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Exercise 3.2 · Q3

Q.Which of the following pairs of sets are equal? Give reasons.

(i) A={−2,3}A = \{-2, 3\}, B={x:x is a solution of x2−x−6=0}B = \{x : x \text{ is a solution of } x^2 - x - 6 = 0\}
(ii) A={x:x is a letter of the word ’FOLLOW’}A = \{x : x \text{ is a letter of the word 'FOLLOW'}\}, B={y:y is a letter of the word ’WOLF’}B = \{y : y \text{ is a letter of the word 'WOLF'}\}
(iii) A={x:x is a letter of the word ’ASSET’}A = \{x : x \text{ is a letter of the word 'ASSET'}\}, B={y:y is a letter of the word ’EAST’}B = \{y : y \text{ is a letter of the word 'EAST'}\}
(iv) A={−1,1}A = \{-1, 1\}; B={x:x is a real number satisfying the equation x2+1=0}B = \{x : x \text{ is a real number satisfying the equation } x^2 + 1 = 0\}
(v) A={1,4,9}A = \{1, 4, 9\}; B={x:x=n2 where ’n’ is a natural number less than 5}B = \{x : x = n^2 \text{ where 'n' is a natural number less than } 5\}
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✓ Free question

Comparing each pair's actual elements: (i), (ii), (iii) turn out identical (equal sets); (iv) and (v) differ (not equal).

[!FORMULA] Two sets AA and BB are equal (A=BA=B) iff they have exactly the same elements, i.e. every element of AA is in BB and every element of BB is in AA.

  1. (i) A={−2,3}A=\{-2,3\}. Solve x2−x−6=0x^2-x-6=0: factor as (x−3)(x+2)=0⇒x=3(x-3)(x+2)=0\Rightarrow x=3 or x=−2x=-2. So B={3,−2}={−2,3}B=\{3,-2\}=\{-2,3\}. Since A={−2,3}=BA=\{-2,3\}=B, A=BA=B, equal.

  2. (ii) A=A= distinct letters of 'FOLLOW' ={F,O,L,W}=\{F,O,L,W\}. B=B= distinct letters of 'WOLF' ={W,O,L,F}={F,O,L,W}=\{W,O,L,F\}=\{F,O,L,W\}. Same set. A=BA=B, equal.

  3. (iii) A=A= distinct letters of 'ASSET' ={A,S,E,T}=\{A,S,E,T\}. B=B= distinct letters of 'EAST' ={E,A,S,T}={A,S,E,T}=\{E,A,S,T\}=\{A,S,E,T\}. Same set. A=BA=B, equal.

  4. (iv) A={−1,1}A=\{-1,1\}. Solve x2+1=0⇒x2=−1x^2+1=0\Rightarrow x^2=-1, which has no real solution. So B=ϕB=\phi (empty set). Since AA has elements and BB has none, A≠BA\ne B, not equal.

  5. (v) A={1,4,9}A=\{1,4,9\}. B={n2:n∈N, n<5}B=\{n^2 : n\in N,\ n<5\}: n=1,2,3,4⇒n2=1,4,9,16n=1,2,3,4\Rightarrow n^2=1,4,9,16, so B={1,4,9,16}B=\{1,4,9,16\}. Since 16∈B16\in B but 16∉A16\notin A, A≠BA\ne B, not equal.

✓Final answer

Equal pairs: (i), (ii), (iii). Not equal: (iv) (B=ϕ≠AB=\phi\ne A), (v) (BB has an extra element 16).

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