Relations And Functions

Given: A = {–1, 2, 3} and B = {1, 3}

To find: (i) A × B (ii) B × A (iii) B × B (iv) A × A

Explanation:

(i) A × B

The sets A = {–1, 2, 3} and B = {1, 3} are given; we need to find the Cartesian product of set A and set B.

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

So, this is the required Cartesian product.

(ii) The sets B = {1, 3} and A = {–1, 2, 3} are given; we need to find the Cartesian product of set B and set A.

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

So this is the required Cartesian product.

(iii) The sets B = {1, 3} and B = {1, 3} are given; we need to find the Cartesian product of set A and set B.

So, B×B = {(1, 1), (1, 3), (3, 1), (3, 3)}

So this is the required Cartesian product.

(iv) The sets A = {–1, 2, 3} and A = {–1, 2, 3} are given; we need to find the Cartesian product of set A and set B.

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

So this is the required Cartesian product.

Given: P = {x: x < 3, x ∈N}, Q = {x : x ≤ 2, x ∈W} where W is the set of whole numbers

To find: (P∪Q) × (P∩Q)

Explanation: Given P = {x: x < 3, x ∈N}

This means set P contains all natural numbers which are less than 3, so

P = {1, 2}

And Q = {x : x ≤ 2, x ∈W}

This means set Q contains all whole numbers which are less than or equal to 2, so

Q = {0, 1, 2}

Now

(P∪Q) is union of set P = {1, 2} and set Q = {0, 1, 2} elements, so

(P∪Q) = {0, 1, 2}

And,

(P∩Q) is intersection of set P = {1, 2} and set Q = {0, 1, 2} elements, so

(P∩Q) = {1, 2}

We need to find the Cartesian product of (P∪Q) = {0, 1, 2} and (P∩Q) = {1, 2}

So,

(P∪Q) × (P∩Q) = {(0, 1), (0, 2), (1, 1), (1, 2), (2, 1), (2, 2)}

This is the required Cartesian product.

Given: A = {x: x ∈W, x < 2} B = {x : x ∈N, 1 < x < 5} C = {3, 5} where W is the set of whole numbers

To find: (i) A × (B∩C) (ii) A × (B∪C)

Explanation: Given A = {x: x ∈W, x < 2}

This means set A contains all whole numbers which are less than 2, so

A = {0, 1}

And B = {x : x ∈N, 1 < x < 5}

This means set B contains all natural numbers which are greater than 1 and less than 5, so

B = {2, 3, 4}

(i) Now

(B∩C) is intersection of set B = {2, 3, 4} and set C = {3, 5} elements, so

(B∩C) = {3}

We need to find the Cartesian product of A = {0, 1} and (B∩C) = {3}

So,

A × (B∩C) = {(0, 3), (1, 3)}

This is the required Cartesian product.

(ii) Now

(B∪C) is union of set B = {2, 3, 4} and set C = {3, 5} elements, so

(B∪C) = {2, 3, 4, 5}

We need to find the Cartesian product of A = {0, 1} and (B∩C) = {2, 3, 4, 5}

So,

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

This is the required Cartesian product.

Given: (2a + b, a – b) = (8, 3)

To find: the value of a and b

Explanation: Now given the ordered pairs are equal, so corresponding elements will be equal, i.e.,

2a + b = 8 and a-b = 3

Now a-b = 3

⇒a = 3 + b

Substituting the value of a in the equation 2a + b = 8, we get

2(3 + b) + b = 8

⇒ 6 + 2b + b = 8

⇒ 3b = 8-6 = 2

Now substituting the value of b in equation (a-b = 3), we get

Hence the value of a and b are and respectively

Given:

To find: the value of a and b

Explanation: Now given the ordered pairs are equal, so corresponding elements will be equal, i.e.,

and a-2b = 6 + b

Now

⇒a = 0

Substituting the value of a in the equation (a-2b = 6 + b), we get

0-2b = 6 + b

⇒ -2b-b = 6

⇒ -3b = 6

⇒ b = -2

Hence the value of a and b are 0 and -2 respectively

Given: A = {1, 2, 3, 4, 5}, S = {(x, y) : x ∈A, y ∈A}

To find: the ordered pairs which satisfy the conditions x + y = 5

Explanation: Given: A = {1, 2, 3, 4, 5}, S = {(x, y) : x ∈A, y ∈A}

We need to find the ordered pair such that x + y = 5, where x and y belongs to set A = {1, 2, 3, 4, 5}

1 + 1 = 2≠5

1 + 2 = 3≠5

1 + 3 = 4≠5

1 + 4 = 5, so one of the ordered pair is (1, 4)

1 + 5 = 6≠5

2 + 1 = 3≠5

2 + 2 = 4≠5

2 + 3 = 5, so one of the ordered pair is (2, 3)

2 + 4 = 6≠5

2 + 5 = 7≠5

3 + 1 = 4≠5

3 + 2 = 5, so one of the ordered pair is (3, 2)

3 + 3 = 6≠5

3 + 4 = 7≠5

3 + 5 = 8≠5

4 + 1 = 5, so one of the ordered pair is (4, 1)

4 + 2 = 6≠5

4 + 3 = 7≠5

4 + 4 = 8≠5

4 + 5 = 9≠5

5 + 1 = 6≠5

5 + 2 = 7≠5

5 + 3 = 8≠5

5 + 4 = 9≠5

5 + 5 = 10≠5

So, the set of ordered pairs satisfying x + y = 5 is {(1,4), (2,3), (3,2), (4,1)}.

Given: A = {1, 2, 3, 4, 5}, S = {(x, y) : x ∈A, y ∈A}

To find: the ordered pairs which satisfy the conditions x + y < 5

Explanation: Given: A = {1, 2, 3, 4, 5}, S = {(x, y) : x ∈A, y ∈A}

We need to find the ordered pair such that x + y<5, where x and y belongs to set A = {1, 2, 3, 4, 5}

1 + 1 = 2<5, so one of the ordered pairs is (1, 1)

1 + 2 = 3<5, so one of the ordered pairs is (1, 2)

1 + 3 = 4<5, so one of the ordered pairs is (1, 3)

1 + 4 = 5

1 + 5 = 6>5

2 + 1 = 3<5, so one of the ordered pairs is (2, 1)

2 + 2 = 4<5, so one of the ordered pairs is (2, 2)

2 + 3 = 5

2 + 4 = 6>5

2 + 5 = 7>5

3 + 1 = 4<5, so one of the ordered pairs is (3, 1)

3 + 2 = 5

3 + 3 = 6>5

3 + 4 = 7>5

3 + 5 = 8>5

4 + 1 = 5

4 + 2 = 6>5

4 + 3 = 7>5

4 + 4 = 8>5

4 + 5 = 9>5

5 + 1 = 6>5

5 + 2 = 7>5

5 + 3 = 8>5

5 + 4 = 9>5

5 + 5 = 10>5

So, the set of ordered pairs satisfying x + y< 5 is {(1,1), (1,2), (1,3), (2, 1), (2,2), (3,1)}.

Given: A = {1, 2, 3, 4, 5}, S = {(x, y) : x ∈A, y ∈A}

To find: the ordered pairs which satisfy the conditions x + y > 8

Explanation: Given: A = {1, 2, 3, 4, 5}, S = {(x, y) : x ∈A, y ∈A}

We need to find the ordered pair such that x + y>8, where x and y belongs to set A = {1, 2, 3, 4, 5}

1 + 1 = 2<8

1 + 2 = 3<8

1 + 3 = 4<8

1 + 4 = 5<8

1 + 5 = 6<8

2 + 1 = 3<8

2 + 2 = 4<8

2 + 3 = 5<8

2 + 4 = 6<8

2 + 5 = 7<8

3 + 1 = 4<8

3 + 2 = 5<8

3 + 3 = 6<8

3 + 4 = 7<8

3 + 5 = 8

4 + 1 = <8

4 + 2 = 6<8

4 + 3 = 7<8

4 + 4 = 8

4 + 5 = 9>8, so one of the ordered pairs is (4, 5)

5 + 1 = 6<8

5 + 2 = 7<8

5 + 3 = 8

5 + 4 = 9>8, so one of the ordered pairs is (5, 4)

5 + 5 = 10>8, so one of the ordered pairs is (5, 5)

So the set of ordered pairs satisfying x + y>8 is {(4, 5), (5, 4),(5,5)}.

Given: R = {(x, y) : x, y ∈W, x² + y² = 25}

To find: the domain and Range of R

Explanation: Given R = {(x, y) : x, y ∈W, x² + y² = 25}

This means set R contains all whole numbers such that the relation between the elements of the set satisfy the condition x² + y² = 25, so

R = {(0,5), (3,4), (4, 3), (5,0)}

Now we need to find the domain and range of set R.

So, the domain of R consists of all the first elements of all the ordered pairs of R, so

Domain of R = {0, 3, 4, 5}

And the range of R contains all the second elements of all the ordered pairs of R, so

Range of R = {5, 4, 3, 0}

Given: R₁ = {(x, y) | y = 2x + 7, where x ∈R and – 5 ≤ x ≤ 5} is a relation

To find: the domain and Range of R₁

Explanation: Given R₁ = {(x, y) | y = 2x + 7, where x ∈R and – 5 ≤ x ≤ 5}

Now we need to find the domain and range of set R₁.

So, the domain of R₁ consists of all the first elements of all the ordered pairs of R₁, i.e., x, and it is also given – 5 ≤ x ≤ 5, so

Domain of R₁ = [-5, 5]

And the range of R contains all the second elements of all the ordered pairs of R₁, i.e., y and it is also given so

y = 2x + 7

Now x ∈ [-5,5]

Multiply both sides with 2, we get

So 2x∈[-10, 10]

Add both sides with 7, we get

2x + 7∈[-3, 17]

Or, y∈[-3, 17]

So,

Range of R₁ = [-3, 17]

Given: R₂ = {(x, y) | x and y are integers and x² + y² = 64} is a relation

To find: R₂

Explanation: Given R₂ = {(x, y) | x and y are integers and x² + y² = 64}

This means set R₂ contains all whole numbers such that the relation between the elements of the set satisfy the condition x² + y² = 64, so

x² + y² = 64

⇒ y² = 64- x²

This is possible only when

64-x²≥0

⇒ 64≥ x²

⇒ x²≤64

Taking square root on both sides, we get

⇒ x≤±√64

⇒ x≤±8

i.e., -8≤x≤8

So, x ∈ [-8,8]

Now for corresponding values of x, y becomes

When x = ±8

So, two of the elements of R₂ = (-8, 0), (8, 0)

When x = ±7

Now as y is an integer, so x cannot take values, ±7

When x = ±6

Now as y is an integer, so x cannot take values, ±6

When x = ±5

Now as y is an integer, so x cannot take values, ±5

When x = ±4

Now as y is an integer, so x cannot take values, ±4

When x = ±3

Now as y is an integer, so x cannot take values, ±3

When x = ±2

Now as y is an integer, so x cannot take values, ±2

When x = ±1

Now as y is an integer, so x cannot take values, ±5

When x = 0

So, two of the element of R₂ = (0, 8), (0, -8)

Hence complete set is

R₂ = {(0, 8), (0, -8), (-8,0), (8, 0)}

Given: R₃ = {(x, |x|) |x is a real number} is a relation

To find: the domain and Range of R₃

Explanation: Given R₃ = {(x, |x|) |x is a real number}

Now we need to find the domain and range of set R₃.

So, the domain of R₃ consists of all the first elements of all the ordered pairs of R₃, i.e., x, and it is also given x is a real number, so

Domain of R₃ = R

And the range of R contains all the second elements of all the ordered pairs of R₃, i.e., |x| and it is also given x is a real number, so

|x| = |R|

⇒ |x|≥0,

i.e., |x| has all positive real numbers including 0

So,

Range of R₃ = [0, ∞)

Given: h = {(4, 6), (3, 9), (– 11, 6), (3, 11)}

To find: whether the given relation is a function with proper reason

Explanation: the given relation is h = {(4, 6), (3, 9), (– 11, 6), (3, 11)}

Here it can be seen that element 3 has two images, i.e., 9 and 11.

And a relation is said to be function if every element of one set has one and only one image in other set.

So, h is not a function

Given: f = {(x, x) | x is a real number}

To find: whether the given relation is a function with proper reason

Explanation: the given relation is f = {(x, x) | x is a real number}

This means the relation f has elements which are real number.

Here it can be seen that element for every x ∈ R there will be unique image.

And a relation is said to be function if every element of one set has one and only one image in other set.

So, f is a function

Given: is a positive integer

To find: whether the given relation is a function with proper reason

Explanation: the given relation is is a positive integer

Here it can be seen that element n is a positive integer and the corresponding will be a unique and distinct number. Therefore, every element in the domain has unique image.

And a relation is said to be function if every element of one set has one and only one image in other set.

So, g is a function

Given: s = {(n, n²) | n is a positive integer}

To find: whether the given relation is a function with proper reason

Explanation: the given relation is s = {(n, n²) | n is a positive integer}

Here it can be seen that element n is a positive integer and the corresponding n² will be a unique and distinct number, as square of any positive integer is unique. Therefore, every element in the domain has unique image.

And a relation is said to be function if every element of one set has one and only one image in other set.

So, s is a function

Given: t = {(x, 3) | x is a real number.

To find: whether the given relation is a function with proper reason

Explanation: the given relation is t = {(x, 3) | x is a real number

Here it can be seen that the domain element x is a real number. And range just has one number i.e., 3 in it. So for every element in the domain has the image 3, it is a constant function.

And a relation is said to be function if every element of one set has one and only one image in other set.

So, t is a function

Given: f and g are real functions such that f (x) = x² + 7 and g (x) = 3x + 5

To find: f (3) + g (– 5)

Explanation: the given

f (x) = x² + 7

Now putting x = 3 in above function, we get

f (3) = 3² + 7 = 9 + 7 = 16……..(i)

and also given

g (x) = 3x + 5

Now putting x = -5 in above function, we get

g (-5) = 3(-5) + 5 = -15 + 5 = -10…………(ii)

Adding equation (i) and (ii), we get

f (3) + g (– 5) = 16-10 = 6

Given: f and g are real functions such that f (x) = x² + 7 and g (x) = 3x + 5

To find:

Explanation: the given

f (x) = x² + 7

Now putting in above function, we get

and also given

g (x) = 3x + 5

Now putting x = 14 in above function, we get

g (14) = 3(14) + 5 = 42 + 5 = 47…………(ii)

Multiplying equation (i) and (ii), we get

Given: f and g are real functions such that f (x) = x² + 7 and g (x) = 3x + 5

To find: f (– 2) + g (– 1)

Explanation: the given

f (x) = x² + 7

Now putting x = -2 in above function, we get

f (-2) = (-2)² + 7 = 4 + 7 = 11……..(i)

and also given

g (x) = 3x + 5

Now putting x = -1 in above function, we get

g (-1) = 3(-1) + 5

= -3 + 5 = 2…………(ii)

Adding equation (i) and (ii), we get

f (– 2) + g (– 1)

= 11 + 2

= 13

Given: f and g are real functions such that f (x) = x² + 7 and g (x) = 3x + 5

To find: f (t) – f (– 2)

Explanation: the given

f (x) = x² + 7

Now putting x = t in above function, we get

f (t) = t² + 7……..(i)

and again, considering the same function

f (x) = x² + 7

Now putting x = -2 in above function, we get

f (-2) = (-2)² + 7 = 4 + 7 = 11…….(ii)

Subtracting equation (i) with (ii), we get

f (t) – f (– 2) = t² + 7-11

= t²-4

Given: f and g are real functions such that f (x) = x² + 7 and g (x) = 3x + 5

To find: , if t ≠ 5

Explanation: the given

f (x) = x² + 7

Now putting x = t in above function, we get

f (t) = t² + 7……..(i)

and again, considering the same function

f (x) = x² + 7

Now putting x = 5 in above function, we get

f (5) = (5)² + 7 = 25 + 7 = 32……..(ii)

We need to find,

Substituting values from equation (i) and (ii), we get

But we know a²-b² = (a + b)(a-b), so above equation becomes,

Cancelling the like terms, we get

Given: f and g be real functions defined by f(x) = 2x + 1 and g(x) = 4x-7

To find: For what real numbers x, f (x) = g (x)

Explanation: to satisfy the condition f(x) = g(x), the given real functions should be equal

i.e., 2x + 1 = 4x-7

⇒ 7 + 1 = 4x-2x

⇒ 8 = 2x

Or, 2x = 8

⇒ x = 4

Hence for x = 4, f (x) = g (x)

Given: f and g be real functions defined by f(x) = 2x + 1 and g(x) = 4x-7

To find: For what real numbers x, f (x) < g (x)

Explanation: to satisfy the condition f(x)<g(x), we should have

i.e., 2x + 1<4x-7

⇒ 7 + 1<4x-2x

⇒ 8<2x

Or, 2x>8

⇒ x>4

Hence for x>4, f (x) > g (x)

Given: f and g be real valued functions defined as f (x) = 2x + 1, g (x) = x² + 1,

To find: f + g

Explanation: this can be obtained by adding functions f(x) and g(x), i.e.,

So, f + g = (f + g)(x)

⇒ f + g = f(x) + g(x)

Substituting the corresponding equation, we get

⇒ f + g = 2x + 1 + x² + 1

⇒ f + g = x² + 2x + 2

Given: f and g be real valued functions defined as f (x) = 2x + 1, g (x) = x² + 1,

To find: f-g

Explanation: this can be obtained by subtracting functions f(x) from g(x), i.e.,

So, f-g = (f-g)(x)

⇒ f-g = f(x)-g(x)

Substituting the corresponding equation, we get

⇒ f-g = 2x + 1-( x² + 1)

⇒ f-g = 2x-x²

Given: f and g be real valued functions defined as f (x) = 2x + 1, g (x) = x² + 1,

To find: fg

Explanation: this can be obtained by multipying functions f(x) and g(x), i.e.,

So, fg = (fg)(x)

⇒ fg = f(x) g(x)

Substituting the corresponding equation, we get

⇒ fg = (2x + 1)( x² + 1)

⇒ fg = 2x(x² ) + 2x(1) + 1(x²) + 1(1)

⇒ fg = 2x³ + 2x + x² + 1

⇒ fg = 2x³ + x² + 2x + 1

Given: f and g be real valued functions defined as f (x) = 2x + 1, g (x) = x² + 1,

To find:

Explanation: this can be obtained by dividing functions f(x) by g(x), i.e.,

So,

Substituting the corresponding equation, we get

Given: a function f : X →R, f (x) = x³ + 1, where X = {–1, 0, 3, 9, 7}

To find: function, f as set of ordered pairs and range of f

Explanation: given f : X →R, f (x) = x³ + 1, where X = {–1, 0, 3, 9, 7}

This means f is a function such that the first elements of all the ordered pair belong to the set X = {–1, 0, 3, 9, 7}. So this is the domain.

Now the second element of all the ordered pair are such that they satisfy the condition f (x) = x³ + 1

When x = -1, f (x) = x³ + 1 becomes

f (-1) = (-1)³ + 1 = -1 + 1 = 0, so ordered pair is (-1, 0)

When x = 0, f (x) = x³ + 1 becomes

f (0) = (0)³ + 1 = 0 + 1 = 1, so ordered pair is (0, 1)

When x = 3, f (x) = x³ + 1 becomes

f (3) = (3)³ + 1 = 27 + 1 = 28, so ordered pair is (3, 28)

When x = 9, f (x) = x³ + 1 becomes

f (9) = (9)³ + 1 = 729 + 1 = 730, so ordered pair is (9, 730)

When x = 7, f (x) = x³ + 1 becomes

f (7) = (7)³ + 1 = 343 + 1 = 344, so ordered pair is (7, 344)

So the given function as a set of ordered pairs is

f = {(-1, 0), (0, 1), (3, 28), (9, 730), (7, 344)}

And the range of f contains all the second elements of all the ordered pairs of f, so

Range of f = {0, 1, 28, 730, 344}

Given: f and g functions defined by f (x) = 3x² – 1 and g (x) = 3 + x

To find: For what x, f (x) = g (x)

Explanation: to satisfy the condition f(x) = g(x), the given real functions should be equal

i.e., 3x² – 1 = 3 + x

⇒ 3x² –x-3-1 = 0

⇒ 3x² –x-4 = 0

We will find the solution by splitting the middle term, i.e.,

⇒ 3x² + 3x-4x-4 = 0

⇒ 3x(x + 1)-4(x + 1) = 0

⇒ (3x-4)(x + 1) = 0

⇒ 3x-4 = 0 or x + 1 = 0

⇒ 3x = 4 or x = -1

Hence for , f (x) = g (x), i.e., given functions are equal.

Given: g = {(1, 1), (2, 3), (3, 5), (4, 7)}, and it is described by relation g (x) = αx + β

To find: whether g is a function, and also to find the values of α and β

Explanation: the given relation is g = {(1, 1), (2, 3), (3, 5), (4, 7)}

Here for every element in the domain has the unique image.

And a relation is said to be function if every element of one set has one and only one image in other set.

So g is a function.

Now given the relation g = {(1, 1), (2, 3), (3, 5), (4, 7)} as

g (x) = αx + β

for ordered pair (1,1), g (x) = αx + β, becomes

g (1) = α(1) + β = 1

⇒ α + β = 1

⇒ α = 1-β………..(i)

Now consider other ordered pair (2, 3), g (x) = αx + β, becomes

g (2) = α(2) + β = 3

⇒ 2α + β = 3

Now substituting value of α from equation (i), we get

⇒ 2(2) + β = 3

⇒ β = 3-4 = -1

Now substituting the value of β in equation (i), we get

α = 1-β = 1-(-1) = 2

Hence the values 2 and -1 should be assigned to α and β to satisfy the given condition g (x) = αx + β, i.e., g (x) = 2x-1

Given:

To find: the domain of function

Explanation: So the domain of a function consists of all the first elements of all the ordered pairs, i.e., x, so we have to find the values of x to get the required domain

Given,

We know the value of cos x lies between -1, 1, so

-1≤ cos x≤ 1

Multiplying by negative sign, we get

Or 1≥- cos x≥ -1

Adding with 1, we get

2≥1- cos x≥0…………(i)

Now for real value of

1- cos x≠0

⇒ cos x≠ 1

Or, x≠2nπ ∀n∈Z

Hence the domain of f = R-{2nπ:n∈Z}

Given:

To find: the domain of function

Explanation: So, the domain of a function consists of all the first elements of all the ordered pairs, i.e., x, so we have to find the values of x to get the required domain

Given,

Now for real value of f,

x + |x|>0

Now when x>0,

x + |x|>0⇒ x + x>0⇒ 2x>0⇒ x>0

Now when x<0,

x + |x|>0⇒ x-x>0⇒ 2x>0⇒ x>0

This is not possible.

So, x>0, to satisfy the given equation.

Hence the domain of f = R ⁺

Given: f(x) = x|x|

To find: the domain of function

Explanation: So, the domain of a function consists of all the first elements of all the ordered pairs, i.e., x, so we have to find the values of x to get the required domain

Given,

f(x) = x|x|

We know x and |x| are defined for all real values.

Hence the domain of f = R

Given:

To find: the domain of function

Explanation: So, the domain of a function consists of all the first elements of all the ordered pairs, i.e., x, so we have to find the values of x to get the required domain

Given,

Now for real value of

x²-1≠0

⇒ (x-1)(x + 1)≠0

⇒ x-1≠0 or x + 1≠0

⇒ x≠1 or x≠-1

Hence the domain of f = R-{-1, 1}

Given:

To find: the domain of function

Explanation: So, the domain of a function consists of all the first elements of all the ordered pairs, i.e., x, so we have to find the values of x to get the required domain

Given,

Now for real value of

28-x≠0

⇒ x≠ 28

Hence the domain of f = R-{28}

Given:

To find: the range of function

Explanation: So, the range of a function consists of all the second elements of all the ordered pairs, i.e., f(x), so we have to find the values of f(x) to get the required range

Given,

Let this be equal to y, so

But x²≥ 0

So

⇒ y>0 and 2y-3≥0

⇒ y>0 and 2y≥3

⇒ y>0 and

Or f(x)>0 and

Hence the range of f =

Given: f(x) = 1-|x-2|

To find: the range of function

Explanation: So, the range of a function consists of all the second elements of all the ordered pairs, i.e., f(x), so we have to find the values of f(x) to get the required range

Given,

f(x) = 1-|x-2|

Now for real value of f,

|x-2|≥ 0

Adding negative sign, we get

Or -|x-2|≤ 0

Adding 1 we get

⇒ 1-|x-2|≤ 1

Or f(x)≤1

⇒ f(x)∈ (-∞, 1]

Hence the range of f = (-∞, 1]

Given: f(x) = |x-3|

To find: the range of function

Explanation: So, the range of a function consists of all the second elements of all the ordered pairs, i.e., f(x), so we have to find the values of f(x) to get the required range

Given,

f(x) = |x-3|

We know |x| are defined for all real values.

And |x-3| will always be greater than or equal to 0.

i.e., f(x)≥0

Hence the range of f = [0, ∞)

Given: f (x) = 1 + 3 cos2x

To find: the range of function

Explanation: So, the range of a function consists of all the second elements of all the ordered pairs, i.e., f(x), so we have to find the values of f(x) to get the required range

Given,

f (x) = 1 + 3 cos2x

We know the value of cos 2x lies between -1, 1, so

-1≤ cos 2x≤ 1

Multiplying by 3, we get

-3≤ 3cos 2x≤ 3

Adding with 1, we get

-2≤ 1 + 3cos 2x≤ 4

Or, -2≤ f(x)≤ 4

Hence f(x)∈ [-2, 4]

Hence the range of f = [-2, 4]

Given: function f(x) = |x-2| + |2 + x|, -3≤ x≤ 3

To find: to redefine the given function

Explanation:

We know

when x>0,

|x-2| is (x-2), x≥2

and |2 + x| is (2 + x), x≥-2

when x>0

|x-2| is -(x-2), x<2

and |2 + x| is -(2 + x), x<-2

Now given, f(x) = |x-2| + |2 + x|, -3≤ x≤ 3

It can be rewritten as,

Or

Or,

Hence the given function can be redefined as shown above.

Given:

To show:

Explanation: given

Now replace x by we get

Hence proved

Given:

To show:

Explanation: given

Now replace x by we get

Hence proved

Given: f(x) = √x and g (x) = x two functions defined in the domain R ⁺ ∪{0},

To find: (f + g)(x)

Explanation: this can be obtained by adding functions f(x) and g(x), i.e.,

⇒ (f + g)(x) = f(x) + g(x)

Substituting the corresponding equation, we get

⇒ (f + g)(x) = √x + x

Given: Given: f(x) = √x and g (x) = x two functions defined in the domain R ⁺ ∪{0},

To find: (f-g)(x)

Explanation: this can be obtained by subtracting functions f(x) from g(x), i.e.,

⇒ (f-g)(x) = f(x)-g(x)

Substituting the corresponding equation, we get

⇒ (f-g)(x) = √x-x

Given: f(x) = √x and g (x) = x two functions defined in the domain R ⁺ ∪{0},

To find: (fg)(x)

Explanation: this can be obtained by multiplying functions f(x) and g(x), i.e.,

⇒ (fg)(x) = f(x) g(x)

Substituting the corresponding equation, we get

⇒ (fg)(x) = (√x)(x)

⇒ (fg)(x) = x√x

Given: f(x) = √x and g (x) = x two functions defined in the domain R ⁺ ∪{0},

To find:

Explanation: this can be obtained by dividing functions f(x) by g(x), i.e.,

So,

Substituting the corresponding equation, we get

Multiply and divide by √x, we get

Given:

To find: the domain and range of function

Explanation: So, the domain of a function consists of all the first elements of all the ordered pairs, i.e., x, so we have to find the values of x to get the required domain

Given,

Now for real value of

x-5≠0 and x-5>0

⇒ x≠5 and x>5

Hence the domain of f = (5, ∞)

And the range of a function consists of all the second elements of all the ordered pairs, i.e., f(x), so we have to find the values of f(x) to get the required range

Now we know for this function

x-5>0

taking square root on both sides, we get

Or

Or

f(x)>0

⇒ f(x)∈(0, ∞)

Hence the range of f = (0, ∞)

Given:

To prove: f(y) = x

Explanation: we have

Now we will replace x with y, we get

From equation (i), we will substitute the value of y in above equation, we get

Cancelling the like terms, we get

f(y) = x

Hence proved

Given: n (A) = m, and n (B) = n

To find: the total number of non-empty relations that can be defined from A to B

Explanation: given n(A) = m and n(B) = n

So n(A×B) = n(A)×n(B) = m×n

And we know a Relation R from a non-empty set A to a non empty set B is a subset of the Cartesian product set A × B.

So total number of relation from A to B = Number of subsets of A×B = 2^mn
So, total number of non-empty relations = 2^mn – 1

Hence the correct option is (D)

Given: [x]² – 5 [x] + 6 = 0, where [ . ] denote the greatest integer function

To find: range of x

Explanation: we have

[x]² – 5 [x] + 6 = 0

We will split the middle term, we get

⇒ [x]² – 3 [x] -2[x] + 6 = 0

⇒ [x]([x]–3)-2([x]-3) = 0

⇒ ([x]-3)([x]-2) = 0

⇒ [x]-3 = 0 or [x]-2 = 0

⇒ [x] = 3 or [x] = 2

⇒ [x] = 2, 3

For [x] = 2, x ∈ [2,3)

For [x] = 3, x ∈ [3,4)

[x] ∈ [2,3) ∪ [3,4)

So, x ∈ [2,4]

Hence the correct answer is option (C).

Given:

To find: the range of the given function

Explanation: So the range of a function consists of all the second elements of all the ordered pairs, i.e., f(x), so we have to find the values of f(x) to get the required range

We know the value of cos x lies between -1, 1, so

-1≤ cos x≤ 1

Multiplying by 2, we get

-2≤ 2cos x≤ 2

Adding a negative sign, we get

2≥ -2cos x≥ -2

Adding with 1, we get

3≥ 1-2cos x≥ -1

Now is defined if

-1≤ 1- 2 cos x < 0 or 0 < 1- 2 cos x ≤ 3

So the correct answer is option (C)

Given:

To find: the relation between f(xy) and f(x).f(y)

Explanation: First we will find f(xy), for this we will replace x with xy in the given equation , we get

Now we will find f(y), for this we will replace x with y in the given equation , we get

Using this we will find the value for f(x).f(y), we get

Comparing equation (i) and (ii), we get

⇒ f(xy)≤ f(x)f(y)

So, the correct answer is option (C)

Given:

To find: the domain of the given function

Explanation: So the domain of a function consists of all the first elements of all the ordered pairs, i.e., x, so we have to find the values of x to get the required domain

let,

Now for real value

a²-x²≥0

⇒ x²≥ a²

⇒ x≥±a

Or –a ≤ x ≤ a

Hence the domain of given function is = [-a, a]

So, the correct answer is option (B)

Given: f (x) = ax + b, where a and b are integers, f (–1) = – 5 and f (3) = 3

To find: the values of a and b

Explanation: we have f (x) = ax + b

Put x = -1 in above equation we get

f (-1) = a(-1) + b

But it is also given f(-1) = -5, so above equation becomes

-5 = -a + b

⇒ b = a-5………(i)

we have f (x) = ax + b

Put x = 3 in above equation we get

f (3) = a(3) + b

But it is also given f(3) = 3, so above equation becomes

3 = 3a + b

Now substituting the value of b from equation (i), we get

3 = 3a + a-5

⇒ 4a = 5 + 3

⇒ 4a = 8

⇒ a = 2

Substituting the value of a in equation (i), we get

b = a-5 = 2-5 = -3

Hence the values of a and b are 2 and -3 respectively.

Hence the correct answer is option (B)

Given:

To find: the domain of the given function

Explanation: So the domain of a function consists of all the first elements of all the ordered pairs, i.e., x, so we have to find the values of x to get the required domain

We have

Now for real value

4-x≥0 and x²-1>0

⇒ 4≥x and x²>1

⇒ x≤4 and -1>x>1

⇒ x≤4 and x>1 and x<-1

⇒ x∈ (-∞, -1)∪(1, 4]

Hence the domain of given function is (-∞, -1)∪(1, 4]

So, the correct answer is option (A)

Given:

To find: the domain and range of function

Explanation: So, the domain of a function consists of all the first elements of all the ordered pairs, i.e., x, so we have to find the values of x to get the required domain

Given,

Now for real value of

x-4≠0

⇒ x≠4

Hence the domain of f = R-{4}

And the range of a function consists of all the second elements of all the ordered pairs, i.e., f(x), so we have to find the values of f(x) to get the required range

Given,

⇒ f(x) = -1

Hence the range of f = {-1}

Hence the correct answer is option (C)

Given:

To find: the domain and range of function

Explanation: So the domain of a function consists of all the first elements of all the ordered pairs, i.e., x, so we have to find the values of x to get the required domain

Given,

Now for real value of

x-1≥0

⇒ x≥1

Hence the domain of f = [1, ∞)

And the range of a function consists of all the second elements of all the ordered pairs, i.e., f(x), so we have to find the values of f(x) to get the required range

We got

Now for real value of

x-1≥0

or

⇒ f(x)≥0

Hence the range of f = [0, ∞)

Hence the correct answer is option (D).

Given:

To find: the domain of the given function

Explanation: So the domain of a function consists of all the first elements of all the ordered pairs, i.e., x, so we have to find the values of x to get the required domain

We have

Now for real value

x²-x-6≠0

Splitting the middle term, we get

⇒ x²-3x + 2x-6≠0

⇒ x(x-3) + 2(x-3)≠0

⇒ (x-3)(x + 2)≠0

⇒ x-3≠ 0 or x + 2≠0

⇒ x≠3 or x≠-2

Hence the domain of given function is R-{-2, 3}

So the correct answer is option (A)

Given: f(x) = 2–|x −5|

To find: the domain and range of function

Explanation: So the domain of a function consists of all the first elements of all the ordered pairs, i.e., x, so we have to find the values of x to get the required domain

Given,

f(x) = 2–|x −5|

Now x is defined for all real numbers

Hence the domain of f is R

And the range of a function consists of all the second elements of all the ordered pairs, i.e., f(x), so we have to find the values of f(x) to get the required range

Now we know

|x-5|≥0

or

-|x-5|≤0

Adding 2 we get

2-|x-5|≤2

⇒ f(x)≤2

Hence the range of f = (-∞, 2]

Hence the correct answer is option (B)

Given: f (x) = 3x² – 1 and g (x) = 3 + x

To find: the domain of the given functions equal

Explanation: So the domain of a function consists of all the first elements of all the ordered pairs, i.e., x, so we have to find the values of x to get the required domain

The two given functions are equal, so

f (x) = g (x)

Substituting the values, we get

3x² – 1 = 3 + x

3x² – 1 - 3 – x = 0

3x²– x-4 = 0

We will find the solution by splitting the middle term, i.e.,

⇒ 3x² + 3x-4x-4 = 0

⇒ 3x(x + 1)-4(x + 1) = 0

⇒ (3x-4)(x + 1) = 0

⇒ 3x-4 = 0 or x + 1 = 0

⇒ 3x = 4 or x = -1

Hence for , f (x) = g (x), i.e., given functions are equal.

Hence the domain is =

Hence the correct answer is option (B)

Given: f and g be two real functions given by

f = {(0, 1), (2, 0), (3, – 4), (4, 2), (5, 1)}

g = {(1, 0), (2, 2), (3, – 1), (4, 4), (5, 3)}

To find: the domain of f.g

Explanation: So the domain of a function consists of all the first elements of all the ordered pairs, i.e., x, so we have to find the values of x to get the required domain

The two given functions are

f = {(0, 1), (2, 0), (3, – 4), (4, 2), (5, 1)}

g = {(1, 0), (2, 2), (3, – 1), (4, 4), (5, 3)}

Domain of f = {0,2, 3, 4, 5}
And Domain of g = {1, 2, 3,4, 5}
Domain of (f . g) = (Domain of f) ∩ (Domain of g) = {2, 3,4, 5}

The correct answer is (a)-(iii), (b)-(iv), (c)-(ii), (d) – (i)

Explanation:

given two functions are

f = {(2, 4), (5, 6), (8, – 1), (10, – 3)}

g = {(2, 5), (7, 1), (8, 4), (10, 13), (11, 5)}

So the domain of f = {2, 5, 8, 10}

And the domain of g = {2, 7, 8, 10, 11}

(i) f-g

Domain of (f - g) = (Domain of f) ∩ (Domain of g) = {2, 8, 10}

So (f-g) = (f-g)(x) = f(x)-g(x)

When x = 2, (f-g) = (f-g)(2) = f(2)-g(2) = 4-5 = -1⇒ (2, -1)

When x = 8, (f-g) = (f-g)(8) = f(8)-g(8) = -1-4 = -5⇒ (8, -5)

When x = 10, (f-g) = (f-g)(10) = f(10)-g(10) = -3-13 = -16⇒ (10, -16)

So f-g = {(2, -1), (8,-5), (10,-16)}

(ii) f + g

Domain of (f + g) = (Domain of f) ∩ (Domain of g) = {2, 8, 10}

So (f + g) = (f + g)(x) = f(x) + g(x)

When x = 2, (f + g) = (f + g)(2) = f(2) + g(2) = 4 + 5 = 9⇒ (2, 9)

When x = 8, (f + g) = (f + g)(8) = f(8) + g(8) = -1 + 4 = 3⇒ (8, 3)

When x = 10, (f + g) = (f + g)(10) = f(10) + g(10) = -3 + 13 = 10⇒ (10, 10)

So f + g = {(2, 9), (8,3), (10,10)}

(iii) f.g

Domain of (f . g) = (Domain of f) ∩ (Domain of g) = {2, 8, 10}

So (f.g) = (f.g)(x) = f(x) g(x)

When x = 2, (f.g) = (f.g)(2) = f(2) g(2) = 4× 5 = 20⇒ (2, 20)

When x = 8, (f.g) = (f.g)(8) = f(8) g(8) = -1× 4 = -4⇒ (8, -4)

When x = 10, (f.g) = (f.g)(10) = f(10) g(10) = -3×13 = -39⇒ (10, -39)

So f.g = {(2, 20), (8,-4), (10,-39)}

(iv)

Domain of = (Domain of f) ∩ (Domain of g) = {2, 8, 10}

So

When x = 2,

When x = 8,

When x = 10,

So

The given statement is false.

Explanation:

given R = {(x, y) : y = x – 5, x, y ∈Z}

This means set R contains numbers such that y = x-5, so

So when x = 5, y becomes

y = x-5 = 5-5 = 0

So corresponding y will be 0,

So (5,2) does not belong to R.

The given statement is false.

Explanation: The set P = {1, 2} is given

⇒ n(P) = 2

Now we need to find P×P×P,

So number of elements in P×P×P, will be

n(P×P×P) = n(P)×n(P)×n(P) = 2× 2× 2 = 8

But given P×P×P set has just 4 elements; hence it is not the set of P×P×P.

The set of P×P×P is

P×P×P = {(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2), (2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)}

The given statement is true.

Explanation:

Now

Cartesian product of set A = {1, 2, 3} and B = {3, 4} is

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

Cartesian product of set A = {1, 2, 3} and C = {4, 5, 6} is

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

Now,

(A×B)∪(A×C) is union of set A×B and set A×C elements, so

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

This is the required Cartesian product.

The given statement is false.

Explanation:

Two ordered pairs are equal, if and only if the corresponding first elements are equal and the second elements are also equal,

So from given criteria

⇒ x-2 = -2 and

⇒ x = -2 + 2 and

⇒ x = 0 and

These are the values of x and y

The given statement is true

Explanation:

Given A × B = {(a, x), (a, y), (b, x), (b, y)}

So

Set A will be set of first element of ordered pairs in A x B

So, A = {a, b}

And B will be set of first element of ordered pairs in A x B

So B = {x, y}