CHSE +2 1st Year Math Chapter 3 Relations and Functions Question Answer

Complete question answer guide for CHSE Odisha +2 First Year Elements of Mathematics Chapter 3 (Relations & Functions). Find detailed solutions for Exercise 3(a), 3(b), and 3(c). plus two first year elements of mathematics chapter 3 question answer, elements of mathematics class 11 chapter 3 relations and functions,class 11 relations and functions elements of mathematics solutions, chse class 11 math chapter 3 relations and functions solutions , elements of mathematics class 11 chapter 3 exercise 3a solutions , elements of mathematics class 11 chapter 3 exercise 3b solutions ,  chse +2 1st year math relations and functions question answer, relations and functions class 11 domain and range problems.

Relations and Functions

Order Pairs

An ordered pair consists of a pair of objects, or elements or numbers or functions in order.
We denote order pairs as (a, b)

• An order pair is not a set of two objects.
• (a, b) = (c, d) ⇒ a = c and b = d
• (a, b) ≠ (b, a)

Cartesian Product Of Sets:

If A and B are non-empty sets, then their Cartesian product, denoted by A × B and defined by A × B = {(a, b): a ∈ A, b ∈ B} = Set of all ordered pairs (a,b) where a ∈ A and b ∈ B
Note:
1. For finite sets A and B |A × B| = |A| . |B|
2. A × B = Φ ⇔ A = Φ or B = Φ
3. A2 = A × A

Properties of Cartesian product:

1. A × B ≠ B × A (Cartesian product is non-commutative)
2. A × (B ∪ C) = (A × B) ∪ (A × C)
3. A × (B ∩ C) = (A × B) ∩ (A × C)
4. A × B = B × A ⇔ A = B
5. A × (B – C) = ( A × B) – (A × C)
6. A ⊂ B ⇒ A × A ⊂ (A × B) ∩ (B × A)
7. A ⊂ B ⇒ A × C ⊂ B × C
8. A ⊂ Band C ⊂ D ⇒ A × C ⊂ B × D
9. (A × B) ∩ (C × D) = ( A ∩ C) × (B ∩ D)

Relation
Let A and B be two arbitrary sets. A binary relation from A to B is a subset of A × B.
OR f is a relation from A to B if f ⊆ A × B
Note:

• If a of A is related to b of B by relation ‘f’ then we write (a,b) ∈ f or a f b
• As Φ ⊂ A × B we have Φ is a relation from A to B. This relation is known as a null of void or empty relation.
• As A × B ⊆ A × B, A × B is also a relation from A to B. This relation is known as universal relation.
• If |A| = m and |B| = n then number of relations from A to B is 2mn

Domain, co-domain, and Range of a relation:
Let f is a relation from A to B. Domain of f = Dom (f) or Df
={x ∈ A : (x, y) ∈ f for some y ∈ B) Co-domain of f = B
Range of ‘f’ = Rng (f) or Rf = {y ∈ B : (x, y) ∈ f for some x ∈ A}

Types Of Relation:
(a) One-many relation: A relation f from A to B is one many if (a, b) and (a, b’) ∈ f ⇒ b ≠ b’
(b) Many-one relations: A relation f from A to B is many-one if (a, b) and (a’, b) ∈ f ⇒ a ≠ a’
(c) One-one relation: A relation f from A to B is one-one if (a, b), (a, b’) ∈ f ⇒ b = b’ and (a, b), (a’, b) ∈ f ⇒ a = a’

Inverse of a relation: 

Let f is a relation from A to B. The inverse of f is denoted by f-1 is a relation from B to A defined as f-1 = {(b, a): (a, b) ∈ f}

Function:
A relation ‘f’ from X to Y is called a function if:
(a) Df = Dom (f) = X and
(b) (x, y) and (x, z) ∈ f ⇒ y = z or A relation from A to B is a function
if ⇒ Domain of f = X i.e All elements of X is engaged in the relation and
⇒ f is not one many.

Note:
(1) If a relation f from X to Y becomes a function then we write f: X → Y.
(2) If f is a function from A to B i.e. f: X → Y and (x, y) ∈ f then we write y = f(x)
(3) Mapping, map, transformation, transform, operator, and correspondence are different synonym terms of function.
(4) If f: X → Y is defined as y = f(x), then

• y is called the value of the function at x or the image of x under f or the dependent variable.
• x is called the independent variable or pre-image of y under f.

Domain, Co-domain or Range of a function:
Let f: X → Y defined as y = f(x)
(a) Domain of ‘f’ = Dom f = Df = {x ∈ X: y = f(x)}
(b) Range of f = Rng f = Rf = f(A) = {f(x) ∈ Y: x ∈ A } Clearly f(A) ⊆ y
(c) If |A| = m, |B| = n then number of functions from A to B = nm

Real valued function :
A function f: A → B is a real-valued function if B ⊆ R.
→ f is a real function if A ⊆ R and B ⊆ R

Techniques to find Domain and Range of a Real function:
(a) Techniques to find Domain: Let the function is defined as y = f(x).
Step -1: Check the values of x for which f(x) is well defined.
Step -2: The set of all values obtained from step -1 is the domain of ‘f.

(b) Techniques to find range: Let the function is y = f(x)

• Method-1 (By inspection):
  → Step -1: Get values of y for all values x ∈ Dom f.
  → Step -2: Set of all these values of y = Rng f
• Method-2:
  → Step -1: Write x in terms of y
  → Step -2: Get values of y for which x is well defined in Dom f.
  → Step -3: Rng (f) = The set of all y obtained from step 2.

Some Real Functions:
(a) Constant function: A function f: A → R defined as f(x) = k, for some k ∈ R is called a constant function.

(b) Identity function: Let A ⊆ R. The function f: A → A defined as f(x) = x, x ∈ A is called the identity function on A. We denote it by IA

(c) Polynomial function: A function f: A → R defined by f(x) = f(x) = a0 + a1x + a2x2 + anxn where a0, a1, a2, ….., an are real numbers and an ≠ 0 is called a polynomial function (polynomial) of degree n.

(d) Rational function: A function of form f(x) = P(x)Q(x) where P(x) and Q(x) are polynomial functions of x is known as a rational function.

(e) Absolute value function OR modulus function: The function f: R → R defined as  f(x) = |x| = {x,−x,x≥0x<0 is called as the modulus function.
→ Rng f = [ 0, ∞] = R+U {0}

Properties Of Modulus Function:
1. For any real number x, we have x2−−√=|x|
2. If a and b are positive real numbers then

• x2 ≤ a2 ⇔ |x| ≤ a
• x2 ≥ a2 ⇔ |x| ≥ a
• a2 ≤ x2 ≤ b2 ⇔ a ≤ |x| ≤ b ⇔ x ∈ [-b, – a] ∪ [a, b]

(f) Signum function: The function f: R → R defined as f(x) = {x|x|,0,x≠0x=0 is called signum function.
→ We denote a signum function as f(x) = sgn(x)
→ Range of a signum function = {-1, 0, 1}

(g) Greatest integer function: The function f: R → R defined by f(x) = [x] is called the greatest integer function. [x] = The greatest among all integers ≤ x. OR [x] = n for n ≤ x < n + 1

Properties of the greatest integer function :
Let n is an integer and x is a real number between n and n + 1
(i) [-n]= -[n]
(ii) [x + k] = [x] + k (for an integer ‘k’)
(iii) [-x] = -[x] – 1
(iv) [x] + [-x] = {−1,0,x∉Zx∈Z
(v) [x] – [-x] = {2[x]+1,2[x],x∉Zx∈Z
(vi) [x] ≥ k ⇒ x ≥ k for k ∈ Z
(vii) [x] ≤ k ⇒ x < k +1 for k ∈ Z
(viii) [x] > k ⇒ x > k + 1 for k ∈ Z
(ix) [x] < k ⇒ x < k for k ∈ Z

(h) Exponential Function: A function f: R → R defined as f(x) = ax where a > 0 and a ≠ 1 is called the exponential function.

Properties Of Exponential Function:
1. ax+y =  ax . ay
2. (ax)y = axy
3. ax = 1 if x = 0
4. If a > 1, ax > ay ⇒ x > y
5. If a < 1, ax > ay ⇒ x < y

Logarithmic Function:
Let a ≠ 1 is a positive real number. The function f: (0, ∞) → R defined by f(x) = logax is called the logarithmic function, where y = logax  ⇔ ay = x
→ Domain of a logarithmic function = (0, ∞) and Range = R

Properties of logarithmic function:
1. loga (xy) = logax + logay
2. loga (x/y) = logax – logay
3. logaa = 1
4. loga(x)y = y logax
5. loga x = 0 ⇔ x = 1
6. logax = logaa , x ≠ 1
7. logab = logab
8. logan(xm) = mn loga|x|

Different Categories of function:

(a) Algebraic Function: A function that can be generated by a variable by a finite number of algebraic operations such as addition, subtraction, multiplication, division, square root, etc. is called an algebraic function (b) Transcendental function: A non-algebraic function is a transcendental function.

⇒ Trigonometric, trigonometric, Exponential, and logarithmic functions are transcendental functions.

Even And Odd Functions:
A function ‘f’ is an even function  if f(-x) = x and is an odd function
if f(-x) = x and is an odd function: if f(-x) = -f(x)
Note:
1. If ‘f’ is any function f(x) + f(-x) is always an even function and f(x) – f(-x) is an odd function.
2. Every function f(x) can be expressed as the sum of an even and an odd function as f(x) = g(x) + h(x), where
g(x) = f(x)+f(−x)2
h(x) = f(x)−f(−x)2

Periodic Function:
A function is called a periodic function with period k if f(x + k) = f(x) for some constant k ≠ 0. The least positive value of k for which f(x + k) = f(x) holds is called the fundamental period of f.

Properties of periodic function :
(1) If k is the period of f then any non-zero integral multiples of k is also a period of f.
(2) If k is the period of f(x) then f(ax + b) is also periodic with period ka
(3) If f1(x) + f2(x) and f3(x) are periodic functions with periods k1, k2, k3, respectively then the function a1f1(x) + a2f2(x) + a3f3(x) is also a periodic function with period, LCM (k1, k2, k3)

Algebra Of Real functions:
(a) Equality of two functions: Two functions f and g are equal iff ‘
(i) Dom f = Dom g
(ii) Co-Dom f = Co-Dom g
(iii) f(x) = g(x) for all x belonging to their common domain.

(b) Addition of two functions: Let f: D1 → R and g: D2 → R be two real functions.
The sum function f + g is defined by f + g: D1 ∩ D2 → R and (f + g)(x) = f(x) + g(x) ∀ x ∈ D1 n D2

(c) Subtraction of two functions: Let f: D1 → R and g: D2 → R. The difference function (f – g) is f – g: D1 ∩ D2) → R defined by (f – g) (x) = f(x) – g(x) ∀ x ∈ D1 ∩ D2

(d) Scalar multiplication: Let f: D → R and c is any scalar. The scalar multiple of f by the scalar c is cf: D → R defined as (cf)(x) = c. f(x) ∀ x ∈ D1.

(e) Multiplication of two functions: Let f: D1 → R and g: D2 → R are two real functions. The product function (fg) is (fg): D1 ∩ D2 → R defined as (fg)(x) = f(x)g(x) ∀ ∈ D1 ∩ D2

(f) The quotient of two functions: Let f: D1 → R and g: D2 → R are two real functions. the quotient function (fg) i,e,. fg: D1 ∩ D2 → R, defined by (fg)(x) = f(x)g(x), ∀ x ∈ D1 ∩ D2

Relations And Functions Ex 3(a)

1. Compute the product A × B when
(i) A = {0} = B
(ii) A = {a, b}, B = {a, b, c}
(iii) A = Z, B = Φ
Solution:
(i) A = {0} = B
∴ A × B = {(0, 0)}

(ii) A = {a, b}, B = {a, b, c}
∴ A × B = {(a, a), (a, b), (a, c), (b, a), (b, b), (b, c)}

(iii) A = Z, B = Φ ∴ AxB = Φ

2.If |A| = m, |B| = n, what can you say about
(i) |A × B| (ii) |P(A) × P(B)|
Solution:
If |A| = m. |B| = n then

(i) lA × B| = mn.

(ii) |P(A)| = 2m . |P(B)| = 2n
∴  |P(A) × P(B)| =2m × 2n = 2m+n

3 . Find x, y if

(i) (x, y) = (-3, 2)

(ii) {x + y, 1) = (1, x – y)

(iii) (2x + y, 1) = (x, 2x + 3y)

Solution:

(i) ∴ x = – 3, y = 2

(ii) ∴ x + y = 1, x – y = 1
∴ 2x = 2 or, x = 1
∴ y=0

(iii) ∴ 2x + y = x, 1 = 2x + 3y
∴ {x + y = 0} × 2
2x + 3y = 1
–      –      –
∴ – y = – 1 or, y = 1
∴ x = – 1

4. If, A × B = B × A then what can you
Solution:

If A × B = B × A then A = B

5. |A × B| = 6. If ( -1, y ), (1, x), (0, y) are in A × B. Write other elements in A × B, where x ≠ y.
Solution:
Let |A × B| = 6 and (-1, y) (1, .x) (0, y) ∈ A × B
⇒ -1, 1, 0 ∈ A and x, y ∈ B
As |A × B| = 6 and 3 × 2 = 6
We have A = {-1, 1, 0} and B = {x, y}
Thus other elements of A × B is (-1, x) , (1, y), (0, x)

Relations And Function Ex 3(b)

1.  Let A = {a, b, c }, |B| = {1, 2}
(a) Determine all the relations from A to B and determine the domain, range, and inverse of each relation.
(b) Determine all the relations from B to A.
(c) Is there any relationship that is both a relation from A to B and B to A? How many?
(d) Of all the relations from A to B, identify which relations are many ones, one-many, and one-one and represent this diagrammatically.
Solution:
(a) A = {a, b, c}, B = {1, 2}
∴ A × B = {(a, 1), (a, 2), (b, 1), (b, 2), (c, 1), (c, 2)}
∴ |A × B| = 6
∴ |P(A × B)| = 26 = 64
∴ There are 64 relations from A to B as any subset of A × B. The domain of these relations is any subset of A. The inverse of these relations is any subset of B × A.
(b) There are 64 relations from B to A as any sub-set of B x A is a relation from B to A.
(c) Φ is the only relation that is from A to B and from B to A.
(d) Some many-one relations are {(a, 1), (b, 1), (c, 1), (b, 2) (c, 2)}, {(a, 2), (b, 2), (c, 2)}.

2.
Are the following sets related?
(i) Φ from A to B.
(ii) A × B from A to B.
(iii) A × Φ from A to Φ.
(iv) Φ × B from Φ to B.
(v) Φ × Φ from Φ to Φ.
(vi) Φ × C from A to B.
(vii) Φ × Φ from A to B.
Determine the domain range and inverse of each of the relations mentioned above
Solution :
(i) Φ from A to B is a relation.
(ii) A × B from A to B is a relation.
(iii) A × Φ from A to Φ is a relation.
(iv) Φ × B from Φ to B is a relation.
(v) Φ × Φ from Φ to Φ is a relation.
(vi) Φ × C from A to B is a relation.
(vii) Φ × Φ from A to B is a relation.
∴ Domain of Φ i.e. DΦ = Φ
Range of Φ i.e., RΦ = Φ
Similarly, DA × B = A, RA × B = β
DA × Φ = Φ, RA × Φ = Φ
D Φ × B = Φ = Φ, R Φ × B = Φ
D Φ × Φ = Φ, R Φ × C = Φ
D Φ × C = Φ, R Φ × C = Φ
D Φ × Φ = Φ, R Φ × Φ = Φ
The inverse of the above relations is Φ, B × A, Φ × A, B × Φ, Φ × Φ, C ×  Φ, and Φ × Φ respectively.

3.
Express the following relations on A to B in each case in tabular form :
(i) A = {n ∈ N : n ≤ 10}, B = N
f = {(x, y) ∈ A × B : y = x2}
Solution:
A = {n ∈ N : n ≤ 10}
= {1, 2, 3,…..10}, B = N
∴ B = {1, 2, 3}
∴ f ={(x, y) ∈ A × B : y = x2}
= {(1, 1), (2, 4), (3, 9)…..(10, 100)}

(ii) A = B = R
∴ f = {(x, y) : x2 + y2 = 1 and |x – y| = 1}
Solution:
A = B = R
∴ f = {(x, y) : x2 + y2 = 1 and |x – y| = 1}
={(0, 1) (1, 0), (-1, 0), (0, -1)}

(iii) (1, 2, 3, 4), B = {1, 2, 3, 4, 5}
f = {x, y) : 2 divides 3x+y}
Solution:
A = {1, 2, 3, 4}, B = {1, 2, 3, 4, 5}
∴ f = {(x, y) : 2 divides 3x+y}
={(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (3, 1), (3, 3), (3,5), (4, 2), (4, 4)}

 4. A and B are non-empty sets such that |A| = m, |B| = n. How many relations can be defined from A to B ? (Remember that the number of relations is the number of subsets of (A × B).
Solution:
|A| = m, |B| = n
⇒ |A × B| = mn
A relation is a subset of A to B
∴ Number of relations from A to B
= Number of subsets of A × B
= 2mn (∴ |A × B| = mn)

5. Give an example of a relation f such that
(i) dom f – rng f (ii) dom f ⊂ rng f
(iii) dom f ⊃ rng f
(iv) f ∪ f-1 = Φ
(v) f = f-1
(vi) f ∪ f-1 ≠ Φ
Solution:
Let A = { 1, 2, 3} = B
(i) Let f = {(x, y) ∈ A × B : x = y}
∴ Dom f = {1, 2, 3} = Range f

(ii) Let f = {(1, 1), (1, 2), (2, 3)}
on A = (1, 2, 3}
∴ Dom f = {1, 2} ⊂ { 1, 2, 3} = Range f

(iii) Do yourself

(iv) Let f = Φ
∴ f-1 = Φ = f ∪ f-1 = Φ

(v) Let f = {(x, y) ∈ A × B; x2 + y2 = 1}, where A = B = {1, – 1, 0}
= {(1, 0), (0, 1), (-1, 0), (0, -1)}
f-1 = {(0, 1) (1, 0), (0, -1), (-1, 0)}
=f

(vi) Let f = {(1, 3), (3, 1)} on A = { 1, 2, 3}
∴ f-1 = {3, 1), (1, 3)},
so that f ∩ f-1 = Φ.

6. Let R = {(a, a3) I a is a prime number less than 10}
Fine (i) R, (ii) dom R, (iii) rng R (iv) R-1 (v) dom R-1 (vi) rng R-1
Solution:
R = {(a, a3)} a is a prime number less than 10}
(i) R = {(2, 8), (3, 27), (5, 125), (7, 343)}
(ii) dom R = {2, 3, 5, 7}
(iii) rng R = {8, 27, 125, 343}
(iv) R-1 = {(8, 2), (27, 3), (125, 5), (343, 7)}
(v) Dom R-1 = {8, 27, 125, 343} = rng R
(vi) rng R-1 = {2, 3, 5, 7} = dom R

7. Let A = {1, 2, 3, 4, 5, 6} and Let R be a relation on A defined by R = {(a, b)} a divides b
Find (i) R, (ii) dom R, (iii) rng R (iv) R-1, (v) Dom R-1 (vi) rng R-1
Solution:
A = {1, 2, 3, 4, 6}
R on A is defined by
R = {(a, b) | a divides b}
(i) R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6) (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)}
(ii) dom R = {1,2, 3, 4, 6} = A
(iii) rng R = {1, 2, 3, 4, 6} = A
(iv) R-1 = {(1, 1), (2, 1), (3, 1), (4, 1), (6, 1), (2, 2) (4, 2), (6, 2), (3, 3), (6, 3), (4, 4), (6, 6)}
(v) dom R-1 = {1, 2, 3, 4, 6} = A
(vi) rng R-1 = {l, 2, 3, 4, 6} = A

Relations And Functions Exercise 3(c)

1. Give an example of a relation that is not a function.
Solution:
Let the relation ‘f’ be defined from
A = {1, 2, 3} to B = {1, 2} as
f = {(1, 1), (1, 2), (2, 1), (3, 2)}
f is not a function because f (1) = 1,
f(1) = 2 but 1 ± 2 i.e. f is one many.

2. If X and Y are sets containing m and n elements respectively then what is the total number of functions from X to Y?
Solution:
If |X| = m and |Y| = n then the number of functions from X to Y = |Y||X| = nm

3Find the domain of the following functions:

(i)9−x2−−−−−√
Solution:
y = f(x) = 9−x2−−−−−√
Clearly, f(x) is well-defined for 9 – x2 ≥ 0 ⇒ x2 ≤ 32 ⇒ – 3 ≤ x ≤ 3
∴ dom f = [ – 3, 3]

(ii) x1+x
Solution:
y = f(x) = x1+x
f(x) is well defined for all x ∈ R
∴ dom f = R

(iii) 1 – |x|
Solution:
y = f(x) = 1 – |x|
f(x) is well defined for all x ∈ R
∴ dom f = R

(iv) 1x2−1
Solution:
y = f(x) = 1x2−1
f(x) is well defined for x2 – 1 ≠ 0
⇒ x2 ≠ 1
⇒ x2 ≠ ± 1
∴ dom f = R – { -1, 1}

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