#### Page No 33:

#### Question 1:

If,
find the values of *x* and *y*.

#### Answer:

It is given that.

Since the ordered pairs are equal, the corresponding elements will also be equal.

Therefore,

and.

∴ *x* = 2 and *y* = 1

#### Question 2:

If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in (A × B)?

#### Answer:

It is given that set A has 3 elements and the elements of set B are 3, 4, and 5.

⇒ Number of elements in set B = 3

Number of elements in (A × B)

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

= 3 × 3 = 9

Thus, the number of elements in (A × B) is 9.

#### Question 3:

If G = {7, 8} and H = {5, 4, 2}, find G × H and H × G.

#### Answer:

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

We know that the Cartesian product P × Q of two non-empty sets P and Q is defined as

P
× Q = {(*p*, *q*): *p*∈
P, *q* ∈
Q}

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

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

#### Question 4:

State whether each of the following statement are true or false. If the statement is false, rewrite the given statement correctly.

(i) If P = {*m*, *n*} and Q = {*n*, *m*}, then P × Q = {(*m*, *n*), (*n*, *m*)}.

(ii) If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (*x*, *y*) such that *x* ∈ A and *y* ∈ B.

(iii) If A = {1, 2}, B = {3, 4}, then A × (B ∩ Φ) = Φ.

#### Answer:

(i) False

If P = {*m*, *n*} and Q = {*n*, *m*}, then

P × Q = {(*m*, *m*), (*m*, *n*), (*n**,* *m*), (*n*, *n*)}

(ii) True

(iii) True

#### Question 5:

If A = {–1, 1}, find A × A × A.

#### Answer:

It is known that for any non-empty set A, A × A × A is defined as

A
× A × A = {(*a*, *b*, *c*): *a*, *b*, *c *∈
A}

It is given that A = {–1, 1}

∴ 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)}

#### Question 6:

If
A × B = {(*a*, *x*),
(*a*, *y*),
(*b*, *x*),
(*b*, *y*)}.
Find A and B.

#### Answer:

It
is given that A × B = {(*a*, *x*),
(*a,* *y*),
(*b*, *x*),
(*b*, *y*)}

We
know that the Cartesian product of two non-empty sets P and Q is
defined as P × Q = {(*p*, *q*): *p* ∈
P, *q* ∈
Q}

∴ A is the set of all first elements and B is the set of all second elements.

Thus,
A = {*a*, *b*}
and B = {*x*, *y*}

#### Question 7:

Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that

(i) A × (B ∩ C) = (A × B) ∩ (A × C)

(ii) A × C is a subset of B × D

#### Answer:

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

We have B ∩ C = {1, 2, 3, 4} ∩ {5, 6} = Φ

∴L.H.S. = A × (B ∩ C) = A × Φ = Φ

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

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

∴ R.H.S. = (A × B) ∩ (A × C) = Φ

∴L.H.S. = R.H.S

Hence, A × (B ∩ C) = (A × B) ∩ (A × C)

(ii) 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)}

We can observe that all the elements of set A × C are the elements of set B × D.

Therefore, A × C is a subset of B × D.

#### Question 8:

Let A = {1, 2} and B = {3, 4}. Write A × B. How many subsets will A × B have? List them.

#### Answer:

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

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

⇒ *n*(A
× B) = 4

We
know that if C is a set with *n*(C)
= *m*,
then *n*[P(C)]
= 2^{m}.

Therefore,
the set A × B has 2^{4} = 16 subsets. 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, 3), (2, 3), (2, 4)},

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

#### 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* and *z* are distinct elements.

#### Answer:

It
is given that *n*(A)
= 3 and *n*(B)
= 2; and (*x*,
1), (*y*,
2), (*z*,
1) are in A × B.

We know that A = Set of first elements of the ordered pair elements of A × B

B = Set of second elements of the ordered pair elements of A × B.

∴ *x*, *y*,
and *z* are the elements of A; and 1 and 2 are the elements of B.

Since *n*(A)
= 3 and *n*(B)
= 2, it is clear that A = {*x*, *y*, *z*}
and B = {1, 2}.

#### Page No 34:

#### Question 10:

The Cartesian product A × A has 9 elements among which are found (–1, 0) and (0, 1). Find the set A and the remaining elements of A × A.

#### Answer:

We
know that if *n*(A)
= *p *and *n*(B)
= *q, *then *n*(A
× B) = *pq*.

∴ *n*(A
× A)
= *n*(A)
× *n*(A)

It
is given that *n*(A
× A)
= 9

∴ *n*(A)
× *n*(A)
= 9

⇒ *n*(A)
= 3

The ordered pairs (–1, 0) and (0, 1) are two of the nine elements of A × A.

We
know that A × A = {(*a,
a*): *a* ∈
A}. Therefore, –1, 0, and 1 are elements of A.

Since *n*(A)
= 3, it is clear that A
= {–1, 0, 1}.

The remaining elements of set A × A are (–1, –1), (–1, 1), (0, –1), (0, 0),

(1, –1), (1, 0), and (1, 1)

#### Page No 35:

#### Question 1:

Let A = {1, 2, 3, … , 14}. Define a relation R from A to A by R = {(*x*, *y*): 3*x* – *y* = 0, where *x*, *y* ∈ A}. Write down its domain, codomain and range.

#### Answer:

The relation R from A to A is given as

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

i.e., R = {(*x*, *y*): 3*x* = *y*, where *x*, *y* ∈ A}

∴R = {(1, 3), (2, 6), (3, 9), (4, 12)}

The domain of R is the set of all first elements of the ordered pairs in the relation.

∴Domain of R = {1, 2, 3, 4}

The whole set A is the codomainof the relation R.

∴Codomain of R = A = {1, 2, 3, …, 14}

The range of R is the set of all second elements of the ordered pairs in the relation.

∴Range of R = {3, 6, 9, 12}

#### Page No 36:

#### Question 2:

Define
a relation R on the set **N** of natural numbers by R = {(*x*, *y*): *y* = *x* + 5, *x* is a natural number less than 4; *x*, *y* ∈ **N**}.
Depict this relationship using roster form. Write down the domain and
the range.

#### Answer:

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

The natural numbers less than 4 are 1, 2, and 3.

∴R = {(1, 6), (2, 7), (3, 8)}

The domain of R is the set of all first elements of the ordered pairs in the relation.

∴ Domain of R = {1, 2, 3}

The range of R is the set of all second elements of the ordered pairs in the relation.

∴ Range of R = {6, 7, 8}

#### Question 3:

A
= {1, 2, 3, 5} and B = {4, 6, 9}. Define a relation R from A to B by
R = {(*x*, *y*):
the difference between *x* and *y* is odd; *x* ∈
A, *y *∈
B}. Write R in roster form.

#### Answer:

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}

∴R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}

#### Question 4:

The given figure shows a relationship between the sets P and Q. write this relation

(i) in set-builder form (ii) in roster form.

What is its domain and range?

#### Answer:

According to the given figure, P = {5, 6, 7}, Q = {3, 4, 5}

(i) R
= {(*x,
y*): *y
= x* – 2; *x* ∈
P} or R = {(*x,
y*): *y
= x* – 2 for *x* = 5, 6, 7}

(ii) R = {(5, 3), (6, 4), (7, 5)}

Domain of R = {5, 6, 7}

Range of R = {3, 4, 5}

#### Question 5:

Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by

{(*a*, *b*): *a*, *b* ∈
A, *b* is exactly divisible by *a*}.

(i) Write R in roster form

(ii) Find the domain of R

(iii) Find the range of R.

#### Answer:

A
= {1, 2, 3, 4, 6}, R = {(*a*, *b*): *a*, *b* ∈
A, *b* is exactly divisible by *a*}

(i) 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)}

(ii) Domain of R = {1, 2, 3, 4, 6}

(iii) Range of R = {1, 2, 3, 4, 6}

#### Question 6:

Determine
the domain and range of the relation R defined by R = {(*x*, *x* + 5): *x* ∈
{0, 1, 2, 3, 4, 5}}.

#### Answer:

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

∴ R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}

∴Domain of R = {0, 1, 2, 3, 4, 5}

Range of R = {5, 6, 7, 8, 9, 10}

#### Question 7:

Write
the relation R = {(*x*, *x*^{3}): *x *is
a prime number less than 10} in roster form.

#### Answer:

R
= {(*x*, *x*^{3}): *x *is
a prime number less than 10}

The prime numbers less than 10 are 2, 3, 5, and 7.

∴R = {(2, 8), (3, 27), (5, 125), (7, 343)}

#### Question 8:

Let
A = {*x*, *y*,
z} and B = {1, 2}. Find the number of relations from A to B.

#### Answer:

It
is given that A = {*x*, *y*,
z} and B = {1, 2}.

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

Since *n*(A
× B) = 6, the number of subsets of A × B is 2^{6}.

Therefore,
the number of relations from A to B is 2^{6}.

#### Question 9:

Let
R be the relation on **Z** defined by R = {(*a*, *b*): *a*, *b* ∈ **Z**, *a *– *b* is an integer}. Find the domain and range of R.

#### Answer:

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

It is known that the difference between any two integers is always an integer.

∴Domain
of R = **Z**

Range
of R = **Z**

#### Page No 44:

#### Question 1:

Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range.

(i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}

(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}

(iii) {(1, 3), (1, 5), (2, 5)}

#### Answer:

(i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}

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

Here, domain = {2, 5, 8, 11, 14, 17} and range = {1}

(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}

Since 2, 4, 6, 8, 10, 12, and 14 are the elements of the domain of the given relation having their unique images, this relation is a function.

Here, domain = {2, 4, 6, 8, 10, 12, 14} and range = {1, 2, 3, 4, 5, 6, 7}

(iii) {(1, 3), (1, 5), (2, 5)}

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

#### Question 2:

Find the domain and range of the following real function:

(i) *f*(*x*)
= –|*x*| (ii)

#### Answer:

(i) *f*(*x*)
= –|*x*|, *x* ∈
R

We
know that |*x*|
=

Since *f*(*x*)
is defined for *x* ∈ **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.

∴The
range of *f* is (–,
0].

(ii)

Sinceis
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 {*x* : –3
≤ *x* ≤
3} or [–3, 3].

For
any value of *x* such that –3 ≤ *x* ≤ 3, the value of *f*(*x*)
will lie between 0 and 3.

∴The
range of *f*(*x*)
is {*x*: 0 ≤ *x* ≤ 3}
or [0, 3].

#### Question 3:

A
function *f* is defined by *f*(*x*)
= 2*x* – 5. Write down the values of

(i) *f*(0), (ii) *f*(7), (iii) *f*(–3)

#### Answer:

The
given function is *f*(*x*)
= 2*x* – 5.

Therefore,

(i) *f*(0)
= 2 × 0 – 5 = 0 – 5 = –5

(ii) *f*(7)
= 2 × 7 – 5 = 14 – 5 = 9

(iii) *f*(–3)
= 2 × (–3) – 5 = – 6 – 5 = –11

#### Question 4:

The function ‘*t*’
which maps temperature in degree Celsius into temperature in degree
Fahrenheit is defined by.

Find (i) *t* (0) (ii) *t* (28) (iii) *t* (–10) (iv) The value of C, when *t*(C)
= 212

#### Answer:

The given function is.

Therefore,

(i)

(ii)

(iii)

(iv) It
is given that *t*(C)
= 212

Thus,
the value of *t*,
when *t*(C)
= 212, is 100.

#### Question 5:

Find the range of each of the following functions.

(i) *f*(*x*)
= 2 – 3*x*, *x* ∈ **R**, *x* > 0.

(ii) *f*(*x*)
= *x*^{2} + 2, *x*, is
a real number.

(iii) *f*(*x*)
= *x*, *x* is a real number

#### Answer:

(i) *f*(*x*)
= 2 – 3*x*, *x* ∈ **R**, *x* > 0

The
values of *f*(*x*)
for various values of real numbers *x* > 0 can be written in the tabular form as

| 0.01 | 0.1 | 0.9 | 1 | 2 | 2.5 | 4 | 5 | … |

| 1.97 | 1.7 | –0.7 | –1 | –4 | –5.5 | –10 | –13 | … |

Thus,
it can be clearly observed that the range of *f* is the set of all real numbers less than 2.

i.e.,
range of *f* = (–,
2)

**Alter:**

Let *x* > 0

⇒
3*x* > 0

⇒
2 –3*x* < 2

⇒ *f*(*x*)
< 2

∴Range
of *f* =
(–,
2)

(ii) *f*(*x*)
= *x*^{2} + 2, *x*, is
a real number

The
values of *f*(*x*)
for various values of real numbers *x* can be written in the tabular form as

| 0 | ±0.3 | ±0.8 | ±1 | ±2 | ±3 | … | |

| 2 | 2.09 | 2.64 | 3 | 6 | 11 | ….. |

Thus,
it can be clearly observed that the range of *f* is the set of all real numbers greater than 2.

i.e.,
range of *f* = [2,)

**Alter:**

Let *x* be any
real number.

Accordingly,

*x*^{2} ≥
0

⇒ *x*^{2} + 2 ≥
0 + 2

⇒ *x*^{2} + 2 ≥
2

⇒ *f*(*x*)
≥
2

∴ Range
of *f* =
[2,)

(iii) *f*(*x*)
= *x, x* is
a real number

It
is clear that the range of *f* is the set of all real numbers.

∴ Range
of *f* = **R**

#### Page No 46:

#### Question 1:

The relation *f* is
defined by

The
relation* g* is defined by

Show
that *f* is
a function and* g *is
not a function.

#### Answer:

The
relation *f* is defined as

It is observed that for

0 ≤ *x* < 3, *f*(*x*)
= *x*^{2}

3 < *x* ≤
10, *f*(*x*)
= 3*x*

Also,
at *x* = 3, *f*(*x*)
= 3^{2} =
9 or *f*(*x*)
= 3 × 3 = 9

i.e.,
at *x* = 3, *f*(*x*)
= 9

Therefore,
for 0 ≤ *x* ≤
10, the images of *f*(*x*)
are unique.

Thus, the given relation is a function.

The
relation* g* is defined as

It
can be observed that for *x* = 2, *g*(*x*)
= 2^{2} =
4 and *g*(*x*)
= 3 × 2 = 6

Hence,
element 2 of the domain of the relation *g* corresponds to two different images i.e., 4 and 6. Hence, this
relation is not a function.

#### Question 2:

If *f*(*x*)
= *x*^{2},
find.

#### Answer:

#### Question 3:

Find the domain of the function

#### Answer:

The given function is.

It
can be seen that function *f* is defined for all real numbers except at *x* = 6 and *x* = 2.

Hence,
the domain of *f* is **R** –
{2, 6}.

#### Question 4:

Find
the domain and the range of the real function *f* defined by.

#### Answer:

The given real function is.

It
can be seen that is defined for (*x* – 1) ≥
0.

i.e.,

is
defined for *x* ≥
1.

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

As *x* ≥
1 ⇒
(*x* –
1) ≥
0 ⇒

Therefore,
the range of *f* is the set of all real numbers greater than or equal to 0 i.e., the
range of *f* = [0,).

#### Question 5:

Find
the domain and the range of the real function *f* defined by *f* (*x*)
= |*x* – 1|.

#### Answer:

The
given real function is* f* (*x*)
= |*x* – 1|.

It
is clear that |*x* – 1| is defined for all real numbers.

∴Domain
of *f* = **R**

Also,
for *x* ∈ **R**,
|*x* – 1| assumes all real numbers.

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

#### Question 6:

Letbe
a function from **R** into **R**.
Determine the range of *f*.

#### Answer:

The
range of *f* is the set of all second elements. It can be observed that all these
elements are greater than or equal to 0 but less than 1.

[Denominator is greater numerator]

Thus,
range of *f* = [0, 1)

#### Question 7:

Let *f*, *g*: **R** → **R** be
defined, respectively by *f*(*x*)
= *x *+ 1, *g*(*x*)
= 2*x* –
3. Find *f* + *g*, *f* – *g* and.

#### Answer:

*f*, *g*: **R** → **R **is
defined as *f*(*x*)
= *x *+ 1, *g*(*x*)
= 2*x* –
3

(*f* + *g*) (*x*)
= *f*(*x*)
+ *g*(*x*)
= (*x* + 1)
+ (2*x* –
3) = 3*x* –
2

∴(*f
+ g*) (*x*)
= 3*x* –
2

(*f
– g*) (*x*)
= *f*(*x*)
– *g*(*x*)
= (*x* + 1)
– (2*x* – 3) = *x* + 1 – 2*x* + 3 = – *x* + 4

∴ (*f
– g*) (*x*)
= –*x* + 4

#### Question 8:

Let *f *=
{(1, 1), (2, 3), (0, –1), (–1, –3)} be a function
from **Z** to **Z** defined by *f*(*x*)
= *ax* + *b*,
for some integers *a*, *b*.
Determine *a*, *b*.

#### Answer:

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

*f*(*x*)
= *ax* + *b*

(1,
1) ∈ *f*

⇒ *f*(1)
= 1

⇒ *a* × 1 + *b* = 1

⇒ *a* + *b* = 1

(0,
–1) ∈ *f*

⇒ *f*(0)
= –1

⇒ *a* × 0 + *b* = –1

⇒ *b* = –1

On
substituting *b* = –1 in *a* + *b* = 1, we obtain *a* + (–1) = 1 ⇒ *a* = 1 + 1 = 2.

Thus,
the respective values of *a* and *b* are 2 and –1.

#### Question 9:

Let R be a relation from **N** to **N** defined by R = {(*a*, *b*): *a*, *b* ∈ **N** and *a* = *b*^{2}}. Are the following true?

(i) (*a*, *a*) ∈ R, for all* a *∈ **N**

(ii) (*a*, *b*) ∈ R, implies (*b*, *a*) ∈ R

(iii) (*a*, *b*) ∈ R, (*b*, *c*) ∈ R implies (*a*, *c*) ∈ R.

Justify your answer in each case.

#### Answer:

R = {(*a*, *b*): *a*, *b* ∈ **N** and *a* = *b*^{2}}

(i) It can be seen that 2 ∈** N**;however, 2 ≠ 2^{2} = 4.

Therefore, the statement “(*a*, *a*) ∈ R, for all* a *∈ **N**” is not true.

(ii) It can be seen that (9, 3) ∈** N **because 9, 3 ∈** N **and 9 = 3^{2}.

Now, 3 ≠ 9^{2} = 81; therefore, (3, 9) ∉** N**

Therefore, the statement “(*a*, *b*) ∈ R, implies (*b*, *a*) ∈ R” is not true.

(iii) It can be seen that (16, 4) ∈ R, (4, 2) ∈ R because 16, 4, 2 ∈** N **and 16 = 4^{2} and 4 = 2^{2}.

Now, 16 ≠ 2^{2} = 4; therefore, (16, 2) ∉** N**

Therefore, the statement “(*a*, *b*) ∈ R, (*b*, *c*) ∈ R implies (*a*, *c*) ∈ R” is not true.

#### Question 10:

Let
A = {1, 2, 3, 4}, B = {1, 5, 9, 11, 15, 16} and *f *=
{(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}. Are the following true?

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

Justify your answer in each case.

#### Answer:

A = {1, 2, 3, 4} and B = {1, 5, 9, 11, 15, 16}

∴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)}

It
is given that *f *=
{(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}

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

It
is observed that *f* is a subset of A × B.

Thus, *f* is a relation from A to B.

(ii) Since
the same first element i.e., 2 corresponds to two different images
i.e., 9 and 11, relation *f *is
not a function.

#### Page No 47:

#### Question 11:

Let *f* be the subset of **Z** × **Z** defined by *f *=
{(*ab*, *a* + *b*): *a*, *b* ∈ **Z**}.
Is *f* a function from **Z** to **Z**:
justify your answer.

#### Answer:

The
relation *f* is defined as *f *=
{(*ab*, *a* + *b*): *a*, *b* ∈ **Z**}

We
know that a relation *f* from a set A to a set B is said to be a function if every element of
set A has unique images in set B.

Since
2, 6, –2, –6 ∈ **Z**,
(2 × 6, 2 + 6), (–2 × –6, –2 + (–6))
∈ *f*

i.e.,
(12, 8), (12, –8) ∈ *f*

It
can be seen that the same first element i.e., 12 corresponds to two
different images i.e., 8 and –8. Thus, relation *f* is not a function.

#### Question 12:

Let
A = {9, 10, 11, 12, 13} and let *f*:
A → **N** be defined by *f*(*n*)
= the highest prime factor of *n*.
Find the range of *f*.

#### Answer:

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

*f*:
A → **N** is defined as

*f*(*n*)
= The highest prime factor of *n*

Prime factor of 9 = 3

Prime factors of 10 = 2, 5

Prime factor of 11 = 11

Prime factors of 12 = 2, 3

Prime factor of 13 = 13

∴*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

The
range of *f* is the set of all *f*(*n*),
where *n* ∈
A.

∴Range
of *f* = {3, 5, 11, 13}

**NCERT Solutions for Class 11 Chemistry Chapters**

- Chapter 1 – Sets
- Chapter 2 – Relations and Functions
- Chapter 3 – Trigonometric Functions
- Chapter 4 – Principle of Mathematical Induction
- Chapter 5 – Complex Numbers and Quadratic Equations
- Chapter 6 – Linear Inequalities
- Chapter 7 – Permutations and Combinations
- Chapter 8 – Binomial Theorem
- Chapter 9 – Sequences and Series
- Chapter 10 – Straight Lines
- Chapter 11 – Conic Sections
- Chapter 12 – Introduction to Three Dimensional Geometry
- Chapter 13 – Limits and Derivatives
- Chapter 14 – Mathematical Reasoning
- Chapter 15 – Statistics
- Chapter 16 – Probability

**NCERT Solutions for Class 11:**