# NCERT Solutions for Class 11 Math Chapter 2 - Relations and Functions

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Maths Class 11 chapter “Relations and Functions” explains Cartesian product of sets, the definition of relation, function as a special kind of relation from one set to another, domain and range of functions; sum, difference, product, and quotients of functions, pictorial representation of a function, ordered pairs; domain, co-domain and range of a relation, number of elements in the Cartesian product of two finite sets, the real-valued function of the real variable, pictorial diagrams, Cartesian product of the reals with itself; domain, co-domain and range of a function; signum and greatest integer functions with their graphs; polynomial, constant, modulus, identity, rational, and much more.

##### Question 1:

If (x + 1, y – 2) = (3, 1), find the values of x and y.

Since the ordered pairs (x + 1, y – 2) and (3, 1)
are equal,
$\therefore$ x + 1 = 3 and y – 2 = 1
$⇒$ x = 2 and y = 3.
Hence, x = 2 and y = 3.

##### Question 2:

If P = {a, b, c} and Q = {r}, form the sets P × Q and Q × P. Are these products equal

We have : P = {a, b, c} and Q = {r}.
$\therefore$ P × Q = {(a, r), (b, r), (c, r)}
and Q × P = {(r, a), (r, b), (r, c)}.
Clearly P × Q$\ne$ Q × P. [$\because$(a, r)$\ne$(r,a);etc.]

##### Question 3:

Let A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}. Find :
A × (B $\cap$ C)

Here B $\cap$ C = {3, 4} $\cap$ {4, 5, 6} = {4}.
$\therefore$ A × (B $\cap$ C) = {1, 2, 3} × {4}
= {(1, 4), (2, 4), (3, 4)}.

##### Question 4:

Let A = {1, 2, 3}, B = {3, 4} and
C = {4, 5, 6}. Find :
(A × B) $\cap$ (A × C)

Here A × B = {(1, 3), (1, 4), (2, 3), (2, 4),
(3, 3), (3, 4) }
and A × C = {(1, 4), (1, 5), (1, 6), (2, 4),
(2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}.
$\therefore$ (A × B) $\cap$ (A × C) = {(1, 4), (2, 4) × (3, 4)}.

##### Question 5:

Let A = {1, 2, 3}, B = {3, 4} and
C = {4, 5, 6}. Find :
A × (B $\mathrm{\upsilon }$ C)

Here B $\mathrm{\upsilon }$ C = {3, 4}$\mathrm{\upsilon }${4, 5, 6} = {3, 4, 5, 6}.
$\therefore$ A × (B $\mathrm{\upsilon }$ 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)}.

##### Question 6:

Let A = {1, 2, 3}, B = {3, 4} and
C = {4, 5, 6}. Find :
(A × B) $\mathrm{\upsilon }$ (A × C)

(A × B) $\mathrm{\upsilon }$ (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)}.

##### Question 7:

If P = {1, 2}, form the set P × P × P.

We have : {1, 2}.
$\therefore$ P × P × P = {(1, 1, 1), (1, 1, 2), (1, 2, 1),
(2, 1, 1), (1, 2, 2), (2, 1, 2), (2, 2, 1), (2, 2, 2)}.

##### Question 8:

If R is the set of all real numbers, what do the Cartesian products R × R and R × R × R represent.

R × R = {(x, y) : x, y $\in$ R}, which represents the co-ordinates of all points in 2-dimensional space.
(ii) R × R × R = {(x, y, z) ; x, y, z $\in$ R}, which represents the co-ordinates of all points in 3-dimensional space.

##### Question 9:

If A × B = {(p, q), (p, r), (m, q), (m, r)}, find A and B.

We have A × B = {(p, q), (p, r),(m, q), (m, r)}.
$\therefore$ A = set of first elements = {p, m}
and B = set of second elements = {q, r}.

##### Question 10:

Let A = {1, 2, 3, 4, 5, 6}. Define a relation R from A to A by
R = {(x, y) : y = x + 1}.
Depict this relation by arrow diagram.

By the question, the relation,
R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}.
The arrow diagram is as shown in the figure. ##### Question 11:

Let A = {1, 2, 3, 4, 5, 6}. Define a relation R from A to A by
R = {(x, y) : y = x + 1}.
Write down the domain, co-domain and range of R.

Domain = {1, 2, 3, 4, 5}
Range = {2, 3, 4, 5, 6}
and Co-domain = {1, 2, 3, 4, 5, 6}.

##### Question 12:

The figure shows a relation between the sets P and Q. Write this relation (i) in set builder form (ii) in roster form What is its domain and range ?

Here the relation R is “x is sqaure of y” (i) In set-builder form :
R = {(x, y) : x is the square of y, x $\in$ P, y $\in$ Q}.
In roster form :
R = {(9, 3), (9, – 3), (4, 2), (4, –2), (25, 5), (25, – 5)}.
(ii) Domain of R = {4, 9, 25}.
(iii) Range of R = {– 2, 2, –3 , 3, – 5, 5}.

##### Question 13:

Set A = {1, 2} and B = {3, 4}. Find the number of relations from A to B.

We have : A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}.
Since n(A × B) = 4,
$\therefore$ the number of subsets of A × B = 24.
Hence, the number of relations from A to B = 24 = 16.

##### Question 14:

Let N be the set of natural numbers and the relation R be defined on N such that :
R = {(x, y) : y = 2x, x, y $\in$ N}.
What is the domain, co-domain and range of R ?
Is this relation a function ?

Domain of R = N, the set of natural numbers.
Co-domain of R = N, the set of natural numbers.
Range of R = set of even natural numbers.
Since each natural number n has one and only one image,
$\therefore$ the relation R is a function.

##### Question 15:

Examine each of the following relations given below and state in each case, giving reasons whether it is a function or not ?
R = {(2, 1), (3, 1), (4, 2)}

Since the elements 2, 3, 4 in the domain of R have unique images,
$\therefore$ R is a function.

##### Question 16:

Examine each of the following relations given below and state in each case, giving reasons whether it is a function or not ?
R = {(2, 2), (2, 4), (3, 3), (4, 4)}

Since the element 2 in the domain of R has two images 2 and 4,
$\therefore$ R is not a function.

##### Question 17:

Examine each of the following relations given below and state in each case, giving reasons whether it is a function or not ?
R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7)}

Since each element in the domain of R has a unique image,
$\therefore$ R is a function.

##### Question 18:

Let N be the set of natural numbers. Define a real valued function f : N $\to$ N by 2x + 1. Using this definition, complete the table given below : The completed table is given by : [$\because$ f(1) = 2 (1) + 1 = 3, f(2) = 2(2) + 1 = 5; etc.]

##### Question 19:

Define the function f : R$\to$R by f (x) = x2, x $\in$ R. Complete the table given below by using this definition. What is the domain and range of this function ?
Draw the graph of f. The complete table is as below : Domain of f = {x : x $\in$ R}.
Range of f = {x : x ≥ 0, x $\in$ R}.
The graph of f is as in the adjoining figure. ##### Question 20:

Draw the graph of the function :
f : R $\to$ R defined by f (x) = x3, x $\in$ R.

We have the following table : ##### Question 21:

Define the real valued function f : R – {0} $\to$ R defined by : Complete the table given below using this defnition.
What is the domain and range of this function ? The complete table is as below. Domain of f = R – {0}.
Range of f = R – {0}.
The graph of f is as shown in the following figure : ##### Question 22:

Let f(x) = x2 and g(x) = 2x + 1 be two real functions. Find :  ##### Question 23:  ##### Question 24:

Let R be the set of real numbers. Define a real function f : R $\to$ R by f (x) = x + 10. Sketch the graph of this function.

We have the following table : The graph of the function is as shown below : ##### Question 25:

Let R be a relation from Q to Q defined by :
R = {(a, b) : a, b $\in$ Q and a – b $\in$ Z}.
Show that : (a, a) $\in$ R for all Q $\in$ Q

Since a – a = 0 $\in$ Z, therefore, (a, a) $\in$ R.

##### Question 26:

Let R be a relation from Q to Q defined by :
R = {(a, b) : a, b $\in$ Q and a – b $\in$ Z}.
Show that : (a, b) $\in$ R implies that (b, a) $\in$ R

(a, b) $\in$ R $⇒$ a – b $\in$ Z $⇒$ b – a $\in$ Z $⇒$ (b, a) $\in$ R.

##### Question 27:

Let R be a relation from Q to Q defined by :
R = {(a, b) : a, b $\in$ Q and a – b $\in$ Z}.
Show that : (a, b) $\in$ R and (b, c) $\in$ R implies that (a, c) $\in$ R

(a, b) $\in$ R and (b, c) $\in$ R
$⇒$ a – b $\in$ Z and b – c $\in$ Z
$⇒$ (a – b) + (b – c) = a – c $\in$ Z.
Hence, (a, c) $\in$ R.

##### Question 28:

Let f = {(1, 1), (2, 3), (0, – 1), (– 1, – 3)} be a linear function from Z to Z. Find f(x).

Since ‘f’ is linear function,
$\therefore$ let f(x) = mx + c ...(1)
Now (1, 1) $\in$ R $⇒$ 1 = m + c $⇒$ m + c = 1 ...(2)
And (0, – 1) $\in$ R $⇒$ – 1 = 0 + c $⇒$ c = – 1.
Putting in (2), m – 1 $⇒$ m = 2.
Putting in (1), f(x) = 2x – 1.

##### Question 29:

Find the domain of the function :  Now x2 – 5x + 4 = (x – 4) (x – 1).
$\therefore$ x2 – 5x + 4 = 0 $⇒$ x = 4, 1.
Thus, that function ‘f’ is defined for all real value except
at x = 4, 1.
Hence, domain of f = R – {1, 4}.

##### Question 30:

The function ‘f’ is defined by : Draw the graph of f (x).

We have : f (x) = 1– x, x < 0.
$\therefore$ f (– 4) = 1 – (– 4) = 1 + 4 = 5
f (–3) = 1 – (–3) = 1 + 3 = 4
f (–2) = 1 – (–2) = 1 + 2 = 3
f (–1) = 1 – (–1) = 1 + 1 = 2.
Also, f (0) = 1.
And f (x) = x + 1, x > 0.
$\therefore$ f (1) = 1 + 1 = 2, f (2) = 2 + 1 = 3,
f (3) = 3 + 1 = 4, f (4) = 4 + 1 = 5 ; .....
Hence, the graph of f is as shown : ##### Question 31:  ##### Question 32:

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

A has 3 elements and B has also 3 elements.
$\therefore$ Number of elements in (A × B) = 3 × 3 = 9.

##### Question 33:

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

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

##### Question 34:

State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly.
If P = {m, n}
Q = {n, m},
then P × Q = {(m, n), (n, m)}.

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

##### Question 35:

State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly.
If A and B are non-empty sets, then A × B is nonempty set of ordered pairs (x, y) such that x $\in$ Β and y $\in$ A.

False.
A × B is a non-empty set of ordered pairs.
(x, y) such that x $\in$ A and y $\in$ B.

##### Question 36:

State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly.
If A = {1, 2}, B = {3, 4}, then :
A × (B $\cap$ $\mathrm{\varphi }$) = $\mathrm{\varphi }$.

True.
[$\because$ B $\cap$ $\mathrm{\varphi }$ = $\mathrm{\varphi }$, $\therefore$ A × (B $\cap$ $\mathrm{\varphi }$) = A × $\mathrm{\varphi }$ = $\mathrm{\varphi }$]

##### Question 37:

If A = {– 1, 1}, find A × A × 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), (1, 1, – 1), (1, 1, 1)}.

##### Question 38:

If A × B = [(a, x), (a, y), (b, x), (b, y)}, find A and B.

We have :
A = {(a, x), (a, y), (b, x), (b, y)}.
$\therefore$ A = {a, b} and B = {x, y}.

##### Question 39:

Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and
D = {5, 6, 7, 8].
Verify that :
A × (B $\cap$ C) = (A × B) $\cap$ (A × C)

B $\cap$ C = {1, 2, 3, 4} $\cap$ {5, 6} = $\mathrm{\varphi }$.
$\therefore$ LHS = A × (B $\cap$ C) = A × $\mathrm{\varphi }$
= {1, 2} × $\mathrm{\varphi }$ = $\mathrm{\varphi }$.
Now A × B = {1, 2} × {1, 2, 3, 4}
= {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}
and A × C = {1, 2} × {5, 6}
= {(1, 5), (1, 6), (2, 5), (2, 6)}.
$\therefore$ RHS = (A × B) $\cap$ (A × C) = $\mathrm{\varphi }$.
Hence, A × (B $\cap$ C) = (A × B) $\cap$ (A × C).

##### Question 40:

Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and
D = {5, 6, 7, 8].
Verify that :
A × C is a subset of B × D.

A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}
[As in part (i)]
B × D = {1, 2, 3, 4} × {5, 6, 7, 8}
= {(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)}.
Clearly each element of A × C is in B × D.
Hence, A × C is a subset of B × D.

##### Question 41:

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

A × B = {1, 2} × {3, 4}
= {(1, 3), (1, 4), (2, 3), (2, 4)}.
No. of subsets of A × B = 24 = 16.
These subsets are :
$\mathrm{\varphi }$, {(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, 4), (2, 3), (2, 4)},
{((2, 3), (2, 4), (1, 3))}, {(1, 3), (1, 4), (2, 4)},
{(1, 3), (1, 4), (2, 3), (2, 4)}.

##### Question 42:

Let A and B be two sets such that n (A) = 3, 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.

Since x, y, z $\in$ A,
$\therefore$ A = {x, y, z}.
Since 1, 2 $\in$ B,
$\therefore$ B = {1, 2}.

##### Question 43:

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.

Here (– 1, 0) $\in$ A × A
and (0, 1) $\in$ A × A.
$\therefore$ – 1, 0 $\in$ A
and 0, 1 $\in$ A.
Thus A = {– 1, 0, 1}.
$\therefore$ A × A = {– 1, 0, 1} × {– 1, 0, 1}
= {(– 1, – 1), (– 1, 0), (– 1, 1), (0, – 1),
(0, 0), (0, 1), (1, – 1), (1, 0), (1, 1)}.
$\therefore$ Remaining elements of A × A are :
(– 1, – 1), (– 1, 1), (0, – 1), (0, 0), (1, – 1), (1, 0), (1, 1).

##### Question 44:

Let A = {1, 2, 3, ........., 14}. Define a relation R from A to A by R = {(x, y) ; 3x – y = 0, where x, y $\in$ A}. Write down its domain, co-domain and range.

Here R = {(x, y) : 3x – y = 0, where x, y $\in$ A}
= {(1, 3), (2, 6), (3, 9), (4, 12)}.
(i) Domain = {1, 2, 3, 4}
(ii) Co-domain = {1, 2, 3, ......., 14}
(iii) Range = {3, 6, 9, 12}. ##### Question 45:

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 $\in$ N}.
Depict this relationship using (i) roster form (ii) an arrow diagram. Write down the domain and range. ##### Question 46:

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 $\in$ A, y $\in$ B}. Write R in roster form.

Here R = {(1, 4), (1, 6), (2, 9), (3, 4),
(3, 6), (5, 4), (5, 6)}
[$\because$ 4 – 1 = 3; odd etc.]

##### Question 47:

Figure given below shows a relationship between the sets P and Q. Write the relation (i) in set builder form (ii) roster form. What is its domain and range ? (i) R = {(x, y) : x – y = 2, 4 < x < 8, x, y $\in$ N}
(ii) R = {(5, 3), (6, 4), (7, 5)}.
Domain = {5, 6, 7}. Range = { 3, 4, 5}.

##### Question 48:

Write the relation ?
R = {(x, x3) : x is a prime number less than 10} in roster form.

Prime numbers less than 10 are 2, 3, 5, 7.
$\therefore$ R = {(2, 8), (3, 27), (5, 123), (7, 343)}.

##### Question 49:

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

We have : A = {x, y, z}
and B = {1, 2}.
$\therefore$ A × B = {(x, 1), (x, 2), (y, 1), (y, 2), (z, 1), (z, 2)}.
$\therefore$ n (A × B) = 6
$⇒$ the number of subsets of n (A × B) = 26.
Hence, the number of relation from A into B = 26.

##### Question 50:

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

We have : R = {(a, b) : a, b $\in$ Z,
a – b is an integer}
$\therefore$ Domain of R = Z.
Range of R = Z.

##### Question 51:

Which of the following relations are functions ? Give reasons. If it is a function, determine its domain and range :
{(2, 1), (5, 1), (8, 1) (11,1), (14, 1), (17, 1)}

We have :
f = {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}.
Here no two ordered pairs have the same first component.
$\therefore$ this relation is a function.
Domain of f = {2, 5, 8, 11, 14, 17}.
Range of f = {1}.

##### Question 52:

Which of the following relations are functions ? Give reasons. If it is a function, determine its domain and range :
{(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}

We have : f = {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}.
Here no two ordered pairs have the same first component. $\therefore$ this relation is a function.
Domain of f = {2, 4, 6, 8, 10, 12, 14}.
Range of f = {1, 2, 3, 4, 5, 6, 7}.

##### Question 53:

Which of the following relations are functions ? Give reasons. If it is a function, determine its domain and range :
{(1, 3), (1, 5), (2, 5)}

We have : f = {(1, 3), (1, 5), (2,5)}. Here 1 appears more than once as a first component in the ordered pairs of f.
Hence, this relation is not a function.

##### Question 54:

Find the domain and range of the following real functions :
f (x) = – | x |

We have : f (x) = – | x |.
Clearly f (x) $\le$ 0 ∀ x $\in$ R.
Domain of f = R.
Range of f = (– $\infty$, 0].

##### Question 55:

Find the domain and range of the following real functions :  ##### Question 56:

A function f is defined by f (x) = 2x – 5. Write down the values of :
f (0)

We have : f (x) = 2x – 5.
f (0) = 2 (0) – 5 = 0 – 5 = – 5.

##### Question 57:

A function f is defined by f (x) = 2x – 5. Write down the values of :
f (7)

We have : f (x) = 2x – 5.
f (7) = 2 (7) – 5 = 14 – 5 = 9.

##### Question 58:

A function f is defined by f (x) = 2x – 5. Write down the values of :
f (– 3)

We have : f (x) = 2x – 5.
f (– 3) = 2 (– 3) – 5 = – 6 – 5 = – 11.

##### Question 59:

The function ‘f’, which maps temperature in degree celsius into temperature degree Fahrenheit is defined by Find :
t (0) Putting C = 0, t (0) = 0 + 32 = 32.

##### Question 60:

The function ‘f’, which maps temperature in degree celsius into temperature degree Fahrenheit is defined by Find :
t (28)  ##### Question 61:

The function ‘f’, which maps temperature in degree celsius into temperature degree Fahrenheit is defined by Find :
t (– 10)  ##### Question 62:

The function ‘f’, which maps temperature in degree celsius into temperature degree Fahrenheit is defined by Find :
The value of C, when t (C) = 212  ##### Question 63:

Find the range of each of the following function :
f (x) = 2 – 3x, x $\in$ R, x > 0 ##### Question 64:

Find the range of each of the following function :
f (x) = x2 + 2, x is a real number ##### Question 65:

Find the range of each of the following function :
f (x) = x, x is a real number

We have f (x) = x $⇒$ y = x.
$⇒$y is a real number.
Hence, Range of f = {y : y $\in$ R}