﻿ Class 11 NCERT Math Solution
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TOPICS
Unit-2(Examples)

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

Solution :-
```   In the Ordered Pairs, the corresponding values are equal.
So, x + 1 = 3
x = 3 - 1
x = 2 and
y - 2 = 1
y = 1 + 2
y = 3
```

Example-2 :-  If P = {a, b, c} and Q = {r}, form the sets P × Q and Q × P. Are these two products equal?

Solution :-
```   Given that P = {a, b, c} and Q = {r}.
P x Q = {(a, r), (b, r), (c, r)}
Q x P = {(r, a), (r, b), (r, c)}
By equality rule, (a, r) is not equal to (r, a).
So, P x Q ≠ Q x P.
```

Example-3 :-  Let A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}. Find
(i) A × (B ∩ C)
(ii) (A × B) ∩ (A × C)
(iii) A × (B ∪ C)
(iv) (A × B) ∪ (A × C)

Solution :-
```   Given that A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}.
(i) A × (B ∩ C)
B ∩ C = {3, 4} ∩ {4, 5, 6} = {4}
A x (B ∩ C) = {1, 2, 3} x {4} = {(1, 4), (2, 4), (3, 4)}
```
```(ii) (A × B) ∩ (A × C)
A x B = {1, 2, 3} x {3, 4} = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}
A x C = {1, 2, 3} x {4, 5, 6} = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
(A × B) ∩ (A × C) = {(1, 4), (2, 4), (3, 4)}
```
```(iii) A × (B ∪ C)
B ∪ C = {3, 4} ∪ {4, 5, 6} = {3, 4, 5, 6}
A × (B ∪ C) = {1, 2, 3} x {3, 4, 5, 6}
= {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)}
```
```(iv)(A × B) ∪ (A × C)
A x B = {1, 2, 3} x {3, 4} = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}
A x C = {1, 2, 3} x {4, 5, 6} = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
(A × B) ∪ (A × C) = {(1, 3), (2, 3), (3, 3), (1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
```

Example-4 :-  If P = {1, 2}, form the set P × P × P.

Solution :-
```   p = {1, 2}
P x P = {1, 2} x {1, 2} = {(1, 1), (1, 2), (2, 1), (2, 2)}
P x P x P = {(1, 1), (1, 2), (2, 1), (2, 2)} x {1, 2}
= {(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2), (2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)}
```

Example-5 :-  If R is the set of all real numbers, what do the cartesian products R × R and R × R × R represent?

Solution :-
```   The Cartesian product R × R represents the set R × R = {(x, y) : x, y ∈ R}
which represents the coordinates of all the points in two dimensional space
and the cartesian product R × R × R represents the set R × R × R ={(x, y, z) : x, y, z ∈ R}
which  represents the coordinates of all the points in three-dimensional space.
```

Example-6 :- If A × B = {(p, q),(p, r), (m, q), (m, r)}, find A and B.

Solution :-
```   Given that A × B = {(p, q),(p, r), (m, q), (m, r)}.
A = set of first elements = {p, m}
B = set of second elements = {q, r}.
```

Example-7 :-  Let A = {1, 2, 3, 4, 5, 6}. Define a relation R from A to A by R = {(x, y) : y = x + 1 }
(i) Depict this relation using an arrow diagram.
(ii) Write down the domain, codomain and range of R.

Solution :-
```   Given that  A = {1, 2, 3, 4, 5, 6} and R = {(x, y) : y =  x + 1 }.
(i) R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}
Now Arrow Diagram is given below :

```
```(ii) Domain of R = {1, 2, 3, 4, 5}
Range of R = {2, 3, 4, 5, 6}
Co-domain of R = {1, 2, 3, 4, 5, 6}
```

Example-8 :-  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?

Solution :-
```(i) In the Set-builder form of R = {(x, y): x is the square of y, x ∈ P, y ∈ Q}.
(ii) In the Set-builder form of R = {(9, 3), (9, –3), (4, 2), (4, –2), (25, 5), (25, –5)}
Domain of R = {9, 4, 25}
Range of R = {3, -3, 2, -2, 5, -5}
Co-domain of R = {5, 3, 2, 1, -1, -2, -3, -5} or set of Q.
```

Example-9 :-  Let A = {1, 2} and B = {3, 4}. Find the number of relations from A to B.

Solution :-
```   Given that A = {1, 2} and B = {3, 4}.
A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}.
Now n(A × B) = 4, the number of subsets of A × B is 2⁴.
Therefore, the number of relations from A into B will be 2⁴.
```

Example-10 :-  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 ∈ N}.
What is the domain, codomain and range of R? Is this relation a function?

Solution :-
```   The domain of R is the set of natural numbers N.
The codomain is also N.
The range is the set of even natural numbers.
Since every natural number n has one and only one image, this relation is a function.
```

Example-11 :-  Examine each of the following relations given below and state in each case, giving reasons whether it is a function or not?
(i) R = {(2, 1), (3, 1), (4, 2)},
(ii) R = {(2, 2), (2, 4), (3, 3), (4, 4)}
(iii) R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7)}.

Solution :-
```(i) Since 2, 3, 4 are the elements of domain of R having their unique images, this relation R is a function.
(ii) Since the same first element 2 corresponds to two different images 2 and 4, this relation is not a function.
(iii) Since every element has one and only one image, this relation is a function.
```

Example-12 :-  Let N be the set of natural numbers. Define a real valued function f : N → à N by f (x) = 2x + 1. Using this definition, complete the table given below.

Solution :-
```
```

Example-13 :-  Define the function f: R → R by y = f(x) = x², x ∈ R. Complete the Table given below by using this definition. What is the domain and range of this function? Draw the graph of f.

Solution :-
```
Domain of f = {x : x ∈ R}. Range of f   = {x²: x ∈ R}. The graph of f is

```

Example-14 :-  Draw the graph of the function f :R → R defined by f (x) = x³, x ∈ R.

Solution :-
```   We have f(0) = 0, f(1) = 1, f(–1) = –1, f(2) = 8, f(–2) = –8,  f(3) = 27; f(–3) = –27, etc.
Therefore,   f = {(x, x³): x ∈ R}. The graph of f is

```

Example-15 :-  Define the real valued function f : R – {0} → R defined by f(x) = 1/x, x ∈ R –{0}. Complete the Table given below using this definition. What is the domain and range of this function?

Solution :-
```
The domain is all real numbers except 0 and its range is also all real numbers except 0. The graph of f is

```

Example-16 :-  Let f(x) = x² and g(x) = 2x + 1 be two real functions.Find (f + g)(x), (f – g)(x), (fg)(x), (f/g)(x).

Solution :-
```   (f + g)(x) = f(x) + g(x) = x² + 2x + 1
(f - g)(x) = f(x) - g(x) = x² - 2x + 1
(f.g)(x) = f(x) . g(x) = (x²)(2x + 1) = 2x³ + x²
(f/g)(x) = f(x) ÷ g(x) = x²/(2x + 1)
```

Example-17 :-  Let f(x) = √x and g(x) = x be two functions defined over the set of non negative real numbers. Find Find (f + g)(x), (f – g)(x), (fg)(x), (f/g)(x).

Solution :-
```   (f + g)(x) = f(x) + g(x) = √x + x
(f - g)(x) = f(x) - g(x) = √x - x
(f.g)(x) = f(x) . g(x) = √x x x = x√x = x3/2
(f/g)(x) = f(x) ÷ g(x) = √x/x = x-1/2
```
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