1 ∈ N
When {y = 0}, {x = y - 1 } = 1 – 1 = 0
As the domain of f is N and as 0 ∉ N, the element 1 ∈ N in the co-domain is not an image of any element in the domain N.
f is not onto.
To show that gof is onto
As f : N → N and g : N → N, we have gof : N → N
Now
gof(x)
{= g(f(x))}
{= g(x + 1)}
{= x + 1 - 1} (∵ xN {⇒ x + 1 > 1)})
{= x}
⇒ Every element yN will be an image of an element xN such that
{y = gof(x) = x}
{∴ gof} is onto.

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.
Consider a set A such that AP(X)
As every set is subset of itself, we can say that
AA
A R A
R is reflexive.
Now, consider two sets A, BP(X) and AB
This does not automatically imply that BA
The relation R is not symmetric.
Now consider three sets A, B, CP(X) such that AB, BC
AC
In otherwords, A R B, B R C implies that A R C
R is transitive.
∴ The relation R is not an equivalence relation as it is not symmetric.
Note: If the question requires us to find whether a relation is an equivalance relation or not, we can conclude that it is not an equivalence relation as soon as we discover that the relation R is not symmetric. In otherwords, we don’t have to check for the relation being transitive, especially when we’re working on multiple choice questions.

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 *.
To Show that X is the identity element of the binary operation *:
For any two sets A, BP(X), the binary operation * is defined as
A * B = AB
Now, for any set AP(X),
We have, A * X = AX = A (∵ X is the universal set and if the set AP(A) then AX)
and X * A = XA = A
X is the identity element of the binary operation *
To show that X is invertible:
For any two sets A, BP(X), the binary operation * is said to be invertible, if
A * B = X = B * A, where X is the identity element under the binary operation *
AB = X = BA
Clearly, this is possible only if A = B = X
So, we have XX = X = XX
X * X = X = X * X
The binary operation * has only one invertible element X

10. Find the number of all onto functions from the set {1, 2, 3, … , n} to itself.
A mapping from a given set to itself will be an onto function only if every element in the domain is mapped to only one element in the co-domain. In otherwords, when a given set is mapped to itself, it has to be a one-one mapping so as to become onto.
Onto function from {1, 2, 3, … , n} to itself is simply a permutation of n symbols 1, 2, 3, … , n.
∴ The total number of onto maps from {1, 2, 3, … , n} to itself is same as total number of permutation on n symbols 1, 2, 3, … , n which is n!.
A more detailed explanation is as follows:
Let’s consider a mapping from a three element set A = {1, 2, 3} to itself
Domain = A= {1, 2, 3}
Co-Domain = A = {1, 2, 3}
i.
Now, the element in the co-domain 1 can be mapped to any one of the three elements 1, 2, 3 in the domain. So, the number of possible mappings for the first element are 3. After this, 2 elements will be left in the domain/co-domain without mapping.
ii.
Now, the second element in the co-domain can be mapped to any of the remaining 2 elements in the domain. So, the number of possible mappings for the second element are 2. After this, the last element will be left in the domain/co-domain without any mapping.
iii.
Now, the third element in the co-domain can be mapped to the only last element left in the domain.
This can be summarised in the following table.
Element
Available mappings
1
3
2
2
3
1
So, the number of onto mappings from given set to itself are 3 × 2 × 1 = 6
The following mapping diagrams and Roster form of the mapping will help further understanding of this.
{(1,1), (2, 2), (3, 3)}
{(1,1), (2, 3), (3, 2)}
{(1,2), (2, 1), (3, 3)}
{(1,2), (2, 3), (3, 1)}
{(1, 3), (2, 1), (3, 2)}
{(1, 3), (2, 2), (3, 1)}
The above procedure can also be applied to a set consisting of n elements (as given in the problem).
Element
Available mappings
1
n
2
{n - 1}
3
{n - 2}
.
.
.
.
{n - 1}
2
n
1
So, the total number of onto mappings from the given set {1, 2, 3, … , n} to itself will be {n × (n - 1) × (n - 2) × ..... × 2 × 1}. This is equal to n!.

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)}
A function F will have an inverse if it is both one-one and onto.
11.i.To find the inverse of F
The function F is given as
F = {(a, 3), (b, 2), (c, 1)}
The function F can also be represented using the mapping diagram depicted below.
The domain of the function is given as
S = {{a, b, c}}
The co-domain of the function is given as
T = {1, 2, 3}
The range of the function is
{1, 2, 3}
To show that F is one-one
As we see, every element in the domain is mapped to a distinct image in the co-domain.
∴ F is one-one
To show that F is onto
As we see, every element in the co-domain T is an image of an element in the domain S
∴ F is onto
As F is both one-one and onto, F is invertible.
⇒ The inverse of F i.e. F-1 exists and is defined as
F-1 = {(3, a), (2, b), (1, c)}
11.ii.To find the inverse of F
The function F is given as
F = {(a, 2), (b, 1), (c, 1)}
The function F can also be represented using the mapping diagram depicted below.
The domain of the function is given as
S = {{a, b, c}}
The co-domain of the function is given as
T = {1, 2, 3}
The range of the function is
{1, 2}
To check whether F is one-one
As we see, two distinct elements in the domain b, cT have the same image 1 in the co-domain S
∴ F is not one-one
To check whether F is onto
As we see, the element 3 ∈ S in the co-domain is not the image of any element in the domain T
∴ F is not onto
As F is neither one-one nor onto, F is not invertible.
⇒ The inverse of F does not exist.
Note: As F is not one-one, in case (ii), we don’t have to check whether it is onto or not. This is because the condition that F is not one-one is enough to conclude that F is not invertible and its inverse does not exist. This strategy can be adopted especially when solving the MCQs. However, when solving the problems, we can go ahead and check for F being onto also.

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.
To show that * is commutative
The binary operation * : R × R → R is defined as
{a * b = |a - b|}
Consider two real numbers a, b ∈ R
Now {a * b = |a - b|}
and {b * a = |b - a|}
We know that {|a - b| = |b - a|}
{⇒ a * b = b * a}
∴ The binary operation * is commutative on the set R.
To show that * is not associative.
Consider three real numbers {a, b, c}R
Now
{(a * b) * c}
{= |a - b| * c}
{= ||a - b| - c|}
{a * (b * c)}
{= a * |b - c|}
{= |a - |b - c||}
So, {(a * b) * c} may not be equal to {a * (b * c)}
∴ The binary operation * is not associative on the set R
Hence, the binary operation * is commutative but not associative.
To show that the binary operation o is not commutative.
The binary operation o : R × R → R is defined as
{a~o~b = a}
Now consider two real numbers {a, b}R
Now {a~o~b = a}
and {b~o~a = b}
Clearly, {a~o~b}{b~o~a}
∴ The binary operation o is not commutative on the set R
To show that o is associative.
Consider three real numbers {a, b, c}R
{(a~o~b)~o~c}
{= a~o~c} {(∵ a~o~b = a)}
{= a}
{a~o~(b~o~c)}
{= a~o~b} {(∵ b~o~c = b)}
{= a}
∴ The binary operation o is associative on the set R
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).
The binary operation * : P(X) × P(X) → P(X) is defined as
A * B = (A - B) ∪ (B - A) A, BP(X)
To show that ϕ is the identity set for the operation *
Consider the set AP(A) and the empty set ϕ,
We have
A * ϕ
= (A - ϕ ) ∪ (ϕ - A)
= Aϕ (∵ A - ϕ = A and ϕ - A = ϕ)
= A
Similarly,
ϕ * A
= (ϕ - A) ∪ (A = ϕ)
= ϕ - A (∵ ϕ - A = ϕ and A - ϕ = A)
= ϕ
A * ϕ = ϕ * A = A AP(X)
∴ Empty set ϕ is the identity for the operation *.
To show that all the elements of A are invertible with A-1 = A
As per the definition, any element a is invertible under the binary operation *, if there exists an element b such that
{a * b = e = b * a}
where e is the identity element under the binary operation *.
Applying this to the sets, the set A is invertible under the binary operation *, if there exists another set B such that
A * B = ϕ = B * A
where ϕ is the empty set.
B = A-1
Now, we have
A * A
= (A - A) ∪ (A - A)
= ϕϕ
= ϕAP(X)
A is invertible, and the inverse of A is A itself.
⇒ B = A-1 = A
∴ All the elements of A of P(X) are invertible with A-1 = A
14. Define a binary operation * on the set {0, 1, 2, 3, 4, 5} as
{a * b = \begin{cases}a + b & \text{if }a + b \lt 6\\a + b - 6 & \text{if }a + b \geq 6\end{cases}}
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.
Under the binary operation *, if for any element aX, there exists an element eX such that
{a * e = e * a = a}
To show that zero is the identity element for the binary operation *
We have
{a * 0}
{= a + 0} (∵ as aX, {a + 0 \lt 6)}
= a
Now,
{0 * a}
{= 0 + a} (∵ as aX, {0 + a \lt 6)}
= a
⇒ There exists an element 0 ∈ X such that
{a * 0 = 0 * a = a}
∴ zero is the identity element for the binary operation *
To show that each element a ≠ 0 of the set is invertible with {6 - a} being the inverse of a.
Any element aX is invertible, if there exists another element bX, such that
{a * b = e = b * a}
where e is the identity element under the binary operation *
As we have seen above, the identity element under the binary operation * is 0.
As the binary operation * is defined as
{a * b = \begin{cases}a + b & \text{if }a + b \lt 6\\a + b - 6 & \text{if }a + b \geq 6\end{cases}}
we have two scenarios.
When {a + b \lt 6}
In this case, we have
{a * b} = a + b = 0
However as a, bX and as {a \gt 0}, and also {b \gt 0}
So, a + b ≠ 0
When {a + b \geq 6}
In this case, we have
{a * b} = a + b - 6 = 0
b = 6 - a
The following will be the table for binary operation * on the set X
a ∈ X
{b = a - 6}
{a * b}~{= a + b - 6}
1
5
0
2
4
0
3
3
0
4
2
0
5
1
0
Thus, every element aX is invertible and {6 - a} is the inverse of a.
{a^{-1} = 6 - a} is the inverse of the element a ∈ {1, 2, 3, 4, 5} (∵ a ≠ 0)
15. Let A = {– 1, 0, 1, 2}, B = {– 4, – 2, 0, 2} and f, g : A → B be functions defined by {f(x) = x^2 – x}, x ∈ A and {g(x) = 2\left|x - \dfrac{1}{2}\right| - 1}, x ∈ A. Are f and g equal? Justify your answer. (Hint: One may note that two functions f : A → B and g : A → B such that {f(a) = g(a)} a ∈ A, are called equal functions).
We know that two functions f : A → B and g : A → B are equal, if aA
{f(a) = g(a)}
Let's now check this by taking each element aA
When a = -1
f(-1)
= (-1)2 - (-1)
= 1 + 1
= 2
g(-1)
{= 2 × \left|-1 - \dfrac{1}{2}\right| - 1}
{= 2 × \left|\dfrac{-2 - 1}{2}\right| - 1}
{= 2 × \left|\dfrac{-3}{2}\right| - 1}
= 3 - 1
= 2
{f(-1) = g(-1)}
When a = 0
f(0)
= 02 - 0
= 0
g(-1)
{= 2 × \left|0 - \dfrac{1}{2}\right| - 1}
{= 2 × \left|\dfrac{0 - 1}{2}\right| - 1}
{= 2 × \left|\dfrac{-1}{2}\right| - 1}
= 1 - 1
= 0
{f(0) = g(0)}
When a = 1
f(1)
= 12 - 1
= 1 - 1
= 0
g(1)
{= 2 × \left|1 - \dfrac{1}{2}\right| - 1}
{= 2 × \left|\dfrac{2 - 1}{2}\right| - 1}
{= 2 × \left|\dfrac{1}{2}\right| - 1}
= 1 - 1
= 0
{f(1) = g(1)}
When a = 2
f(1)
= 22 - 2
= 4 - 2
= 2
g(1)
{= 2 × \left|2 - \dfrac{1}{2}\right| - 1}
{= 2 × \left|\dfrac{4 - 1}{2}\right| - 1}
{= 2 × \left|\dfrac{3}{2}\right| - 1}
= 3 - 1
= 2
{f(2) = g(2)}
Thus we see that for all aA, we have
{f(a) = g(a)}
∴ Both f and g are equal.
Justification: aA, {f(a) = g(a)}
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

Let R be the relation that contains (1, 2) and (1, 3).
To be reflexive on the set A = {1, 2, 3}, R must include (1, 1), (2, 2) and (3, 3)
To be symmetric on the set A = {1, 2, 3}, R must include (2, 1) and (3, 1) (∵ (1, 2) ∈ R, (2, 1) should also belong to R, also as (1, 3) ∈ R, (3, 1) should also belong to R so that R is symmetric)
The only possible mappings that R does not include so far are (2, 3) and (3, 2).
However, adding any one of these mappings to R i.e. (2, 3) or (3, 2), it will break down the symmetry. To retain symmetry, we should either add both of them or not add any of them.
However, adding both of them would make the relation R transitive.
So, only when R = {(1, 2), (1, 3), (1, 1), (2, 2), (3, 3), (2, 1), (3, 1)}, the relation R satisfies the condition of being both reflexive and symmetric but not transitive.
So, the correct option is A.
17. Let A = {1, 2, 3}. Then number of equivalence relations containing (1, 2) is
(A) 1
(B) 2 ✔
(C) 3
(D) 4

A relation R on the set A = {1, 2, 3} will be an equivalent relation if it is
i.
Reflexive
ii.
Symmetric
iii.
and Transive
To be reflexive on the set A, it must include the mappings (1, 1), (2, 2) and (3, 3)
To be symmetric while containing (1, 2), it must also include (2, 1)
So, when R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}, we can see that it is transitive too.
So, this is the smallest mapping that R must satisfy to be an equivalent relation.
The other elements that we can include in R are (1, 3), (3, 1), (2, 3), (3, 2)
If we add (1, 3) to R, then we must also add (3, 1) so that it remains symmetric. We should also add (2, 3) so that it continues to be transitive.
However, adding just (2, 3) will make R no more symmetric. So, we should also add (3, 2) so that it continues to be symmetric.
To summarise, the next combination of mappings to be added to R to keep it equivalent is to add all the four remaining mappings (1, 3), (3, 1), (2, 3) and (3, 2).
Just to re-iterate, we should add all of them to the relation R. If we just add either 1 or 2 or 3 of them, R will no longer be equivalent. So, we should add all of them.
So, R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2)} is the next possible combination for R so that it remains as an equivalent relation. As, this is the largest combination that we can attain (R is the universal set in this case) for the relation R, the total number of possible mappings, which are equivalent, from A to A is 2. These are repeated below for your reference.
R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}
R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2)}
So, the option B is the correct answer.
18. Let f : R → R be the Signum Function defined as
f(x) = \begin{cases}1 & x \gt 0\\0 & x = 0\\-1 & x \lt 0\end{cases}
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]?
The functions f : R → R and g : R → R are defined as
f(x) = \begin{cases}1 & x \gt 0\\0 & x = 0\\-1 & x \lt 0\end{cases}
and {g(x) = [x]}
The coincidence of fog and gof need to be verified in the domain (0, 1]
{⇒ x \gt 0} and {x \leq 1}
Now,
fog(x)
{= f(g(x))}
{= \begin{cases}f(0) & \text{when }0 \lt x \lt 1\\f(1) & \text{when }x = 1\end{cases}}
{= \begin{cases}0 & \text{when }0 \lt x \lt 1\\1 & \text{when }x = 1\end{cases}}
and
gof(x)
{= g(f(x))}
{= g(1)} {(∵ x \gt 0}, {f(x) = 1)}
= [1]
= 1
19. Number of binary operations on the set {a, b} are
(A) 10
(B) 16 ✔
(C) 20
(D) 8

The binary operation * on the set {a, b} is given by
{ * : \{a, b\} × \{a, b\} → \{a, b\}}
Now, the cross product of the given set will be equal to
{\{a, b\} × \{a, b\} = \{(a, a)}, {(a, b)}, {(b, a)}, {(b, b)\}}
As we know, the number of elements in the cross product are 2 × 2 = 4
Now, the binary operation * can be re-defined as
{* : \{(a, a), (a, b), (b, a), (b, b)\} → \{a, b\}}
The following are the various possibilities in which the binary operation * can be re-defined
{ \left.\begin{aligned}* (a, a) = a\\* (a, a) = b \end{aligned}\right\} = \text{2 possibilities}}
{ \left.\begin{aligned}* (a, b) = a\\* (a, b) = b \end{aligned}\right\} = \text{2 possibilities}}
{ \left.\begin{aligned}* (b, a) = a\\* (b, a) = b \end{aligned}\right\} = \text{2 possibilities}}
{ \left.\begin{aligned}* (b, b) = a\\* (b, b) = b \end{aligned}\right\} = \text{2 possibilities}}
The total number of possibilities are
2 × 2 × 2 × 2 = 24 = 16
For, instance, one of the possibilities of defining the binary operation * will be
{* (a, a) = a}, {* (a, b) = a}, {* (b, a) = b}, {* (b, b) = a}
The other possible combination is {* (a, a) = b}, {* (a, b) = a}, {* (b, a) = a}, {* (b, b) = a}
Similary, we can have 14 more ways (overall, a total of 16 ways) of defining the binary operation *
For your information, so far in this chapter, we have seen only the Signum function that has a maximum of 3 possibilities. In this case, we're dealing with all the possibilities for the binary operation *
So, the option B is the correct answer.
To be specific, the number of binary operations on a two element set are {2^{2^2}}
Similarly, the number of binary operations on a three element set are {3^{3^2}}
So, in general, the number of binary operations on an n element set are {n^{n^2}}