JEE Mathematics Sets MCQs Set 05

Answer: (b) Set of even prime numbers

Explanation:
1. No odd number is divisible by 2, so (a) is empty.
2. The only even prime is 2, so (b) is \(\{2\}\) and is not empty.
3. No number is both less than 5 and greater than 7, and parallel lines have no common point, so (c) and (d) are empty.

Teacher's Note:
1. A null set has no elements. Test each description for at least one example.

Answer: (b) 7, 4

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

Teacher's Note:
1. Factorise 112 as a power of 2 times an odd number.

Answer: (a) 16

Explanation:
1. For \(n=2,3,4,5\), \(x=9,13,17,21\), so \(A\) has 4 elements.
2. The number of subsets is \(2^4=16\).

Teacher's Note:
1. A set with k elements has \(2^k\) subsets.

Answer: (c) 7

Explanation:
1. \(A=\{1,4,9\}\) has 3 elements, so it has \(2^3=8\) subsets.
2. Proper subsets exclude the set itself: \(8-1=7\).

Teacher's Note:
1. Proper subsets number \(2^k-1\).

Answer: (d) None of these

Explanation:
1. Every subset of A or of B is a subset of \(A\cup B\), so \(P(A)\cup P(B)\subseteq P(A\cup B)\).
2. Equality fails in general: if \(A=\{1\}\), \(B=\{2\}\), then \(\{1,2\}\in P(A\cup B)\) but not in \(P(A)\cup P(B)\).
3. Since \(P(A)\cup P(B)\ne P(A\cup B)\) generally, and the other options are also wrong, the answer is none of these.

Teacher's Note:
1. Use a small example to test set identities.

Answer: (b) \(N\)

Explanation:
1. \((A\cup B)\cap B'=A\cap B'\), since \(B\cap B'=\phi\).
2. \(A'\cup(A\cap B')=(A'\cup A)\cap(A'\cup B')=A'\cup B'=(A\cap B)'\).
3. \(A\) has odd numbers and \(B\) has even numbers, so \(A\cap B=\phi\) and \((A\cap B)'=N\).

Teacher's Note:
1. Simplify with the distributive law, then use that A and B are disjoint.

Answer: (a) \(\{3,4,10\}\)

Explanation:
1. \(A\cap B=\{3,4,10\}\) and \(A\cap C=\{4\}\).
2. Their union is \(\{3,4,10\}\).

Teacher's Note:
1. Compute each intersection and then join them.

Answer: (d) \(A\cap B^c\)

Explanation:
1. \((A\cap B)^c=A^c\cup B^c\) (De Morgan).
2. \(A\cap(A^c\cup B^c)=(A\cap A^c)\cup(A\cap B^c)=A\cap B^c\).

Teacher's Note:
1. Apply De Morgan's law and distribute.

Answer: (c) odd prime number

Explanation:
1. \(A\cap B\) contains numbers that are both odd and prime.
2. These are the odd primes: 3, 5, 7, 11, ... (the number 2 is prime but even, and 1 is odd but not prime).

Teacher's Note:
1. Describe the intersection by combining both properties.

Answer: (b) 12

Explanation:
1. A number in both sets is a common multiple of 4 and 6.
2. The LCM of 4 and 6 is 12, so \(A\cap B\) is the set of multiples of 12.

Teacher's Note:
1. The intersection of multiples is the multiples of the LCM.

Answer: (a) \(\{i,-i\}\)

Explanation:
1. \(A=\{1,-1\}\) and \(B=\{1,-1,i,-i\}\).
2. The symmetric difference contains elements in exactly one of the sets: \(\{i,-i\}\).

Teacher's Note:
1. Solve the equations over the complex numbers.

Answer: (d) I and II

Explanation:
1. I: elements of A not in B are exactly those of A not in \(A\cap B\), so I is true.
2. II: splitting A into the part in B and the part not in B gives II, which is true.
3. III: the correct identity is \(A-(B\cup C)=(A-B)\cap(A-C)\), so III is false. Hence I and II are correct.

Teacher's Note:
1. Test III with elements that lie in only one of B and C.

Answer: (b) \(A'\cup B\)

Explanation:
1. \((A\cap B')'=A'\cup B\).
2. \(A'\cup B\cup(B\cap C)=A'\cup B\), because \(B\cap C\subseteq B\).

Teacher's Note:
1. The absorption law removes a subset of a set already in the union.

Answer: (b) 700

Explanation:
1. \(n(A\cup B)=200+300-100=400\).
2. \(n(A'\cap B')=n(X)-n(A\cup B)=300\), so \(n(X)=700\).

Teacher's Note:
1. The complement of a union is the intersection of the complements.

Answer: (b) 41

Explanation:
1. The multiples of 3 below 100 are \(3,6,\dots,99\): \(n(S)=33\).
2. The primes below 20 are 2, 3, 5, 7, 11, 13, 17, 19: \(n(P)=8\).
3. The sum is 41.

Teacher's Note:
1. Count carefully: the last multiple of 3 below 100 is 99.

Answer: (b) 60

Explanation:
1. \(n(F\cup S)=n(F)+n(S)-n(F\cap S)\).
2. \(=50+20-10=60\).

Teacher's Note:
1. Subtract the overlap so that people who speak both are not counted twice.

Answer: (b) 25

Explanation:
1. Students playing at least one game: \(25+20-10=35\).
2. Neither: \(60-35=25\).

Teacher's Note:
1. Subtract the union from the total to get those playing neither.

Answer: (c) exactly 25

Explanation:
1. Total readings \(=300\times5=1500\).
2. Each newspaper has 60 readers, so the number of newspapers is \(\frac{1500}{60}=25\).

Teacher's Note:
1. Count the same total in two ways: by students and by newspapers.

Answer: (b) 290

Explanation:
1. \(n(H\cup E)=450+300-200=550\).
2. Neither: \(840-550=290\).

Teacher's Note:
1. Use inclusion-exclusion and subtract from the total.

Answer: (a) \(\{x: |x|<1,\ x\in I\}\)

Explanation:
1. (a) Integers with \(|x|<1\): only 0, so it is a singleton.
2. (b) \(\{5,-5\}\) has two elements; (c) \(\{1,-1\}\) has two; (d) has no real roots, so it is empty.

Teacher's Note:
1. Solve each condition over the stated number system.

Answer: (c) \(A=B\)

Explanation:
1. \(A\in P(A)=P(B)\), so \(A\subseteq B\). Similarly \(B\in P(B)=P(A)\), so \(B\subseteq A\).
2. Hence \(A=B\).

Teacher's Note:
1. A set is always an element of its own power set.

Answer: (a) \(D\subset A,\ D\subset B\) and \(D\subset C\)

Explanation:
1. \(x^2-8x+12=(x-2)(x-6)\), so \(A=\{2,6\}\).
2. \(D=\{6\}\) is contained in \(A=\{2,6\}\), \(B=\{2,4,6\}\) and \(C\) (all even numbers from 2).
3. So \(D\subset A\), \(D\subset B\), \(D\subset C\) are all true.

Teacher's Note:
1. List the elements of each set first.

Answer: (a) 20

Explanation:
1. \(n(A\cup B\cup C)=40+30+30-10-8-5+3=80\).
2. Families buying none: \(100-80=20\).

Teacher's Note:
1. Apply the three-set inclusion-exclusion formula.

Answer: (b) 20

Explanation:
1. Students studying at least one subject: \(200-20=180\).
2. \(180=120+90+70-40-30-50+x\), so \(180=160+x\).
3. Hence \(x=20\).

Teacher's Note:
1. Use inclusion-exclusion backwards to find the triple overlap.

Answer: (d) 160

Explanation:
1. \(n(C\cup H\cup B)=224+240+336-64-80-40+24=640\).
2. Boys who played no game: \(800-640=160\).

Teacher's Note:
1. Add the three sets, subtract the pairs and add back the triple overlap.

Answer: (b) \(A\cap B\) is a singleton set

Explanation:
1. Intersection: \(\frac4x=x\), so \(x^2=4\) and \(x=\pm2\).
2. B requires \(x>0\), so only \(x=2\) works, giving the point \((2,2)\).
3. So \(A\cap B\) is a singleton.

Teacher's Note:
1. Check the extra condition \(x>0\) before counting solutions.