Miscellaneous Exercise — Sets

1. 1.M.13 marksNCERT Class 11 Mathematics, Sets, Reprint 2026-27

   Decide which of the following sets are subsets of one another: A={x in R : x^2-8x+12=0}, B={2,4,6}, C={2,4,6,8,...}, D={6}.

   Hint. Solve the quadratic defining A first, then compare every pair of sets for the subset relation.

   x^2-8x+12=0 factors as (x-2)(x-6)=0, so A={2,6}. Now D={6} is a subset of A, since 6 is A's only relevant... checking directly: A={2,6} subset B={2,4,6} (both 2,6 are in B); B subset C (2,4,6 are all even naturals); D={6} subset A, B and C.

   ✦ Working through each part gives: the chain is D subset A subset B subset C: every element of D is in A, every element of A is in B, and every element of B is in C.

2. 1.M.24 marksNCERT Class 11 Mathematics, Sets, Reprint 2026-27

   Determine true or false, proving true statements and giving a counterexample for false ones: (i) If x in A and A in B, then x in B. (ii) If A subset B and B in C, then A in C. (iii) If A subset B and B subset C, then A subset C. (iv) If A not-subset B and B not-subset C, then A not-subset C. (v) If x in A and A not-subset B, then x in B. (vi) If A subset B and x not-in B, then x not-in A.

   Hint. Statements (iii) and (vi) are genuine, provable properties of subsets; the rest fail because membership (in) and subset-hood (subset) do not chain together automatically.

   (i) False: A={1}, B={{1},2}; x=1 is in A and A is in B, but 1 is not an element of B (B's elements are the set {1} and the number 2). (ii) False: A={1}, B={1,2}, C={{1,2},3}; A subset B and B in C, but A itself is not an element of C. (iii) True: if x in A then x in B (since A subset B), and then x in C (since B subset C), so every element of A is in C. (iv) False: A={1,2}, B={1,3}, C={1,2,4}; A not-subset B and B not-subset C, yet A actually IS a subset of C, contradicting the claim. (v) False: A={1,2}, B={2,3}; A not-subset B (1 is missing from B), x=1 in A, but 1 is not in B. (vi) True: this is the contrapositive of A subset B -- if x were in A, it would have to be in B, so x not-in B forces x not-in A.

   ✦ Working through each part gives: true: (iii) and (vi). False: (i),(ii),(iv),(v) -- each disproved by a specific counterexample.

3. 1.M.33 marksNCERT Class 11 Mathematics, Sets, Reprint 2026-27

   Let A, B, C be sets such that A cup B = A cup C and A cap B = A cap C. Show that B = C.

   Hint. Express B using the absorption identity B = B cap (A cup B), then substitute the two given equalities and use the distributive law to reduce everything to C.

   B = B cap (A cup B) = B cap (A cup C) [given] = (B cap A) cup (B cap C) [distributive law] = (A cap B) cup (B cap C) = (A cap C) cup (B cap C) [given] = (A cup B) cap C [distributive law, reversed] = (A cup C) cap C [given] = C, since (A cup C) cap C always simplifies to C.

   ✦ Working through each part gives: b=C, since the given equalities force the same chain of substitutions to collapse both sides onto C.

4. 1.M.44 marksNCERT Class 11 Mathematics, Sets, Reprint 2026-27

   Show that the following four conditions are equivalent: (i) A subset B (ii) A-B = phi (iii) A cup B = B (iv) A cap B = A.

   Hint. Show a full cycle (i) implies (ii) implies (iii) implies (iv) implies (i), which proves all four are equivalent to each other.

   (i) to (ii): if A subset B, no element of A can be missing from B, so A-B has nothing in it. (ii) to (iii): if A-B=phi, every element of A is already in B, so adding A to B changes nothing, giving A cup B=B. (iii) to (iv): if A cup B=B, then A is contained in A cup B=B, so A subset B, and intersecting A with its superset B just returns A, so A cap B=A. (iv) to (i): if A cap B=A, then since A cap B is always a subset of B, A itself is a subset of B.

   ✦ Working through each part gives: all four conditions imply each other in a closed cycle, so they are logically equivalent.

5. 1.M.52 marksNCERT Class 11 Mathematics, Sets, Reprint 2026-27

   Show that if A subset B, then C-B subset C-A.

   Hint. Take an arbitrary element of C-B and show it must also belong to C-A.

   Let x be any element of C-B, so x is in C and x is not in B. If x were in A, then since A subset B, x would have to be in B -- but it is not. So x is not in A either. Since x is in C and not in A, x is in C-A.

   ✦ Working through each part gives: every element of C-B is shown to lie in C-A, so C-B subset C-A whenever A subset B.

6. 1.M.62 marksNCERT Class 11 Mathematics, Sets, Reprint 2026-27

   Show that for any sets A and B: A = (A cap B) cup (A-B), and A cup (B-A) = (A cup B).

   Hint. Split A's own elements into those that are also in B and those that are not -- these two pieces are disjoint and together make up all of A.

   Every element of A either lies in B (contributing to A cap B) or does not lie in B (contributing to A-B); these two cases are mutually exclusive and cover A completely, so their union is exactly A. For the second identity: A union (B-A) collects everything in A, plus whatever part of B lies outside A -- together this is precisely everything in A or B, i.e. A cup B.

   ✦ Working through each part gives: both identities hold by directly matching which set each element belongs to.

7. 1.M.72 marksNCERT Class 11 Mathematics, Sets, Reprint 2026-27

   Using properties of sets, show that (i) A cup (A cap B) = A, (ii) A cap (A cup B) = A.

   Hint. These are the absorption laws -- use the fact that A cap B is always a subset of A, and A is always a subset of A cup B.

   (i) since A cap B subset A, taking the union of A with one of its own subsets adds nothing new, leaving A unchanged. (ii) since A subset A cup B, intersecting A with a set that already contains it entirely just returns A.

   ✦ Working through each part gives: both simplify to A directly from the subset relations A cap B subset A and A subset A cup B.

8. 1.M.82 marksNCERT Class 11 Mathematics, Sets, Reprint 2026-27

   Show that A cap B = A cap C need not imply B = C.

   Hint. One counterexample is enough to disprove the general claim.

   Take A={1}, B={1,2}, C={1,3}. Then A cap B={1} and A cap C={1}, so the two intersections are equal, but B={1,2} is not equal to C={1,3}.

   ✦ Working through each part gives: the counterexample A={1}, B={1,2}, C={1,3} shows A cap B=A cap C does not force B=C.

9. 1.M.93 marksNCERT Class 11 Mathematics, Sets, Reprint 2026-27

   Let A and B be sets. If A cap X = B cap X = phi and A cup X = B cup X for some set X, show that A = B.

   Hint. Use the hint: write A as A cap (A cup X), substitute the given equality A cup X = B cup X, then expand with the distributive law.

   A = A cap (A cup X) = A cap (B cup X) [given] = (A cap B) cup (A cap X) [distributive law] = (A cap B) cup phi [given A cap X=phi] = A cap B. By the identical argument with A and B swapped, B = B cap (B cup X) = B cap (A cup X) = (B cap A) cup (B cap X) = (A cap B) cup phi = A cap B. So both A and B equal A cap B, hence A = B.

   ✦ Working through each part gives: both A and B reduce to exactly A cap B through the same chain of substitutions, so A=B.

10. 1.M.102 marksNCERT Class 11 Mathematics, Sets, Reprint 2026-27

    Find sets A, B, C such that A cap B, B cap C and A cap C are each non-empty, but A cap B cap C = phi.

    Hint. Arrange three sets so each pair shares exactly one element, but no single element is common to all three at once.

    Take A={1,2}, B={2,3}, C={1,3}. Then A cap B={2}, B cap C={3}, A cap C={1} -- all non-empty -- but there is no element common to all three sets simultaneously, so A cap B cap C=phi.

    ✦ Working through each part gives: the sets A={1,2}, B={2,3}, C={1,3} satisfy all the required conditions.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 11 Mathematics textbook, Reprint 2026-27 (kemh101.pdf) — five numbered exercises (1.1-1.5, 39 questions total) plus the chapter's Miscellaneous Exercise (10 questions); every question solved directly from the given sets rather than from memory, including the multi-part true/false and subset-vs-membership questions in Exercise 1.3 Q3 and the Miscellaneous Exercise Q2, which require care to avoid conflating 'element of' with 'subset of'. Questions are referenced from the NCERT textbook for identification.