TN 11th, Maths Chapter 1, Exercise 1.1

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Class       : XI

Subject   : Maths

Chapter  : 1

Exercise : 1.1

Que 1 :

Write the following in roster form:

(i)  {x ∈ N : x² < 121 and x is a prime};

Solution :

x is a Prime Number; x² < 121

Let 2² = 4, 3² = 9 , 5² = 25, 7² = 49, (i.e) x²<121

So We Take 2,3,5,7

A = {2, 3, 5, 7}

(ii) The set of all positive roots of the equation

(x – 1)(x + 1)(x2 – 1) = 0;

Solution :

Let

x-1= 0;        x+1 = 0;        x²+1 = 0

x = 1;              x = -1;             x = ±1

Let "B"  as the the set of all positive roots of the equation (x – 1)(x + 1)(x2 – 1) 

B = { 1, -1, ±1} Only Positive Roots

So We Take B = {1}

B = {1}

(iii) {x ∈ N : 4x + 9 < 52};

Solution :

             4x + 9 < 52

Add (-9) On Both Side

       4x + 9 – 9 < 52 – 9

                    4x < 43              

                      x < 43/4                  

              (i.e.) x < 10.75

But x ∈ N

∴ A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

(iv)  {x : (x−4 ÷ x+2) = 3, x ∈ R – {-2}}.

Solution :

                   x-4/x+2 = 3

Divided By ( x + 2 ) On Both Side

  [(x-4)(x+2)]/(x+2) = 3(x+2) {∴ x≠2}

               (i.e.) x – 4 = 3(x + 2)

                        x – 4 = 3x + 6

                    – 4 – 6 = 3x – x

                             2x = -10

                               x = -5 

                           ∴ A = {-5} 

Que 2 :

Write the set {-1,1} in set builder form.

Solution :

Let A = { -1, 1 }

A { x ∈ ℝ, x Is The Root Of Equation x²-1=0}

A = {x: x²= 1}

Que 3 :

State whether the following sets are finite or infinite.

(i) {x ∈ N : x is an even prime number}

Solution :

A = { 2 } 

A is a Finite Set

(ii) {x ∈ N : x is an odd prime number}

Solution :

A = { 3, 5, 7, 11, 13, ......}

A Is a Infinite Set

(iii) {x ∈ Z : x is even and less than 10}

Solution :

A = { ....., -4, -2, 0, 2, 4, 6, 8 }

A Is a Infinite Set

(iv) {x ∈ R : x is a rational number}

Solution :

A = { All Rational Numbers }

A Is a Infinite Set

(v) {x ∈ N : x is a rational number}

Solution :

A = { 1, 2, 3, 4, 5, ...... }

A Is a infinite Set

Que 4 :

By taking suitable sets A, B, C, verify the following results:

Solution :

To prove the following results let us take

U = {1, 2, 5, 7, 8, 9, 10}

A = {1, 2, 5, 7}

B = {2, 7, 8, 9}

C = {1, 5, 8, 7}

(i) To prove:

A × (B ∩ C) = (A × B) ∩ (A × C)

B ∩ C = {8};

A = {1, 2, 5, 7};

A × (B ∩ C) = {1, 2, 5, 7} × {8}

                    = {(1, 8), (2, 8), (5, 8), (7, 8)}

A x B = {(1, 2), (1, 7), (1, 8), (1, 9), (2, 2), (2, 7), (2, 8), (2, 9), (5, 2), (5, 7), (5, 8), (5, 9), (7, 2), (7, 7), (7, 8), (7, 9)}

A × C = {(1, 1), (1, 5),(1, 8), (1, 10), (2, 1), (2, 5), (2, 8), (2, 10), (5, 1), (5, 5), (5, 8), (5, 10), (7, 1), (7, 5), (7, 8), (7, 10)}

(A × B) ∩ (A × C) = {(1, 8), (2, 8), (5, 8), (7, 8)

Hence Proved

A × (B ∩ C) = (A × B) ∩ (A × C)

(ii) To prove :

A × (B ∪ C) = (A × B) (A × C)

B ∪ C = {1, 2, 5, 7, 8, 9, 10}

A = {1, 2, 5, 7};

A × (B ∪ C) = {(1, 1), (1, 2), (1, 5), (1, 7), (1, 8), (1, 9), (1, 10), (2, 1), (2, 2), (2, 5), (2, 7), (2, 8), (2, 9), (2, 10), (5, 1), (5, 2), (5, 5), (5, 7), (5, 8), (5, 9), (5, 10), (7, 1), (7, 2), (7, 5), (7, 7), (7, 8), (7, 9), (7, 10))

A × B = {(1, 2), (1, 7), (1, 8), (1, 9), (2, 2), (2, 7), (2, 8), (2, 9), (5, 2), (5, 7), (5, 8), (5, 9),

(7, 2), (7, 7), (7, 8), (7, 9)}

A × C = {(1, 1), (1, 5), (1, 8), (1, 10), (2, 1), (2, 5), (2, 8), (2, 10), (5, 1), (5, 5), (5, 8), (5, 10), (7, 1), (7, 5), (7, 8), (7, 10)}

(A × B) ∪ (A × C) = (1, 1), (1, 2), (1, 5), (1, 7), (1, 8), (1,9), (1, 10), (2, 1), (2, 2), (2, 5), (2, 7), (2, 8), (2, 9), (2, 10), (5, 1), (5, 2), (5, 5), (5, 7), (5, 8), (5, 9), (5, 10), (7, 1), (7, 2), (7, 5), (7, 7), (7, 8), (7, 9), (7, 10)}

Hence Proved

A × (B ∪ C) = (A × B) ∪ (A × C)

(iii) To Prove 

(A × B) ∩ (B × A) = (A ∩ B) × (B ∩ A)

A × B = {(1, 2), (1, 7), (1, 8), (1, 9) (2, 2), (2, 7), (2, 8), (2, 9) (5, 2), (5, 7), (5, 8), (5, 9) (7, 2), (7, 7), (7, 8), (7, 9)}

B × A = {(2, 1), (2, 2), (2, 5), (2, 7) (7, 1), (7, 2), (7, 5), (7, 7) (8, 1), (8, 2), (8, 5), (8, 7) (9,1), (9, 2), (9, 5), (9, 7)}

(A × B) ∩ (B × A) = {(2, 2), (2, 7), (7, 2), (7, 7)}

(A ∩ B) = {2, 7}

(B ∩ A) = {2, 7}

(A ∩ B) × (B ∩ A) = {2, 7} × {2, 7}

                               = {(2, 2), (2, 7), (7, 2), (7, 7)}

Hence Proved

(A × B) ∩ (B × A) = (A ∩ B) × (B ∩ A)

(iv) To prove

C – (B – A) = (C ∩ A) ∪ (C ∩ B)

B – A = {8, 9};

C = {1, 5, 8, 10};             

C – (B – A) = {1, 5, 10}

C ∩ A = {1}

C ∩ B = {1, 5, 10}

(C ∩ A) ∪ (C ∩ B) = {1} ∪ {1, 5, 10}

                               = {1, 5, 10}

Hence Proved

C – (B – A) = (C ∩ A) ∪ (C ∩ B)

(v) To prove

(B – A) ∩ C = (B ∩ C) – A = B ∩ (C – A)

B – A = {8, 9}

(B – A) ∩ C = {8}

B ∩ C = {8}

A = {1, 2, 5, 7}

So (B ∩ C) – A = {8}

C – A = {8, 10}

B = {2, 7, 8, 9}

B ∩ (C – A) = {8}

Hence Proved

(B – A) ∩ C = (B ∩ C) – A = B ∩ (C – A)

(vi) To prove

(B – A) ∪ C = (B ∪ C) – (A – C)

B – A = {8, 9},

C = {1, 5, 8, 10}

(B – A) ∪ C = {1, 5, 8, 9, 10} ……. (1)

B ∪ C = {1, 2, 5, 7, 8, 9, 10}

A – C = {2, 7}

(B ∪ C) – (A – C) = {1, 5, 8, 9, 10} ……… (2)

(1) = (2)

⇒ (B – A) ∪ C = (B ∪ C) – (A – C)

Que 5 :

Justify the trueness of the statement.

“An element of a set can never be a subset of itself.”

Solution :

Let A = { a, b, c, d }

Each and every element of the set A can be a subset of a set itself

Example  A = {a}, {b}, {c}, {d}

Hence The Given Statement Is Not True.

A set itself can be a subset of itself

(i.e.) A ⊆ A.

But it cannot be a proper subset.

Que 6 :

If n(P(A)) = 1024, n(A ∪ B) = 15 and n(P(B)) = 32, then find n(A ∩ B).

Solution :

n(P(A)) = 1024

               = 2¹⁰

      n(A) = 10

n(A ∪ B) = 15                      

n(P(B)) = 32

              = 2⁵

      n(B) = 5

We Know That

         n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

                    15 = 10 + 5 – n(A ∩ B)

         n(A ∩ B) = 15 – 15

                         = 0

Que 7 :

If n(A ∩ B) = 3 and n(A ∪ B) = 10, then find n(P(A(A ∆ B))

Solution :

n(A ∪ B) = 10

n(A ∩ B) = 3

n(A ∆ B) = 10 – 3 = 7

and n(P(A ∆ B)) = 27

                             = 128

Que 8 :

For a set A, A × A contains 16 elements and two of its elements are (1, 3) and (0, 2). Find the elements of A.

Solution :

A × A = 16 elements = 4 × 4

⇒ A has 4 elements

∴ A = {0, 1, 2, 3}

Que 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, z are distinct elements.

Solution :

n(A) = 3 ⇒ set A contains 3 elements,

n(B) = 2 ⇒ set B contains 2 elements,

we are given (x, 1), (y, 2), (z, 1) are elements in A × B ⇒ A = {x, y, z} and B = {1, 2}

Que 10 :

If A × A has 16 elements,

S = {(a, b) ∈ A × A : a < b}; (-1, 2) and (0, 1) are two elements of S, then find the remaining elements of S.

Solution :

n(A × A) = 16 ⇒ n( A) = 4

S ={(-1, 0), (-1, 1), (0, 2), (1, 2)}

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