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

Answer:

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

Answer:

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.]

Answer:

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)}.

Answer:

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)}.

Answer:

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)}.

Answer:

(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)}.

Answer:

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)}.

Answer:

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.

Answer:

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}.

Answer:

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.

Answer:

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

Answer:

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}.

Answer:

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.

Answer:

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.

Answer:

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

Answer:

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

Answer:

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

Answer:

The completed table is given by :

[$\because$ f(1) = 2 (1) + 1 = 3, f(2) = 2(2) + 1 = 5; etc.]

Answer:

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.

Answer:

We have the following table :

Answer:

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 :

Answer:

Answer:

Answer:

We have the following table :

The graph of the function is as shown below :

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

Now x² – 5x + 4 = (x – 4) (x – 1).
$\therefore$ x² – 5x + 4 = 0 $\Rightarrow$ 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}.

Answer:

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 :

Answer:

Answer:

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

Answer:

(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)}.

Answer:

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

Answer:

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

Answer:

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

Answer:

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)}.

Answer:

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

Answer:

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).

Answer:

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.

Answer:

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)}.

Answer:

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

Answer:

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).

Answer:

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}.

Answer:

Answer:

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

Answer:

(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}.

Answer:

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

Answer:

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
$\Rightarrow$ the number of subsets of n (A × B) = 2⁶.
Hence, the number of relation from A into B = 2⁶.

Answer:

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

Answer:

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}.

Answer:

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}.

Answer:

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.

Answer:

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

Answer:

Answer:

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

Answer:

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

Answer:

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

Answer:

Putting C = 0, t (0) = 0 + 32 = 32.

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Answer:

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