Where Mindful Learning Begins
Where Mindful Learning Begins
Free, step-by-step Class 11 Maths NCERT Solutions for Chapter 1 Ex 1.5 — all 7 questions solved, covering the complement of a set, De Morgan's laws, Venn diagram representations, and the standard complement identities.
Questions 1–3 build direct computational fluency — finding A′ by removing A's elements from the universal set U, including compound expressions like (A∪C)′ and (B−C)′, and complements taken with respect to the natural numbers in Question 3. Question 4 verifies De Morgan's laws numerically before Question 5 asks you to see the same idea visually through Venn diagrams — (A∪B)′ and A′∩B′ shade the exact same region, and so do (A∩B)′ and A′∪B′. Question 6 applies complements to a geometric set, and Question 7 closes the exercise with four fill-in-the-blank identities worth memorising cold: A∪A′=U, A∩A′=φ, and the two involving φ′ and U′.
Given U=\{1,2,3,4,5,6,7,8,9\}.
(i) A'=U-A — remove 1, 2, 3, 4 from U.
(ii) B'=U-B — remove 2, 4, 6, 8 from U.
(iii) A\cup C=\{1,2,3,4,5,6\}, so (A\cup C)' removes these from U.
(iv) A\cup B=\{1,2,3,4,6,8\}, so (A\cup B)' removes these from U.
(v) The complement of a complement returns the original set.
(vi) B-C: elements of B not in C. B=\{2,4,6,8\}, C=\{3,4,5,6\}; removing 4 and 6 (which are in C) leaves B-C=\{2,8\}. Then (B-C)'=U-\{2,8\}.
Given U=\{a,b,c,d,e,f,g,h\}. Each complement is found by removing the given set's elements from U.
(i) Remove a, b, c from U.
(ii) Remove d, e, f, g from U.
(iii) Remove a, c, e, g from U.
(iv) Remove f, g, h, a from U.
The universal set here is \mathbb{N}=\{1,2,3,\ldots\}.
(i) Every natural number is either even or odd. The complement of the even naturals is exactly the odd naturals.
(ii) Likewise, the complement of the odd naturals is the even naturals.
(iii) The complement consists of every natural number that is NOT a multiple of 3.
(iv) The complement consists of every natural number that is NOT prime (this includes 1, which is neither prime nor composite).
(v) Divisible by both 3 and 5 means divisible by 15. The complement is every natural number not divisible by 15.
(vi) The complement is every natural number that is not a perfect square.
(vii) The complement is every natural number that is not a perfect cube.
(viii) Solving x+5=8 gives x=3, so the set is {3}. Its complement is every natural number except 3.
(ix) Solving 2x+5=9 gives x=2, so the set is {2}. Its complement is every natural number except 2.
(x) The complement of {x : x ≥ 7} is every natural number less than 7.
(xi) 2x+1\gt10\Rightarrow x\gt4.5\Rightarrow x\ge5 (since x is a natural number). The complement is every natural number less than 5.
First find the individual complements: A'=U-A=\{1,3,5,7,9\} and B'=U-B=\{1,4,6,8,9\}.
A\cup B=\{2,3,4,5,6,7,8\}, so (A\cup B)'=U-(A\cup B)=\{1,9\}.
A'\cap B'=\{1,3,5,7,9\}\cap\{1,4,6,8,9\}=\{1,9\}.
A\cap B=\{2\}, so (A\cap B)'=U-\{2\}=\{1,3,4,5,6,7,8,9\}.
A'\cup B'=\{1,3,5,7,9\}\cup\{1,4,6,8,9\}=\{1,3,4,5,6,7,8,9\}.
In each diagram, U is the rectangle, and A and B are the two circles inside it. The shaded region is the answer.
(i) (A\cup B)' — first find A\cup B (everything inside A or B), then shade everything outside it.
(ii) A'\cap B' — A′ is everything outside A, B′ is everything outside B; their intersection is everything outside both.
(iii) (A\cap B)' — first find A\cap B (the lens-shaped overlap of A and B), then shade everything except it.
(iv) A'\cup B' — the region outside A, together with the region outside B.
A' consists of every triangle in U that does NOT have at least one angle different from 60° — that is, every angle must equal 60°.
A triangle in which all three angles are 60° is, by definition, an equilateral triangle.
(i) A set together with its complement covers the entire universal set.
(ii) Since \phi'=U, this becomes U\cap A, which is simply A.
(iii) A set and its complement share no elements.
(iv) Since U'=\phi, this becomes \phi\cap A, which is empty.
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