Relations And Functions

Given:

Since the ordered pairs are equal, the corresponding elements are equal.

Therefore,

Hence by solving we get, x = 2 and y = 1.

Given: Set A has 3 elements and the set B = {3, 4, 5}

As we see the number of elements in set B = 3.

Number of elements in (A× B) = (Number of elements in set A) × (Number of elements in B)

⇒ n(A × B) = n(A) × n(B)

⇒ n(A × B) = 3 × 3

⇒ n(A × B) = 9

Hence, the number of elements in (A×B) = 9.

Given: G = {7, 8} and H = {5, 4, 2}

By definition of Cartesian product of two non-empty Set P and Q:

P × Q = {(p, q): p Є P, q Є Q}

Therefore, G × H = {(7,5), (7,4), (7,2), (8,5), (8,4), (8,2)}.

And H × G = {(5,7), (5,8), (4,7), (4,8), (2,7), (2,8)}.

(i) Given: P = {m, n} and Q = {n, m}

By definition of Cartesian product of two non-empty Set P and Q:

P × Q = {(p, q): p Є P, q Є Q}

Therefore, P × Q = {(m, n), (m, m), (n, m), (n, n)}.

⇒ P × Q ≠ {(m, n), (n, m)}.

Hence, the statement is false.

(ii) Given: A and B are non-empty sets and x Є A and y Є B.

By definition of Cartesian product of two non-empty Set P and Q:

P × Q = {(p, q): p Є P, q Є Q}

⇒ A × B = {(x, y): x Є A, y Є B}

Hence, the statement is true.

(iii) Given: A = {1, 2}, B = {3, 4}

To Prove:

As

By definition if either of the two set P and Q is null set then P × Q will also be a null set. i.e. P × Q = ϕ.

⇒ .

Hence, the statement is true.

Given: A = {–1, 1}
cartesian product of a two sets A and B is given by set of multiplication of every member of the set with every other member of the set. Thus pairs of members of set are formed.
For given A,
A × A = {-1, 1} × {-1, 1}

A × A = {(-1, -1), (-1, 1), (1, -1), (1, 1)}
Now,

By definition A × A × A = {(a,b,c) : a,b,c Є A}.

⇒ A × A × A = {(-1, -1, -1), (-1, -1, 1), (-1, 1, -1), (1, -1, -1), (-1, 1, 1), (1, -1, 1), (1, 1, -1), (1, 1, 1)}

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

By definition of Cartesian product of two non-empty Set P and Q:

P × Q = {(p, q): p Є P, q Є Q}

Hence, we see A = set of all first elements.

B = set of all second elements.

⇒ A = {a,b} and B = {x,y}.

(i) Given: A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}

To verify: A × (B ∩ C) = (A × B) ∩ (A × C)

As we see,

By definition if either of the two set P and Q is null set then P × Q will also be a null set. i.e. P × Q = ϕ .

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

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

From (1) and (2)

(ii) Given: A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}

To verify: A × C is a subset of B × D.

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

(B × D) = {(1,5), (1,6), (1,7), (1,8), (2,5), (2,6), (2,7), (2,8), (3,5), (3,6), (3,7), (3,8), (4,5), (4,6), (4,7), (4,8)}

As we see all the elements of set A × B are there in set B × D.

Hence, A × C is a subset of B × D.

Given: A = {1, 2} and B = {3, 4}

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

Number of elements in (A × B) = 4

⇒ n(A × B) = 4

Then number of subsets of set (A × B) = 2ⁿ = 2⁴ = 16

⇒ number of subsets of set (A × B) = 16

These are:

{ϕ}, {(1,3)}, {(1,4)}, {(2,3)}, {(2,4)}, {(1,3),(1,4)}, {(1,3),(2,3)}, {(1,3),(2,4)}, {(1,4),(2,3)}, {(1,4),(2,4)}, {(2,3),(2,4)}, {(1,3),(1,4),(2,3)}, {(1,3),(1,4),(2,4)}, {(1,4),(2,3),(2,4)}, {(1,3),(2,3), (2,4)}, {(1,3),(1,4),(2,3),(2,4)}.

Given: n(A) = 3 and n(B) = 2 and If (x, 1), (y, 2), (z, 1) are in A × B.

By definition of Cartesian product of two non-empty Set P and Q:

P × Q = {(p, q): p Є P, q Є Q}

Hence, we see P = set of all first elements.

Q = set of all second elements.

⇒ (x, y, z) are elements of A and (1,2) are elements of B.

⇒ As n(A) = 3 and n(B) = 2 so, A = {x, y, z} and B = {1, 2}.

Given: Cartesian product A × A having 9 elements among which are found (–1, 0) and (0,1).

Number of elements in (A× B) = (Number of elements in set A) × (Number of elements in B)

⇒ n(A × A) = n(A) × n(A)

⇒ n(A × A) = 9 (given)

⇒ n(A) × n(A) = 9

⇒ n(A) = 3

By definition A × A = {(a, a): a Є A}.

Therefore, -1, 0 and 1 are the elements of set A.

Because, n(A) = 3 therefore, A = {-1, 0, 1}.

Hence the remaining elements of set (A × A) are:

(-1,-1), (-1,1), (0,0), (0, -1), (1,1), (1, -1) and (1, 0).

Given: A = {1, 2, 3, ..., 14} and R = {(x, y) : 3x – y = 0, where x, y ∈ A}

As the relation R from A to A is given as:

R = {(x, y) : 3x – y = 0, where x, y A}

⇒ R = {(x, y) : 3x = y, where x, y A}

Hence the relation in roaster form, R = {(1, 3), (2, 6), (3, 9), (4, 12)}

As Domain of R = set of all first elements of the order pairs in the relation.

⇒ Domain of R = {1, 2, 3, 4}

Co-domain of R = the whole set A

⇒ Co-domain of R = {1, 2, 3, ..., 14}

Range of R = set of all second elements of the order pairs in the relation.

⇒ range of R = {3, 6, 9, 12}.

Given: R = {(x, y) : y = x + 5, x is a natural number less than 4; x, y ∈ N}.

As x is a natural number which is less than 4.

Hence the relation in roaster form, R = {(1,6), (2,7), (3,8)}

As Domain of R = set of all first elements of the order pairs in the relation.

⇒ Domain of R = {1, 2, 3}

Range of R = set of all second elements of the order pairs in the relation.

⇒ range of R = {6, 7, 8}.

Given: A = {1, 2, 3, 5} and B = {4, 6, 9}

R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B}

Hence the relation in roaster form, R = {(1,4), (1,6), (2,9), (3,4), (3,6), (5,4), (5,6)}

Given:

As we see in given figure: P = {5,6,7}, Q = {3,4,5}

(i) Hence the relation in set builder form, R = {(x, y): y = x-2; x Є P}

or R = {(x, y): y = x-2; for x = 5, 6, 7}

(ii) And the relation in roaster form, R = {(5,3), (6,4), (7,5)}

As Domain of R = set of all first elements of the order pairs in the relation.

⇒ Domain of R = {5, 6, 7}

Range of R = set of all second elements of the order pairs in the relation.

⇒ range of R = {3, 4, 5}.

Given: A = {1, 2, 3, 4, 6}

R = {(a, b): a, b ∈ A, b is exactly divisible by a}

Hence the relation in roaster form, R = {(1,1), (1,2), (1,3), (1,4), (1,6), (2,2), (2,4), (2,6), (3,3), (3,6), (4,4), (6,6)}

As Domain of R = set of all first elements of the order pairs in the relation.

⇒ Domain of R = {1, 2, 3, 4, 6}

Range of R = set of all second elements of the order pairs in the relation.

⇒ range of R = {1, 2, 3, 4, 6}.

Given: R = {(x, x + 5) : x ∈ {0, 1, 2, 3, 4, 5}}

Hence the relation in roaster form, R = {(0,5), (1,6), (2,7), (3,8), (4,9), (5,10)}

As Domain of R = set of all first elements of the order pairs in the relation.

⇒ Domain of R = {0, 1, 2, 3, 4, 5}

Range of R = set of all second elements of the order pairs in the relation.

⇒ range of R = {5, 6, 7, 8, 9, 10}.

Given: R = {(x, x³) : x is a prime number less than 10}

As we know the prime number less than 10 are 2, 3, 5 and 7.

Hence the relation in roaster form, R = {(2,8), (3,27), (5,125), (7,343)}

Given: A = {x, y, z} and B = {1, 2}

∴ A × B = {(x,1), (x,2), (y,1), (y,2), (z,1), (z,2)}

⇒ n(A × B) = 6

Then number of subsets of set (A × B) = 2ⁿ = 2⁶

⇒ number of relations from A to B = 2⁶.

Given: R = {(a, b): a, b ∈ Z, a – b is an integer}

As we know that the difference between any two integers is always an integer.

As Domain of R = set of all first elements of the order pairs in the relation.

⇒ Domain of R = Z

Range of R = set of all second elements of the order pairs in the relation.

⇒ range of R = Z

Since, 2, 5, 8, 11, 14 and 17 are the elements of domain R having their unique images. Hence, this relation R is a function.

As Domain of R = set of all first elements of the order pairs in the relation.

⇒ Domain of R = {2, 5, 8, 11, 14, 17}

Range of R = set of all second elements of the order pairs in the relation.

⇒ range of R = {1}.

Since, 2, 5, 8, 11, 14 and 17 are the elements of domain R having their unique images. Hence, this relation R is a function.

As Domain of R = set of all first elements of the order pairs in the relation.

⇒ Domain of R = {2, 4, 6, 8, 10, 12, 14}

Range of R = set of all second elements of the order pairs in the relation.

⇒ range of R = {1, 2, 3, 4, 5, 6, 7}.

Since, the same first element 1 corresponds to two different images 3 and 5. Hence, this relation is not a function.

Given: f(x) = –|x|

_{As we know,}

Since, f(x) is defined for x Є R, the domain of f is R.

It can be observed that the range of f(x) = -|x| is all real numbers except positive real numbers. Because will will always get a negative number when we put a value from domain.

Thus, the range of function is f(x) is (-∞, 0].

Given:

Domain: These are the values of x for which f(x) is defined.

from the given f(x) we can say that, f(x) should be real and for that,

9 - x² ≥ 0 [Since a value less than 0 will give an imaginary value]

(3 + x)(3 - x) ≥ 0

Now there are two critical points, x = + 3 and x = - 3

Taking a value less than - 3 and putting in the expression we get,

(3 - 5)(3 + 5) = -ve value and thus

Plotting these on number line we get,

Since, f(x) is defined for all real numbers that are greater than or equal to -3 and less than or equal to 3, the domain of f(x) is [-3, 3].

Given: f (x) = 2x –5

(i) f(0) = 2× 0 – 5

= 0 -5 = -5

(i) f(7) = 2× 7 – 5

= 14 -5 = 9

(i) f(-3) = 2× (-3) – 5

= -6 -5 = -11

Given:

(i) t(0)

⇒ t(0) = 32

(ii) t(28)

(iii) t(-10)

⇒ t(-10) = -18 + 32 = 14

(iv) t(C) = 212

⇒ Value of C, when t(C) is 212 = 100.

Given: f (x) = 2 – 3 x, x ∈ R, x > 0

As x > 0

⇒ 3x > 0

⇒ -3x < 0

⇒ -3x + 2 < 0 + 2

⇒ 2 – 3x < 2

⇒ f(x) < 2

Hence, range of f(x) = (-∞, 2).

Given: f(x) = x² + 2, x is a real number.

As x² > 0

⇒ x² + 2 > 0 + 2

⇒ x² + 2 > 2

⇒ f(x) > 2

Hence, range of f(x) = [2, ∞).

Given: f(x) = x, x is a real number.

It is clear that range of f(x) is set of all real numbers.

Hence, range of f(x) = R.

Given:

As

⇒ f(x) = x² for 0≤x<3

And f(x) = 3xfor 3≤x<10

At x = 3, f(x) = x² = 3² = 9

Also, at x = 3, f(x) = 3x = 3×3 = 9

Hence, we see for 0≤x≤10, f(x) has unique images. Thus, by definition of a function the given relation is function.

Now,

As

⇒ g(x) = x² for 0≤x<2

And g(x) = 3xfor 2≤x<10

At x = 2, g(x) = x² = 2² = 4

Also, at x = 2, g(x) = 3x = 3×2 = 6

Hence, element 2 of the domain of relation g(x) corresponds to two different images i.e. 4 and 6.

because for 0≤x≤10, f(x) does not have unique images. Thus, by definition of a function the given relation is not a function.

Given: f(x) = x²

Then

Given:

Hence, we see f(x) can be defined at all real numbers except at x = 2 and x = 6. Hence the domain of f(x) is (R - {2, 6}).

Given: f(x) = √(x - 1)

To Find: Domain and Range of f(x)

Now for real values of f(x), x≥1.

Thus,

√(x - 1) is defined for x≥1.

Hence, the domain of f is the set of all real numbers greater than or equal to 1 i.e., the domain of f = [1, ∞).

Now for Range,

As, x ≥ 1, x - 1 ≥ 0

And thus √x - 1 ≥ 0

Hence, the Range of f is the set of all real numbers greater than or equal to 0 i.e., the Range of f = [0, ∞).

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

As we see |x-1| is defined for all real numbers (R).

Hence the domain of f = R.

Also, for x Є R, |x-1| assumes all real numbers.

Hence, the range of f is set of all non-negative real numbers.

Given:

To Find:

Range of any function is the set of values obtained after mapping is done in domain of the function

So every value of the co domain that are being mapped is Range of the function

For finding the range of any function

Let y = f(x)

and then find the values of y for which x exists.

Given: f(x) = x + 1, g(x) = 2x – 3.

To find: f + g, f - g,

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

(f + g)x = x + 1 + 2x - 3

⇒ (f + g)x = 3x - 2

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

(f - g)x = (x + 1) - (2x - 3)

⇒ (f - g)x = x +1 - 2x + 3

⇒ (f - g)x = 4 - x

Given: f = {(1, 1), (2, 3), (0, –1), (–1, –3)}

f(x) = ax + b

(1,1) Є f ⇒ for x = 1, f(x) = 1

⇒ 1 = a(1) + b

⇒ a+b = 1 …. (1)

Similarly, (0, -1) Є f ⇒ for x = 0, f(x) = -1

⇒ -1 = a(0) + b

⇒ b = -1

⇒ a-1 = 1 (from 1)

⇒ a = 2.

Hence a = 2 and b = -1.

Given: R = {(a,b): a, b Є N and a = b²}

(i) (a,a) Є R, for all a Є N

As we can see that 3 Є N but 3 ≠ 3² = 9.

Hence, the statement is not true.

(ii) (a,b) Є R, implies (b,a) Є R

As we can see that (4,2) Є N and 4 = 2² = 4 but 2 ≠ 4² = 16 hence, (2,4) does not belong to N.

Hence, the statement is not true.

(iii) (a,b) Є R, (b,c) R implies (a,c) Є R.

As we see, (9, 3) Є R, (16, 4) Є R because 3,4,9,16 Є N and 9 = 3² and 16 = 4².

Now, 9 ≠ 4² = 16; therefore, (9, 4) does not belong to N.

Hence, the statement is not true.

Given: A = {1,2,3,4}, B = {1,5,9,11,15,16} and f = {(1,5), (2,9), (3,1), (4,5), (2,11)}

Now, A × B = {(1,1), (1,5), (1,9), (1,11), (1,15), (1,16), (2,1), (2,5), (2,9), (2,11), (2,15), (2,16), (3,1), (3,5), (3,9), (3,11), (3,15), (3,16), (4,1), (4,5), (4,9), (4,11), (4,15), (4,16).

(i) f is a relation from A to B.

f = {(1,5), (2,9), (3,1), (4,5), (2,11)}.

A relation from a non empty set A to a non empty set B is a subset of the Cartesian product A × B.

And we can see f is a subset of A × B.

Hence f is a relation from A to B statement is true.

(ii) f is a function from A to B.

f = {(1,5), (2,9), (3,1), (4,5), (2,11)}.

As we observe that same first element i.e. 2 corresponds to two different images that is 9 and 11. Thus f is not a function from A to B.

Given: f = {(ab, a + b): a, b Є Z}

For value 2, 6, -2, -6 Є Z,

f = {(2 × 6, 2 + 6), (-2 × -6, -2 - 6), (2 × -6, 2 - 6), (-2 × 6, -2 + 6).

⇒ f = {(12, 8), (12, -8), (-12, -4), (-12, 4).

Here we observe that same first element i.e. 12 corresponds to two different images that is 8 and -8. Thus f is not a function.

Given: A = {9,10,11,12,13}

f : A→N be defined by f(n) = the highest prime factor of n.

Prime factor of 9 = 3

Prime factor of 10 = 2,5

Prime factor of 11 = 11

Prime factor of 12 = 2,3

Prime factor of 13 = 13

f(n) = the highest prime factor of n.

Hence,

f(9) = the highest prime factor of 9 = 3

f(10) = the highest prime factor of 10 = 5

f(11) = the highest prime factor of 11 = 11

f(12) = the highest prime factor of 12 = 3

f(13) = the highest prime factor of 13 = 13

As the range of f is the set of all f(n), where n Є A.

Thus, the range of f is: {3, 5, 11, 13}.