Question 1. Show that the function f : R → {x ∈ R : – 1 < x < 1} defined by f(x) = [Tex]\frac{x}{1+|x|} [/Tex], x ∈ R is one one and onto function.
Solution:
As, it is mentioned here
f : R → {x ∈ R : – 1 < x < 1} defined by [Tex]f(x) = \frac{x}{1+|x|} [/Tex], x ∈ R
As, we know f is invertible, if and only if f is one-one and onto.
ONE-ONE
For the pair of number, we will deal with three cases:
Case 1: When both numbers p and p are positive numbers.
The function f is defined as
Case 1: When both numbers p and q are positive numbers.
[Tex]f(p) = \frac{p}{1+|p|}[/Tex]
[Tex]f(q) = \frac{q}{1+|q|}[/Tex]
f(p) = f(q)
[Tex]\frac{p}{1+|p|} = \frac{q}{1+|q|}[/Tex]
[Tex]\frac{p}{1+p} = \frac{q}{1+q}[/Tex]
p(1+q) = q(1+p)
p = q
Case 2: When number p and q are negative numbers.
[Tex]f(p) = \frac{p}{1+|p|}[/Tex]
[Tex]f(q) = \frac{q}{1+|q|}[/Tex]
f(p) = f(q)
[Tex]\frac{p}{1+|p|} = \frac{q}{1+|q|}[/Tex]
[Tex]\frac{p}{1-p} = \frac{q}{1-q}[/Tex]
p(1-q) = q(1-p)
p = q
Case 3: When p is positive and q is negative
[Tex]f(p) = \frac{p}{1+|p|}[/Tex]
[Tex]f(q) = \frac{q}{1+|q|}[/Tex]
f(p) = f(q)
[Tex]\frac{p}{1+|p|} = \frac{q}{1+|q|}[/Tex]
[Tex]\frac{p}{1+p} = \frac{q}{1-q}[/Tex]
p(1-q) = q(1+p)
p + q = 2pq
Here, RHS will be negative and LHS will be positive. Hence, the case 3 result is impossible.
So, the function f is one-one, for case 1 and case 2.
ONTO
Case 1: When p>0.
[Tex]f(p) = \frac{p}{1+|p|}[/Tex]
[Tex]y = \frac{p}{1+p}[/Tex]
[Tex]p = \frac{y}{1-y} (y≠1)[/Tex]
Case 2: When p <0
[Tex]f(p) = \frac{p}{1+|p|}[/Tex]
[Tex]y = \frac{p}{1-p}[/Tex]
[Tex]p = \frac{y}{1+y} (y≠-1)[/Tex]
Hence, p is defined for all the values of y, p∈ R
Hence f is onto.
As, f is one-one and onto. This f is an invertible function.
Question 2. Show that the function f : R → R given by f(x) = x3 is injective.
Solution:
As, it is mentioned here
f : R → R defined by f(x) = x3, x ∈ R
To prove f is injective (or one-one).
ONE-ONE
The function f is defined as
f(x) = x3
f(y) = y3
f(x) = f(y)
x3 = y3
x = y
The function f is one-one, so f is injective.
Question 3. 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.
Solution:
Given, A and B are the subsets of P(x), A⊂ B
To check the equivalence relation on P(X), we have to check
As, we know that every set is the subset of itself.
Hence, A⊂ A and B⊂ B
ARA and BRB is reflexive for all A,B∈ P(X)
As, it is given that A⊂ B. But it doesn’t make sure that B⊂ A.
To be symmetric it has to be A = B
ARB is not symmetric.
When A⊂ B and B⊂ C
Then of course, A⊂ C
Hence, R is transitive.
So, as R is not symmetric.
R is not an equivalence relation on P(X).
Question 4. Find the number of all onto functions from the set {1, 2, 3, … , n} to itself.
Solution:
Onto function from the set {1,2,3,…..,n} to itself is just same as the permutations of n.
1×2×3×4×…….×n
Which is n!.
Question 5. 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 [Tex]g(x) = 2|x-\frac{1}{2}|-1 [/Tex] 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).
Solution:
Given, f, g : A → B be functions defined by f(x) = x2 – x, x ∈ A and g(x) = [Tex]2|x-\frac{1}{2}|-1 [/Tex] x ∈ A
At x = -1
f(0) = (-1)2 – (-1) = 2
g(0) = [Tex]2|-1-\frac{1}{2}|-1 [/Tex] = 2
Here, f(-1) = g(-1) and 2=2
At x = 0
f(0) = 02 – 0 = 0
g(0) = [Tex]2|0-\frac{1}{2}|-1 [/Tex] = 0
Here, f(0) = g(0) and 0=0
At x = 1
f(1) = 12 – 1 = 0
g(1) = [Tex]2|1-\frac{1}{2}|-1 [/Tex] = 0
Here, f(1) = g(1) and 1=1
At x = 2
f(1) = 22 – 2 = 2
g(1) = [Tex]2|2-\frac{1}{2}|-1 [/Tex] = 2
Here, f(2) = g(2) and 2=2
For, every c∈ A, f(c) = g(c)
Hence, f and g are equal functions.
Question 6. 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:
R = {(1,1),(1,2),(1,3),(2,1),(2,2),(3,1),(3,3)}
Reflexive : (1,1), (2,2), (3,3) ∈ R
Symmetric: (1,2), (2,1)∈ R and (1,3), (3,1) ∈ R
R is not Transitive because, (1,2), (1,3) ∈ R but (3,2) ∉R
So, if we will add (3,2) and (2,3) or both, then R will become transitive.
New, R = {(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)}
Hence, A is the correct option.
Question 7. Let A = {1, 2, 3}. Then number of equivalence relations containing (1, 2) is
(A) 1
(B) 2
(C) 3
(D) 4
Solution:
Smallest equivalence relations containing (1, 2):
R = {(1,1),(2,2),(1,2),(2,1),(3,3)}
or R = {(1,1),(2,2),(1,2),(2,1),(3,2),(2,3)(3,3)}
Hence, B is the correct option.
Deleted Questions
Let f : R → R be defined as f(x) = 10x + 7. Find the function g : R → R such that g o f = f o g = 1R.
Solution:
As, it is mentioned here
f : R → R be defined as f(x) = 10x + 7
To, prove the function one-one
Let’s take f(x) = f(y)
10x + 7 = 10y + 7
x = y
Hence f is one-one.
To, prove the function onto
y ∈ R, y = 10x+7
[Tex]x = \frac{y-7}{10} ∈ R[/Tex]
So, it means for y ∈ R, there exists [Tex]x = \frac{y-7}{10}[/Tex]
[Tex]f(x) = f(\frac{y-7}{10}) = 10(\frac{y-7}{10})+7 = y – 7 + 7 = y[/Tex]
Hence f is onto.
As, f is one-one and onto. This f is invertible function.
Let’s say g : R → R be defined as [Tex]g(y) = \frac{y-7}{10}[/Tex]
[Tex]g o f = g(f(x)) = g(10x+7) = \frac{(10x+7)-7}{10} = \frac{10x}{10} = x[/Tex]
[Tex]f o g = f(g(x)) = f(\frac{x-7}{10}) = 10(\frac{x-7}{10})+7 = x – 7 + 7 = x[/Tex]
Hence, g : R → R such that g o f = f o g = 1R.
g : R → R is defined as [Tex]g(y) = \frac{y-7}{10}[/Tex]
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:
The function f is defined as
[Tex]f(n)= \begin{cases} n-1, \hspace{0.2cm}n \hspace{0.2cm}is\hspace{0.2cm} odd\\ n+1,\hspace{0.2cm}n \hspace{0.2cm}is\hspace{0.2cm} even \end{cases}[/Tex]
As, we know f is invertible, if and only if f is one-one and onto.
ONE-ONE
For the pair of number, we will deal with three cases:
Case 1: When both numbers p and q are odd numbers.
f(p) = p-1
f(q) = q-1
f(p) = f(q)
p-1 = q-1
p – q = 0
Case 2: When both numbers p and q are even numbers.
f(p) = p+1
f(q) = q+1
f(p) = f(q)
p+1 = q+1
p – q = 0
Case 3: When p is odd and q is even
f(p) = p-1
f(q) = q+1
f(p) = f(q)
p-1 = q+1
p – q = 2
Subtracting an odd number and even always gives a odd number, not even. Hence, the case 3 result is impossible.
So, the function f is one-one, for case 1 and case 2 only.
ONTO
Case 1: When p is odd number
f(p) = p-1
y = p-1
p = y+1
Hence, when p is odd y is even.
Case 2: When p is even number
f(p) = p+1
y = p+1
p = y-1
Hence, when p is even y is odd.
So, it means for y ∈ W, there exists p = y+1 and y-1 for odd and even value of p respectively.
Hence f is onto.
As, f is one-one and onto. This f is an invertible function.
Let’s say g : W → W be defined as [Tex]g(y)= \begin{cases} y-1, \hspace{0.2cm}y \hspace{0.2cm}is\hspace{0.2cm} odd\\ y+1,\hspace{0.2cm}y \hspace{0.2cm}is\hspace{0.2cm} even \end{cases}[/Tex]
f = g
Hence, The inverse of f is f itself
If f : R → R is defined by f(x) = x2– 3x + 2, find f (f(x)).
Solution:
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
f(f(x)) = x4 – 6x3 + 10x2 – 3x
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 |).
Solution:
Two functions, f : N → Z and g : Z → Z
Taking f(x) = x and g(x) = |x|
Let’s check, whether g is injective or not
g(5) = |5| = 5
g(-5) = |-5| = 5
As, we can see here that
Taking two integers, 5 and -5
g(5) = g(-5)
but, 5 ≠-5
So, g is not an injective function.
Now, g o f: N → Z is defined as
g o f = g(f(x)) = g(x) = |x|
Now, as x,y∈ N
g(x) = |x|
g(y) = |y|
g(x) = g(y)
|x| = |y|
x = y (both x and y are positive)
Hence, g o f is an injective.
Give examples of two functions f : N → N and g : N → N such that g o f is onto but f is not onto.
(Hint : Consider f(x) = x + 1 and [Tex]g(x)= \begin{cases} x-1, \hspace{0.2cm}x>1\\ 1,\hspace{0.2cm}x=1 \end{cases}[/Tex]
Solution:
Two functions, f : N → N and g : N → N
Taking f(x) = x+1 and [Tex]g(x)= \begin{cases} x-1, \hspace{0.2cm}x>1\\ 1,\hspace{0.2cm}x=1 \end{cases}[/Tex]
As, f(x) = x+1
y = x+1
x = y-1
But, when y=1, x = 0. Which doesn’t satiny this relation f : N → N.
Hence. f is not an onto function.
Now, g o f: N → N is defined as
g o f = g(f(x)) = g(x+1)
When x+1=1, we have
g(x+1) = 1 (1∈ N)
And, when x+1>1, we have
g(x+1) = (x+1)-1 = x
y = x, which also satisfies x,y∈ N
Hence, g o f is onto.
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:
Given, P(X) × P(X) → P(X) is defined as A*B = A∩B ∀ A, B ∈ P(X)
This implies, A⊂ X and B ⊂ X
So, A∩X = A and B∩X = B ∀ A, B ∈ P(X)
⇒ A*X = A and B*X = B
Hence, X is the identity element for intersection of binary operator.
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)}
Solution:
As, F = {(a, 3), (b, 2), (c, 1)} and S = {a,b,c} and T={1,2,3}
F: S→T is defined as
F(a) = 3, F(b) = 2 and F(c) = 1
F is one-one and onto.
Taking F-1, so F-1: T→S
a = F-1(3), b = F-1(2) and c = F-1(1)
F-1 = {(3,a),(2,b),(1,c)}
(ii) F = {(a, 2), (b, 1), (c, 1)}
Solution:
As, F = {(a, 2), (b, 1), (c, 1)}
F: S→T is defined as
F(a) = 2, F(b) = 1 and F(c) = 1
Here, F(b) = F(c) but b ≠c
Hence, F is not one-one.
So, F is not invertible and F-1 doesn’t exists.
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.
Solution:
Binary operations ∗ : R × R → R defined as a ∗b = |a – b|
a*b = |a-b|
b*a = |b-a| = |-(a-b)| = |a-b|
a*b = b*a
Hence, ∗ is commutative.
Now, let’s take a=1, b=2 and c=3 for better understanding
a*(b*c) = a*|b-c| = |a-|b-c|| = |1-|2-3|| = 0
(a*b)*c = |a-b|*c = ||a-b|-c| = ||1-2|-3| = 2
a*(b*c) ≠(a*b)*c
Hence, ∗ is not associative.
Binary operations o : R × R → R defined as a o b = a, ∀ a, b ∈ R
a o b = a
b o a = b
a o b ≠b o a
Hence, o is not commutative.
a o (b o c) = a o b = a
(a o b) o c = a o c = a
a o (b o c) ≠(a o b) o c
Hence, o is associative.
Let’s check for a ∗ (b o c) = (a ∗ b) o (a ∗ c) a, b, c ∈ R
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, let’s check for a o (b * c) = (a o b) * (a o c)
a o (b * c) = a
(a o b) * (a o c) = a * a = |a-a| = 0
Hence, a o (b * c) ≠(a o b) * (a o c)
o does not distribute over ∗
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:
Set X, such that P(X) × P(X) → P(X) be defined as A * B = (A – B) ∪ (B – A), ∀ A, B ∈ P(X)
φ*A = (φ-A) U (A-φ) = φ U A = A
A*φ = (A-φ) U (φ-A) = A U φ = A
Hence, φ is the identity element for the operation * on P(X)
A*A = (A-A) U (A-A) = φ U φ = φ
⇒ A = A-1
Hence, all the elements A of P(X) are invertible with A–1 = A.
Define a binary operation ∗ on the set {0, 1, 2, 3, 4, 5} as
[Tex]a*b= \begin{cases} a+b, \hspace{0.2cm}a+b<6\\ a+b-6,\hspace{0.2cm}a+b\geq6 \end{cases}[/Tex]
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:
Let the set x = {0, 1, 2, 3, 4, 5}
Let’s take i as identity element, where a*i = a = i*a ∀ a ∈ x
a*0 = a
0*a = a, when (a+0<6)
Hence, zero is the identity for this operation
An element a ∈ x is invertible if there exists b ∈ x such that a*b = b*a = 0
[Tex]i.e.,\begin{cases} a+b=0=b+a, \hspace{0.2cm}a+b<6\\ a+b-6=0=b+a-6,\hspace{0.2cm}a+b\geq6 \end{cases}[/Tex]
From above equations, we have
a = -b or b = 6-a
But, as x = {0, 1, 2, 3, 4, 5} and a,b∈ x. Then a≠-b
Hence, b = 6-a is the inverse of an element a∈ x
a≠0
a-1 = 6-a
Let f : R → R be the Signum Function defined as
[Tex]f(x)= \begin{cases} 1, \hspace{0.2cm}x<0\\ 0,\hspace{0.2cm}x=0\\ -1\hspace{0.2cm}x>0 \end{cases}[/Tex]
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:
Given, f : R → R and g : R → R
when x ∈ (0,1]
[x] = 1, when x=1
[x] = 0, when 0<x<1
[Tex]g(x)= \begin{cases} 1, \hspace{0.2cm}x=1\\ 0,\hspace{0.2cm}0<x<1 \end{cases}[/Tex]
Now, fog(x)=f(g(x)) = f([x])
[Tex]f([x])= \begin{cases} f(1), \hspace{0.2cm}x=1\\ f(0),\hspace{0.2cm}0<x<1 \end{cases} = [/Tex] [Tex]\begin{cases} 1, \hspace{0.2cm}x=1\\ 1,\hspace{0.2cm}0<x<1 \end{cases}[/Tex]
And, Now gof(x) = g(f(x))
g(1) = [1] = 1
g(0) = [0] = 0
g(-1) = [-1] = -1
When x ∈ (0,1), fog = 0 and gof = 1. fog(1) ≠gof(1)
Hence, fog and gof do not coincide in (0, 1].
Number of binary operations on the set {a, b} are
(A) 10
(B) 16
(C) 20
(D ) 8
Solution:
Let A = {a,b}
A x A = {a,b} x {a,b}
R = {(a,a),(a,b),(b,a),(b,b)}
Number of elements are 4.
Hence, the number of binary operations on the set will be 24 = 16
Hence, B is the correct option.
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