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 |x^{2} + y^{2}| ≤ 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}

|x^{2} + y^{2}| ≤ 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) 2^{3}

(2) 3^{2}

(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) 2^{17}

(2) 17^{2}

(3) 34

(4) Insufficient data

Answer:

(2) 17^{2}

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 17^{2}

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 2^{n2}. Here n = 3

∴ Required number = 2^{32} = 2^{9}

= 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) = x^{2} 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) = x^{2}

The image of x under f is always positive since x^{2} is positive for x = 1 and x = – 1 ∈ R

f(1) = 1^{2} = 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 ) = x^{2} 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) = x^{2}

f(-3) = (-3)^{2} = 9

f(0) = 0^{2} = o

f(3) = 3^{2} = 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) = x^{2} if 1 ≤ x ≤ 4 —– (2)

Put y = x^{2} ⇒ 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 ⇒ y^{2} = 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: