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Exercise 3.3 · Q1

Q.For the following sets, find their union and intersection.

(i) A={x:x is the letter of the word ’MATHEMATICS’}A = \{x : x \text{ is the letter of the word 'MATHEMATICS'}\}.
B={x:x is the letter of the word ’TRIGONOMETRY’}B = \{x : x \text{ is the letter of the word 'TRIGONOMETRY'}\}.
(ii) A={x:x is a natural number less than 6}A = \{x : x \text{ is a natural number less than } 6\}.
B={x:x is a multiple of 2 from 1 to 10}B = \{x : x \text{ is a multiple of } 2 \text{ from } 1 \text{ to } 10\}
(iii) A={x:x=2n−1,n∈N}A = \{x : x = 2n - 1, n \in N\} and B={x:x=2n,n∈N}B = \{x : x = 2n, n \in N\}
(iv) A={2,4,6,8,10}A = \{2, 4, 6, 8, 10\} and B={2,4}B = \{2, 4\}
(v) A={x:x=sin⁡θ where 0≤θ≤π/2}A = \{x : x = \sin\theta \text{ where } 0 \leq \theta \leq \pi/2\}
B={x:x=cos⁡θ where 0≤θ≤π/2}B = \{x : x = \cos\theta \text{ where } 0 \leq \theta \leq \pi/2\}
Sikkim CbseNCERTSubjective· 3mImportance★★★★★est
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✓ Free question

Working out A∪BA\cup B and A∩BA\cap B for all five given pairs by listing elements and comparing.

[!FORMULA] A∪B={x:x∈A or x∈B}A\cup B=\{x : x\in A \text{ or } x\in B\}; A∩B={x:x∈A and x∈B}A\cap B=\{x : x\in A \text{ and } x\in B\}.

  1. (i) Distinct letters: A=A= letters of MATHEMATICS ={M,A,T,H,E,I,C,S}=\{M,A,T,H,E,I,C,S\} (8 letters); B=B= letters of TRIGONOMETRY ={T,R,I,G,O,N,M,E,Y}=\{T,R,I,G,O,N,M,E,Y\} (9 letters). Common letters: M,T,I,EM,T,I,E. Combined distinct letters: A,C,E,G,H,I,M,N,O,R,S,T,YA,C,E,G,H,I,M,N,O,R,S,T,Y (13 letters).

A∪B={A,C,E,G,H,I,M,N,O,R,S,T,Y}A\cup B=\{A,C,E,G,H,I,M,N,O,R,S,T,Y\}; A∩B={E,I,M,T}A\cap B=\{E,I,M,T\}.

  1. (ii) A={x∈N:x<6}={1,2,3,4,5}A=\{x\in N : x<6\}=\{1,2,3,4,5\}; B={2,4,6,8,10}B=\{2,4,6,8,10\} (multiples of 2 from 1 to 10). Common: 2,42,4.

A∪B={1,2,3,4,5,6,8,10}A\cup B=\{1,2,3,4,5,6,8,10\}; A∩B={2,4}A\cap B=\{2,4\}.

  1. (iii) A={x=2n−1:n∈N}={1,3,5,7,…}A=\{x=2n-1 : n\in N\}=\{1,3,5,7,\ldots\} (odd naturals); B={x=2n:n∈N}={2,4,6,8,…}B=\{x=2n : n\in N\}=\{2,4,6,8,\ldots\} (even naturals). Every natural number is either odd or even, so together they cover NN, and no number is both.

A∪B=NA\cup B=N (the set of all natural numbers); A∩B=ϕA\cap B=\phi (disjoint sets).

  1. (iv) A={2,4,6,8,10}A=\{2,4,6,8,10\}; B={2,4}B=\{2,4\}. Since every element of BB is already in AA, B⊂AB\subset A.

A∪B=A={2,4,6,8,10}A\cup B=A=\{2,4,6,8,10\}; A∩B=B={2,4}A\cap B=B=\{2,4\}.

  1. (v) A={x=sin⁡θ:0≤θ≤π/2}A=\{x=\sin\theta : 0\le\theta\le\pi/2\}: as θ\theta runs from 00 to π/2\pi/2, sin⁡θ\sin\theta runs from 00 to 11, so A=[0,1]A=[0,1]. B={x=cos⁡θ:0≤θ≤π/2}B=\{x=\cos\theta : 0\le\theta\le\pi/2\}: as θ\theta runs from 00 to π/2\pi/2, cos⁡θ\cos\theta runs from 11 down to 00, so B=[0,1]B=[0,1] too — the same interval.

A∪B=[0,1]A\cup B=[0,1]; A∩B=[0,1]A\cap B=[0,1].

✓Final answer

(i) A∪B={A,C,E,G,H,I,M,N,O,R,S,T,Y}A\cup B=\{A,C,E,G,H,I,M,N,O,R,S,T,Y\}, A∩B={E,I,M,T}A\cap B=\{E,I,M,T\}.

(ii) A∪B={1,2,3,4,5,6,8,10}A\cup B=\{1,2,3,4,5,6,8,10\}, A∩B={2,4}A\cap B=\{2,4\}.

(iii) A∪B=NA\cup B=N, A∩B=ϕA\cap B=\phi.

(iv) A∪B={2,4,6,8,10}A\cup B=\{2,4,6,8,10\}, A∩B={2,4}A\cap B=\{2,4\}.

(v) A∪B=A∩B=[0,1]A\cup B=A\cap B=[0,1].

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