ML Aggarwal Class 11 Maths Solutions Section A Chapter 2 Relations and Functions Ex 2.3

Effective ML Aggarwal Class 11 Solutions ISC Chapter 2 Relations and Functions Ex 2.3 can help bridge the gap between theory and application.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 2 Relations and Functions Ex 2.3

Question 1.
Which of the following relations are functions ? Give reasons. If it is a function, determine its domain and range.
(i) {(1, 1), (1, 2), (1, 3), (1, 4)}
(ii) {(a, b), (b, e), (e, d), (d, e)}
(iii) {(1, 2), (3, 1), (1, 3), (4, i)}
(iv) {(1, 2), (2, 2), (3, 2), (4, 2)}
(v) {(2, 1), (0, – 1), (3, 1), (5, 4), (- 1, 0), (3, 4), (1, 0)}
(vi) ((2, 1), (4, 2), (6, 3), (8, 4), (10, 3), (14, 7)}.
Solution:
(i) Given R = {(1, 1), (1, 2), (1, 3), (1, 4)}
Since element I has distinct images 1, 2, 3 and 4 respectively.
Thus element 1 has no unique image.
Thus given relation is not a function.

(ii) Given R = {(a, b), (b, e), (c, d), (d, e)}
Since every element has a unique image and hence R be a function,
∴ Domain of R = {a, b, e, d}
and Range of R = {b, e, d, e}

(iii) Given Relation R = {(1, 2), (3, 1), (1, 3), (4, 1)}
Since element 1 has two images 2 and 3.
Hence relation R is not a function.

(iv) Given relation R = {(1, 2), (2, 2), (3, 2), (4, 2)}
Clearly different elements have same image i.e. 2.
and no element has more than one image.
Thus given relation is a function.
∴ domain = {1, 2, 3, 4)
and Range = {2}

(v) Given relation R = {(2, 1), (0, – 1), (3, 1), (5, 4), (- 1, 0), (3, 4), (1, 0)}
Here, element 3 has two images 1 and 4.
∴ given relation is not a function.

(vi) Given relation R = {(2, 1), (4, 2), (6, 3), (8, 4), (10, 3), (14, 7)}.
Thus, different elements have unique ¡mage
and hence given relation be a function.
∴ domain = {2, 4, 6, 8, 10, 14}
and Range = {1, 2, 3, 4, 7}.

Question 2.
If X = {- 4, 1, 2, 3) and Y = {a, b, c), which of the following relations ¡s a function from X to Y?
(j) {(- 4, a), (1, a), (2, b)}
(ii) {(- 4, b), (1, b), (2, a), (3, c)}
(iii) {(- 4, a), (1, a), (2, b), (3, c), (1, b)}
Solution:
Given X = {- 4, 1, 2, 3}
and Y = {a, b, c}
(i) Let R = {(- 4, a), (1, a), (2, b)}
but element 3 has no image
∴ given relation is not a function from X to Y.

(ii) Given R = {(- 4, b), (1, b), (2, a), (3, c)}
Since every element has a unique image.
∴ given relation be a function from X to Y.

(iii) Let R {(- 4, a), (t. a), (2, b), (3, c), (1, b))
Since element 1 have two images a and b in Y
∴ R is not a function from X to Y.

Question 3.
Let A = {1, 2, 3, 4}, B = {1, 5,9, 11, 15, 16} and f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}.
Are the following true?
(i) f is a relation from A to B.
(ii) f is a function from A to B?
Justify your answer in each case.
Solution:
(i) Given A = {1, 2, 3, 4},
B = {1, 5, 9, 11, 15, 16}
andf {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}
Clearly f ⊆ A × B and hence f be a relation from A to B.

(ii) Since element 2 has two images 9 and 11.
∴ every element in A does not have unique image in B.
∴ f is not a function from A to B.

Question 4.
Is the given relation a function ? Justify your answer.
(1) f = {(x, x) : x ∈ R}
(ii) g = {(n, n²) : n is a positive integer}
(iii) h = {(x, 3) : x ∈ R)
Solution:
(i) Given relation R’ = {(x, x) : x ∈ R}
So every element of R has a unique image x in R.
Thus R’ be a function.

(ii) Given g = {(n, n²) : n is a positive integer}
since every element n ∈ N has a unique
image n² ∈ N s.t. (n, n²) ∈ g
Thus g be a function from N to N.

(iii) Given h = {(x, 3) : x ∈ R)
Since different elements of R has a unique image 3.
∴ given relation be a function.

Short and Long answer questions (5 to 24) :

Question 5.
(i) A function ‘‘ is defined by f(x) = x² + 1 where x ∈ {- 1, 0, 1, 3). List the elements of f.
(ii) Is a func(ionf is defined byf(x) = 2x – 1 where x ∈ {- 2, 0, 3, 5}, then find its range.
Solution:
(i) Given a function f is defined by
f(x) = x² + 1
where x ∈ {- 1, 0, 1, 3}
When x = – 1
∴ f(x) = (- 1)² + 1 = 2
⇒ (- 1, 2) ∈ f
When x = 0
⇒ (0, 1) ∈ f
When x = 1
∴ f(x) = 1² + 1 = 2
⇒ (1, 2) ∈ f
When x = 3
∴ f(x) = 3² + 1 = 10
⇒ (3, 10) ∈ f
∴ f = {(- 1, 2), (0, 1), (1, 2), (3, 10)}.

(ii) Given a functionf is defined by
f(x) = 2x – 1
where x ∈ {- 2, 0, 3, 5}
When x = – 2
∴ f(x) = 2 × (- 2) -1 = – 5
⇒ (- 2, – 5) ∈ f
When x = 0
∴ f(x) = 0 – 1 = – 1
⇒ (0, – 1) ∈ f
When x = 3
∴ f(x) = 6 – 1 = 5
⇒ (3, 5) ∈ f
When x = 5
∴ f(x) = 10 – 1 = 9
⇒ (5, 9) ∈ f
∴ f = {(- 2, – 5), (0,- 1), (3, 5), (5, 9)}
Thus range of f = {- 5, – 1, 5, 9}.

Question 6.
Does the adjacent arrow diagram represent a function? If so, write its range.
Solution:
Since every element of (domain) set A has a unique image in (codomain) set B.
Thus given arrow diagram represents a function.
Range = {0, 2, 7}.

Question 7.
The adjacent arrow diagram represents a relation. Represent the relation in roster form. Is this relation a function ? Give reasons for your answer.

Solution:
Since 1, 3 and 5 has image 4.
∴ (1, 4), (3, 4), (5, 4) ∈ R
elements 2 and 4 has images 1 and 3 respectively.
∴ (2, 1), (4, 3) ∈ R.
Thus, R = {(1, 4), (3, 4), (5, 4), (2, 1),(4, 3)}
Thus every element in domain has a unique image in codomain hence the given relation be a function.

Question 8.
The adjacent arrow diagram represents a relation. List the pairs that satisfy the relation. Is this relation a function ? Also find the rule (relation) for the above correspondence.

Clearly the relation of from A to B is given by {(- 1, 1), (0, 0), (1, 1), (2, 4), (- 2, 4)}
Solution:
Clearly every element in domain A has a unique image in codomain B.
∴ Given relation be a function.
Clearly f : A → B is defined by
f(x) = x² ∀ x ∈ A

Question 9.
Write the relation represented by the adjoining diagram, by listing the ordered pairs. State the domain, the codomain and the range of the relation. Is the relation a function ?

Solution:
Clearly the set of ordered pairs {(p, 3), (q, 7), (r, 7), (s, 1)}
represents a relation R since R ⊆ A × B.
domain (R) = {p, q, r s} ;
codomain (R) = {1, 3, 5, 7}
and range (R) = {1, 3, 7}

Question 10.
A = {- 2, – 1, 1, 2} and f = {(x, \(\frac{1}{x}\)) : x ∈ A}.
(i) List the domain of f.
(ii) List the range of f.
(iii) Is f a function ?
Solution:
Given A = {- 2, – 1, 1, 2}
and f = {(x, \(\frac{1}{x}\)) : x ∈ A}
∴ f = {(- 2, – \(\frac{1}{2}\)), (- 1, – 1), (1, 1), (2, \(\frac{1}{2}\))}

(i) Thus domain of f = {- 2, – 1, 1, 2)

(ii) Range of f = (- \(\frac{1}{2}\), – 1, 1, \(\frac{1}{2}\)}

(iii) Since every element of A has a unique image in A.
Thus given relationf represents a function.

Question 11.
Express the following function as a set of ordered pairs and find its range :
f : X → R defined by f(x) = x³ + 1, where X = {- 1, 0, 3, 9, 7}.
Solution:
Given f : X → R defined by
f(x) = x³ + 1, where x = {- 1, 0, 3, 9, 7}
When x = – 1
∴ f(x) = (- 1)³ + 1 = 0
⇒ (- 1, 0) ∈ f
When x = 0
∴ f(x) = 0 + 1 = 1
⇒ (0, 1) ∈ f
When x = 3
∴ f(x) = 3³ + 1 = 28
⇒ (3, 28) ∈ f
When x = 9
∴ f(x) = 9³ + 1 = 730
⇒ (9, 730) ∈ f
When x = 7
∴ f(x) = 7³ + 1 = 344
⇒ (7, 344) ∈ f
∴ f = {(- 1, 0), (0, 1), (3, 28), (9, 730), (7, 344)}
∴ range of f = {0, 1, 28, 730, 344}.

Question 12.
If f(x) = ax + b, where a and b are integers, f (- 1) – 5 and f (3) = 3, find a and b.
Solution:
Given f(x) = ax + b
Now f (- 1) = – 5
⇒ – 5 = – a + b ……………….(1)
Now f (3) = 3
⇒ 3 = 3a + b ………………….(2)
eqn. (2) – eqn. (1) gives ;
8 = 4a
⇒ a = 2
∴ from (1) ;
b = – 5 + a = – 3
Hence f(x) = 2x – 3.

Question 13.
If (a, 8) and (2, b) are ordered pairs which belong to the mapping f : x → 3x + 4 where x ∈ R, find a and b.
Solution:
Given a function f : R → R defined by
f(x) = 3x + 4 ∀ x ∈ R
Now (a, 8) ∈ f
∴ 8 = 3a + 4
⇒ 3a = 4
⇒ a = 4/3
and (2, b) ∈ f
∴ b = 6 + 4b = 10

Question 14.
If a function f from R to R is defined by f = {(x, 3x – 5) : x ∈ R}, find the values of a and b given that (a, 4) and (1, b) belong to f.
Solution:
Given a function f : R → R defined by
f(x) = 3x – 5 ∀ x ∈ R
Since (a, 4) ∈ f
∴ 4 = 3a – 5
⇒ 3a = 9
⇒ a = 3
and (1, b) ∈ f
∴ b = 3 x 1 – 5 = – 2

Question 15.
Let A = {1, 2, 3}, B = {4, 5, 6, 7} and f : {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show thatf is one-one but not onto.
Solution:
Given A = {1, 2, 3} ;
B = {4, 5, 6,7}
and f = {(1, 4), (2, 5), (3, 6)} be a function from A to B
since different elements 1, 2, 3 in A have different images 4, 5 and 6 in B respectively.
∴ f is one-one.
Since the element 7 in B has no pre-image in A.
∴ f is not onto.
Moreover codoinain B ≠ R_f.

Question 16.
Show that the function f : Q → Q defined by f(x) = 3x – 2 ¡s one-one.
Solution:
Given a function f : Q → Q defined by
f(x) = 3x – 2 ∀ x ∈ Q
∀ x, y ∈ Q he such that
f(x) = f(y)
⇒ 3x – 2 = 3y – 2
⇒ x = y
∴ f is one-one.

Question 17.
Show that the function f : N → N defined by f(x) = 2x – 1 is one-one but not onto.
Solution:
Given the function f : N → N defined by
f (x) = 2x – 1 ∀ x ∈ N
∀ x, y ∈ N be such that
f(x) = f(y)
⇒ 2x – 1 = 2y – 1
⇒ 2x = 2y
⇒ x = y
∴ f is one-one.
Let n ∈ N be any arbitrary element such that f (n) = 2
⇒ 2n – 1 = 2
⇒ 2n = 3
⇒ n = \(\frac{3}{2}\) ∉ N
Hence element 2 in codoniain of f has no pre-imag in N.
∴ f is not onto.
Hencef is one-one and not onto.

Question 18.
Show that the function f : N → N defined by f (n) = n² is injective.
Solution:
Given, f : N → N defined by
f(x) = n² ∀ n ∈ N
∀ m, n ∈ N be s.t.
f(n) = f(m)
⇒ n² = m²
⇒ (n – m) (n + m) = 0
⇒ n – m = 0
[∵ n, m ∈ N
∴ n + m ≠ 0]
⇒ n = m
Hence f is one-one i.e. injective.

Question 19.
Show that the function f : N → N defined by f(m) = m² + 2m + 3 is one-one but not onto.
Solution:
Given function f : N → N defined by
f (m) = m² + 2m + 3 ∀ m ∈ N
∀ m, n ∈ N be s.t
f(m) = f(n)
⇒ m² + 2m + 3 = n² + 2n + 3
⇒ m² – n² + 2 (m – n) = 0
⇒ (m – n) (mn + 2) = 0
⇒ m – n = 0
[∵ m + n + 2 ≠ 0 as m, n ∈ N]
Thus, f is one-one.
Let n ∈ N be any arbitrary element such that f(n) = 2
⇒ n² + 2n + 3 = 2
⇒ n² + 2n + 1 = 0
⇒ (n + 1)² = 0
⇒ n = – 1, – 1 ∉ N
Hence element 2 in codomain of f has no pre-image in N (domain of f).
∴ f is not onto.
Thus, f is one-one but not onto.

Question 20.
In each of the following cases, state whether the function is bijective or not. Justify your answer.
(i) f : R → R defined by f(x) = 2x + 1
(ii) f : R → R defined by f(x) = 3 – 4x
Solution:
(i) Given f : R → R defined by
f(x) = 2x + 1 ∀ x ∈ R
∀ x, y ∈ R s.t. f(x) = f (y)
⇒ 2x + 1 = 2y + 1
⇒ 2x = 2y
⇒ x = y
Thus, f is one-one.

Onto:
Let y ∈ R be any arbitrary element
and let y = f(x) = 2x + 1
⇒ x = \(\frac{y-1}{2}\) ∈ R
[∵ y ∈ R ∃x ∈ R]
∀ y ∈ R ∃ x ∈ R s.t.
f(x) = f \(\left(\frac{y-1}{2}\right)\)
= 2 \(\left(\frac{y-1}{2}\right)\) + 1
= y – 1 + 1 = y
∴ f is onto.
Hence f is 1 – 1 and onto.
∴ f is bijective.

(ii) Given f : R → R defined by
f(x) = 3 – 4x ∀ x ∈ R
One – one:
∀ x, y ∈ R s.t. f(x) = f(y)
⇒ 3 – 4x = 3 – 4y
⇒ – 4x = – 4y
⇒ x = y
∴ f is one – one.
Onto :
Consider any y ∈ R (codomain of f)
and let y = f(x) = 3 – 4x
⇒ 4x = 3 – y
⇒ x = \(\frac{3-y}{4}\)
since y ∈ R = 3 – y ∈ R
⇒ \(\frac{3-y}{4}\) ∈ R
⇒ x ∈ R
Thus, ∀ y ∈ R ∃
x = \(\frac{3-y}{4}\) ∈ R s.t.
f(x) = f(\(\frac{3-y}{4}\))
= 3 – 4 (\(\frac{3-y}{4}\)) = y
∴ f is onto.
Thus f is one-one, onto and therefore f is bijective.

Question 21.
Show that the function f : R – {3} → R defined by f(x) = \(\frac{x-2}{x-3}\) is one-one but not onto.
Solution:
Given a function f : R – {3) → R defined by
f(x) = \(\frac{x-2}{x-3}\)
one – one:
∀ x, y ∈ R – {3} s.t.f(x) = f(y)
⇒ \(\frac{x-2}{x-3}=\frac{y-2}{y-3}\)
⇒ (x – 2) (y – 3) = (x – 3) (y – 2)
⇒ xy – 3x – 2y + 6 = xy – 2x – 3y + 6
⇒ – x + y = 0
⇒ x = y
∴ f is one – one
Since I ∈ R
Let x ∈ R – {3) s.t.f(x) = 1
⇒ \(\frac{x-2}{x-3}\) = 1
⇒ x – 2 = x – 3
⇒ 2 = 3
which is impossible
Thus element I ∈ codomain of f has no pre-image.
∴ f is not onto.
Hence f is 1 – 1 but not onto.

Question 22.
Show that the function f : R → R defined by f(x) = |x + 2| is neither one – one nor onto.
Solution:
Given the function f : R → R defined by
f(x) = |x + 2| ∀ x ∈ R
one-one :
since f(0) = |0 + 2| = 2
and f (- 4) = |- 4 + 2| = 2
Thus elements 0 and – 4 have same image 2.
∴ f is many one i.e. not one-one.
Now, |x + 2| ≥ 0 ∀ x ∈ R
∴ range of f = [0, ∞) which is a proper subset of R.
Thus codomain of f ≠ R
Thus, f is not onto.
Hence, f is neither 1 – 1 nor onto.

Question 23.
Show that the function f : R → R defined by f(x) = 3x² – 2 is neither one- one nor onto.
Solution:
Given, f : R → R defined by
f(x) = 3x² – 2 ∀ x ∈ R
∀ x, y ∈ R s.t. f(x) = f(y)
⇒ 3x² – 2 = 3y² – 2
⇒ 3x² = 3y²
⇒ x = ± y
∴ x ≠ y (always)
∴ f is not one – one.
Moreover,
f (1) = 3 – 2 = 1 ;
f (- 1) = 3 – 2 = 1
Hence, elements 1 and – 1 have same image 1.
∴ f is not one – one.
Since – 5 ∈ R.
Let x ∈ R be any arbitrary element s.t. f(x) = – 5
⇒ 3x² – 2 = – 5
⇒ 3x² = – 3
Thus, element – 5 has no pre-image in R.
∴ f is not onto.
Hence f is neither one-one nor onto.

Question 24.
If A = R – {- 2}, B = R – {1}and f : A → B is a function defined by f(x) = \(\frac{x-3}{x+2}\) for all x ∈ A, then show that f is a one- one correspondence.
Solution:
Given A = R – {- 2} ;
B = R – {1}
and f : A → B defined by
f(x) = \(\frac{x-3}{x+2}\) ∀ x ∈ A
One – one:
∀ x, y ∈ A s.t. f(x) = f(y)
⇒ \(\frac{x-3}{x+2}=\frac{y-3}{y+2}\)
⇒ (x – 3) (y + 2) = (y – 3) (x + 2)
⇒ xy + 2x – 3y – 6 = xy + 2y – 3x – 6
⇒ 5x = 5y
⇒x = y
∴ f is one-one.
Onto:
Let y ∈ B = R – { 1 } be any arbitrary element
and let y = \(\frac{x-3}{x+2}\)
⇒ xy + 2y = x – 3
⇒ x (y – 1) = – 2y – 3
⇒ x = \(\frac{2 y+3}{1-y}\)
since y ≠ 1
\(\frac{2 y+3}{1-y}\) ∈ R, as y ∈ R
Also \(\frac{2 y+3}{1-y}\) ≠ – 2
if \(\frac{2 y+3}{1-y}\) = – 2
⇒ 2y + 3 = – 2 + 2y
⇒ 3 = – 22
which is impossible.
Thus x = \(\frac{2 y+3}{1-y}\) ∈ R – {- 2} = A
∴ ∀ y ∈ B ∃x ∈ A s.t.
f(x) = f \(\left(\frac{2 y+3}{1-y}\right)\)
= \(\frac{\frac{2 y+3}{1-y}-3}{\frac{2 y+3}{1-y}+2}\)
= \(\frac{2 y+3-3+3 y}{2 y+3+2-2 y}\) = y
∴ f is onto.
Thus f is one-one onto.
∴ f is one-one correspondence.