Class 11 Maths Ncert Solutions Chapter 2 Ex 2.2 Relations & Functions PDF

# Class 11 Maths Ncert Solutions Ex 2.2

## Class 11 Maths Ncert Solutions Chapter 2 Ex 2.2

Q.1: Let X = {1, 2, 3, 4, . . . . . 14}. Define a relation Z from X to X by Z= {(a, b): 3a – b = 0, where a, b $\in$ X}. Find its co – domain, domain and range.

Sol:

The relation ‘Z’ from ‘X to X’ is:

Z = {(a, b): 3a – b = 0, where a, b $\in$ X}

Z = {(a, b): 3a = b, where a, b $\in$ X}

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

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

Domain of Z = {1, 2, 3, 4}

The set X is the co – domain of the relation Z.

Therefore, co – domain of Z = X = {1, 2, 3, 4, . . . . . . 14}

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

Therefore, Range of Z = {3, 6, 9, 12}

Q.2: Define a relation Z on the set N of natural no. by Z = {(a, b): b = a + 5, a is a natural no less than 4; a, b $\in$ N}. Give this relationship in the roaster form. Find the domain and the range.

Sol:

Z = {(a, b): b = a + 5, a is a natural number less than 4; a, b $\in$ N}.

Natural numbers less than 4 are 1, 2 and 3.

Z = {(1, 6), (2, 7), (3, 8)}

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

Domain of Z = {1, 2, 3}

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

Therefore, Range of Z = {6, 7, 8}

Q.3: M = {1, 2, 3, 5} and N = {4, 6, 9}. Define a relation Z from M to N by Z = {(a, b): the difference between a and b is odd; a $\in$ M, b $\in$ N}. Find Z in roster form.

Sol:

M = {1, 2, 3, 5}

N = {4, 6, 9}

Z = {(a, b): the difference between a and b is odd; a $\in$ M, b $\in$ N}

Therefore, Z = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}

Q.4: The figure given below shows a relationship between the sets A and B. Find the following relation:

(i) In set-builder form

(ii) In roster form.

What is its range and domain?

Sol:

According to the information given in the figure:

A = {5, 6, 7}

B = {3, 4, 5}

(i).   Z = {(a, b): b = a – 2; a $\in$ A}

(or),  Z = {(a, b): b = a – 2 for a = 5, 6, 7}

(ii).   Z = {(5, 3), (6, 4), (7, 5)}

Domain of Z = {5, 6, 7}

Range of Z = {3, 4, 5}

Q.5: Let X = {1, 2, 3, 4, 6}. Let Z be the relation on X defined by {(p, q): p, q $\in$ X, q is divisible by p}.

(i) Write Z in the roster form

(ii) Find domain of Z

(iii) Find range of Z

Sol:

X = {1, 2, 3, 4, 6}

Z = {(p, q): p, q $\in$ X, q is divisible by p}

(i) Z = {(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 Z = {1, 2, 3, 4, 6}

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

Q.6: Find the range and domain of the relation Z defined by Z = {(a, a + 5): a $\in$ {0, 1, 2, 3, 4, 5}}.

Sol:

Z = {(a, a + 5): a $\in$ {0, 1, 2, 3, 4, 5}}

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

Domain = {0, 1, 2, 3, 4, 5}

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

Q.7: Find the relation Z = {(a, a3$a^{3}$): a is a prime number less than 10} in the roster form.

Sol:

Z = {(a, a3$a^{3}$): a is a prime no. less than 10}

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

Therefore, Z = {(2, 8), (3, 27), (5, 125), (7, 343)}

Q.8: Let X = {a, b, c} and Y = {11, 12}. Find the no. of relations from X to Y.

Sol:

It is given that X = {a, b, c} and Y = {11, 12}.

X × Y = {(a, 11), (a, 12), (b, 11), (b, 12), (c, 11), (c, 12)}

As n(X × Y) = 6, the no of subsets of X × Y = 26$2^{6}$.

Therefore, the number of relations from X to Y is 26$2^{6}$.

Q.9: Let Z be the relation on P defined by Z = {(x, y): x, y $\in$ P, x – y is an integer}. Find the range and domain of Z.

Sol:

Z = {(x, y): x, y $\in$ P, x – y is an integer}

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

Domain of Z = P

Range of Z = P