Ncert Solutions Maths class 11th

26. (i) X ∩ Y = {1,3,5} ∩ {1,2,3} = {1,3}

(ii) A = {a, e, i, o, u} ∩ {a, b, c} = {a}

(iii) A ∩ B = {3,6,9,12 …} ∩ {1,2,3,4,5}

= {3}

(iv) A ∩ B = {2,3,4,5,6} ∩ {7,8,9} = $?$

(v) A ∩ B = {1,2,3} ∩ $?$ = $?$

25. (i) A$\cup$ B = {1, 2, 3, 4}$\cup$ {3, 4, 5, 6}

= {1, 2, 3, 4, 5, 6}

(ii) A $\cup$C = {1, 2, 3, 4}$\cup$ {5, 6, 7, 8}

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

(iii) B$\cup$ C = {3, 4, 5, 6} $\cup$ {5, 6, 7, 8}

= {3, 4, 5, 6, 7, 8}

(iv) B $\cup$D = {3, 4, 5, 6} $\cup$ {7, 8, 9, 10}

= {3, 4, 5, 6, 7, 8, 9, 10}

(v) A$\cup$ B $\cup$C = (A $\cup$B)$\cup$ C = {1,2,3,4,5,6} $\cup$ {5,6,7,8}

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

(vi) A $\cup$B$\cup$ D = (A$\cup$ B)$\cup$ D = {1,2,3,4,5,6}$\cup$ {7,8,9,10}

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

(vii) B$\cup$ C$\cup$ D = (B$\cup$ C)$\cup$ D = {3,4,5,6,7,8}$\cup$ {7,8,9,10}

= {3, 4, 5, 6, 7, 8, 9, 10}

24.If A? B then let a $\in$ A and also a $\in$ B.

but b $\in$ B and b $\notin$ A i.e, A = {a} and B = {a, b}

So, A$\cup$ B = {x : x $\in$ A or x $\in$ B}

= {a, b}

= B

23. Given, A = {a, b}

B = {a, b, c}

Yes A $\subset$ B as a, b $\in$ A and a, b $\in$ B.

And A$\cup$ B = {a, b} $\cup$ {a, b, c} = {a, b, c} = B

22. (i) X∪ Y = {1,3,5}∪ {1,2,3} = {1,2,3,5}.

(ii) A ∪B = {a, e, i, o, u} ∪ {a, b, c} = {a, b, c, e, i, o, u}

(iii) A = {3, 6, 9, 12 …}

B = {1, 2, 3, 4, 5}

So, A∪ B = {3, 6, 9, 12, …}∪ {1, 2, 3, 4, 5}

= {1, 2, 3, 4, 5, 6, 9, 12 …}

(iv) A = {2, 3, 4, 5, 6}

B = {7, 8, 9}

So, A∪ B = {2, 3, 4, 5, 6}∪ {7, 8, 9} = {2, 3, 4, 5, 6, 7, 8, 9}

(v) A ∪B = {1, 2, 3} ∪$?$ = {1, 2, 3}.

21. Universal set of A, B and C must includes all elements of A, B and C, i.e. 0,1,2,3,4,5,6,8.

So, the universal set of A, B and C is (iii) {0,1,2,3,4,5,6,7,8,9,10}.

20. (i) The set of all triangles. U= {x : x is a triangle in a plane}

(ii) The set of all triangles. U= {x : x is a triangle in a plane}

19. (i) [–3, 0] = {x : x such that x $\in$ R, –3 < x <0}

(ii) [6, 12] = {x : x $\in$ R, 6 ≤ x ≤ 12}

(iii) (6, 12] = {x : x $\in$ R, 6 < x 12}

(iv) [–23, 5) = {x : x $\in$ R, –23 ≤ x < 5}

18. (i) As x does not include –4 while 6 is included.

(–4, 6].

(ii) As x does not include both –12 and –10.

(–12, –10)

(iii) As x includes 0 but does not includes 7.

[0, 7)

(iv) As includes both 3 and 4

[3, 4]

17. As A= $?$ , no elements

n(A) = 0 = m

So, n[P(A)] = 2^m = 2⁰ = 1.