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

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

Question 1.
If A = (1, 2, 3), B = {4, 5} and C = {5, 6}, then verify that
(i) A × (B ∪ C) = (A × B) ∪ (A × C)
(ii) A × (B ∩ C) = (A × B) ∩ (A × C)
(iii) A × (B – C) = (A × B) – (A × C).
Solution:
Given A = {1, 2, 3},
B = {4, 5}
and C = {5, 6)
(i) Now, B ∪ C = {4, 5, 6}
L.H.S. = A × (B ∪ C)
= {1, 2, 3} × {4, 5, 6}
= {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
NowA × B = ((1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}
A × C = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6)}
∴ R.H.S.= (A × B) ∪ (A × C)
= {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
Thus L.H.S = R.H.S
∴ A × (B ∪ C) = (A × B) ∪ (A × C)

(ii) Here, B ∩ C = {5}
L.H.S. = A × (B C)
= {1, 2, 3} × (5)
= {(1, 5), (2, 5), (3, 5)}
Now, A × B = {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}
A × C = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6)}
R.H.S. = (A × B) ∩ (A × C)
= {(1, 5), (2, 5), (3, 5)}
∴ L.H.S = R.H.S.
Thus, A × (B ∩ C) = (A × B) ∩ (A × C)

(iii) Now, B – C = {4}
L.H.S = A × (B – C) = {1, 2, 3}
= {(1, 4), (2, 4), (3, 4)}
Now A × B = {(1, 4), (1, 5), (2. 4), 2, 5), 2,, 4), (3. 5)}
A × C = {(1, 5), (1, 6), (2, 5), (2, 6). (3, 5), (3, 6)}
R.H.S = (A × B) – (A × C)
= {(1, 4), (2, 4), (3, 4)}
∴ A × (B – C) = (A × B) – (A × C)

Question 2.
Let A = {2, 4, 6, 8} and B = {0, 6, 8, 9, 10}. Find the elements of (A ∩ B) × (A – B) corresponding to the relation ‘is a multiple of’.
Solution:
Given A = (2, 4, 6, 8}
and B = {0, 6, 8, 9, 10}
∴ A ∩ B = {6, 8}
and A – B = {2, 4}
(A ∩ B) × (A – B) = {(6, 2), (6, 4), (8, 2), (8, 4)}
Now elements corresponding to relation is ‘a multiple of’
since 6 is a multiple of 2
⇒ (6, 2) ∈ R
8 is a muÌ!p!e of 2
⇒ (8, 2) ∈ R
b a multiple of 4
⇒ (8, 4) ∈ R
but 6 is not a multiple of 4
⇒ (6, 4) ∈ R
∴ required elements are (6, 2),(8, 2), (8, 4).

Question 3.
Let A = {6,7, 8, 10}, B = {2, 4, 5}, a ∈ A, b ∈ B and R be the relation from A to B defined by a R b if and only if a is divisible by b. Write R in the roster form.
Solution:
Given A = {6, 7,8, 10}
and B = {2, 4, 5}
and R be the relation A to B defined by a R b if f a is divisible by b.
since 6 is divisible by 2
⇒ 6 R 2 ⇒ (6, 2) ∈ R
6 is not divisible by 4 and 5
⇒ (6, 4), (6, 5) ∉ R
Also, 8 is divisible by 2
⇒ (8, 2) ∈ R
8 is divisible by 4
⇒ (8, 4) ∈ R
10 is divisible by 2
⇒ (0, 2) ∈ R
and 10 is divisible by 5
⇒ (10, 5) ∈ R
Hence, R = {(6. 2), (8. 2), (8, 4), (10, 2), (10, 5)}.

Question 4.
Let R = {(x, y) : x + 2y < 6, x, y ∈ N}
(i) Find the domain and the range of R
(ii) Write R as a set of ordered pairs.
Solution:
Given R = {(x, y) : x + 2y < 6. x, y ∈ N}
since, x, y ∈ N, x + 2y < 6
⇒ y < \(\frac{6-x}{2}\)
When x = 1 ;
y < \(\frac{6-1}{2}=\frac{5}{2}\)
⇒ y = 1, 2 (∵ y ∈ N)
Thus, (1, 1), (1, 2) ∈ R
When x = 2 ;
y < \(\frac{6-2}{2}=\frac{4}{2}\) = 2
⇒ y = 1 (∵ y ∈ N)
∴ (2, 1) ∈ R
When x = 3 ;
y < \(\frac{6-3}{2}=\frac{3}{2}\)
⇒ y = 1 (∵ y ∈ N)
∴ (3, 1) ∈ R
When x = 4 ;
y < \(\frac{6-4}{2}=\frac{2}{2}\) = 1
since there is no natural number which is less than 1.
So for other values of x, we donot get y ∈ N
Thus R = {(1, 1), (1, 2), (2, 1), (3, 1)}

(i) Domain of R = {1, 2, 3}
and Range of R = {1, 2}

Question 5.
Let R = {(x, y) ; y = x + 1 and y ∈ {0, 1, 2, 3, 4, 5)}.
(i) List the elements of R..
(ii) Represent R by an arrow diagram.
Solution:
Given R = {(x, y) : y = x + 1 and y ∈ {0, 1, 2, 3, 4, 5}}
(i) Since y = x + 1, y ∈ {0, 1, 2, 3, 4, 5}
When y = 0
⇒ 0 = x + 1
⇒ x = – 1
∴ (- 1, 0) ∈ R
When y = 1
⇒ 1 = x + 1
⇒ x = 0
∴ (0, 1) ∈ R
When y = 2
⇒ 2 = x + 1
⇒ x = 1
∴ (1, 2) ∈ R
When y = 3
⇒ 3 = x + 1
⇒ x = 2
∴ (2, 3) ∈ R
When y = 4
⇒ 4 = x + 1
⇒ x = 3
(3, 4) ∈ R
When y = 5
⇒ 5 = x + 1
⇒ x = 3
∴ (4, 5) ∈ R
Thus, R = {(- 1, 0), (0, 1), (1, 2), (2, 3), (3, 4), (4, 5)}

(ii)

Question 6.
Let f be the subset of Q × Z defined by f = {(\(\frac{m}{n}\), m) : m, n ∈ Z, n ≠ 0} function from Q to Z ? Justify your answer.
Solution:
Given f ⊂ Q × Z defined by
f = {(\(\frac{m}{n}\), m) : m, n ∈ Z, n ≠ 0}
Here f (\(\frac{1}{2}\)) = 1 ;
f (\(\frac{2}{4}\)) = 2
since \(\frac{1}{2}=\frac{2}{4}\)
So element \(\frac{1}{2}\) ∈ Q have two images 1 and 2 in Z.
∴ f is not a function from Q to Z.
since each element in Q does not have unique image in Z.

Question 7.
Let f : X → Y be defined byf(x) = x² for all x ∈ X where X = {- 2, – 1, 0, 1, 2, 3} and Y = {0, 1, 4, 7, 9, 10}.
Write the relation f in the roster form. Is f a function ?
Solution:
Given f : X → Y be defined by
f(x) = x² ∀ x ∈ X
where X = {- 2, – 1, 0, 1, 2, 3}
and Y = {0, 1, 4, 7, 9, 10}
When x = – 2,
f (- 2) = (- 2)² = 4
⇒ (- 2, 4) ∈ R
When x = 1,
f (- 1) = (- 1)² = 1
⇒ (- 1, 1) ∈ R
When x = 0,
f (0) = 0² = 0
⇒ (0, 0) ∈ R
When x = 1,
f (1) = 1² = 1
⇒ (1, 1) ∈ R
When x = 2,
f (2) = 2² = 4
⇒ (2, 4) ∈ R
When x = 3,
f (3) = 3² = 9
⇒ (3, 9) ∈ R
Thus, R {(- 2, 4), (- 1, 1), (0, 0), (1, 1), (2, 4), (3, 9)}
since elements 1, – 1 have same image 1.
∴ f is many one function since every element in X has unique image in Y.

Question 8.
Is g = {(1, 1), (2, 3), (3, 5), (4, 7)} a function ? If this is described by the relation g (x) = αx + β, then what values should be assigned to α and β.
Solution:
Given g = {(1, 1), (2, 3), (3, 5), (4, 7)}
Since every element in domain of g has a unique image in codomain of g.
Thus g be a function.
Since g (x) = αx + β ……………….(1)
Since (1, 1) ∈ g
∈ when x = 1, y = 1
∴ from (1) ; we have
1 = α + β ………………(2)
Since (2, 3) ∈ g i.e. when x = 2, y = 2
∴ 3 = 2α + β
∴ from (2) and (3) ;
α = 2 ; β = – 1

Question 9.
Consider the function f (x) = x + \(\frac{1}{x}\), x ∈ R, x ≠ 0. Is f one – one?
Solution:
Given f (x) = x + \(\frac{1}{x}\)
Here f (2) = 2 + \(\frac{1}{2}\) = \(\frac{5}{2}\)
and f (\(\frac{1}{2}\)) = \(\frac{1}{2}+\frac{1}{\frac{1}{2}}\)
= \(\frac{1}{2}+2=\frac{5}{2}\)
Since element 2 and \(\frac{1}{2}\) in domain of f have same image ï in codomain of f.
Thus different elements have same image.
∴ f is not one-one.

Question 10.
Prove that the function f : N → N defined by f (x) = 3x – 2 is one – one but not onto.
Solution:
Given f : N → N defined by
f (x) = 3x – 2 ∀ x ∈ N
∀ x, y ∈ N s.t. f (x) = f (y)
⇒ 3x – 2 = 3y – 2
⇒ 3x = 3y
⇒ x = y
∴ f is one – one.
Since 2 ∈ N (codomain of f)
Let x ∈ N be any element such that f (x) = 2
⇒ 3x – 2 = 2
Hence the element 2 in codomain off has no pre-image in N (domain of f)
∴ f is not onto.

P.Q. Find the domain of the function f given by f (x) = \(\frac{1}{\sqrt{|x|-x}}\).
Solution:
Given f (x) = \(\frac{1}{\sqrt{|x|-x}}\)
For D_f : f (x) must be a real number .
⇒ \(\frac{1}{\sqrt{|x|-x}}\) must be a real number
⇒ |x| – x > 0
⇒ |x| > x
⇒ x < 0 [∵ |x| = x if x = 0 and |x| = x if x > 0
⇒ |x| – x = 0 if x ≥ 0]
∴ D_f = (- ∞, 0).

Question 11.
Determine a quadratic function ‘f’ defined by f (x) = ax² + bx + c if f (0) = 6, f (2) = 11 and f (- 3) = 6.
Solution:
Given f (x) = ax² + bx + c ……………..(1)
Now f (0) = 6
⇒ 6 = 0 + 0 + c
⇒ c = 6
f (2) = 11
⇒ 11 = 4a + 2b + c
⇒ 4a + 2b = 5 ……………(2)
and f (- 3) = 6
⇒ 6 = 9a – 3b + c
⇒ 9a – 3b = 0
⇒ 3a – b = 0 …………………..(3)
From (2) and (3) ; we have
4a + 2 (3a) = 5
⇒ 10a = 5
⇒ a = \(\frac{1}{2}\)
∴ from (3) ;
b = 3a = \(\frac{3}{2}\)
∴ from (1) ;
f (x) = \(\frac{1}{2}\) x² + \(\frac{3}{2}\) x + 6

Question 12.
Find the domain and the range of the functionf(x) = 2 – 3x². Also find f (- 2) and the numbers which are associated with the number – 25 in its range.
Solution:
Given f (x) = 2 – 3x²
For D_f : (x) must be a real number
⇒ 2 – 3x² must be a real number
which is a real number for all x ∈ R
For R_f :
Let y = f(x)
= 2 – 3x² ∀ x ≤ R
⇒ 3x² – y
⇒ x² = \(\frac{2-y}{3}\)
since x² ≥ 0 ∀ x ∈ R
⇒ \(\frac{2-y}{3}\) ≥ 0
⇒ 2 – y ≥ 0
⇒ y ≤ 2
∴ R_f = (- ∞, 2]
since – 2 ∈ R
:. f (- 2) = 2 – 3 (- 2)²
= 2 – 12
= – 10
since – 25 ∈ R_f,
let x ∈ D_f s.t. f (x) = 25
⇒ 2 – 3x² = – 25
⇒ 3x² = 27
⇒ x² = 9
⇒ x = ± 3 ∈ R

Question 13.
Find the domain and the range of the following functions :
(i) \(\sqrt{x-3}\)
(ii) \(\sqrt{25-x^2}\)
(iii) 5 – |x + 1|
Solution:
(i) Given, f (x) = \(\sqrt{x-3}\)
For D_f : f (x) must be a real number
⇒ \(\sqrt{x-3}\) must be a real number
⇒ x – 3 ≥ 0
⇒ x ≥ 3
Thus D = [3, ∞)
For R_f :
Let y = f (x)
= \(\sqrt{x-3}\) ∀ x ∈ D_f
as x ≥ 3
⇒ x – 3 ≥ 0
⇒ 4x – 3 ≥ 0
∴ R_f = [0, ∞)

(ii) Given f (x) = \(\sqrt{25-x^2}\)
For D_f : f (x) must be a real number
⇒ \(\sqrt{25-x^2}\) must be a real number.
⇒ 25 – x² ≥ 0
⇒ x² ≤ 25
⇒ |x| ≤ 5
⇒ – 5 ≤ x ≤ 5
∴ D_f = [- 5, 5]
For R_f :
Let y = f (x)
= \(\sqrt{25-x^2}\)
Since square root of real number is always non-negative
∴ y ≥ 0
On squaring ; we get
y² = 25 – x²
⇒ x² = 25 – y²
since x² ≥ 0
⇒ 25 – y² ≥ 0
⇒ 25 ≥ y²
⇒ y² ≤ 25
⇒ |y| ≤ 5
⇒ – 5 ≤ y ≤ 5 and y ≥ 0
⇒ 0 ≤ y ≤ 5
∴ R_f = [0, 5]

(iii) Given f (x) = 5 – \x + 1|
For D_f : f (x) must be a real number
⇒ 5 – |x + 1| must be a real number
which is a real number for all x ∈ R
∴ D_f = R
For R_f :
Let y = f (x) = 5 . |x + 1|
since x ≥ 0 ∀ x ∈ R
⇒ |x + 1| ≥ 0 ∀ x ∈ R
⇒ – |x + 1| ≤ 0
⇒ 5 – |x + 1| ≤ 5
⇒ y ≤ 5
∴ R_f = (- ∞, 5].

Question 14.
Draw the graph of the function f (x) = \(\begin{cases}1+2 x, & x<0 \\ 3+5 x, & x \geq 0\end{cases}\). Hence, find its range.
Solution:
Given f (x) = \(\begin{cases}1+2 x, & x<0 \\ 3+5 x, & x \geq 0\end{cases}\)
The graph consists of two parts.
Case – I :
When x < 0, f (x) = 1 + 2x

Table of values are given as under :

x	– 1	– 2	– 3
y	– 1	– 3	– 5

The part of graph is a straight line as shown in adjoining figure.

Case – II :
When x ≥ 0, f (x) = 3 + 5x
Table of values is given as under:

x	0	1	\(\frac{1}{5}\)
y	3	8	4

The part of graph is a straight line as shown in adjoining figure.
From graph it is clear that R_f = (- ∞, 0) ∪ [3, ∞).

Question 15.
If f (x) = 2x + 5 an g (x) = x² – 1 are two real valued functions, find tht following functions :
(i) f + g
(ii) f – g
(iii) fg
(iv) \(\frac{f}{g}\)
(v) \(\frac{g}{f}\)
(vi) 3g + 2f²
Solution:
Given f (x) = 2x + 5
and g (x) = x² – 1
Here D_f = R ; D_g = R
So f and g have same domains.
(i) (f + g) (x) = f (x) + g (x)
= 2x + 5 + x² – 1
= x² + 2x + 4 ∀ x ∈ R

(ii) (f – g) (x) = f (x) – g (x)
= 2x + 5 – x² + 1
= 2x + – x t I
= 2x – x² + 6 ∀ x ∈ R

(iii) (fg) (x) = f (x) g (x)
= (2x + 5) (x² – 1)
= 2x³ + 5x² – 2x – 5 ∀ x ∈ R

(iv) \(\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{2 x+5}{x^2-1}\)
with domain D
where D = R except g (x) = 0
i.e. x² – 1 = 0
⇒ x = ± 1
⇒ D = R – {- 1}

(v) \(\left(\frac{g}{f}\right)\) (x) = \(\frac{g(x)}{f(x)}\)
= \(\frac{x^2-1}{2 x+5}\)
with domain D₁
where D₁ = R exccpt f (x) = 0
i.e. 2x + 5
i.e. x = – \(\frac{5}{2}\)

(vi) (3g + 2f²) (x) = 3 g (x) – 2f² (x)
= 3g (x) + 2 (f (x))²
= 3 (x² – 1) + 2 (2x + 5)²
= 3x² – 3 + 2 (4x² + 20x + 25)
= 11x² + 40x + 47 ∀ x ∈ R