In NCERT Solutions for Class 11 maths chapter 2 relations and functions, we learn about ordered pair, a cartesian product of sets, relations, representation of a relation, function as a special kind of relation, function as a correspondence, equal functions, real functions, domain of real functions, some standard real functions and their graphs, operations on real functions.

Section Name | Topic Name |

2 | Relations and Functions |

2.1 | Introduction |

2.2 | Cartesian Product of Sets |

2.3 | Relations |

2.4 | Functions |

2.5 | Summary |

**Contents**show

### Class 11 Maths NCERT Solutions Chapter 2

**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

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

f(x) | 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

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

f(x) | 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.

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}

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