Class 11 Maths Ncert Solutions Ex 1.3

Class 11 Maths Ncert Solutions Chapter 1 Ex 1.3

Q.1: Fill in the blanks properly using and ⊄.

(i).  {3, 4, 5} ____ {2, 3, 4, 5, 6}

(ii).  {a, b, c} ____ {d, c, d}

(iii).  {y: y is a pupil of Class 11 of the school} ____ {y: y is students of the school}

(iv).  {y: y is a circle in the plane} ____ {y: y is a circle in the same plane with radius 2 unit}

(v).  {y: y is an equilateral triangle in a plane} ____ {y: y is a rectangle in the same plane}

(vi).  {y: y is an equilateral triangle in a plane} ____ {y: y is a triangle in the plane}

(vii).  {y: y is an odd natural number} ____ {x: x is an integers}

 

Answer:

(i).  {3, 4, 5} {2, 3, 4, 5, 6}

(ii).  {a, b, c} ⊄ {d, c, d}

(iii).  {y: y is a pupil of Class 11 of the school} {y: y is students of the school}

(iv).  {y: y is a circle in the plane} ⊄ {y: y is a circle in the same plane with radius 2 unit}

(v).  {y: y is an equilateral triangle in a plane} ⊄ {y: y is a rectangle in the same plane}

(vi).  {y: y is an equilateral triangle in a plane} ⊄ {y: y is a triangle in the plane}

(vii).  {y: y is an odd natural number} {x: x is an integers}

 

 

Q.2: State whether the given statements are true or false:

(i).  {b, c} ⊄ {c, d, e}

(ii).  {a, e, i} {x: x is a vowel in the English alphabets}

(iii).  {1, 2, 3} {1, 2, 4, 5}

(iv).  {c} {b, c, d}

(v).  {b} {a, b, c, d}

(vi).  {y: y is an even natural no. less than 6} {y: y is a natural no. which can divide 36}

 

Answer:

(i).  False

Since, Each element of {d, c} is present in {c, d, e}

 

(ii).  True

Since, a, e and i are the three vowels of the English alphabet.

 

(iii).  False

Since, 3 {1, 2, 3} but 3 {1, 2, 4, 5}

 

(iv).  True

Since, Element c of {c} is also present in {b, c, d}.

 

(v).  False

Since, Element b of {b} is also present in {a, b, c, d}.

 

(vi).  True

Since, {y: y is an even natural number less than 6} = {2, 4}

{y: y is a natural no. which can divide 36} = {1, 2, 3, 4, 6, 9, 12, 18, 36}

Each elements of {2, 4} are present in {1, 2, 3, 4, 6, 9, 12, 18, 36}

 

 

Q.3: Let X = {11, 12, {13, 14}, 15}. According to the given set which of the given statements are false? Explain why.

(i).  {13, 14} X

(ii).  {13, 14} X

(iii).  {{13, 14}} X

(iv).  11 X

(v).  11 X

(vi).  {11, 12, 15} X

(vii).  {11, 12, 15} X

(viii).  {11, 12, 13} X

 (ix).  Ø X

(x).  Ø X

(xi).  {Ø} X

 

Answer:

Given:

X = {11, 12, {13, 14}, 15}

(i).  The Statement {13, 14} X is False

Since, 13 {13, 14}

But, 13 X

 

(ii). The Statement {13, 14} X is True

Since, {13, 14} is an element of X

 

(iii). The Statement {{13, 14}} X is True

Since, {13, 14} {{13, 14}} and {13, 14} X

 

(iv). The Statement 11 X is True

Since, 11 is an element of X

 

(v). The Statement 11 X is False

Since, an element of a set can never be subset of itself.

 

(vi). The Statement {11, 12, 15} X is True

Since, each element of {11, 12, 15} is present in X

 

(vii). The Statement {11, 12, 15} X is False

Since, {11, 12, 13} is not an element of X.

 

(viii). The Statement {11, 12, 13} X is False

Since, 13 {11, 12, 13}

But, 13 X

 

 (ix). The Statement Ø X is False

Since, X does not contain element Ø

 

(x).  The Statement Ø X is True

Since, Ø is subset of every set.

 

(xi).  The Statement {Ø} X is False

Since, Ø {Ø}

Ø X.

 

 

Q.4: Write all the subsets of the given sets:

(i).  {b}

(ii).  {b, c}

(iii).  {2, 3, 4}

(iv).  Ø

 

Answer:

(i).  {b}:

Subsets are as given: Ø and {b}

 

(ii).  {b, c}:

Subsets are as given: Ø, {b}, {c} and {b, c}

 

(iii).  {2, 3, 4}:

Subsets are as given: Ø, {2}, {3}, {4}, {2, 3}, {3, 4}, {4, 2} and {2, 3, 4}

 

(iv).  Ø:

Subsets are as given: Ø

 

 

Q.5: How many elements has P(X), if X = Ø ?

 

Answer:

If X has m elements that is n(X) = m, then n[P(X)] = 2m

If X = Ø,  then n(X) = 0

Therefore, n[P(X)] = 20 = 1

Therefore, P(X) has one element.

 

 

Q.6: Write the given in the form of intervals:

(i).  {y: y R, –5 < y ≤ 7}

(ii).  {y: y R, –13 < y < –11}

(iii).  {y: y R, 1 ≤ y < 8}

(iv).  {y: y R, 4 ≤ y ≤ 5}

 

Answer:

(i).  {y: y R, –5 < y ≤ 7} = (-5, 7]

 

(ii).  {y: y R, –13 < y < –11} = (-13, -11)

 

(iii).  {y: y R, 1 ≤ y < 8} = [1, 8)

 

(iv).  {y: y R, 4 ≤ y ≤ 5} = [4, 5]

 

 

Q.7:  Write the given intervals in the form of set – builder:

(i).  (–4, 1)

(ii).  [7, 13]

(iii).  (7, 13]

(iv).  [–24, 6)

 

Answer:

(i).  (–4, 1) = {y: y R, -4 < y < 1}

 

(ii).  [7, 13] = {y: y R, 7 y 13}

 

(iii).  (7, 13] = {y: y R, 7 < y 13}

 

(iv).  [–24, 6) = {y: y R, –24 x < 6}

 

 

Q.8: What universal set/ sets would you propose for the given sets?

(i).  The set of right triangles

(ii).  The set of isosceles triangles

 

Answer:

(i).  The set of right triangles

Proposed universal sets are:

  • The set of triangles
  • The set of polygons

 

(ii).  The set of isosceles triangles

Proposed universal sets are:

  • The set of triangles
  • The set of polygons
  • The set of two – dimensional figures

 

 

Q.9: X = {1, 3, 5}, Y = {2, 4, 6} and Z = {0, 2, 4, 6, 8}

 

Which of the given sets can be considered as the universal set for the given sets X, Y and Z?

(i).  {0, 1, 2, 3, 4, 5, 6}

(ii).  Ø

(iii).  {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

(iv).  {1, 2, 3, 4, 5, 6, 7, 8}

 

Answer:

(i).  {0, 1, 2, 3, 4, 5, 6}

X {0, 1, 2, 3, 4, 5, 6}

Y {0, 1, 2, 3, 4, 5, 6}

Z ⊄ {0, 1, 2, 3, 4, 5, 6}

Therefore, the set {0, 1, 2, 3, 4, 5, 6} cannot be the universal set for the sets X, Y and Z.

 

(ii).  Ø

X ⊄ Ø

Y ⊄ Ø

Z ⊄ Ø

Therefore, the set Ø cannot be the universal set for the sets X, Y and Z.

 

(iii).  {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

X {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Y {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Z {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Therefore, the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} is the universal set for the sets X, Y and Z.

 

(iv).  {1, 2, 3, 4, 5, 6, 7, 8}

X {1, 2, 3, 4, 5, 6, 7, 8}

Y {1, 2, 3, 4, 5, 6, 7, 8}

Z ⊄ {1, 2, 3, 4, 5, 6, 7, 8}

Therefore, the set {1, 2, 3, 4, 5, 6, 7, 8} cannot be the universal set for the sets X, Y and Z.