Category: Class 11

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 1 Sets, Relations and Functions Ex 1.5 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets, Relations and Functions Ex 1.5

Choose the correct or the most suitable answer.

Question 1.
If A = {(x,y): y = e^x, x ∈ R} and B = {(x, y): y = e^-x, x ∈ R},then n(A ∩ B) is
(1) Infinity
(2) 0
(3) 1
(4) 2
Answer:
(3) 1

Explaination:
Given
A = {(x,y): y = e^x, x ∈ R}
B = {(x,y): y = e^-x, x ∈ R}
Consider the curve y = e^x
When x = 0 ⇒ y = e^-0 = 1
When x = ∞ ⇒ y = e^-∞ = ∞
When x = -∞ ⇒ y = e^∞ = 0

Consider the curve y = e^-x.
When x = 0 ⇒ y = e^-0 = 1
When x = ∞ ⇒ y = e^-∞ = 0
When x = -∞ ⇒ y = e^∞ = ∞


(0, 1) is the only point common to y = e<sup.x and y = e^-x
∴ A ∩ B = {(0,1)} ⇒ n (A ∩ B ) = 1

Question 2.
If A = { ( x, y): y = sin x, x ∈ R } and B = { ( x, y): y = cos x, x ∈ R} then A ∩ B contains
(1) no element
(2) infinitely many elements
(3) only one element
(4) cannot be determined
Answer:
(2) infinitely many elements

Explaination:
Given A = { (x, y): y = sin x, x ∈ R}
B = {(x, y): y = cos x, x ∈ R }
Consider the equations y = sin x and y = cos x
sin x = cos x ⇒ \(\frac{\sin x}{\cos x}\) = 1
tan x = 1 ⇒ x = nπ + \(\) for n ∈ z
There are infinite number of common points for the sets A and B

Question 3.
The relation R defined on a set A = {0, -1, 1, 2} by x R y if |x² + y²| ≤ 2,
then which one of the following is true.
(1) R = {(0,0), (0,-1), (0,1), (-1,0), (-1,1), (1,2), (1,0)}
(2) R^-1 = {(0, 0), (0, -1), (0, 1), (-1, 0), (1, 0)}
(3) Domain of R is {0,- 1, 1, 2}
(4) Range of R is {0, -1, 1}
Answer:
(4) Range of R is {0, -1, 1}

Explaination:
A= {0, -1, 1, 2}
|x² + y²| ≤ 2
The values of x and y can be 0, -1 or 1
So range = {0, -1, 1}

Question 4.
If f(x) = |x – 2 | + |x + 2| x ∈ R then
(1)
(2)
(3)
(4)

Answer:
(1)

Explaination:
f(x) = |x – 2| + |x + 2|, x ∈ R
Divide the Real line into three intervals

In the interval (2, ∞), both the factors x – 2 and x + 2 are positive.
∴ f(x) = x – 2 + x + 2 = 2x
f(x) = 2x for all x ∈ (2, ∞)

In the interval (- ∞, – 2 ] both the factors x – 2 and x + 2 are negative.
∴ f(x) = – (x – 2) – (x + 2)
= – x + 2 – x – 2 = – 2x
∴ f(x) = – 2x for all x ∈ (- ∞,- 2]

In the interval (—2, 2], the factor x – 2 is negative and the factor x + 2 is positive.
∴ f(x) = – (x – 2) + (x + 2)
f(x) = – x + 2 + x + 2 = 4
Thus f(x) = 4 for all x ∈ (- 2, – 2]

Question 5.
Let R be the set of all real numbers. Consider the following subsets of the plane R × R:
S = { (x, y): y = x + 1 and 0 < x < 27 and
T = {(x, y): x – y is an integer} Then which of the following is true?
(1) T is an equivalence relation but S Is not an equivalence relation
(2) Neither S nor T is an equivalence relation
(3) Both S and T are equivalence relation
(4) S is an equivalence relation but T is not an equivalence relation.
Answer:
(1) T is an equivalence relation but S Is not an equivalence relation

Explanation:
(0, 1), (1, 2) it is not an equivalence relation
T is an equivalence relation

Question 6.
Let A and B be subsets of the universal set N, the set of natural numbers. Then A’ ∪ [(A ∩ B) ∪ B’] is
(1) A
(2) A’
(3) B
(4) N
Answer:
(4) N

Explaination:
Let N = {1, 2, 3, ……….. 10}
A = { 1, 2, 3, 4, 5 }
B = {6, 7, 8, 9, 10}
A’ = {6, 7, 8, 9, 10 }
B’ = { 1 , 2, 3, 4, 5 }
A ∪ B = {1, 2, 3, 4, 5} ∪ {6, 7, 8, 9, 10}
A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(A ∪ B) ∩ B’ = {1,2, 3, 4, 5, 6,7, 8, 9,10} ∩ { 1, 2, 3, 4 , 5 }
(A ∪ B) ∩ B’= {1,2, 3,4, 5}
A’ ∪ [(A ∪ B ) ∩ B’] = { 6, 7, 8, 9 , 10 } ∪ {1, 2, 3, 4, 5 }
= {1, 2, 3, 4,5, 6, 7, 8, 9, 10}
A’ ∪ [(A ∪ B) ∩ B’] = N

Question 7.
The number of students who take both the subjects Mathematics and Chemistry is 70. This represents 10% of the enrollment in Mathematics and 14% of the enrollment in Chemistry. The number of students takes atleast one of these two subjects, is
(1) 1120
(2) 1130
(3) 1100
(4) Insufficient data
Answer:
(2) 1130

Explanation:
n(M ∪ C) = n(M) + n(C) – n(M ∩ C)
= 700 + 500 – 70
= 1130

Question 8.
If n[ (A × B) n (A × C) ] = 8 and n(B ∩ C) = 2 then n(A) is
(1) 6
(2) 4
(3) 8
(4) 16
Answer:
(2) 4

Explaination:
Given n[(A × B) n (A × C)] = 8
n(B ∩ C) = 2
n[(A × B) ∩ (A × C)] = 8
A × (B ∩ C) = (A × B) ∩ (A × C) ]
⇒ n [A × (B ∩ C)] = 8
⇒ n(A) . n (B ∩ C) = 8
⇒ n(A). 2 = 8
⇒ n(A) = \(\frac{8}{2}\) = 4

Question 9.
If n(A) = 2 and n(B ∪ C) = 3,then n[(A × B) ∪ (A × C)] is
(1) 2³
(2) 3²
(3) 6
(4) 5
Answer:
(3) 6

Explaination:
n[(A × B) ∪ (A × C)] = n[ A × (B ∪ C)] = n(A) × n(B ∪ C)
= 2 × 3
= 6

Question 10.
If two sets A and B have 17 elements in common, then the number of elements common to the set A × B and B × A is
(1) 2¹⁷
(2) 17²
(3) 34
(4) Insufficient data
Answer:
(2) 17²

Explanation:
If two sets A and B have 17 elements in common, then the number of elements common to the set A × B
and B × A is 17²

Question 11.
For non-empty sets A and B, if A ⊂ B then (A × B) ∩ (B × A) is equal to
(1) A ∩ B
(2) A × A
(3) B × B
(4) None of these
Answer:
(2) A × A

Explanation:
When A ⊂ B, (A × B) ∩ (B × A) = A × A

Question 12.
The number of relations on a set containing 3 elements is
(1) 9
(2) 81
(3) 512
(4) 1024
Answer:
(3) 512

Explanation:
The number of relations on a set containing n elements is 2ⁿ². Here n = 3
∴ Required number = 2³² = 2⁹
= 512

Question 13.
Let R be the universal relation on a set X with more than one element. Then R is
(1) not reflexive
(2) not symmetric
(3) transitive
(4) none of the above
Answer:
(3) transitive

Explanation:
Given R is a universal relation on the set X.
The universal relation is always an equivalence relation.
R is reflexive, symmetric, and transitive.

Question 14.
Let X = { 1, 2, 3, 4 } and R = { ( 1, 1 ), (1, 2), (1, 3), (2, 2), (3, 3), (2, 1), (3, 1), (1, 4), ( 4, 1) } . Then R is
(1) Reflexive
(2) Symmetric
(3) Transitive
(4) Equivalence
Answer:
(2) Symmetric

Explanation:
(4, 4} ∉ R ⇒ R is not reflexive
(1, 4), (4, 1) ∈ R ⇒ R is symmetric
(1, 4), (4, 1) ∈ R but (4, 4) ∉ R
So R is not transitive

Question 15.
The range of the function \(\frac{1}{1-2 \sin x}\) is
(1) (- ∞, – 1) ∪ (\(\frac{1}{3}\), ∞)
(2) (-1, \(\frac{1}{3}\))
(3) [-1, \(\frac{1}{3}\)]
(4) (- ∞, – 1] ∪ [\(\frac{1}{3}\), ∞)
Answer:
(4) (- ∞, – 1] ∪ [\(\frac{1}{3}\), ∞)

Explaination:

Question 16.
The range of the function
f(x) = |[x] – x|, x ∈ R is
(1) [0, 1]
(2) [0, ∞)
(3) [0, 1)
(4) (0, 1)
Answer:
(3) [0, 1)

Explaination:
f(x) = |[x] – x|
When x = 1 ,
we have [x] = [1] = 1
f(1) = |1 – 1| = o

When x = 1.5
we have [x} = [1.5] = 1
f(1.5) = |1 – 1.5| = |- 0.5| = 0.5

When x = 2.5
we have [x] = [2.5] = 2
f (2.5) = |2 – 2.5| = |- 0.5| = 0.5

When x = – 2.5
we have [x] = [- 2.5] = – 3
f (-2.5) = |- 3 – (-2.5| = |- 3 + 0.5| = |- 0.5| = 0.5
∴ Range of f(x) = |[x] – x| is [0, 1)

Question 17.
The rule f(x) = x² is a bijection if the domain and the co – domain are given by
(1) R, R
(2) R,(0, ∞)
(3) (0, ∞)R
(4) [0, ∞), [0, ∞)
Answer:
(4) [0, ∞), [0, ∞)

Explaination:
Let x ∈ R, then x can be negative or zero or positive.
Given f(x) = x²
The image of x under f is always positive since x² is positive for x = 1 and x = – 1 ∈ R
f(1) = 1² = 1
f(-1) = (-1)² = 1
∴ 1, – 1 have the same image
∴ f is not one – one if the domain is R.
Suppose the domain is [0, ∞) then f is one – one and onto.
Domain = [0, ∞)
Co-domain = [0, ∞)

Question 18.
The number of constant functions from a set containing m elements to a set containing n elements is
(1) mn
(2) m
(3) n
(4) m + n
Answer:
(3) n

Explanation:
Let A be a set having m elements and B be a set having n elements.
When all the elements of A mapped onto the first element of B we get the first constant function. When all the elements of A mapped onto the second element of B we get the second constant function.

When all the elements of A mapped on to the n^th element of B, we get the n^th constant function.
∴ The number of constant functions possible is n.

Question 19.
The function f:[0, 2π] → [- 1, 1] defined by f(x) = sin x is
(1) one to one
(2) onto
(3) bijection
(4) cannot be defined
Answer:
(2) onto

Explaination:
f : [0, 2π] → [- 1, 1]
Defined by f (x) = sin x
f(0) = sin 0 = 0

Question 20.
If the function f : [-3, 3] → S defined by f(x ) = x² is onto, then S is
(1) [-9, 9]
(2) R
(3) [-3, 3]
(4) [0, 9]
Answer:
(4) [0, 9]

Explaination:
f: [-3, 3] → S defined by f(x) = x²
f(-3) = (-3)² = 9
f(0) = 0² = o
f(3) = 3² = 9
∴ S = [0, 9]

Question 21.
Let X = {1, 2, 3, 4}, Y = { a, b , c, d } and f = {(1, a), (4, b), (2, c), (3, d), (2, d)}. Then f is
(1) an one – to – one function
(2) an onto function
(3) a function which is not one to one
(4) not a function
Answer:
(4) not a function

Explaination:
X = {1, 2, 3, 4}, Y = {a, b, c, d>
f = {(1, a), (4, b), (2, c), (3, d), (2, d)}

f is not a function since 2 ∈ X has two images c and d.

Question 22.
The inverse of

(1)
(2)
(3)
(4)
Answer:
(1)

Explaination:

Let f(x) = x if x < 1 —— (1)
Put y = x then
(1) ⇒ f(x) = y
⇒ x = f^-1(y) if y < 1
⇒ y = f^-1(y) if y < 1
⇒ f^-1(x) = x if x < 1
Let f(x) = x² if 1 ≤ x ≤ 4 —– (2)
Put y = x² ⇒ x = √y, if 1 ≤ y ≤ 16
then (2) ⇒ f(x) = y
⇒ x = f^-1(y) if 1 ≤ y ≤ 16
⇒ √(y) = f^-1(y) if 1 ≤ y ≤ 16
⇒ √x = f^-1(y) if 1 ≤ x ≤ 16
Let f(x) = 8√x if x > 4 ———– (3)
Put y = 8√x ⇒ y² = 64 x
⇒ x = \(\frac{y^{2}}{64}\) if y>16
then (3) ⇒ f(x) = y
⇒ x = f^-1(y) if y > 16
⇒ √y = f^-1(y) if y > 16
⇒ \(\frac{y^{2}}{64}\) = f^-1(x) if x > 16

Question 23.
Let f: R → R be defined by f(x) = 1 – |x|. Then the range of f is
(1) R
(2) (1, ∞)
(3) (-1, ∞)
(4) (- ∞, 1)
Answer:
(4) (- ∞, 1)

Explaination:
f: R ➝ R defined by
f(x) = 1 – |x|
For example,
f(1) = 1 – 1 = 0
f(8) = 1 – 8 = -1
f(-9) = 1 – 9 = -8
f(-0.2) = 1 – 0.2 = 0.8
so range = (-∞, 1]

Question 24.
The function f : R → R be defined by f(x) = sin x + cos x is
(1) an odd function
(2) neither an odd function nor an even function
(3) an even function
(4) both odd function and even function
Answer:
(2) neither an odd function nor an even function

Explaination:
f : R → R is defined by f(x) = sin x + cos x
f(-x) = sin (-x) + cos (-x) = -sin x + cos x ≠ f (x)
If f(-x) = -f(x) then f(x) is an odd function.
If f(-x) = f(x) then f(x) is an even function.
∴ f (x) is neither odd function nor an even function.

Question 25.
The function f : R → R be defined by

(1) an odd function
(2) neither an odd function nor an even function
(3) an even function
(4) both odd function and even function.
Answer:
(3) an even function

Explanation:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 1 Sets, Relations and Functions Ex 1.4 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets, Relations and Functions Ex 1.4

Question 1.
For the curve y = x³ given in figure below, draw
(i) y = – x³
(ii) y = x³ + 1
(iii) y = x³ – 1
(iv) y = (x + 1)³ with the same scale.

Answer:
(i) y = – x³

The graph y = – x³ is the reflection of the graph y = x³ about x-axis.
The graph of y = – f( x) is the reflection of the graph of y = f( x) about x – axis.

(ii) y = x³ + 1

The graph of y = x³ + 1, causes the graph y = x³ a shift to the upward by 1 unit.
The graph of y = f(x) + d, d > 0 causes the graph y = f(x) a shift to the upward by d units.

(iii) y = x³ – 1

The graph of y = x³ – 1, causes the graph y = x³ a shift to the downward by 1 unit.
The graph of y = f(x) – d, d > 0 causes the graph y = f(x) a shift to the downward by d units.

(iv) y = (x + 1)⁰⁰³³

The graph y = (x + 1)³ causes the graph of y = x³ a shift to the left by 1 unit.
The graph of y = f(x + c), c > 0 causes the graph y = f(x) a shift to the left by c units.

Question 2.
For the given curve y = x^1/3 given in figure draw
(i) y = – x^1/3
(ii) y = x^1/3 + 1
(iii) y = x^1/3 – 1
(iv) y = (x + 1)^1/3
Answer:

(i) y = – x^1/3
-y = x^1/3
(-y)³ = x
-y³ = x
When y = 0 ⇒ – 0³ ⇒ x = 0
y = 1 ⇒ – 1³ = x ⇒ x = – 1
y = 2 ⇒ – 2³ = x ⇒ x = – 8
y = 3 ⇒ – 3³ = x ⇒ x = – 27
y = -1 ⇒ – (-1)³ = x ⇒ x = 1
y = -2 ⇒ – (-2)³ = x ⇒ x = 8
y = -3 ⇒ – (- 3)³ = x ⇒ x = 27

The graph of y = – x^1/3 is the reflection of the graph of y = x^1/3 about the x-axis.
The graph of y = – f(x) is the reflection of the graph of y = f(x) about x – axis.

(ii) y = x^1/3 + 1
y – 1 = x^1/3
⇒ (y – 1)³ = x

When
y = 0 ⇒ (0 – 1 )³ = x ⇒ x = – 1
y = 1 ⇒ (1 – 1)³ = x ⇒ x = 0
y = 2 ⇒ ( 2 – 1 )³ = x ⇒ x = 1
y = 3 ⇒ (3 – 1)³ = x ⇒ x = 8
y = -1 ⇒ (-1 – 1)³ = x ⇒ x = – 8
y = -2 ⇒ (-2 – 1)³ = x ⇒ x = – 27

The graph of y = x^1/3 + 1 causes the graph y = x^1/3 a shift to the upward by 1 unit.
The graph of y = f(x) + d, d > 0 causes the graph y = f(x) a shift to the upward by d units.

(iii) y = x^1/3 – 1
y + 1 = x^1/3
( y + 1)³ = x
When
y = 0 ⇒ (0 + 1)³ = x ⇒ x = 1
y = 1 ⇒ ( 1 + 1)³ = x ⇒ x = 8
y = 2 ⇒ (2 + 1)³ = x ⇒ x = 27
y = – 1 ⇒ (-1 + 1)³ = x ⇒ x = 0
y = – 2 ⇒ (-2 + 1)³ = x ⇒ x = – 1
y = – 3 ⇒ (-3 + 1)³ = x ⇒ x = – 8

The graph of y = x^1/3 – 1 causes the graph y = x^1/3 a shift to the downward by 1 unit.
The graph of y = f(x) – d, d > 0 causes the graph y = f(x) a shift to the downward by d units.

(iv) (x + 1)^1/3
y³ = x + 1
When
y = 0 ⇒ 0³ = x + 1 ⇒ x = -1
y = 1 ⇒ 1³ = x + 1 ⇒ x = 0
y = 2 ⇒ 2³ = x + 1 ⇒ x = 8 – 1 = 7
y = 3 ⇒ 3³= x + 1 ⇒ x = 27 – 1 = 26
y = – 1 ⇒ (-1)³ = x + 1 ⇒ x = – 1 – 1 = – 2
y = -2 ⇒ (-2)³ = x + 1 ⇒ x = – 8 – 1 = – 9

The graph of y = (x + 1)³ causes the graph y = x^1/3 a shift to the left by 1 unit.
The graph of y = f(x + c), c > 0 causes the graph y = f(x) a shift to the left by c units.

Question 3.
Graph the functions f(x) = x³ and g (x) = \(\sqrt[3]{x}\) on the same coordinate plane. Find fog and the graph it on the plane as well. Explain your results.
Answer:
Given functions are f(x) = x3 and g(x) = x^1/3
fog (x) = f(g(x))
= f\(\left(x^{\frac{1}{3}}\right)\)
= \(\left(x^{\frac{1}{3}}\right)^{3}\) = x
f(x) = x³

g(x) = x^1/3

Graph of fog(x) = x

Since fog(x) = x is symmetric about the line y = x, g(x) is the inverse image of f(x).
∴ g(x) = f^-1(x)

Question 4.
Write steps to obtain the graph of the function y = 3 (x – 1 )² + 5 from the graph y = x²
Answer:
Step 1:
Draw the graph y = x²

Step 2:
The graph of y = (x – 1)² shifts to the right for one unit.

The graph of y = (x – 1 )² shifts the graph
y = x² to the right by 1 unit.
The graph of y = f(x – c), c > 0 causes the graph y = f(x) a shift to the right by c units.

Step 3:
The graph of y = 3 (x – 1)² compresses towards y – axis that is moves away from the x – axis since the multiplying factor is which is greater than 1.

The graph of y = 3 (x – 1)² compresses the graph y = (x – 1)² towards the y-axis that is moving away from the x-axis since the multiplying factor is greater than 1.

For the graph y = kf(x), If k is a positive constant greater than one, the graph moves away from the x-axis. If k is a positive constant less than one, the graph moves towards the x-axis.

Step 4:
The graph of y = 3(x – 1)² + 5 causes the shift to the upward for 5 units.


The graph of y = 3(x – 1)² + 5 causes the graph y = 3(x – 1)² shifts to the upward for 5 units.

The graph of y = f(x) + d, d > 0 causes the graph y = f(x) a shift to the upward by d units.

Question 5.
From the curve y = sin x, graph the functions.
(i) y = sin (- x)
(ii) y = -sin(-x)
(iii) y = sin\(\left(\frac{\pi}{2}+x\right)\) which is cos x
(iv) y = sin\(\left(\frac{\pi}{2}-x\right)\) which is also cos x
(Refer Trigonometry )
Answer:
y = sin x

(i) y = sin(-x)
y = – sin x

The graph of y = sin (- x) is the reflection of the graph of y = sin x about y-axis.
The graph of y = f(- x) is the reflection of the graph of y = f(x) about y – axis.

(ii) y = – sin (-x)
y = sin x


y = – sin (-x) is the reflection of y = sin (-x) about the x – axis.
The graph of y = – f( x) is the reflection of the graph of y = f( x) about x – axis.

(iii) y = sin\(\left(\frac{\pi}{2}+x\right)\)


The graph of y = sin \(\left(\frac{\pi}{2}+x\right)\) causes y = sin x a shift to the left by \(\frac{\pi}{2}\) units.
The graph of y = f(x + c), c > 0 causes the graph y = f(x) a shift to the left by c units.

(iv) y = sin\(\left(\frac{\pi}{2}-x\right)\)



The graph of sin \(\left(\frac{\pi}{2}-x\right)\) causes the graph y = sin x a shift to the right by \(\frac{\pi}{2}\) unit.

The graph of y = f(x – c), c > 0 causes the graph y = f(x) a shift to the right by c units.

Question 6.
From the curve y = x draw
(i) y = -x
(ii) y = 2x
(iii) y = x + 1
(iv) y = \(\frac{1}{2}\)x + 1
(v) 2x + y + 3 = 0
Answer:

(i) y = -x

Graph of y = – x is the reflection of the graph of y = x about the x – axis.
The graph of y = – f(x) is the reflection of the graph of y = f(x) about x – axis.

(ii) y = 2x

The graph of y = 2x compresses the graph y = x towards the y-axis that is moving away from the x-axis since the multiplying factor is 2 which is greater than 1.

The graph of y = k f(x), k > 0 moves away from the x-axis if k is greater than 1.

(iii) y = x + 1


The graph of y = x + 1 causes the graph y = x shift to upward by 1 unit.

The graph of y = f(x) + d, d > 0 causes the graph y = f(x) a shift to the upward by d units.

(iv) y = \(\frac{1}{2}\)x + 1
When
x = 0 ⇒ y = \(\frac{1}{2}\) × 0 + 1 = 1
x = 2 ⇒ y = \(\frac{1}{2}\) × 2 + 1 = 2
x = 4 ⇒ y = \(\frac{1}{2}\) × 4 + 1 = 2 + 1 = 3
x = 6 ⇒ y = \(\frac{1}{2}\) × 6 + 1 = 3 + 1 = 4
x = – 2 ⇒ y = \(\frac{1}{2}\) × – 2 +1= – 1 + 1 = 0
x = – 4 ⇒ y = \(\frac{1}{2}\) × – 4 + 1 = – 2 + 1 = – 1
x = – 6 ⇒ y = \(\frac{1}{2}\) × – 6 + 1 = – 3 + 1 = – 2


The graph of y = \(\frac{1}{2}\)x + 1 stretches the graph y = x towards the x – axis since the multiplying factor is \(\frac{1}{2}\) which is less than 1 and shifts to the upward by 1 unit.

The graph of y = kf(x), k > 0 moves towards the x-axis if k is less than 1.
The graph of y = f(x) + d, d >0 causes the graph y = f(x) a shift to the upward by d units.

(v) 2x + y + 3 = 0
y = -2x – 3
When
x = 0 ⇒ y = -2 × 0 – 3 = -3
x = 1 ⇒ y = -2 × 1 – 3 = -5
x = \(\frac{1}{2}\) ⇒ y = – 2 × \(\frac{1}{2}\) – 3 = – 1 – 3 = – 4
x = 2 ⇒ y = -2 × 2 – 3 = – 4 – 3 = – 7
x = – 1 ⇒ y = -2 × – 1 – 3 = 2 – 3 = – 1
x = – 2 ⇒ y = 2 × – 2 – 3 = 4 – 3 = 1
x = – 3 ⇒ y = -2 × -3 – 3 = 6 – 3 = 3


The graph of y = – 2x – 3 stretches the graph y = x towards the x-axis since the multiplying factor is – 2 which is less than 1 and causes the shift to the downward by 3 units.5

The graph of y = kf(x), k > 0 moves towards the x-axis if k is less than 1.
The graph of y = f(x) – d, d >0 causes the graph y = f(x) a shift to the downward by d units.

Question 7.
From the curve y = |x| draw
(i) y = |x – 1| + 1
(ii) y = |x + 1| – 1
(iii) y = |x + 2| – 3
Answer:
y = |x|

(i) y = |x – 1| + 1
(a) Consider y = |x – 1|

x = 0 ⇒ y = – x + 1 ⇒ y = 1
x = 1 ⇒ y = x – 1 ⇒ y = 0
x = 2 ⇒ y = x – 1 ⇒ y = 1
x = 3 ⇒ y = x – 1 ⇒ y = 2
x = – 1 ⇒ y = – x + 1 ⇒ y = 2
x = – 2 ⇒ y = – x + 1 ⇒ y = 3

The graph of y = |x – 1| causes the graph y = |x| a shift to the right by 1 unit.

The graph of y = f(x – c), c > 0 causes the graph y = f(x) a shift to the right by c units.

(b) Consider y = |x – 1| + 1


x = 0 ⇒ y = – x + 2 ⇒ y = 2
x = 1 ⇒ y = x ⇒ y = 1
x = 2 ⇒ y = x ⇒ y = 2
x = 3 ⇒ y = x ⇒ y = 3
x = – 1 ⇒ y = – x + 2 ⇒ y = 3
x = – 2 ⇒ y = – x + 2 ⇒ y = 4

The graph of y = |x – 1| + 1 shift the graph y = |x| to the right by 1 unit and causes a shift to the upward by 1 unit.

The graph of y = f( x – c), c > 0 causes the graph y = f(x) a shift to the right by c units.
The graph of y = f(x) + d, d > 0 causes the graph y = f(x) a shift to the upward by d units.

(ii) y = |x + 1| – 1
(a) Consider y = |x + 1|

x = 0 ⇒ y = x + 1 ⇒ y = 1
x = 1 ⇒ y = x + 1 ⇒ y = 2
x = 2 ⇒ y = x + 1 ⇒ y = 3
x = 3 ⇒ y = x + 1 ⇒ y = 4
x = – 1 ⇒ y = x + 1 ⇒ y = 0
x = – 2 ⇒ y = – (x + 1) ⇒ y = 1
x = – 3 ⇒ y = – (x + 1) ⇒ y = 2

The graph of y = |x + 1| shifts the graph y = |x| to the left by 1 unit.

The graph of y = f( x + c), c > 0 causes the graph y = f(x) a shift to the left by e units.

(b) Consider y = |x + 1| – 1

x = 0 ⇒ y = x ⇒ y = 0
x = 1 ⇒ y = x ⇒ y = 1
x = 2 ⇒ y = x ⇒ y = 2
x = 3 ⇒ y = x ⇒ y = 3
x = – 1 ⇒ y = x ⇒ y = – 1
x = – 2 ⇒ y = – x – 2 ⇒ y = 0
x = – 3 ⇒ y = – x – 2 ⇒ y = 1
x = – 4 ⇒ y = – x – 5 ⇒ y = -1

The Graph of y = |x + 1 | – 1 shift the graph y = |x| to the left by 1 unit and causes a shift downward by 1 unit.

The graph of y = f(x + c), c > 0 causes the graph y = f(x) a shift to the left by c units.

The graph of y = f(x) – d, d > 0 causes the graph y = f(x) a shift to the downward by d units.

(iii) y = |x + 2| – 3
(a) Consider the curve y = |x + 2|

x = 0 ⇒ y = x + 2 ⇒ y = 2
x = 1 ⇒ y = x + 2 ⇒ y = 3
x = 2 ⇒ y = x + 2 ⇒ y = 4
x = 3 ⇒ y = x + 2 ⇒ y = 5
x = – 1 ⇒ y = x + 2 ⇒ y = 1
x = – 2 ⇒ y = x + 2 ⇒ y = 0
x = – 3 ⇒ y = – x – 2 ⇒ y = 1

The graph of y = |x + 2| shifts the graph y = |x| to the left by 2 units.

The graph of y = f(x + c), c > 0 causes the graph y = f(x) a shift to the left by c units.

(b) Consider the curve y = |x + 2| – 3

x = 0 ⇒ y = x – 1 ⇒ y = – 1
x = 1 ⇒ y = x – 1 ⇒ y = 0
x = 2 ⇒ y = x – 1 ⇒ y = 1
x = 3 ⇒ y = x – 1 ⇒ y = 2
x = – 1 ⇒ y = x – 1 ⇒ y = – 2
x = – 2 ⇒ y = x – 1 ⇒ y = – 3
x = – 3 ⇒ y = – x – 5 ⇒ y = – 2


The graph of y = |x + 2| – 3 shifts the graph y = |x| to the left by 2 units and causes a shift downward by 3 units.

The graph of y = f(x + c), c > 0 causes the graph y = f(x) a shift to the left by c units.
The graph of y = f(x) – d, d > 0 causes the graph y = f(x) a shift to the down ward by d units.

Question 8.
From the curve y = sin x, draw y = sin |x| (Hint: sin(- x) = – sin x)
Answer:
y = sin |x|
(a) y = sin x

(b) Consider y = sin |x|

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 1 Sets, Relations and Functions Ex 1.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets, Relations and Functions Ex 1.3

Question 1.
Suppose that 120 students are studying in 4 sections of eleventh standard in a school. Let A denote the set of students and B denote the set of the sections. Define a relation from A to B as “x related to y if the student x belongs to the section y”. Is this relation a function? What can you say about the inverse relation? Explain your answer.
Answer:
Given A denotes the set of students and B denotes the set of sections. Aslo given there 120 students and 4 sections.

Let f be a relation from A to B as “x related to y if the student x belongs to the section y”

Two are more students in A may belong to same section in B. But one student in A cannot belong to two or more sections in B. Every student in A can belong to any one of the section in B. Therefore / is a function.

In B we can have sections without students. Every element in B need not have preimage in A.
∴ f need not be onto.
Thus, f is a function and inverse relation for f need not exist.

Question 2.
Write the values of f at – 4, 1, -2, 7, 0 if

Answer:

When x = -4
f(x) = – x + 4
f(-4) = – (-4) + 4
= 4 + 4 = 8

When x = 1
f(x) = x – x²
f(1) = 1 – 1²
= 1 – 1 = 0

When x = -2
f(x) = x² – x
f(-2) = (-2)² – (-2)
= 4 + 2 = 6

When x – 7
f(x) = 0
⇒ f(7) = 0

When x = 0
f(x) = x² – x
⇒ f(0) = 0² – 0 = 0

Question 3.
Write the values of f at – 3, 5, 2, – 1, 0 if

Answer:

When x = – 3
f(x) = x² + x – 5
f(-3) = (-3)² + (-3) – 5
= 9 – 3 – 5
= 9 – 8 = 1

When x = 5
f(x) = x² + 3x – 2
f(5) = 5² + 3(5) – 2
= 25 + 15 – 2
= 40 – 2 = 38

When x = 2
f(x) = x² – 3
f(2) = 2² – 3 = 4 – 3 = 1

When x = – 1
f(x) = x² + x – 5
f(-1) = (-1)² – 1 – 5 = 1 – 1 – 5 = – 5

When x = 0
f(x) = x² – 3
f(0) = 0² – 3

Question 4.
State whether the following relations are functions or not. If it is a function, check for one – to – oneness and ontoness. If it is not a function, state why?
(i) If A = { a, b, c } and f = { (a, c), (b, c), (c, b) }; (f : A → A)
Answer:
A = { a, b, c }
f = {(a, c), (b, c), (c, b)}; f : A → A

f is a function since every element in the domain has a unique image in the codomain.
f is not one-one.
a, b belonging to the domain A has the same image in the codomain A. f is not onto since belonging to the codomain A does not have preimage in the domain A Thus the relation / is a function from A to A and it is neither one-one nor onto.

(ii) If X = { x, y, z } and f = { (x, y), (x, z), (z, x) }; (f: X → X)
Answer:
X = { x, y, z }
f = {(x, y), (x, z), (z , x) } f : X → X

The relation f: X → X is not a function since the element x in the domain has two images in the co-domain.

Question 5.
Let A = {1, 2, 3, 4} and B = { a, b , c, d } Give a function from A → B for each of the following.
(i) neither one-to-one nor onto.
Answer:

f = { (1, b) , (2, c) , (3, d) , (4, d)
f is a function, it not one to one and not onto.

(ii) not one – to – one but onto
Answer:
Does not exists

(iii) one – to – one but not onto
Answer:
Does not exist

(iv) one – to – one and onto
Answer:

f = { (1, a) , (2, b) , (3, c) , (4, d) }
f is a function which is one – to – one and onto.

Question 6.
Find the domain of \(\frac{1}{1-2 \sin x}\)
Answer:
Let f(x) = \(\frac{1}{1-2 \sin x}\)
When 1 – 2 sin x = 0
⇒ 1 = 2 sin x
sin x = \(\frac{1}{2}\)
⇒ sin x = sin \(\left(\frac{\pi}{6}\right)\)
x = nπ + (- 1)ⁿ\(\frac{\pi}{6}\), n ∈ Z
sin x = sin α ⇒ x = nπ + (-1)ⁿd, n ∈ Z
∴ Domain of f(x) is

Question 7.
Find the largest possible domain of the real valued function f(x) = \(\frac{\sqrt{4-x^{2}}}{\sqrt{x^{2}-9}}\)
Answer:

∴ For no real values of x, f (x) is defined.
∴ Domain of f(x) = { }

Question 8.
Find the range of the function \(\frac{1}{2 \cos x-1}\)
Answer:
Let f(x) = \(\frac{1}{2 \cos x-1}\)
Range of cosine function is
– 1 ≤ cos x ≤ 1
– 2 ≤ 2 cos x ≤ 2
– 1 ≤ 2 cos x – 1 ≤ 2 – 1
– 3 ≤ 2 cos x – 1 ≤ 1
\(-\frac{1}{3}\) ≥ \(\frac{1}{2 \cos x-1}\) ≥ 1

Question 9.
Show that the relation xy = – 2 is a function for a suitable domain. Find the domain and the range of the function.
Answer:
xy = – 2 ⇒ y = -2/x
which is a function
The domain is (-∞, 0) ∪ (0, ∞) and range is R – {0}

Question 10.
If f, g : R → R are defined by f(x) = |x| + x and g(x) = |x| – x find gof and fog.
Answer:
Given

Question 11.
If f, g, h are real-valued functions defined on R, then prove that (f + g)oh = foh + goh what can you say about fo(g + h )? Justify your answer.
Answer:
Given f : R → R , g : R → R and h : R → R (f + g) oh: R → R and (f o h + g o h) : R → R for any x ∈ R.
[(f + g)oh] (x) = (f + g) h(x)
= f(h(x)) + g(h(x))
= foh(x) + goh(x)
∴ (f + g)oh = foh + goh
Also fo(g + h)(x) = f((g + h)(x)) for any x ∈ R
= f(g(x) + h(x))
= f(g(x)) + f(h(x))
= fog (x) + foh(x)
∴ fo(g + h) = fog +foh

Question 12.
If f : R → R is defined by f( x ) = 3x – 5, Prove that f is a bijection and find its inverse.
Answer:
Given f(x) = 3x – 5
Let y = 3x – 5
y + 5 = 3x
⇒ \(\frac{y+5}{3}\) = x
Let g(y) = \(\frac{y+5}{3}\)
gof (x) = g(f(x))
= g(3x – 5)
= \(\frac{3 x-5+5}{3}\) = \(\frac{3 x}{3}\) = x
gof (x) = x
fog (y) = f(g(y))
= f\(\left(\frac{y+5}{3}\right)\)
= 3\(\left(\frac{y+5}{3}\right)\) – 5
= y + 5 – 5
fog(y) = y
∴ gof = I_x and fog = I_Y
Hence f and g are bijections and inverses to each ot1er.
Hence f is a bijection and f^-1(y) = \(\frac{y+5}{3}\)
Replacing y by x we get f^-1(x) = \(\frac{x+5}{3}\)

Question 13.
The weight of the muscles of a man is a function of his bodyweight x and can be expressed as W ( x ) = 0.35x. Determine the domain of this function.
Answer:
W(x) = 0.35x
Since bodyweight x is positive and if it increases then W(x) also increases.
Domain is (0, ∞) i.e.,x > 0

Question 14.
The distance of an object falling is a function of time t and can be expressed as s ( t) = – 16t². Graph the function and determine if it is one – to – one.
Answer:
Given s (t) = – 16t²
s (t₁) = s (t₂) ⇒ – 16t₁² = – 16t₂²
⇒ t₁² = t₂²
⇒ ± t₁ = ± t₂
Since s (t₁) = s (t₁) 14 t₁ = t₂
∴ The function s(t) is not one-one
Graph of s(t) = – 16t²
Take the time along x – axis and distance along y – axis.

Question 15.
The total cost of airfare on a given route is comprised of the base cost C and the fuel Surcharge S in rupee. Both C and S are functions of the mileage m; C ( m ) = 0.4 m + 50 and S ( m ) = 0.03m. Determine a function for the total cost of a ticket in terms of the mileage and find the airfare for flying 1600 miles.
Answer:
C – base cost,
S = fuel surcharge,
m = mileage
C(m) = 0.4 m + 50
S(m) = 0.03 m
Total cost = C(m) + S(m)
= 0.4 m + 50 + 0.03 m
= 0.43 m + 50
for 1600 miles
T(c) = 0.43 (1600) + 50 = 688 + 50 = ₹ 738

Question 16.
A salesperson whose annual earnings can be represented by the function A (x) = 30,000 + 0.04 x, where x is the rupee value of the merchandise, he sells. His son is also in sales and his earnings are represented by the function S(x) = 25,000 + 0.05 x. Find (A + S)(x) and determine the total family income if they each sell Rs. 1,50,00,000 worth of merchandise.
Answer:
Given A (x) = 30,000 + 0.04 x
S (x) = 25,000 + 0.05x
A(x) + S(x) = 30,000 + 0.04 x + 25,000 + 0.05x
(A + S)(x) = 55,000 + 0.09 x
Given x = 1,50,00,000
Then (A + S)(x) = 55000 + 0.09 × 1,50,00,000
= 55000 + 1,35000000
Total family income = Rs. 14,05,000

Question 17.
The function for exchanging American dollars for Singapore Dollar on a given day is f (x) = 1.23x, where x represents the number of American dollars. On the same day, the function for exchanging Singapore Dollar to Indian Rupee is g(y) = 50.50y, where y represents the number of Singapore dollars. Write a function which will give the exchange rate of American dollars in terms of Indian rupee.
Answer:
Given f(x) = 1.23x
where x represents the number of American dollars
g(y) = 50.50y
where y represents the number of Singapore dollars.

To convert American dollars to Indian rupees, we must find
gof (x) = g(f(x))
= g (1.23x)
= 50.50 (1.23x)
= 62.115x
∴ The function for the exchange rate of American can dollars in terms of Indian rupees is
gof (x) = 62.1 15x

Question 18.
The owner of a small restaurant can prepare a particular meal at a cost of Rs. 100. He estimates that if the menu price of the meal is x rupees, then the number of customers who will order that meal at that price in an evening is given by the function D (x) = 200 – x. Express his day revenue total cost and profit on this meal as functions of x.
Answer:
cost of one meal = ₹ 100
Total cost = ₹ 100 (200 – x)
Number of customers = 200 – x
Day revenue = ₹ (200 – x) x
Total profit = day revenue – total cost
= (200 – x) x – (100) (200 – x)

Question 19.
The formula for converting from Fahrenheit to Celsius temperature is y = \(\). Find the inverse of this function and determine whether the inverse is also a function?
Answer:

Question 20.
A simple cipher takes a number and codes it, using the function f( x) = 3x – 4. Find the inverse of this function, determine whether the inverse is also a function and verify the symmetrical property about the line y = x (by drawing the lines)
Answer:
Given f(x) = 3x – 4
Let y = 3x – 4
⇒ y + 4 = 3x
⇒ x = \(\frac{y+4}{3}\)
Let g(y) = \(\frac{y+4}{3}\)
gof (x) = g (f(x) )
= g(3x – 4)
= \(\frac{3 x-4+4}{3}=\frac{3 x}{3}\)
gof(x) = x
and fog(y) = f(g(y))
= f\(\left(\frac{y+4}{3}\right)\)
= 3\(\left(\frac{y+4}{3}\right)\)
= y + 4 – 4 = y
fog (y) = y
Hence g of = I_x and fog = I_y
This shows that f and g are bijections and inverses of each other.
Hence f is bijection and f^-1(y) = \(\frac{y+4}{3}\)
Replacing y by x we get f^-1(x) = \(\frac{x+4}{3}\)
The line y = x

f(x) =
The line y =3x-4
When x = 0 ⇒ y = 3 × 0 – 4 = -4
When x = 1 ⇒ y = 3 × 1 – 4 = -1
When x = -1 ⇒ y = 3 × -1 – 4 = -7
When x = 2 ⇒ y = 3 × 2 – 4 = 2
When x = -2 ⇒ y = 3 × -2 – 4 = -10
When x = 3 ⇒ y = 3 × 3 – 4 = 5
When x = -3 ⇒ y = 3 × -3 – 4 = -13

The line y = \(\frac{x+4}{3}\)

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 1 Sets, Relations and Functions Ex 1.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets, Relations and Functions Ex 1.2

Question 1.
Discuss the following relations for reflexivity, symmetricity and transitivity:
(i) The relation R defined on the set of all positive integers by “m R n if m divides n”.
(ii) Let P denote the set of all straight lines in a plane. The relation R defined by “ l R m if l is perpendicular to m”.
(iii) Let A be the set consisting of all the members of a family. The relation R defined by “a R b if a is not a sister of b”.
(iv) Let A be the set consisting of all the female members of a family. The relation R defined by “a R b if a is not a sister of b”.
(v) On the set of natural numbers the relation R defined by “x R y if x + 2y = 1”

(i) The relation R defined on the set of all positive integers by “m R n if m divides n”.
Answer:
S = {set of all positive integers}

(a) mRm ⇒ ‘m’ divides’m’ ⇒ reflexive

(b) mRn ⇒ m divides n but
nRm ⇒ n does not divide m
(i.e.,) mRn ≠ nRm
It is not symmetric

(c) mRn ⇒ nRr as n divides r
It is transitive

(ii) Let P denote the set of all straight lines in a plane. The relation R defined by“ l R m if l is perpendicular to m”.
Answer:
Let P denote the set of all straight lines in a plane. The relation R is defined by l R m if l is perpendicular to m.
R = {(l, m): l is perpendicular to m}

(a) Reflexive:
Let l be any line in the plane P. Then line l is not perpendicular to itself.
{1, 1) ∉ R
∴ R is not reflexive.

(b) Symmetric:
Let (1, m) ∉ R ⇒ l is perpendicular to m
∴ m is perpendicular to l.
Hence (m, l) ∈ R
∴ R is symmetric.

(c) Transitive;
Let (l, m), (m, n) ∈ R
⇒ l is perpendicular to m.
∴ l is parallel to n. (l , n) ∉ R
Hence R is not transitive.

(iii) Let A be the set consisting of all the members of a family. The relation R defined by “a R b if a is not a sister of b”.
Answer:
A = {set of all members of the family}
aRb is a is not a sister of b

(a) aRa ⇒ a is not a sister of a It is reflexive

(b) aRb ⇒ a is not a sister of b.
bRa ⇒ b is not a sister of a.
It is symmetric

(c) aRb ⇒ a is not a sister of b.
bRc ⇒ b is not a sister of c.
⇒ aRc ⇒ a can be a sister of c
It is not transitive.

(iv) Let A be the set consisting of all the female members of a family. The relation R defined by “a R b if a is not a sister of b”.
Answer:
Given A is the set containing female members of the family.
Let M = Mother
H = Female child
A = { M, H }
The relation R on A is defined by aRb if a is not a sister of b.
R = {(M, M), (M, H), (H, H), (H, M)}

(a) Reflexive:
Clearly (M, M) and (H, H) ∈ R.
∴ R is reflexive.

(b) Symmetric:
For (M, H) ∈ R, we have (H, M ) ∈ R
∴ R is symmetric.

(c) Transitive:
For (M,M),(M,H) ∈ R ⇒ (M, H) ∈ R
(M,H),(H,H) ∈ R ⇒ (M, H) ∈ R
(H,H),(H,M) ∈ R ⇒ (H, M) ∈ R
(H,M),(M,M) ∈ R ⇒ (H, M) ∈ R
(H,M),(M,H) ∈ R (H, H) ∈ R
∴ R is transitive.

(v) On the set of natural numbers the relation R defined by “x R y if x + 2y = 1”
Answer:
N= {1, 2, 3, 4, 5,….}
xRy if x + 2y = 1 R is an empty set

(a) xRx ⇒ x + 2x = 1 ⇒ x = \(\frac{1}{3}\) ∉ N. It is not reflexive
xRy = yRx ⇒ x + 2y = 1 It does not imply that y + 2x = 1 as y = \(\frac{1-x}{2}\) It is not symmetric.

(b) -x = y ⇒ (-1, 1) ∉ N
It is not transitive.

Question 2.
Let X = { a , b , c , d } and R = { (a, a ) , (b, b ), (a, c)}. Write down the minimum number of ordered pairs to be included to R to make it
(i) reflexive
(ii) symmetric
(iii) transitive
(iv) equivalence
Answer:
Given X = { a, b, c, d }
R = { (a, a), (b, b), (a, c) }
(i) The minimum ordered pairs to be included to R in order to make R to be reflexive is (c, c) and (d, d)
(ii) The minimum ordered pairs to be included to R in order to make R to be symmetric is (c, a).
(iii) R is transitive. We need not add any pair.
(iv) After adding the ordered pairs (c, c),(d, d), (c, a) the new relation becomes
R, = {(a, a), (b, b), (c, c), (d, d), (a, c), (c, a)}
The new relation satisfies, reflexive, symmetric and transitive property.
∴ R₁ is an equivalence relation.

Question 3.
Let A = { a, b , c } and R = { (a, a ) , (b, b ), (a, c ) }. Write down the minimum number of ordered pairs to be included to R to make it
(i) reflexive
(ii) symmetric
(iii) transitive
(iv) equivalence
Answer:
Given A = {a, b, c }
R = { (a, a), (b, b),(a, c) }
(i) The minimum ordered pair to be included to R in order to make it reflexive is (c, c).
(ii) The minimum ordered pair to be included to R in order to make it symmetrical is (c, a).
(iii) R is transitive. We need not add any pair.
(iv) After including the ordered pairs (c, c),(c, a) to R the new relation becomes
R₁ = { (a, a), (b, b), (c, c) , (a, c) , (c, a) }
R₁ is reflexive symmetric and transitive.
∴ R₁ is an equivalence relation.

Question 4.
Let P be the set of all triangles in a plane and R be the relation defined on P as a R b if a is similar to b. Prove that R is an equivalence relation.
Answer:
Given P = the set of all triangles in a plane.
R is the relation defined by a R b if a is similar to b.
R = {(a, b) : a is similar to b for a, b ∈ p }

(a) Reflexive:
(a, a) ⇒ a is similar to a for all a ∈ P
∴ R is reflexive.

(b) Symmetric: .
Let (a,b) ∈ R ⇒ a is similar to b
⇒ b is similar to a
∴ (b, a) ∈ R
Hence R is symmetric.

c) Transitive:
Let (a, b) and ( b, c) ∈ R
(a, b) ∈ R ⇒ a is similar to b
(b, c) ∈ R ⇒ b is similar to c
∴ a is similar to c.
Hence R is transitive.
∴ R is an equivalence relation on P.

Question 5.
On the set of natural numbers let R be the relation defined by a R b if 2a + 3b = 30. Write down the relation by listing all the pairs. Cheek whether it is
(i) reflexive
(ii) symmetric
(iii) transitive
(iv) equivalence
Answer:
Given N = set of natural numbers
R is the relation defined by a R b if 2a + 3b = 30


When a > 15, b negative and does not belong to N.
∴ R = { (3,8),(6,6), (9,4), (12,2)}.
(i) R is not reflexive since (a, a) ∉ R for all a ∈ N.
(ii) R is not symmetric since for (3, 8) ∈ R, (8, 3) ∉ R
(iii) Clearly R is transitive since we cannot find elements (a, b), (b, c) in R such that (a, c) ∉ R
∴ R is not an equivalence relation.

Question 6.
Prove that the relation ‘friendship’ is not an equivalence relation on the set of all people in Chennai.
Answer:
(a) S = aRa (i.e. ) a person can be a friend to himself or herself.
So it is reflective.

(b) aRb ⇒ bRa so it is symmetric

(c) aRb, bRc does not ⇒ aRc so it is not transitive
⇒ It is not an equivalence relation

Question 7.
On the set of natural numbers let R be the relation defined by a R b if a + b < 6. Write down the relation by listing all the pairs. Check whether it is
(i) reflexive
(ii) symmetric
(iii) transitive
(iv) equivalence.
Answer:
N = the set of natural numbers.
R is the relation defined on N by
a R b if a + b ≤ 6
R = { (a, b), a, b ∈ N / a + b ≤ 6}
a + b ≤ 6 ⇒ b ≤ 6 – a

a = 1,
b ≤ 6 – 1 = 5
b is 1, 2, 3, 4, 5
∴ (1, 1), (1, 2), (1, 3), (1, 4), (1, 5) ∈ R

a = 2,
b ≤ 6 – 2 = 4
b is 1, 2, 3, 4
∴ (2, 1), (2, 2),(2, 3), (2, 4) ∈ R

a = 3,
b < 6 – 3 = 3
b is 1, 2, 3
∴ (3, 1), (3, 2), (3, 3) ∈ R

a = 4 ,
b < 6 – 4 = 2
b is 1, 2
∴ (4, 1), (4, 2) ∈ R

a = 5,
b < 6 – 5 = 1
b is 1
∴ (5, 1) ∈ R
∴ R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (5, 1)}

(i) Reflexive:
R is not reflexive since (4, 4), (5, 5) ∈ R

(ii) Symmetric:
Cleary R is symmetric forever (x, y) ∈ R, we have (y, x) ∈ R.

(iii) Transitive:
(3, 1), (1, 5) ∈ R ⇒ (3,5) ∉ R
∴ R is not transitive.

(iv) R is not an equivalence relation.

Question 8.
Let A = { a, b, c }. What is the equivalence relation of smallest cardinality on A? What is the equivalence relation of largest cardinality on A?
Answer:
R = {{a, a), (b, b), (c, c)} is this smallest cardinality of A to make it equivalence relation n(R) = 3

(i) R = {(a, a), {a, b), (a, c), (b, c), (b, b), {b, c), (c, a), (c, b), (c, c)}
n(R) = 9 is the largest cardinality of R to make it equivalence.

Question 9.
In the set Z of integers, define m Rn if m – n is divisible by 7. Prove that R is an equivalence relation.
Answer:
Z = set of all integers
Relation R is defined on Z by m R n if m – n is divisible by 7.
R = {(m, n), m, n ∈ Z/m – n divisible by 7}
m – n divisible by 7
∴ m – n = 7k where k is an integer.

a) Reflexive:
m – m = 0 = 0 × 7
m – m is divisible by 7
∴ (m, m ) ∈ R for all m ∈ Z
Hence R is reflexive.

b) Symmetric:
Let (m, n ) ∈ R ⇒ m – n is divisible by 7
m – n = 7k
n – m = – 7k
n – m = (-k)7
∴ n – m is divisible by 7
∴ (n, m) ∈ R.

c) Transitive:
Let (m, n) and (n , r) ∈ R
m – n is divisible by 7
m – n = 7k ——— (1)
n – r is divisible by 7
n – r = 7k₁ ——— (2)
(m – n) + (n – r) = 7k + 7k₁
m – r = ( k + k₁) 7
m – r is divisible by 7.
∴ (m, r) ∈ R
Hence R is transitive.
R is an equivalence relation.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 1 Sets, Relations and Functions Ex 1.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets, Relations and Functions Ex 1.1

Question 1.
Write the following in roaster form.
(i) {x ∈ N : x² < 121 and x is a prime}
Answer:
Let A = { x ∈ N : x² < 121 and x is a prime }
A = { 2, 3, 5, 7 }

(ii) The set of positive roots of the equation (x – 1) ( x + 1) (x – 1 ) = 0
Answer:
The set of positive roots of the equations
(x – 1) (x + 1) (x² – 1) = 0
(x – 1 ) (x + 1 ) (x + 1) (x – 1) = 0
(x + 1 )² (x – 1)² = 0
(x + 1)² = 0 or (x – 1)² = 0
x + 1 = 0 or x – 1 = 0
x = -1 or x = 1
A = { 1 }

(iii) {x ∈ N : 4x + 9 < 52}
Answer:
4x + 9 < 52
4x + 9 – 9 < 52 – 9
4x < 43
x < \(\frac{43}{4}\) (i.e.) x < 10.75 4
But x ∈ N
∴ A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

(iv)
Answer:
Let A =
⇒ \(\frac{x-4}{x+2}\) = 3
⇒ x – 4 = 3(x + 2)
⇒ x – 4 = 3x + 6
⇒ 3x – x = – 4 – 6
2x = – 10
⇒ x = \(-\frac{10}{2}\) = -5
A = { -5 }

Question 2.
Write the set {-1, 1} in set builder form.
Answer:
A = {x : x² – 1 = 0, x ∈ R}

Question 3.
State whether the following sets are finite or infinite.
(i) {x ∈ N : x is an even prime number }
Answer:
Let A = { x ∈ N : x is an even prime number )
A = {2}
A is a finite set.

(ii) {x ∈ N: x is an odd prime number }
Answer:
Let B = {x ∈ N : x is an odd prime number}
B = {1, 3, 5, 7, 11, …………….. }
B is an infinite set.

(iii) {x ∈ Z : x is even and < 10 }
Answer:
C = {x ∈ Z : x is even and< 10}
C = { ……….. -8, -6, -4, -2, 0, 2, 4, 6, 8}
C is an infinite set.

(iv) {x ∈ R : x is a rational number }
Answer:
D = { x ∈ R : x is a rational number }
D is an infinite set.

(v) {x ∈ N: x is a rational number }
Answer:
E = { x ∈ N : x is a rational number )
E = {1, 2, 3, 4, 5, 6, …………..)
Every integer is a rational number.
∴ E is an infinite set.

Question 4.
By taking suitable sets A, B, C, verify the following results.
(i) A × (B ∩ C) = (A × B) ∩ (A × C)
Answer:
To prove: A × (B ∩ C) = (A × B) ∩ (A × C)
B ∩ C = {8}; A = {1, 2, 5, 7}
So A × (B ∩ C) = {1, 2, 5, 7} × {8}
= {(1, 8), (2. 8), (5, 8), (7, 8)}
Now A x B = {(1, 2), (1, 7), (1, 8), (1, 9), (2, 2), (2, 7), (2, 8), (2, 9), (5, 2), (5, 7), (5, 8), (5, 9), (7, 2), (7, 7), (7, 8), (7, 9)} …. ( 1)
A × C = {(1, 1), (1, 5),(1, 8), (1, 10), (2, 1), (2, 5), (2, 8), (2, 10), (5, 1), (5, 5), (5, 8), (5, 10), (7, 1), (7, 5), (7, 8), (7, 10)}
(A × B) ∩ (A × C) = {(1, 8), (2, 8), (5, 8), (7, 8)} ……… (2)
(1) = (2)
⇒ A × (B ∩ C) = (A × B) ∩ (A × C)

(ii) A × (B ∪ C) = (A × B) ∪ (A × C)
Answer:
Let A = {1, 2} , B = {3, 4}, C = {4, 5}
B ∪ C = {3, 4} ∪ {4, 5}
B ∪ C = {3, 4, 5)
A × (B ∪ C) = {1, 2} × {3, 4, 5}
A × (B ∪ C) = { (1, 3),( 1, 4),(1, 5),(2, 3), (2, 4),(2,5)} ——– (1)
A × B = {1, 2} × {3, 4}
A × B = { (1, 3), (1, 4), (2, 3), (2, 4) }
A × C = {1, 2} × {4, 5}
A × C = { (1, 4), (1, 5), (2, 4), (2, 5 )}
(A × B) ∪ (A × C) = {(1, 3), (1, 4), (2, 3), (2, 4)} ∪ {(1, 4 ), (1, 5 ), ( 2, 4 ), (2, 5)}
(A × B) ∪ (A × C) = { (1, 3) (1, 4), (1, 5), (2, 3), (2, 4), (2, 5)} —— (2)
From equations (1) and (2)
A × (B U C) = (A × B) U (A × C)

(iii) (A × B) ∩ (B × A) = (A ∩ B) × ( B ∩ A)
Answer:
A × B = {(1, 2), (1, 7), (1, 8), (1, 9) (2, 2), (2, 7), (2, 8), (2, 9) (5, 2), (5, 7), (5, 8), (5, 9) (7, 2), (7, 7), (7, 8), (7, 9)}
B × A = {(2, 1), (2, 2), (2, 5), (2, 7) (7, 1), (7, 2), (7, 5), (7, 7) (8, 1), (8, 2), (8, 5), (8, 7) (9,1), (9, 2), (9, 5), (9, 7)}
L.H.S. (A × B) ∩ (B × A) = {(2, 2), (2, 7), (7, 2), (7, 7)} …. (1)
R.H.S. A ∩ B = {2, 7}
B ∩ A = {2, 7}
(A ∩ B) × (B ∩ A) = {2, 7} × {2, 7}
= {(2, 2), (2, 7), (7, 2), (7, 7)} ……… (2)
(1) = (2) ⇒ LHS = RHS

(iv) C – (B – A) = (C ∩ A) ∪ (C ∩ B’)
Answer:
Let A = {1, 2, 3) , B = {2, 3, 4) , C = {3, 4, 5}

B – A = { 2, 3, 4 ) – {1, 2, 3}
B – A = {4}
C – (B – A) = {3, 4, 5} – {4}
C – (B – A) = {3, 5} —- (1)
C ∩ A = {3, 4, 5} ∩ { 1, 2, 3)
C ∩ A = {3}
B’ = {1, 5}
C ∩ B’ = {3, 4, 5} ∩ {1, 5}
C ∩ B’ = {5}
(C ∩ A) ∪ (C∩B’) = {3} ∪ {5}
(C∩A) ∪ (C ∩ B’) = {3, 5} —— (2)
From equations (1) and (2)
C – (B – A) = (C ∩ A) u (C ∩ B’)

(v) (B – A) ∩ C = (B ∩ C) – A = B ∩ (C – A)
Answer:
To prove (B – A) ∩ C = (B ∩ C) – A = B ∩ (C – A)
A= {1, 2, 5, 7}, B = {2, 7, 8, 9}, C = {1, 5, 8, 10}
Now B – A = {8, 9}
(B – A) ∩ C = {8} ……. (1)
B ∩ C = {8}
A = {1, 2, 5, 7}
So (B ∩ C) – A = {8} …… (2)
C – A = {8, 10}
B = {2, 7, 8, 9}
B ∩ (C – A) = {8} …. (3)
(1) = (2) = (3)

(vi) (B – A) ∪ C = (B ∪ C) – (A – C)
Answer:
Let A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8}
B – A = {3, 4, 5, 6} – {1, 2, 3, 4}
B – A = {5, 6}
(B – A) ∪ C = {5, 6} ∪ {5, 6, 7, 8}
(B – A) ∪ C = { 5 , 6 , 7 , 8 } ——- (1)
B ∪ C = { 3, 4, 5, 6 } ∪ { 5, 6, 7,8 }
B ∪ C = { 3, 4, 5, 6, 7, 8 }
A – C = { 1, 2, 3, 4 } – { 5, 6, 7, 8 }
A – C = { 1 , 2, 3 , 4 }
(B ∪ C) – (A – C) = {3, 4, 5, 6, 7, 8} – {1, 2, 3, 4}
(B ∪ C) – (A – C) = { 5, 6, 7, 8 } —-(2)
From equations (1) and (2)
(B – A) ∪ C = (B ∪ C) – (A – C)

Question 5.
Justify the trueness of the statement: “An element of a set can never be a subset of itself”.
Answer:
A set itself can be a subset of itself (i.e.) A ⊆ A.
But it cannot be a proper subset.

Question 6.
If n(P(A)) = 1024, n(A ∪ B) = 15 and n(P(B)) = 32, find n(A ∩ B).
Answer:
Given n(P(A)) = 1024 , n(A ∪ B) = 15, n(P(B)) = 32
n(P(A)) = 1024 = 2¹⁰ n(A) = 10
n(P(B)) = 32 = 2⁵ = n(B) = 5
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
15 = 10 + 5 – n(A ∩ B)
15 = 15 – n (A ∩ B)
n(A ∩ B) = 0

Question 7.
If n (A ∩ B ) = 3 and n(A ∪ B ) = 10 , then find n(P(A ∆ B)).
Answer:

n(A ∪ B) = 10; n(A ∩ B) = 3
n(A ∆ B) = 10 – 3 = 7
and n(P(A ∆ B)) = 27 = 128

Question 8.
For a set A, A × A contains 16 elements and two of its elements are (1, 3) and (0, 2). Find the elements of A.
Answer:
Given A × A contains 16 elements.
∴ A contains 4 elements.
Also, (1, 3) and (0, 2) are two elements of A × A.
∴ A = { 0, 1, 2, 3 }

Question 9.
Let A and B be two sets such that n (A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1) are in A × B , find A and B, where x , y , z are distinct elements.
Answer:
n(A) = 3 ⇒ set A contains 3 elements
n(B) = 2 ⇒ set B contains 2 elements –
we are given (x, 1), (y, 2), (z, 1) are elements in A × B ⇒ A = {x, y, z} and B = {1, 2}

Question 10.
If A × A has 16 elements, S = { ( a, b ) ∈ A × A: a < b } ; (-1, 2) and (0, 1) are two elements of S , then find the remaining elements of S.
Answer:
Given A × A has 16 elements.
∴ A has 4 elements.
Also S = {(a, b) ∈ A × A; a < b}
Given (-1, 2) and (0, 1) ∈ S
A = {-1, 0 , 1 , 2 }
The elements of S are
S = { (-1, 0), (-1, 1) ,(-1, 2),(0, 1), (0, 2), (1, 2)}
∴ The other elements of the sets are
(-1, 0), (-1, 1) , (0, 2), (1, 2)

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Samacheer Kalvi 11th Business Maths Operations Research Ex 10.3 Text Book Back Questions and Answers

Choose the correct answer.

Question 1.
The critical path of the following network is:

(a) 1-2-4-5
(b) 1-3-5
(c) 1-2-3-5
(d) 1-2-3-4-5
Answer:
(d) 1-2-3-4-5
Hint:
1-2-4-5 ⇒ EFT = 20 + 12 + 10 = 42
1-3-5 ⇒ EFT = 25 + 8 = 33
1-2-3-5 ⇒ EFT = 20 + 10 + 8 = 38
1-2-3-4-5 ⇒ EFT = 20 + 10 + 5 + 10 = 45

Question 2.
Maximize: z = 3x₁ + 4x₂ subject to 2x₁ + x₂ ≤ 40, 2x₁ + 5x₂ ≤ 180, x₁, x₂ ≥ 0. In the LPP, which one of the following is feasible comer point?
(a) x₁ = 18, x₂ = 24
(b) x₁ = 15, x₂ = 30
(c) x₁ = 2.5, x₂ = 35
(d) x₁ = 20.5, x₂ = 19
Answer:
(c) x₁ = 2.5, x₂ = 35
Hint:
z = 3x₁ + 4x₂
Let us solve the equations
2x₁ + x₂ = 40 ………(1)
2x₁ + 5x₂ = 180 ……….(2)
(1) – (2) ⇒ -4x₂ = -140
x₂ = 35
We have 2x₁ + x₂ =40
2x₁ + 35 = 40
2x₁ = 5
x₁ = 2.5

Question 3.
One of the conditions for the activity (i, j) to lie on the critical path is:
(a) E_j – E_i = L_j – L_i = t_ij
(b) E_i – E_j = L_j – L_i = t_ij
(c) E_j – E_i = L_i – L_j = t_ij
(d) E_j – E_i = L_j – L_i ≠ t_ij
Answer:
(a) E_j – E_i = L_j – L_i = t_ij

Question 4.
In constructing the network which one of the following statement is false?
(a) Each activity is represented by one and only one arrow, (i.e) only one activity can connect any two nodes.
(b) Two activities can be identified by the same head and tail events.
(c) Nodes are numbered to identify an activity uniquely. Tail node (starting point) should be lower than the head node (end point) of an activity.
(d) Arrows should not cross each other.
Answer:
(b) Two activities can be identified by the same head and tail events.

Question 5.
In a network while numbering the events which one of the following statement is false?
(a) Event numbers should be unique.
(b) Event numbering should be carried out on a sequential basis from left to right.
(c) The initial event is numbered 0 or 1.
(d) The head of an arrow should always bear a number lesser than the one assigned at the tail of the arrow.
Answer:
(d) The head of an arrow should always bear a number lesser than the one assigned at the tail of the arrow.

Question 6.
A solution which maximizes or minimizes the given LPP is called:
(a) a solution
(b) a feasible solution
(c) an optimal solution
(d) none of these
Answer:
(a) a solution

Question 7.
In the given graph the coordinates of M₁ are

(a) x₁ = 5, x₂ = 30
(b) x₁ = 20, x₂ = 16
(c) x₁ = 10, x₂ = 20
(d) x₁ = 20, x₂ = 30
Answer:
(c) x₁ = 10, x₂ = 20
Hint:
4x₁ + 2x₂ = 80 (or) 2x₁ + x₂ = 40
2x₁ + x₂ = 40 ……(1)
2x₁ + 5x₂ = 120 ……(2)
(1) – (2) ⇒ -4x₂ = -80
x₂ = 20
But, 2x₁ + x₂ = 40
2x₁ + 20 = 20
x₁ = 10

Question 8.
The maximum value of the objective function Z = 3x + 5y subject to the constraints x > 0, y > 0 and 2x + 5y ≤ 10 is:
(a) 6
(b) 15
(c) 25
(d) 31
Answer:
(b) 15
Hint:
Let 2x + 5y = 10


∴ Maximum Value = 15

Question 9.
The minimum value of the objective function Z = x + 3y subject to the constraints 2x + y ≤ 20, x + 2y ≤ 20, x > 0 and y > 0 is:
(a) 10
(b) 20
(c) 0
(d) 5
Answer:
(c) 0
Hint:
O(0, 0) is a comer point.
So Z = 0 + 3(0) = 0
∴ Minimum value is 0

Question 10.
Which of the following is not correct?
(a) Objective that we aim to maximize or minimize
(b) Constraints that we need to specify
(c) Decision variables that we need to determine
(d) Decision variables are to be unrestricted
Answer:
(d) Decision variables are to be unrestricted

Question 11.
In the context of network, which of the following is not correct?
(a) A network is a graphical representation.
(b) A project network cannot have multiple initial and final nodes
(c) An arrow diagram is essentially a closed network
(d) An arrow representing an activity may not have a length and shape
Answer:
(d) An arrow representing an activity may not have a length and shape

Question 12.
The objective of network analysis is to:
(a) Minimize total project cost
(b) Minimize total project duration
(c) Minimize production delays, interruption and conflicts
(d) All the above
Answer:
(b) Minimize total project duration

Question 13.
Network problems have advantage in terms of project:
(a) Scheduling
(b) Planning
(c) Controlling
(d) All the above
Answer:
(d) All the above

Question 14.
In critical path analysis, the word CPM mean:
(a) Critical path method
(b) Crash projecf management
(c) Critical project management
(d) Critical path management
Answer:
(a) Critical path method

Question 15.
Given an L.P.P maximize Z = 2x₁ + 3x₂ subject to the constrains x₁ + x₂ ≤ 1, 5x₁ + 5x₂ ≥ 0 and x₁ ≥ 0, x₂ ≥ 0 using graphical method, we observe:
(a) No feasible solution
(b) unique optimum solution
(c) multiple optimum solution
(d) none of these
Answer:
(a) No feasible solution

Tamilnadu State Board New Syllabus Samcheer Kalvi 11th Business Maths Guide Pdf Chapter 10 Operations Research Ex 10.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 10 Operations Research Ex 10.2

Samacheer Kalvi 11th Business Maths Operations Research Ex 10.2 Text Book Back Questions and Answers

Question 1.
Draw the network for the project whose activities with their relationships are given below:
Activities A, D, E can start simultaneously; B, C > A; G, F > D, C; H > E, F.
Solution:
Given: (i) A, D, E can start simultaneously.
(iii) A < B, C; C, D < G, F; E, F < H
Working rule:
A < B, C implies activity A is the immediate predecessor of activities B and C.
i.e., for activities B and C to occur, activity ‘A’ has to be completed.
Similarly for activities G, F to occur D and C has to completed for activity H to occur E and F has to be completed.
Step-1:

(∵ A, D and E are independents events)
Step-2:

Note: Activities B, G and H are not a part of any activities immediate predecessor. So they have to merge into the lost node 5.

Question 2.
Draw the event oriented network for the following data:

Solution:
Step-1:

Step-2:

Step-3:

Question 3.
Construct the network for the projects consisting of various activities and their precedence relationships are as given below:
A, B, C can start simultaneously: A < F, E; B < D, C; E, D < G
Solution:
Step-1:

Step-2:

Step-3:

Question 4.
Construct the network for each the projects consisting of various activities and their precedence relationships are as given below:

Solution:
Step-1:

Step-2:

Step-3:

Question 5.
Construct the network for the project whose activities are given below.

Calculate the earliest start time, earliest finish time, latest start time and latest finish time of each activity. Determine the critical path and the project completion time.
Solution:

Forward pass:
E₁ = 0 + 3 = 3
E₂ = E₁ + t₁₂ = 8 + 3 = 11
E₃ = 3 + 12 = 15
E₄ = E₂ + 6 (or) E₃ + 3 = 11 + 6 (or) 15 + 3 = 18
[We must select a maximum value for forwarding pass]
E₄ = 15 + 3 = 18
E₂ = E₂ + 3 = 11 + 3 = 14
E₆ = E₃ + 8 = 15 + 8 = 23
E₇ = E₆ + 8 = 23 + 8 = 31
Backward pass:
L₇ = 31
L₆ = L₇ – 8 = 31 – 8 = 23
L₅ = L₇ – 3 = 31 – 3 = 28
L₄ = L₇ – 5 = 31 – 5 = 26
L₃ = L₆ – 8 = 23 – 8 = 15
L₂ = L₅ – 3 (or) L₄ – 6 = (28 – 3) (or) (26 – 6) = 25 (or) 20
[We must select a minimum value for a backward pass and maximum value for forwarding pass]
L₁ = L₂ – 8 (or) L₃ – 12 = 20 – 8 (or) 15 – 12 = 12 (or) 3 = 3
L₀ = 0

Critical path 0-1-3-6-7 and the duration is 31 works.

Question 6.
A project schedule has the following characteristics:

Construct the network and calculate the earliest start time, earliest finish time, latest start time, and latest finish time of each activity and determine the Critical path of the project and duration to complete the project.
Solution:

Forward pass:
E₁ = 0
E₂ = 0 + 4 = 4
E₃ = 0 + 1 = 1
E₄ = (E₂ + 1) (or) (E₃ + 1) = (4 + 1) (or) (1 + 1) = 5 (or) 2
[whichever is maximum we must select forward pass]
∴ E₄ = 5
E₅ = 1 + 6 = 7
E₆ = 7 + 4 = 11
E₇ = 8 + 7 = 15
E₈ = (E₇ + 2) (or) (E₆ + 1)
= (15 + 2) or (11 + 1)
= 17 or 12 [whichever is maximum]
∴ E₈ = 17
E₉ = 5 + 5 = 10
E₁₀ = (E₉ + 7) (or) (E₈ + 5)
= (10 + 7) (or) (17 + 5)
= 17 (or) 22 [Which is maximum]
E₁₀ = 22
Backward pass:
L₁₀ = 22
L₉ = 22 – 7 = 15
L₈ = 22 – 5 = 17
L₇ = 17 – 2 = 15
L₆ = 17 – 1 = 16
L₅ = (16 – 4) (or) (15 – 8) [whichever is minimum]
L₅ = 7
L₄ = 15 – 5 = 10
L₃ = (10 – 1) (or) (7 – 6) [whichever is minimum]
L₃ = 1
L₂ = 10 – 1 = 9
L₁ = 0
We can also find the critical path by this method, which also helps us to counter check the solution obtained by the table method.

So critical path is 1-3-5-7-8-10 as it takes 22 units to complete the project.

Question 7.
Draw the network and calculate the earliest start time, earliest finish time, latest start time, and latest finish time of each activity and determine the Critical path of the project and duration to complete the project.

Solution:

Forward pass:
E₁ = 0
E₂ = 0 + 6 = 6
E₃ = 0 + 5 = 5
E₄ = 6 + 10 = 16
E₅ = (E₄ + 6) (or) (E₃ + 4)
E₅ = (16 + 6) (or) (5 + 4) [whichever is maximum]
E₅ = 22
E₆ = (E₅ + 9)(or) (E₄ + 2)
E₆ = (22 + 9) (or) (4 + 2) [whichever is maximum]
E₆ = 31
Backward pass:
L₆ = 31
L₅ = 31 – 9 = 22
L₄ = 22 – 6 (or) 31 – 2 [whichever is minimum]
L₄ = 16
L₃ = 22 – 4 = 18
L₂ = 16 – 10 = 6
L₁ = 6 – 6 = 0


∴ EFT and LFT are the same on 1-2, 2-4, 4-5, 5-6, the vertical path is 1-2-4-5-6 and duration is 31 days to complete.

Question 8.
The following table gives the activities of a project and their duration in days.

Construct the network and calculate the earliest start time, earliest finish time, latest start time, and latest finish time of each activity and determine the critical path of the project and duration to complete the project.
Solution:

Forward pass:
E₁ = 0
E₂ = 0 + 5 = 5
E₃ = (0 + 8) (or) (5 + 6) [whichever is maximum]
E₃ = 11
E₄ = (11 + 5) (or) (5 + 7) [whichever is maximum]
E₄ = 16
E₅ = (11 + 4) (or) (16 + 8) [whichever is maximum]
E₅ = 24
Backward pass:
L₅ = 24
L₄ = 24 – 8 = 16
L₃ = (24 – 4) (or) (16 – 5) [whichever is minimum]
L₃ = 11
L₂ = (11 – 6) (or) (16 – 7) [whichever is minimum]
L₂ = 5
L₁ = (5 – 5) (or) (11 – 8) = 0 [whichever is minimum]


EFT and LFT are the same on 1-2, 2-3, 3-4, and 4-5
So critical path is 1-2-3-4-5 and the time taken is 24 days.

Question 9.
A project has the following time schedule

Construct the network and calculate the earliest start time, earliest finish time, latest start time, and latest finish time of each activity and determine the critical path of the project and duration to complete the project.
Solution:

Forward pass:
E₁ = 0
E₂ = 0 + 7 = 7
E₃ = 7 + 14 = 21
E₄ = 7 + 5 = 12
E₅ = (21 + 11) (or) (2 + 7) = 32 [whichever is maximum]
E₆ = 0 + 6 = 6
E₇ = 6 + 11 = 17
E₈ = 32 + 4 = 36
Backward pass:
L₈ = 36
L₇ = 36 – 18 = 18
L₆ = 18 – 11 = 7
L₅ = 35 – 4 = 31
L₄ = 32 – 7 = 25
L₃ = 32 – 11 = 21
L₂ = (21 – 14) (or) (25 – 5) [whichever is minimum]
L₂ = 7
L₁ = 7 – 7 = 0

EFT and LFT are same is 1-2, 2-3, 3-5, 5-8.
Hence the critical path is 1-2-3-5-8 and the time to complete is 36 days.

Question 10.
The following table uses the activities in construction projects and relevant information.

Draw the network for the project, calculate the earliest start time, earliest finish time, latest start time, and latest finish time of each activity, and find the critical path. Compute the project duration.
Solution:

Forward pass:
E₁ = 0
E₂ = 0 + 22 = 22
E₃ = (0 + 27) (or) (22 + 12) = 34 [Maximum of both]
E₄ = (22 + 14) (or) (34 + 16) = 40
E₅ = 40 + 12 = 52
Backward pass:
L₅ = 32
L₄ = 52 – 12 = 40
L₃ = 40 – 6 = 34
L₂ = (40 – 14) (or) (34 – 12) = 22 [Minimum of both]
L₁ = 22 – 22 = 0

EFT and LFT are the same for 1-2, 2-3, 3-4, and 4-5.
So critical path is 1-2-3-4-5 time to complete is 52 days.

Tamilnadu State Board New Syllabus Samcheer Kalvi 11th Business Maths Guide Pdf Chapter 10 Operations Research Ex 10.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 10 Operations Research Ex 10.1

Samacheer Kalvi 11th Business Maths Operations Research Ex 10.1 Text Book Back Questions and Answers

Question 1.
A company produces two types of pens A and B. Pen A is of superior quality and pen B is of lower quality. Profits on pens A and B are ₹5 and ₹3 per pen respectively. Raw materials required for each pen A is twice as that of pen B. The supply of raw material is sufficient only for 1000 pens per day. Pen A requires a special clip and only 400 such clips are available per day. For pen B, only 700 clips are available per day. Formulate this problem as a linear programming problem.
Solution:
(i) Variables: Let x₁ and x₂ denotes the number of pens in type A and type B.

(ii) Objective function:
Profit on x₁ pens in type A is = 5x₁
Profit on x₂ pens in type B is = 3x₂
Total profit = 5x₁ + 3x₂
Let Z = 5x₁ + 3x₂, which is the objective function.
Since the B total profit is to be maximized, we have to maximize Z = 5x₁ + 3x₂

(iii) Constraints:
Raw materials required for each pen A is twice as that of pen B.
i.e., for pen A raw material required is 2x₁ and for B is x₂.
Raw material is sufficient only for 1000 pens per day
∴ 2x₁ + x₂ ≤ 1000
Pen A requires 400 clips per day
∴ x₁ ≤ 400
Pen B requires 700 clips per day
∴ x₂ ≤ 700

(iv) Non-negative restriction:
Since the number of pens is non-negative, we have x₁ > 0, x₂ > 0.
Thus, the mathematical formulation of the LPP is
Maximize Z = 5x₁ + 3x₂
Subj ect to the constrains
2x₁ + x₂ ≤ 1000, x₁ ≤ 400, x₂ ≤ 700, x₁, x₂ ≥ 0

Question 2.
A company produces two types of products say type A and B. Profits on the two types of product are ₹ 30/- and ₹ 40/- per kg respectively. The data on resources required and availability of resources are given below.

Formulate this problem as a linear programming problem to maximize the profit.
Solution:
(i) Variables: Let x₁ and x₂ denote the two types products A and B respectively.

(ii) Objective function:
Profit on x₁ units of type A product = 30x₁
Profit on x₂ units of type B product = 40x₂
Total profit = 30x₁ + 40x₂
Let Z = 30x₁ + 40x₂, which is the objective function.
Since the profit is to be maximized, we have to maximize Z = 30x₁ + 40x₂

(iii) Constraints:
60x₁ + 120x₂ ≤ 12,000
8x₁ + 5x₂ ≤ 600
3x₁ + 4x₂ ≤ 500

(iv) Non-negative constraints:
Since the number of products on type A and type B are non-negative, we have x₁, x₂ ≥ 0
Thus, the mathematical formulation of the LPP is
Maximize Z = 30x₁ + 40x₂
Subject to the constraints,
60x₁ + 120x₂ ≤ 12,000
8x₁ + 5x₂ ≤ 600
3x₁ + 4x₂ ≤ 500
x₁, x₂ ≥ 0

Question 3.
A company manufactures two models of voltage stabilizers viz., ordinary and autocut. All components of the stabilizers are purchased from outside sources, assembly and testing is carried out at company’s own works. The assembly and testing time required for the two models are 0.8 hour each for ordinary and 1.20 hours each for auto-cut. Manufacturing capacity 720 hours at present is available per week. The market for the two models has been surveyed which suggests maximum weekly sale of 600 units of ordinary and 400 units of auto-cut. Profit per unit for ordinary and auto-cut models has been estimated at ₹ 100 and ₹ 150 respectively. Formulate the linear programming problem.
Solution:
(i) Variables : Let x₁ and x₂ denote the number of ordinary and auto-cut voltage stabilized.

(ii) Objective function:
Profit on x₁ units of ordinary stabilizers = 100x₁
Profit on x₂ units of auto-cut stabilized = 150x₂
Total profit = 100x₁ + 150x₂
Let Z = 100x₁ + 150x₂, which is the objective function.
Since the profit is to be maximized. We have to
Maximize, Z = 100x₁ + 15x₂

(iii) Constraints: The assembling and testing time required for x₁ units of ordinary stabilizers = 0.8x₁ and for x₂ units of auto-cut stabilizers = 1.2x₂
Since the manufacturing capacity is 720 hours per week. We get 0.8x₁ + 1.2x₂ ≤ 720
Maximum weekly sale of ordinary stabilizer is 600
i.e., x₁ ≤ 600
Maximum weekly sales of auto-cut stabilizer is 400
i.e., x₂ ≤ 400

(iv) Non-negative restrictions:
Since the number of both the types of stabilizer is non-negative, we get x₁, x₂ ≥ 0.
Thus, the mathematical formulation of the LPP is,
Maximize Z = 100x₂ + 150x₂
Subject to the constraints
0.8x₁ + 1.2x₂ ≤ 720, x₁ ≤ 600, x₂ ≤ 400, x₁, x₂ ≥ 0

Question 4.
Solve the following linear programming problems by graphical method.
(i) Maximize Z = 6x₁ + 8x₂ subject to constraints 30x₁ + 20x₂ ≤ 300; 5x₁ + 10x₂ ≤ 110; and x₁, x₂ ≥ 0.
(ii) Maximize Z = 22x₁ + 18x₂ subject to constraints 960x₁ + 640x₂ ≤ 15360; x₁ + x₂ ≤ 20 and x₁, x₂ ≥ 0.
(iii) Minimize Z = 3x₁ + 2x₂ subject to the constraints 5x₁ + x₂ ≥ 10; x₁ + x₂ > 6; x₁+ 4x₂ ≥ 12 and x₁, x₂ ≥ 0.
(iv) Maximize Z = 40x₁ + 50x₂ subject to constraints 3x₁ + x₂ ≤ 9; x₁ + 2x₂ ≤ 8 and x₁, x₂ ≥ 0.
(v) Maximize Z = 20x₁ + 30x₂ subject to constraints 3x₁ + 3x₂ ≤ 36; 5x₁ + 2x₂ ≤ 50; 2x₁ + 6x₂ ≤ 60 and x₁, x₂ ≥ 0.
(vi) Minimize Z = 20x₁ + 40x₂ subject to the constraints 36x₁ + 6x₂ ≥ 108; 3x₁ + 12x₂ ≥ 36; 20x₁ + 10x₂ ≥ 100 and x₁, x₂ ≥ 0.
Solution:
(i) Given that 30x₁ + 20x₂ ≤ 300
Let 30x₁ + 20x₂ = 300

Therefore
3x₁ + 2x₂ = 30

Also given that 5x₁ + 10x₂ ≤ 110
Let 5x₁ + 10x₂ = 110
x₁ + 2x₂ = 22

To get point of intersection, (i.e., the to get eo-ordinates of B)
3x₁ + 2x₂ = 30 …….(1)
x₁ + 2x₂ = 22 ……..(2)
(1) – (2) ⇒ 2x₁ = 8
x₁ = 4
x₁ = 4 substitute in (1),
x₁ + 2x₂ = 22
4 + 2x₂ = 22
2x₂ = 18
x₂ = 9
i.e., B is (4, 9)
The feasible region satisfying all the given conditions is OABC.
The co-ordinates of the points are O(0, 0), A(10, 0), B(4, 9), C(0, 11).

The maximum value of Z occurs at B.
∴ The optimal solution is x₁ = 4, x₂ = 9 and Z_max = 96

(ii) Given that 960x₁ + 640x₂ ≤ 15360
Let 960x₁ + 640x₂ = 15360
3x₁ + 2x₂ = 48

Also given that x₁ + x₂ ≤ 20
Let x₁ + x₂ = 20

To get point of intersection
3x₁ + 2x₂= 48 …..(1)
x₁ + x₂ = 20 ……(2)
(2) × -2 ⇒ -2x₁ – 2x₂ = -40 …..(3)
(1) + (3) ⇒ x₁ = 8
x₁ = 8 substitute in (2),
8 + x₂ = 20
x₂ = 12

The feasible region satisfying all the given conditions is OABC.
The co-ordinates of the comer points are O(0, 0), A(16, 0), B(8,12) and C(0, 16).

The maximum value of Z occurs at B(8, 12).
∴ The optimal solution is x₁ = 8, x₂ = 12 and Z_max = 392

(iii) Given that 5x₁ + x₂ ≥ 10
Let 5x₁ + x₂ = 10

Also given that x₁ + x₂ ≥ 6
Let x₁ + x₂ = 6

Also given that x₁ + 4x₂ ≥ 12
Let x₁ + 4x₂ = 12

To get C
5x₁ + x₂ = 10 ……..(1)
x₁ + x₂ = 6 ………(2)
(1) – (2) ⇒ 4x₁ = 4
⇒ x₁ = 1
x = 1 substitute in (2)
⇒ x₁ + x₂ = 6
⇒ 1 + x₂ = 6
⇒ x₂ = 5
∴ C is (1, 5)
To get B
x₁ + x₂ = 6
x₁ + 4x₂ = 12
(1) – (2) ⇒ -3x₂ = -6
x₂ = 2
x₂ = 2 substitute in (1), x₁ = 4
∴ B is (4, 2)

The feasible region satisfying all the conditions is ABCD.
The co-ordinates of the comer points are A(12, 0), B(4, 2), C(1, 5) and D(0, 10).

The minimum value of Z occours at C(1, 5).
∴ The optimal solution is x₁ = 1, x₂ = 5 and Z_min = 13

(iv) Given that 3x₁ + x₂ ≤ 9
Let 3x₁ + x₂ = 9

Also given that x₁ + 2x₂ ≤ 8
Let x₁ + 2x₂ = 8

3x₁ + x₂ = 9 ………(1)
x₁ + 2x₂ = 8 ……..(2)
(1) × 2 ⇒ 6x₁ + 2x₂ = 18 ……..(3)
(2) + (3) ⇒ -5x₁ = -10
x₁ = 2
x₁ = 2 substitute in (1)
3(2) + x₂ = 9
x₂ = 3
The feasible region satisfying all the conditions is OABC.
The co-ordinates of the corner points are O(0, 0), A(3, 0), B(2, 3), C(0, 4)

The maximum value of Z occurs at (2, 3).
∴ The optimal solution is x₁ = 2, x₂ = 3 and Z_max = 230

(v) Given that 3x₁ + 3x₂ ≤ 36
Let 3x₁ + 3x₂ = 36

Also given that 5x₁ + 2x₂ ≤ 50
Let 5x₁ + 2x₂ = 50

x₁ + x₂ = 12 ……(1)
5x₁ + 2x₂ = 50 ……….(2)
(1) × 2 ⇒ 2x₁ + 2x₂ = 24 ………(3)
(2) – (3) ⇒ 3x₁ = 26
x₁ = \(\frac{26}{3}\) = 8.66
put x₁ = \(\frac{26}{3}\) substitute in (1)
x₁ + x₂ = 12
x₂ = 12 – x₁
x₂ = 26 – \(\frac{26}{3}\) = \(\frac{10}{3}\) = 3.33
Also given that 2x₁ + 6x₂ ≤ 60
Let 2x₁ + 6x₂ = 60
x₁ + 3x₂ = 30

x₁ + x₂ = 12 …….(1)
x₁ + 3x₂ = 30 …….(2)
(1) – (2) ⇒ -2x₂ = -18
x₂ = 9
x₂ = 9 substitute in (1) ⇒ x₁ = 3

The feasible region satisfying all the given conditions is OABCD.
The co-ordinates of the comer points are O(0, 0), A(10, 0), B(\(\frac{26}{3}, \frac{10}{3}\)), and C = (3, 9) and D (0, 10)

The maximum value of Z occurs at C(3, 9)
∴ The optimal solution is x₁ = 3, x₂ = 9 and Z_max = 330

(vi) Given that 36x₁ + 6x₂ ≥ 108
Let 36x₁ + 6x₂ = 108
6x₁ + x₂ = 18

Also given that 3x₁ + 12x₂ ≥ 36
Let 3x₁ + 12x₂ = 36
x₁ + 4x₂ = 12

Also given that 20x₁ + 10x₂ ≥ 100
Let 20x₁ + 10x₂ = 100
2x₁ + x₂ = 10

The feasible region satisfying all the conditions is ABCD.
The co-ordinates of the comer points are A(12, 0), B(4, 2), C(2, 6) and D(0, 18).

The minimum value of Z occurs at B(4, 2)
∴ The optimal solution is x₁ = 4, x₂ = 2 and Z_min = 160

Tamilnadu State Board New Syllabus Samcheer Kalvi 11th Business Maths Guide Pdf Chapter 9 Correlation and Regression Analysis Ex 9.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 9 Correlation and Regression Analysis Ex 9.3

Samacheer Kalvi 11th Business Maths Correlation and Regression Analysis Ex 9.3 Text Book Back Questions and Answers

Choose the correct answer.

Question 1.
An example of a positive correlation is:
(a) Income and expenditure
(b) Price and demand
(c) Repayment period and EMI
(d) Weight and Income
Answer:
(a) Income and expenditure

Question 2.
If the values of two variables move in the same direction then the correlation is said to be:
(a) Negative
(b) Positive
(c) Perfect positive
(d) No correlation
Answer:
(b) Positive

Question 3.
If the values of two variables move in the opposite direction then the correlation is said to be:
(a) Negative
(b) Positive
(c) Perfect positive
(d) No correlation
Answer:
(a) Negative

Question 4.
Correlation co-efficient lies between:
(a) 0 to ∞
(b) -1 to +1
(c) -1 to 0
(d) -1 to ∞
Answer:
(b) -1 to +1

Question 5.
If r(X, Y) = 0 the variables X and Y are said to be:
(a) Positive correlation
(b) Negative correlation
(c) No correlation
(d) Perfect positive correlation
Answer:
(c) No correlation

Question 6.
The correlation coefficient from the following data N = 25, ΣX = 125, ΣY = 100, ΣX² = 650, ΣY²= 436, ΣXY = 520:
(a) 0.667
(b) -0.006
(c) -0.667
(d) 0.70
Answer:
(a) 0.667
Hint:

Question 7.
From the following data, N = 11, ΣX = 117, ΣY = 260, ΣX² = 1313, ΣY² = 6580, ΣXY = 2827. the correlation coefficient is:
(a) 0.3566
(b) -0.3566
(c) 0
(d) 0.4566
Answer:
(a) 0.3566
Hint:

Question 8.
The correlation coefficient is:

Answer:
(b) r(X, Y) = \(\frac{cov(x, y)}{\sigma_{x} \sigma_{y}}\)

Question 9.
The variable whose value is influenced or is to be predicted is called:
(a) dependent variable
(b) independent variable
(c) regressor
(d) explanatory variable
Answer:
(a) dependent variable

Question 10.
The variable which influences the values or is used for prediction is called:
(a) Dependent variable
(b) Independent variable
(c) Explained variable
(d) Regressed
Answer:
(b) Independent variable

Question 11.
The correlation coefficient:

Answer:
(a) \(r=\pm \sqrt{b_{x y} \times b_{y x}}\)

Question 12.
The regression coefficient of X on Y:

Answer:
(a) \(b_{x y}=\frac{\mathrm{N} \Sigma d x d y-(\Sigma d x)(\Sigma d y)}{\mathrm{N} \Sigma d y^{2}-(\Sigma d y)^{2}}\)

Question 13.
The regression coefficient of Y on X:

Answer:
(c) \(b_{y x}=\frac{\mathrm{N\Sigma} d x d y-(\Sigma d x)(\Sigma d y)}{\mathrm{N} \Sigma d x^{2}-(\Sigma d x)^{2}}\)

Question 14.
When one regression coefficient is negative, the other would be:
(a) Negative
(b) Positive
(c) Zero
(d) None of these
Answer:
(a) Negative

Question 15.
If X and Y are two variates, there can be at most:
(a) one regression line
(b) two regression lines
(c) three regression lines
(d) more regression lines
Answer:
(b) two regression lines

Question 16.
The lines of regression of X on Y estimates:
(a) X for a given value of Y
(b) Y for a given value of X
(c) X from Y and Y from X
(d) none of these
Answer:
(a) X for a given value of Y

Question 17.
Scatter diagram of the variate values (X, Y) give the idea about:
(a) functional relationship
(b) regression model
(c) distribution of errors
(d) no relation
Answer:
(a) functional relationship

Question 18.
If regression co-efficient of Y on X is 2, then the regression co-efficient of X on Y is:
(a) ≤ \(\frac{1}{2}\)
(b) 2
(c) > \(\frac{1}{2}\)
(d) 1
Answer:
(a) ≤ \(\frac{1}{2}\)

Question 19.
If two variables move in a decreasing direction then the correlation is:
(a) positive
(b) negative
(c) perfect negative
(d) no correlation
Answer:
(a) positive

Question 20.
The person suggested a mathematical method for measuring the magnitude of the linear relationship between two variables say X and Y is:
(a) Karl Pearson
(b) Spearman
(c) Croxton and Cowden
(d) Ya Lun Chou
Answer:
(a) Karl Pearson

Question 21.
The lines of regression intersect at the point:
(a) (X, Y)
(b) \((\overline{\mathrm{X}}, \overline{\mathrm{Y}})\)
(c) (0, 0)
(d) (σ_x, σ_y)
Answer:
(b) \((\overline{\mathrm{X}}, \overline{\mathrm{Y}})\)

Question 22.
The term regression was introduced by:
(a) R.A Fisher
(b) Sir Francis Galton
(c) Karl Pearson
(d) Croxton and Cowden
Answer:
(b) Sir Francis Galton

Question 23.
If r = -1, then correlation between the variables:
(a) perfect positive
(b) perfect negative
(c) negative
(d) no correlation
Answer:
(b) perfect negative

Question 24.
The coefficient of correlation describes:
(a) the magnitude and direction
(b) only magnitude
(c) only direction
(d) no magnitude and no direction
Answer:
(a) the magnitude and direction

Question 25.
If Cov(x, y) = -16.5, \(\sigma_{x}^{2}\) = 2.89, \(\sigma_{y}^{2}\) = 100. Find correlation coeffient.
(a) -0.12
(b) 0.001
(c) -1
(d) -0.97
Answer:
(d) -0.97