Welcome to Our Tamil Crowd Website..You Will Get All Job news, Tamil breaking news, Educational news, Health Tips here..

## Chapter 1 Relations and Functions Ex 1.1

Question 1.

Find A × B, A × A and B × A

(i) A = {2,-2,3} and B = {1,-4}

(ii) A = B = {p,q]

(iii) A= {m,n} ; B = (Φ)

Solution:

(i) A = {2,-2,3}, B = {1,-4}

A × B = {(2, 1), (2, -4), (-2, 1), (-2, -4), (3,1) , (3,-4)}

A × A = {(2, 2), (2,-2), (2, 3), (-2, 2), (-2, -2), (-2, 3), (3, 2), (3, -2), (3,3) }

B × A = {(1, 2), (1, -2), (1, 3), (-4, 2), (-4, -2), (-4,3)}

(ii) A = B = {(p,q)]

A × B = {(p, p), {p, q), (q, p), (q, q)}

A × A = {(p, p), (p, q), (q, p), (q, q)}

B × A = {(p,p), {p, q), (q,p), (q, q)}

(iii) A = {m,n} × Φ

A × B = { }

A × A = {(m,m), (m,n), (n, m), (n, n)}

B × A = { }

Question 2.

Let A = {1, 2, 3} and B = {x | x is a prime number less than 10}. Find A × B and B × A.

Solution:

A = {1, 2, 3}, B = {2,3, 5,7}

A × B = {(1,2), (1,3), (1,5), (1,7), (2,2), (2.3) , (2,5), (2,7), (3,2), (3,3), (3, 5), (3,7)}

B × A = {(2,1), (2,2), (2,3), (3,1), (3,2), (3.3) , (5,1), (5,2), (5,3), (7,1), (7,2) , (7, 3)}

Question 3.

If B × A = {(-2, 3),(-2, 4),(0, 3),(0, 4),(3, 3), (3, 4)} find A and B.

Solution:

B × A ={(-2,3), (-2,4), (0,3), (0,4), (3,3), (3,4)}

A = {3, 4), B = { -2, 0, 3}

Question 4.

If A ={5, 6}, B = {4, 5, 6} , C = {5, 6, 7}, Show that A × A = (B × B) ∩ (C × C).

Solution:

A = {5,6}, B = {4,5,6},C = {5,6, 7}

A × A = {(5, 5), (5, 6), (6, 5), (6, 6)} …(1)

B × B = {(4, 4), (4, 5), (4, 6), (5, 4),

(5,5), (5,6), (6,4), (6,5), (6,6)} …(2)

C × C = {(5,5), (5,6), (5,7), (6,5), (6,6),

(6, 7), (7, 5), (7, 6), (7, 7)} …(3)

(B × B) ∩ (C × C) = {(5, 5), (5,6), (6, 5), (6,6)} …(4)

(1) = (4)

A × A = (B × B) ∩ (C × C)

It is proved.

Question 5.

Given A ={1, 2, 3}, B = {2, 3, 5}, C = {3, 4} and D = {1, 3, 5}, check if (A ∩ C) x (B ∩ D) = (A × B) ∩ (C × D) is true?

Solution:

LHS = {(A∩C) × (B∩D)

A ∩C = {3}

B ∩D = {3,5}

(A ∩ C) × (B ∩ D) = {(3, 3) (3, 5)} …(1)

RHS = (A × B) ∩ (C × D)

A × B = {(1,2), (1,3), (1,5), (2,2), (2,3), (2, 5), (3, 2), (3,3), (3,5)}

C × D = {(3,1), (3,3), (3,5), (4,1), (4, 3), (4,5)}

(A × B) ∩ (C × D) = {(3, 3), (3, 5)} …(2)

∴ (1) = (2) ∴ It is true.

Question 6.

Let A = {x ∈ W | x < 2}, B = {x ∈ N |1 < x < 4} and C = {3, 5}. Verify that

(i) A × (B ∪ C) = (A × B) ∪ (A × C)

(ii) A × (B ∩ C) = (A × B) ∩ (A × C)

(iii) (A ∪ B) × C = (A × C) ∪ (B × C)

(i) A × (B ∪ C) = (A × B) ∪ (A × C)

Solution:

A = {x ∈ W|x < 2} = {0,1}

B = {x ∈ N |1 < x < 4} = {2,3,4}

C = {3,5}

LHS =A × (B ∪ C)

B ∪ C = {2,3,4} ∪ {3,5}

= {2, 3, 4, 5}

A × (B ∪ C) = {(0, 2), (0, 3), (0,4), (0, 5), (1.2) , (1,3), (1,4),(1,5)} …(1)

RHS = (A × B) ∪ (A × C)

(A × B) = {(0,2), (0,3), (0,4), (1,2), (1,3), (1,4)}

(A × C) = {(0,3), (0,5), (1,3), (1,5)}

(A × B) ∪ (A × C)= {(0, 2), (0, 3), (0,4), (1, 2), (1.3), (1,4), (0, 5), (1,5)} ….(2)

(1) = (2), LHS = RHS

Hence it is proved.

(ii) A × (B ∩ C) = (A × B) ∩ (A × C)

LHS = A × (B ∩ C)

(B ∩ C) = {3}

A × (B ∩ C) = {(0, 3), (1, 3)} …(1)

RHS = (A × B) ∩ (A × C)

(A × B) = {(0,2),(0,3),(0,4),(1,2), (1,3),(1,4)}

(A × C) = {(0,3), (0,5), (1,3), (1,5)}

(A × B) ∩ (A × C) = {(0, 3), (1, 3)} …(2)

(1) = (2) ⇒ LHS = RHS.

Hence it is verified.

(iii) (A ∪ B) × C = (A × C) ∪ (B × C)

LHS = (A ∪ B) × C

A ∪ B = {0,1,2,3,4}

(A ∪ B) × C = {(0,3), (0,5), (1,3), (1,5), (2, 3), (2, 5), (3, 3), (3, 5), (4, 3), (4, 5)} (1)

RHS = (A × C) ∪ (B × C)

(A × C) = {(0,3), (0,5), (1,3), (1,5)}

(B × C) = {(2, 3), (2, 5), (3, 3), (3, 5), (4, 3), (4, 5)}

(A × C) ∪ (B × C) = {(0, 3), (0, 5), (1, 3), (1, 5), (2, 3), (2, 5), (3, 3), (3, 5), (4, 3), (4, 5)} …(2)

(1) = (2)

∴ LHS = RHS. Hence it is verified.

Question 7.

Let A = The set of all natural numbers less than 8, B = The set of all prime numbers less than 8, C = The set of even prime number. Verify that

(i) (A ∩ B) × c = (A × C) ∩ (B × C)

(ii) A × (B – C ) = (A × B) – (A × C)

A = {1,2, 3, 4, 5, 6, 7}

B = {2, 3, 5, 7}

C = {2}

Solution:

(i)(A ∩ B) × C = (A × c) ∩ (B × C)

LHS = (A ∩ B) × C

A ∩ B = {2, 3, 5, 7}

(A ∩ B) × C = {(2, 2), (3, 2), (5, 2), (7, 2)} …(1)

RHS = (A × C) ∩ (B × C)

(A × C) = {(1,2), (2,2), (3,2), (4,2), (5,2), (6,2), (7,2)}

(B × C) = {2,2), (3,2), (5,2), (7,2)}

(A × C) ∩ (B × C) = {(2,2), (3,2), (5,2), (7,2)} …(2)

(1) = (2)

∴ LHS = RHS. Hence it is verified.

(ii) A × (B – C) = (A × B) – (A × C)

LHS = A × (B – C)

(B – C) = {3,5,7}

A × (B – C) = {(1,3), (1, 5), (1,7), (2,3), (2,5), (2.7) , (3,3), (3,5), (3,7), (4,3), (4,5), (4,7), (5,3), (5,5), (5,7), (6,3) , (6,5), (6,7), (7,3), (7,5), (7.7)} …(1)

RHS = (A × B) – (A × C)

(A × B) = {(1,2), (1,3), (1,5), (1,7),

(2, 2), (2, 3), (2, 5), (2, 7),

(3, 2), (3, 3), (3, 5), (3, 7),

(4, 2), (4, 3), (4, 5), (4, 7),

(5, 2), (5, 3), (5, 5), (5, 7),

(6, 2), (6, 3), (6, 5), (6, 7),

(7, 2), (7, 3), (7, 5), (7,7)}

(A × C) = {(1,2), (2,2),(3,2),(4,2), (5,2), (6,2), (7,2)}

(A × B) – (A × C) = {(1, 3), (1, 5), (1, 7), (2, 3), (2, 5), (2, 7), (3, 3), (3, 5), (3, 7), (4, 3), (4, 5), (4, 7), (5, 3), (5, 5), (5, 7), (6, 3), (6, 5), (6, 7), (7, 3), (7, 5), (7,7) } …(2)

(1) = (2) ⇒ LHS = RHS.

Hence it is verified.

## Chapter 1 Relations and Functions Ex 1.3

Question 1.
Let f = {(x,y)|x,y ∈ N and y = 2x} be a relation on N. Find the domain, co-domain and range. Is this relation a function?

Solution:

F = {(x, y)|x, y ∈ N and y = 2x}
x = {1,2,3,…}
y = {1 × 2, 2 × 2, 3 × 2, 4 × 2, 5 × 2 …}
R = {(1,2), (2,4), (3, 6), (4, 8), (5, 10),…}
Domain of R = {1, 2, 3, 4,…},
Co-domain = {1,2,3…..}
Range of R = {2, 4, 6, 8, 10,…}
Yes, this relation is a function.

Question 2.
Let X = {3, 4, 6, 8}. Determine whether the relation R = {(x,fx))|x ∈ X, f(x) = x2 + 1} is a function from X to N ?

Solution:

x = {3,4, 6, 8}
R = ((x,f(x))|x ∈ X,f(x) = X2 + 1}
f(x) = x2 + 1
f(3) = 32 + 1 = 10
f(4) = 42 + 1 = 17
f(6) = 62 + 1 = 37
f(8) = 82 + 1 = 65

R = {(3, 10), (4,17), (6,37), (8, 65)}
R = {(3, 10), (4,17), (6,37), (8, 65)}
Yes, R is a function from X to N.

Question 3.
Given the function f: x → x2 – 5x + 6, evaluate
(i) f(-1)
(ii) A(2a)
(iii) f(2)
(iv) f(x – 1)

Solution:

Give the function f: x → x2 – 5x + 6.
(i) f(-1) = (-1)2 – 5(1) + 6 = 1 + 5 + 6 = 12
(ii) f(2a) = (2a)2 – 5(2a) + 6 = 4a2 – 10a + 6
(iii) f(2) = 22 – 5(2) + 6 = 4 – 10 + 6 = 0
(iv) f(x- 1) = (x – 1)2 – 5(x – 1) + 6
= x2 – 2x + 1 – 5x + 5 + 6
= x2 – 7x + 12

Question 4.
A graph representing the function f(x) is given in figure it is clear that f (9) = 2.

(i) Find the following values of the function
(a) f(0)
(b) f(7)
(c) f(2)
(d) f(10)
(ii) For what value of x is f (x) = 1?
(iii) Describe the following
(i) Domain
(ii) Range.
(iv) What is the image of 6 under f?

Solution:

From the graph
(a) f(0) = 9
(b) f(7) = 6
(c) f(2) = 6
(d) f(10) = 0
(ii) Atx = 9.5, f(x) = 1
(iii) Domain = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
= {x |0 < x < 10, x ∈ R}
Range = {x|0 < x < 9, x ∈ R}
= {0, 1,2, 3,4, 5,6, 7, 8,9}
(iv) The image of 6 under f is 5.

Question 5.
Let f(x) = 2x + 5. If x ≠ 0 then find

Solution:

Given f(x) = 2x + 5, x ≠ 0.

Question 6.
A function fis defined by f(x) = 2x – 3
(i) find f(0)+f(1)2
(ii) find x such that f(x) = 0.
(iii) find x such that f(x) = x.
(iv) find x such that f(x) = f(1 – x).

Solution:

Given f(x) = 2x – 3

Question 7.
An open box is to be made from a square piece of material, 24 cm on a side, by cutting equal squares from the corners and turning up the sides as shown in figure. Express the volume V of the box as a function of x.

Solution:

Volume of the box = Volume of the cuboid
= l × b × h cu. units
Here l = 24 – 2x
b = 24 – 2x
h = x
∴ V = (24 – 2x) (24 – 2x) × x
= (576 – 48x – 48x + 4×2)x
V = 4x3 – 96x2 + 576x

Question 8.
A function f is defined bv f(x) = 3 – 2x . Find x such that f(x2) = (f(x))2.

Solution:

f(x) = 3 – 2x
f(x2) = 3 – 2x2

Question 9.
A plane is flying at a speed of 500 km per hour. Express the distance d travelled by the plane as function of time t in hours.

Solution:

Question 10.
The data in the adjacent table depicts the length of a woman’s forehand and her corresponding height. Based on this data, a student finds a relationship between the height (y) and the forehand length(x) as y = ax + b, where a, b are constants.

(i) Check if this relation is a function.
(ii) Find a and b.
(iii) Find the height of a woman whose forehand length is 40 cm.
(iv) Find the length of forehand of a woman if her height is 53.3 inches.

Solution:

(i) Given y = ax + b …(1)
The ordered pairs are R = {(35, 56) (45, 65) (50, 69.5) (55, 74)}
∴ Hence this relation is a function.

Substituting a = 0.9 in (2) we get
⇒ 65 = 45(.9) + b
⇒ 65 = 40.5 + b
⇒ b = 65 – 40.5
⇒ b = 24.5
∴ a = 0.9, b = 24.5
∴ y = 0.9x + 24.5
(iii) Given x = 40 , y = ?
∴ (4) → y = 0.9 (40) + 24.5
⇒ y = 36 + 24.5
⇒ y = 60.5 inches
(iv) Given y = 53.3 inches, x = ?
(4) → 53.3 = 0.9x + 24.5
⇒ 53.3 – 24.5 = 0.9x
⇒ 28.8 = 0.9x

∴ When y = 53.3 inches, x = 32 cm

## Chapter 1 Relations and Functions Ex 1.4

Question 1.
Determine whether the graph given below represent functions. Give reason for your answers concerning each graph.

Solution:

(i) It is not a function. The graph meets the vertical line at more than one points.
(ii) It is a function as the curve meets the vertical line at only one point.
(iii) It is not a function as it meets the vertical line at more than one points.
(iv) It is a function as it meets the vertical line at only one point.

Question 2.
Let f :A → B be a function defined by f(x) = x2 – 1, Where A = {2,4,6,10,12},
B = {0,1, 2, 4, 5, 9}. Represent/by
(i) set of ordered pairs;
(ii) a table;
(iii) an arrow diagram;
(iv) a graph

Solution:

f: A → B
A = {2,4, 6, 10, 12},

B = {0,1, 2, 4,5,9}

(i) Set of ordered pairs
= {(2, 0), (4, 1), (6, 2), (10, 4), (12, 5)}
(ii) a table

(iii) an arrow diagram;

(iv) a graph

Question 3.
Represent the function f = {(1, 2),(2, 2),(3, 2), (4,3), (5,4)} through
(i) an arrow diagram
(ii) a table form
(iii) a graph

Solution:

f = {(1,2), (2, 2), (3, 2), (4, 3), (5, 4)}
(i) An arrow diagram.

(ii) a table form

(iii) A graph representation.

Question 4.
Show that the function f : N → N defined by f{x) = 2x – 1 is one – one but not onto.

Solution:

N = {1,2, 3, 4, 5,…}
f(1) = 2(1) – 1 = 1
f(2) = 2(2) – 1 = 3
f(3) = 2(3) – 1 = 5
f(4) = 2(4) – 1 = 7
f(5) = 2(5) – 1 = 9

Hence f : N → N is a one-one function.
A function f: N → N is said to be onto function if the range of f is equal to the co-domain of f
Range = {1,3, 5, 7, 9,…}
Co-domain = {1, 2,3,..}
But here the range is not equal to co-domain. Therefore it is one-one but not onto function.

Question 5.
Show that the function f: N → N defined by f (m) = m2 + m + 3 is one – one function.

Solution:

f: N → N
f(m) = m2 + m + 3
N = {1,2, 3,4, 5…..}m ∈ N
f{m) = m2 + m + 3
f(1) = 12 + 1 + 3 = 5
f(2) = 22 + 2 + 3 = 9
f(3) = 32 + 3 + 3 = 15
f(4) = 42 + 4 + 3 = 23

In the figure, for different elements in the (X) domain, there are different images in f(x). Hence f: N → N is a one to one but not onto function as the range of f is not equal to co-domain.
Hence it is proved.

Question 6.
Let A = {1, 2,3,4) and B = N. Letf: A → B be
defined by f(x) = x3 then,
(i) find the range off
(ii) identify the tpe of function

Solution:

A = {1,2,3,4}
B = N
f: A → B,f(x) = x3
(i) f(1) = 13 = 1
f(2) = 23 = 8
f(3) = 33 = 27
f(4) = 43 = 64
(ii) Therange of f = {1,8,27,64 )
(iii) It is one-one and into function.

Question 7.
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 – 4x2

Solution:

(i) f : R → R
f(x) = 2x + 1
f(1) = 2(1) + 1 = 3
f(2) = 2(2) + 1 = 5
f(-1) = 2(-1) + 1 = -1
f(0) = 2(0) + 1 = 1
It is a bijective function. Distinct elements of A have distinct images in B and every element in B has a pre-image in A.
(ii) f: R → R; f(x) = 3 – 4x2
f(1) = 3 – 4(12) = 3 – 4 = -1
f(2) = 3 – 4(22) = 3 – 16 = -13
f(-1) = 3 – 4(-1)2 = 3 – 4 = -1
It is not bijective function since it is not one-one

Question 8.
Let A = {-1, 1} and B = {0, 2}. If the function f: A → B defined by f(x) = ax + b is an onto function? Find a and b.

Solution:

A= {-1, 1},B = {0,2}
f: A → B, f(x) = ax + b
f(-1) = a(-1) + b = -a + b
f(1) = a(1) + b = a + b
Since f(x) is onto, f(-1) = 0
⇒ -a + b = 0 …(1)
& f(1) = 2
⇒ a + b = 2 …(2)
-a + b = 0

Question 9.
If the function f is defined by

(i) f(3)
(ii) f(0)
(iii) f(-1.5)
(iv) f(2) + f(-2)

Solution:

(i) f(3) ⇒ f(x) = x + 2 ⇒ 3 + 2 = 5
(ii) f(0) ⇒ 2
(iii) f (- 1.5) = x – 1
= -1.5 – 1 = -2.5

Question 10.
A function f: [-5,9] → R is defined as follows:

Solution:

f : [-5, 9] → R
(i) f(-3) + f(2)
f(-3) = 6x + 1 = 6(-3) + 1 = -17
f(2) = 5×2 – 1 = 5(22) – 1 = 19
∴ f(-3) + f(2) = -17 + 19 = 2
(ii) f(7) – f(1)
f(7) = 3x – 4 = 3(7) – 4 = 17
f(1) = 6x + 1 = 6(1) + 1 = 7
f(7) – f(1) = 17 – 7 = 10
(iii) 2f(4) + f(8)
f(4) = 5x2 – 1 = 5 × 42 – 1 = 79
f(8) = 3x – 4 = 3 × 8 – 4 = 20
∴ 2f(4) + f(8) = 2 × 79 + 20 = 178

Question 11
The distance S an object travels under the influence of gravity in time t seconds is given by S(t) = 12 gt2 + at + b where, (g is the 2 acceleration due to gravity), a, b are constants. Check if the function S (t)is one-one.

Solution:

Yes, for every different values of t, there will be different values as images. And there will be different preimages for the different values of the range. Therefore it is one-one function.

Question 12.
The function ‘t’ which maps temperature in Celsius (C) into temperature in Fahrenheit (F) is defined by ty(C)= F where F = 95 C +32 . Find,
(i) t(0)
(ii) t(28)
(iii) t(-10)
(iv) the value of C when t(C) = 212
(v) the temperature when the Celsius value is equal to the Farenheit value.

Solution:

## Chapter 1 Relations and Functions Ex 1.5

Question 1
Using the functions f and g given below, find fog and gof. Check whether fog = gof.
(i) f(x) = x – 6, g(x) = x2
(ii) f(x) = 2x, g(x) = 2x2– 1
(iii) f(x) = x+63g(x) = 3 – x
(iv) f(x) = 3 + x, g(x) = x – 4
(v) f(x) = 4x2– 1,g(x) = 1 + x

Solution:

(i) f(x) = x – 6, g(x) = x2
fog(x) = f(g(x)) = f(x2) = x2 – 6 …(1)
gof(x) = g(f(x)) = g(x – 6) = (x – 6)2
= x2 + 36 – 12x = x2 – 12x + 36…(2)
(1) ≠ (2)
∴ fog(x) ≠ gof(x)

(ii) f(x) = 2x,g(x) = 2x2 – 1

(iii) f(x) = x+63,g(x) = 3 – x

(iv) f(x) = 3 + x, g(x) = x – 4
fog(x) = f(g(x)) = f(x -4) = 3 + x – 4
= x – 1 …(1)
gof(x) = g(f(x)) = g(3 + x) = 3 + x – 4
= x – 1 …(2)
Here fog(x) = gof(x)

(v) f(x) = 4x2 – 1, g(x) = 1 + x
fog(x) = f(g(x)) = f(1 + x) = 4(1 + x)2 – 1
= 4(1 + x2 + 2x) – 1 = 4 + 4x2 + 8x – 1
= 4x2 + 8x + 3 …(1)
gof(x) = g(f(x)) = g(4x2 – 1)
= 1 + 4x2 – 1 = 4x2 …(2)
(1) ≠ (2)
∴ fog(x) ≠ gof(x)

Question 2.
Find the value of k, such that fog = gof
(i) f(x) = 3x + 2, g(x) = 6x – k
(ii) f(x) = 2x – k, g(x) = 4x + 5

Solution:

(i) f(x) = 3x + 2, g(x) = 6x – k
fog(x) = f(g(x)) = f(6x – k) = 3(6x – k) + 2
= 18x – 3k + 2 …(1)
gof(x) = g(f(x)) = g(3x + 2) = 6(3x + 2) – k
= 18x + 12 – k …(2)
(1) = (2)

2k = -10
k = -5

(ii) f(x) = 2x – k, g(x) = 4x + 5
fog(x) = f(g(x)) = f(4x + 5) = 2(4x + 5) – k
= 8x + 10 – k …(1)
gof(x) = g(f(x)) = g(2x – k) = 4(2x – k)+ 5
= 8x – 4k + 5 ….(2)
(1) = (2)

3k = -5
k = −53

Question 3.
if f(x) = 2x – 1, g(x) = x+12, show that fog = gof = x

Solution:

f(x) = 2x – 1, g(x) = x+12, fog = gof = x

Question 4.
(i) If f (x) = x2 – 1, g(x) = x – 2 find a, if gof(a) = 1.
(ii) Find k, if f(k) = 2k – 1 and fof (k) = 5.

Solution:

(i) f(x) = x2 – 1, g(x) = x – 2
Given gof(a) = 1
gof(x) = g(f(x)
= g(x2 – 1) = x2 – 1 – 2
= x2 – 3
gof(a) ⇒ a2 – 3 = 1 =+ a2 = 4
a = ± 2
(ii) f(k) = 2k – 1
fo f(k) = 5
f(f(k)m = f(2k – 1) = 5
⇒ 2(2k – 1) – 1 = 5
4 k – 2 – 1 = 5 ⇒ 4k = 8
k = 2

Question 5.
Let A,B,C ⊂ N and a function f: A → B be defined by f(x) = 2x + 1 and g : B → C be defined by g(x) = x2. Find the range of fog and gof

Solution:

f(x) = 2x + 1
g(x) = x2
fog(x) = fg(x)) = f(x2) = 2x2 + 1
gof(x) = g(f(x)) = g(2x + 1) = (2x + 1)2
= 4x2 + 4x + 1
Range of fog is
{y/y = 2x2 + 1, x ∈ N}
Range of gof is
{y/y = (2x + 1)2, x ∈ N}.

Question 6.
If f(x) = x2 – 1. Find (i)f(x) = x2 – 1, (ii)fofof

Solution:

(i) f(x) = x2 – 1
fof(x) = f(fx)) = f(x2 – 1)
= (x2 – 1 )2 – 1;
= x4 – 2x2 + 1 – 1
= x4 – 2x2
(ii) fofof = f o f(f(x))
= f o f (x4 – 2x2)
= f(f(x4 – 2x2))
= (x4 – 2x2)2 – 1 =
= x8 – 4x6 + 4x4 – 1

Question 7.
If f: R → R and g : R → R are defined by f(x) = x5 and g(x) = x4 then check if f,g are one-one and fog is one-one?

Solution:

f(x) = x5
g(x) = x4
fog = fog(x) = f(g(x)) = f(x4)
= (x4)5 = x20
f is one-one, g is not one-one.
∵ g(1) = 14 = 1
g(-1) = ( -1)4 = 1
Different elements have same images
fog is not one-one. [∵ fog (1) = fog (-1) = 1]

Question 8.
Consider the functions f(x), g(x), h(x) as given below. Show that (fog)oh = fo(goh) in each case.
(i) f(x) = x – 1, g(x) = 3x + 1 and h(x) = x2
(ii) f(x) = x2, g(x) = 2x and h(x) = x + 4
(iii) f(x) = x – 4, g(x) = x2 and h(x) = 3x – 5

Solution:

(i) f(x) = x – 1, g(x) = 3x + 1 and h(x) = x2
f(x) = x – 1
g(x) = 3x + 1
f(x) = x2
(fog)oh = fo(goh)
LHS = (fog)oh
fog = f(g(x)) = f(3x + 1) = 3x + 1 – 1 = 3x
(fog)oh = (fog)(h(x)) = (fog)(x2) = 3×2 …(1)
RHS = fo(goh)
goh = g(h(x)) = g(x2) = 3x2 + 1
fo(goh) = f(3x2 + 1) = 3x2 + 1 – 1= 3x2 …(2)
LHS = RHS Hence it is verified.

(ii) f(x) = x2, g(x) = 2x, h(x) = x + 4
(fog)oh = fo(goh)
LHS = (fog)oh
fog = f(g(x)) = f(2x) = (2x)2 = 4x2
(fog)oh = (fog) h(x) = (fog) (x + 4)
= 4(x + 4)2 = 4(x2 + 8x+16)
= 4x2 + 32x + 64 …(1)
RHS = fo(goh) goh = g(h(x)) = g(x + 4)
= 2(x + 4) = (2x + 8)
fo(goh) = f(goh) = f(2x + 8) = (2x + 8)2
= 4x2 + 32x + 64 …(2)
(1) = (2)
LHS = RHS
∴ (fog)oh = fo(goh) It is proved.

(iii) f(x) = x – 4, g(x) = x2, h(x) = 3x – 5
(fog)oh = fo(goh)
LHS = (fog)oh
fog = f(g(x)) = f(x2) = x2 – 4
(fog)oh = (fog)(3x – 5) = (3x – 5)2 – 4
= 9x2 – 30x + 25 -4
= 9x2 – 30x + 21 …(1)
∴ RHS = fo(goh)
(goh) = g(h(x)) = g(3x – 5) = (3x – 5)2
= 9x2 – 30x + 25
fo(goh) = f(9x2 – 30 x + 25)
= 9x2 – 30x + 25 – 4
= 9x2 – 30x + 21 …(2)
(1) = (2)
LHS = RHS
∴ (fog)oh = fo(goh)
It is proved.

Question 9.
Let f ={(-1, 3),(0, -1),(2, -9)} be a linear function from Z into Z . Find f(x).

Solution:

f ={(-1,3), (o,-1), 2,-9)
f(x) = (ax) + b ….(1)
is the equation of all linear functions.
∴ f(-1) = 3
f(0) = -1
f(2) = -9
f(x) = ax + b
f(-1) = -a + b = 3 …(2)
f(0) = b = -1
-a – 1 = 3 [∵ substituting b = – 1 in (2)]
-a = 4
a = -A
The linear function is -4x – 1. [From (1)]

Question 10.
In electrical circuit theory, a circuit C(t) is called a linear circuit if it satisfies the superposition principle given by C(at1 + bt2) = aC(t1) + bC(t2), where a,b are constants. Show that the circuit C(t) = 31 is linear.

Solution:

Given C(t) = 3t. To prove that the function is linear
C(at1) = 3a(t1)
C(bt2) = 3 b(t2)
C(at1 + bt2) = 3 [at1 + bt2] = 3 at1 + 3 bt2
= a(3t1) + b(3t2) = a[C(t1) + b(Ct62)]
∴ Superposition principle is satisfied.
Hence C(t) = 3t is linear function.

## Chapter 1 Relations and Functions Ex 1.6

Question 1.
If n(A × B) = 6 and A = {1, 3} then n(B) is
(1) 1
(2) 2
(3) 3
(4) 6

(3) 3
Hint:
If n(A × B) = 6
A = {1, 1}, n(A) = 2
n(B) = 3

Question 2.
A = {a, b, p}, B = {2, 3}, C = {p, q, r, s} then
n[(A ∪ C) × B] is
(1) 8
(2) 20
(3) 12
(4) 16

(3) 12
Hint:
A = {a, b,p}, B = {2,3}, C = {p, q, r, s}
n (A ∪ C) × B
A ∪C = {a, b,p, q, r,s}
(A ∪C) × B = {{a, 2), (a, 3), (b, 2), (b, 3), (p, 2), (p, 3), (q, 2), (q, 3), (r, 2), (r, 3), (s, 2), (s, 3)
n [(A ∪ C) × B] = 12

Question 3.
If A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5,6,7,8} then state which of the following statement is true.
(1) (A × C) ⊂ (B × D)
(2) (B × D) ⊂ (A × C)
(3) (A × B) ⊂ (A × D)
(4) (D × A) ⊂ (B × A)

(1) (A × C) ⊂ (B × D)]
Hint:
A = {1, 2}, B = {1, 2, 3, 4},
C = {5, 6}, D ={5,6, 7, 8}
A × C ={(1,5), (1,6), (2, 5), (2, 6)}
B × D = {(1,5),(1,6),(1,7),(1,8),(2,5),(2,6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8)}
∴ (A × C) ⊂ B × D it is true

Question 4.
If there are 1024 relations from a set A = {1, 2, 3, 4, 5} to a set B, then the number of elements in B is
(1) 3
(2) 2
(3) 4
(4) 6

(2) 2
Hint:
n(A) = 5
n(B) = x
n(A × B) = 1024 = 210
25x = 210
⇒ 5x = 10
⇒ x =2

Question 5.
The range of the relation R = {(x, x2)|x is a prime number less than 13} is
(1) {2,3,5,7}
(2) {2,3,5,7,11}
(3) {4,9,25,49,121}
(4) {1,4,9,25,49,121}

(3) {4, 9, 25, 49,121}]
Hint:
R = {(x, x2)/X is a prime number < 13}
The squares of 2, 3, 5, 7, 11 are
{4, 9, 25,49,121}

Question 6.
If the ordered pairs (a + 2,4) and (5,2a+b)are equal then (a,b) is
(1) (2,-2)
(2) (5,1)
(3) (2,3)
(4) (3,-2)

(4) (3,-2)
Hint:
(a + 2, 4), (5, 2a + b)
a + 2 = 5
a = 3
2a + b = 4
6 + b = 4
b = -2

Question 7.
Let n(A) = m and n(B) = n then the total number of non-empty relations that can be defined from A to B is
(1) mn
(2) nm
(3) 2mn-1
(4) 2mn

(4) 2mn
Hint:
n(A) = m,n(B) = n
n(A × B) = 2mn

Question 8.
If {(a,8),(6,b)}represents an identity function, then the value of a and b are respectively
(1) (8,6)
(2) (8,8)
(3) (6,8)
(4) (6,6)

(1) (8,6)
Hint:
{{a, 8), (6,b)}
a = 8
b = 6

Question 9.
Let A = {1,2,3,4} and B = {4,8,9,10}. A function f : A → B given by f = {(1,4),(2,8),(3,9),(4,10)} is a
(1) Many-one function
(2) Identity function
(3) One-to-one function
(4) Into function
(3) One-to one function
Hint:
A = {1, 2, 3, 4), B = {4, 8, 9,10}

Question 10.
If f(x) = 2x2 and g (x) = 13x, Then fog is

Hint:

Queston 11.
If f: A → B is a bijective function and if n(B) = 7, then n(A) is equal to
(1) 7
(2) 49
(3) 1
(4) 14

(1) 7
Hint:
In a bijective function, n(A) = n(B)
⇒ n(A) = 7

Question 12.
Let f and g be two functions given by f = {(0,1), (2,0), (3, -4), (4,2), (5,7)} g = {(0,2), (1, 0), (2, 4), (-4, 2), (7,0)} then the range of fog is
(1) {0,2,3,4,5}
(2) {-4,1,0,2,7}
(3) {1,2,3,4,5}
(4) {0,1,2}

(4) {0,1,2}
Hint:
gof = g(f(x))
fog = f(g(x))
= {(0,2),(1,0),(2,4),(-4,2),(7,0)}
Range of fog = {0,1,2}

Question 13.
Let f(x) = 1+x2−−−−−√ then
(1) f(xy) = f(x),f(y)
(2) f(xy) > f(x),f(y)
(3) f(xy) < f(x).f(y)
(4) None of these

(3) f(xy) < f(x).f(y)
Hint:

Question 14.
If g = {(1, 1),(2, 3),(3, 5),(4, 7)} is a function given by g(x) = ∝x + β then the values of ∝ and β are
(1) (-1,2)
(2) (2,-1)
(3) (-1,-2)
(4) (1,2)

(2) (2,-1)
Hint:
g(x) = αx + β
α = 2
β = -1
g(x) = 2x – 1
g(1) = 2(1) – 1 = 1
g(2) = 2(2) – 1 = 3
g(3) = 2(3) – 1 = 5
g(4) = 2(4) – 1 = 7

Question 15.
f(x) = (x + 1)3 – (x – 1)3 represents a function which is
(1) linear
(2) cubic
(3) reciprocal