JEE Mathematics Set Operations Relations and Mappings MCQs Set 09

Answer: (a) \(B'-A'=A-B\)

Explanation:
1. Since \(B\subset A\), every element outside \(A\) is also outside \(B\), so \(A'\subset B'\).
2. \(B'-A'=B'\cap A=A\cap B'=A-B\).
3. \(A'-B'=A'\cap B=\phi\) and \(A'\cap B'=A'\), so the other options fail.

Teacher's Note:
1. Rewrite a difference as an intersection with a complement: X - Y = X and Y'.
2. When one set contains another, complements reverse the inclusion.

Answer: (b) \(A\not\subset B\)

Explanation:
1. \(\sin\theta=\tan\theta\) gives \(\sin\theta(1-\sec\theta)=0\), so \(\sin\theta=0\) or \(\cos\theta=1\).
2. Hence \(A=\{n\pi:n\in Z\}\) (note \(\cos\theta=\pm1\) when \(\sin\theta=0\), and \(\cos\theta\ne0\) there).
3. \(B=\{2n\pi:n\in Z\}\) is a proper subset of \(A\), so \(A\not\subset B\).

Teacher's Note:
1. Solve each defining equation fully before comparing sets.
2. Check the excluded values (here cos = 0) so that no solution is lost or added.

Answer: (b) \(P(S)\)

Explanation:
1. \(2^n=256\) gives \(n=8\).
2. \(S\) has 3 elements, \(P(S)\) has \(2^3=8\) elements.
3. So the set with exactly 8 elements is \(P(S)\).

Teacher's Note:
1. A set with n elements has 2^n subsets.
2. The power set of a 3-element set always has 8 elements.

Answer: (c) exactly 25

Explanation:
1. Count student-newspaper readings in two ways: students times papers per student gives \(300\times5=1500\).
2. Each newspaper accounts for 60 readings, so number of newspapers \(=1500/60=25\).

Teacher's Note:
1. Double counting is the standard trick for such problems.
2. Count the same total from the student side and the newspaper side.

Answer: (a) \(A'\)

Explanation:
1. \((A\cup B)'=A'\cap B'\) by De Morgan's law.
2. So the set is \((A'\cap B')\cup(A'\cap B)=A'\cap(B'\cup B)=A'\cap U=A'\).

Teacher's Note:
1. De Morgan: the complement of a union is the intersection of complements.
2. Factor the common set A' out, then B' union B is the universal set.

Answer: (b) \((A\cup B)\setminus(A\cap B)\)

Explanation:
1. The set is the symmetric difference: elements in exactly one of \(A\), \(B\).
2. Elements in at least one but not in both: \((A\cup B)\setminus(A\cap B)\).
3. Option (a) is \((A\setminus B)\cap(B\setminus A)=\phi\) and (c) is only \(A\setminus B\).

Teacher's Note:
1. Symmetric difference = union minus intersection.
2. Test each option with a small example such as A={1,2}, B={2,3}.

Answer: (a) \(d=bc\)

Explanation:
1. A number lies in both bN and cN exactly when it is a multiple of both b and c.
2. Since b and c are coprime, their LCM is bc, so the intersection is the set of multiples of bc.
3. Thus \(d=bc\).

Teacher's Note:
1. The intersection of multiple sets corresponds to the LCM.
2. For coprime numbers LCM equals product.

Answer: (a) \(B\cap C'\)

Explanation:
1. Inside \(C'\), \((A\cap B'\cap C')'=A'\cup B\cup C\) reduces to \(A'\cup B\).
2. Also \((A\cup B\cup C)\cap C'=(A\cup B)\cap C'\).
3. So we need \((A\cup B)\cap(A'\cup B)\cap C'=B\cap C'\).

Teacher's Note:
1. Use De Morgan on the middle bracket first.
2. (A or B) and (not A or B) simplifies to B.

Answer: (a) 20

Explanation:
1. Fish lovers \(=100-50=50\); chicken lovers 60; people who like at least one \(=100-10=90\).
2. Both \(=50+60-90=20\).

Teacher's Note:
1. Venn diagrams or the inclusion-exclusion formula turn most word problems on sets into simple arithmetic.
2. Always check the answer by substituting back into the given counts.

Answer: (c) \(A\cap B\cap C\)

Explanation:
1. Every element of U lies in at least one of A, B, C.
2. An element in exactly one set, say A only, lies in A-B; similarly for B only (in B-C) and C only (in C-A).
3. An element in exactly two sets also falls in one of the three differences (A,B only in B-C; A,C only in A-B; B,C only in C-A).
4. Only elements in all three sets are in none of them, so the complement is \(A\cap B\cap C\).

Teacher's Note:
1. Check each Venn region against each difference in turn.
2. The complement picks out the regions not covered.

Answer: (d) \(1000\le p\le1499\)

Explanation:
1. \(p=1000+500-n(A\cap B)=1500-n(A\cap B)\).
2. \(1\le n(A\cap B)\le500\) (it cannot exceed the smaller set).
3. So \(1000\le p\le1499\).

Teacher's Note:
1. The overlap is at most the size of the smaller set.
2. Maximum union needs minimum overlap, and the reverse.

Answer: (a) 10

Explanation:
1. Percentages who did not lose each organ: 30, 20, 25, 15; at most \(30+20+25+15=90\) percent escaped at least one loss.
2. So at least \(100-90=10\) percent lost all four.

Teacher's Note:
1. For minimum overlap, spread the people who missed each loss as far apart as possible.
2. Subtract the sum of the complements from 100.

Answer: (c) \(X=Y\)

Explanation:
1. \(X=(X\cup A)-A\) because X has nothing in common with A.
2. Likewise \(Y=(Y\cup A)-A\).
3. As \(X\cup A=Y\cup A\), the two right sides are equal, so \(X=Y\).

Teacher's Note:
1. If a union and a disjoint part are known, the other part is fixed.
2. This is a cancellation law for sets.

Answer: (c) \(39\le x\le63\)

Explanation:
1. Union \(=63+76-x=139-x\le100\), so \(x\ge39\).
2. Intersection cannot exceed the smaller set, so \(x\le63\).

Teacher's Note:
1. Use union at most 100 for the lower bound.
2. Use the smaller set for the upper bound.

Answer: (d) (1) and (2)

Explanation:
1. (1): both sides are the part of A outside B, true.
2. (2): A splits into the part inside B and the part outside B, true.
3. (3): the correct law is \(A-(B\cup C)=(A-B)\cap(A-C)\), so it is false (take A={1,2}, B={1}, C={2}: LHS empty, RHS {1,2}).

Teacher's Note:
1. Test doubtful identities with small sets.
2. A difference with a union becomes an intersection of differences.

Answer: (b) 26, 27 and 28 only

Explanation:
1. Let \(x=n(A\cap B\cap C)\) and \(t=n(A\cap C)\). Then \(n(A\cup B\cup C)=45-8-9-t+x=28-(t-x)\).
2. Since A has 10 elements, \(8+t-x\le10\), so \(t-x\le2\), and \(t\ge x\).
3. So the value is 28, 27 or 26 (each is attainable), and 29 is impossible.

Teacher's Note:
1. Express the union by inclusion-exclusion using the unknown triple overlap.
2. Bound the unknowns using the sizes of the individual sets.

Answer: (b) (ii) and (iii) only

Explanation:
1. Own at least one: \(100-65=35\) percent, so (ii) holds.
2. Both \(=25+15-35=5\) percent, so (i) is wrong.
3. 5% of the town is 2000 families, so the town has 40,000 families and (iii) holds.

Teacher's Note:
1. Venn diagrams or the inclusion-exclusion formula turn most word problems on sets into simple arithmetic.
2. Always check the answer by substituting back into the given counts.

Answer: (b) (ii) and (iii) only

Explanation:
1. Students playing at least one game \(=60-15=45\).
2. Inclusion-exclusion: \(45=23+15+20-7-5-4+x\) so \(x=3\) play all three and (i) is wrong.
3. Hockey but not Cricket \(=23-4=19\), so (ii) holds.
4. Hockey and Cricket but not Basketball \(=4-3=1\), so (iii) holds.

Teacher's Note:
1. Venn diagrams or the inclusion-exclusion formula turn most word problems on sets into simple arithmetic.
2. Always check the answer by substituting back into the given counts.

Answer: (b) 3

Explanation:
1. \(2^m-2^n=112\) gives \(2^n(2^{m-n}-1)=112=16\times7\).
2. So \(2^n=16\), \(n=4\) and \(2^{m-n}-1=7\), so \(m-n=3\) (with m = 7).

Teacher's Note:
1. Factor out the smaller power of 2.
2. The odd factor must equal 2^k - 1 for some k.

Answer: (b) 3

Explanation:
1. \(A=\{1,3,5,15\}\), \(B=\{2,3,5,7\}\), \(C=\{2,4,6,8\}\).
2. A union C is \(\{1,2,3,4,5,6,8,15\}\); its intersection with B is \(\{2,3,5\}\), which has 3 elements.

Teacher's Note:
1. List the sets explicitly for such small problems.
2. Take the union first, then intersect.

Answer: (b) 1

Explanation:
1. \(2a^2=35-3b^2\) must be even, so b is odd; b = 1 gives \(a^2=16\), b = 3 gives \(a^2=4\) (b = 5 is too large).
2. Ratios a/b: b = 1 gives 4 or -4; b = 3 gives 2/3 or -2/3.
3. Only the value 4 is a positive integer, so the set has 1 element.

Teacher's Note:
1. Restrict by parity and size first.
2. Remember the ratio must be a positive integer.