NCERT Solutions: Class 11 Maths Chapter 1 - Sets

A set is a well-defined collection. (i) is well-defined: January, June and July. (iv) depends on the given class but is definite. (v), (vi), (vii) and (viii) are also definite collections. In (ii), "most talented" is subjective; in (iii), "best-cricket batsmen" is subjective; in (ix), "most dangerous" is subjective. Hence those three are not well-defined collections.

The elements of $A$ are exactly $1,2,3,4,5,6$. Therefore the numbers $5,4,2$ belong to $A$, while $8,0,10$ do not belong to $A$.

List each distinct element satisfying the given condition. In (iii), the two-digit numbers with digit sum 8 are $17,26,35,44,53,62,71,80$. In (iv), the prime divisors of $60 = 2^2 \times 3 \times 5$ are $2,3,5$. Repeated letters are written only once in a set.

Identify the pattern in each roster form: the first is the first four multiples of 3, the second is powers of 2 from $2^1$ to $2^5$, the third is powers of 5 from $5^1$ to $5^4$, the fourth is all positive even integers, and the fifth is the squares from $1^2$ to $10^2$.

For (ii), the integers strictly between $-\frac12$ and $\frac92$ are $0,1,2,3,4$. For (iii), $x^2\le4$ gives integer values from $-2$ to $2$. Months not having 31 days are February, April, June, September and November. The consonants before $k$ are $b,c,d,f,g,h,j$.

The natural number divisors of 6 are $1,2,3,6$; the prime divisors of 6 are $2,3$; the distinct letters of MATHEMATICS are $M,A,T,H,E,I,C,S$; and the odd natural numbers less than 10 are $1,3,5,7,9$.

No odd natural number is divisible by 2, so (i) is empty. The even prime numbers set is $\{2\}$, so (ii) is not empty. No natural number can satisfy both $x\lt5$ and $x\gt7$, so (iii) is empty. Parallel lines have no common point, so (iv) is empty.

A set with a definite number of elements is finite; otherwise it is infinite. There are 12 months, so (i) is finite. The set $\{1,2,3,\ldots\}$ and the positive integers greater than 100 continue without end, so (ii) and (iv) are infinite. The set from 1 to 100 and the set of primes less than 99 have finitely many elements, so (iii) and (v) are finite.

There are infinitely many lines parallel to the $x$-axis, one for each possible $y$-coordinate. The English alphabet has 26 letters, so it is finite. Multiples of 5 continue indefinitely, so that set is infinite. The animals living on the earth at a given time form a finite set. Infinitely many circles can pass through the origin, since their centres and radii can vary while still including $(0,0)$.

The order of elements does not matter, so (i) is equal. In (ii), $12\in A$ but $12\notin B$, and $18\in B$ but $18\notin A$. In (iii), both sets are exactly $\{2,4,6,8,10\}$. In (iv), $A$ contains all multiples of 10, while $B$ includes numbers such as 15 and 25 that are not multiples of 10.

(i) $x^2+5x+6=0$ gives $(x+2)(x+3)=0$, so $B=\{-2,-3\}$, not $\{2,3\}$. Therefore $A\ne B$. (ii) The distinct letters in FOLLOW are $F,O,L,W$, and the distinct letters in WOLF are $W,O,L,F$. Hence both sets contain the same elements, so $A=B$.

Sets are equal when they have exactly the same elements, regardless of order. $B=\{1,2,3,4\}$ and $D=\{3,1,4,2\}$ contain the same elements. $E=\{-1,1\}$ and $G=\{1,-1\}$ contain the same elements. The remaining sets differ in at least one element.

A set is a subset of another set only if every element of the first is in the second. In (i), all of $2,3,4$ occur in the second set. In (ii), $a$ is missing from the second set. In (iii), every Class XI student of the school is a student of the school. In (iv), not every circle has radius 1 unit. In (v), triangles are not rectangles. In (vi), every equilateral triangle is a triangle. In (vii), every even natural number is an integer.

(i) Since $a,b$ are both in $\{b,c,a\}$, $\{a,b\}\subset\{b,c,a\}$, so the given statement is false. (ii) $a$ and $e$ are vowels. (iii) $2\notin\{1,3,5\}$. (iv) $a\in\{a,b,c\}$, so $\{a\}$ is a subset. (v) The elements of $\{a,b,c\}$ are $a,b,c$, not the set $\{a\}$. (vi) The even natural numbers less than 6 are $\{2,4\}$, and both divide 36.

The elements of $A$ are $1,2,\{3,4\},5$. (i) is incorrect because 3 and 4 are not separate elements of $A$. (ii) is correct because $\{3,4\}$ is an element of $A$. (iii) is correct because its only element $\{3,4\}$ belongs to $A$. (iv) is correct. (v) is incorrect because 1 is an element, not a set being compared as a subset. (vi) is correct because $1,2,5$ all belong to $A$. (vii) is incorrect because $\{1,2,5\}$ is not an element of $A$. (viii) is incorrect because $3\notin A$. (ix) is incorrect because $\phi$ is not an element of $A$. (x) is correct because $\phi$ is a subset of every set. (xi) is incorrect because it would require $\phi\in A$.

A set with $n$ elements has $2^n$ subsets. Thus the one-element set has 2 subsets, the two-element set has 4 subsets, the three-element set has 8 subsets, and the empty set has only itself as a subset.

A round bracket means the endpoint is excluded, and a square bracket means the endpoint is included. Apply this rule to each inequality.

Translate each interval endpoint into the corresponding strict or non-strict inequality. Parentheses give strict inequalities; square brackets give inclusive inequalities.

A universal set must contain every element of the set being discussed. Every right triangle and every isosceles triangle is a triangle, so the set of all triangles in a plane is the natural choice. Since every triangle is also a polygon, the set of all polygons in a plane is also possible.

A universal set for $A,B,C$ must contain every element in $A\cup B\cup C=\{0,1,2,3,4,5,6,8\}$. Set (i) misses 8. Set (ii) misses all non-empty elements. Set (iii) contains all required elements. Set (iv) misses 0. Therefore only (iii) works.

The union contains all elements that are in either set, with repetitions removed. In (iii), $A=\{3,6,9,12,\ldots\}$ and $B=\{1,2,3,4,5\}$, so the union is $\{1,2,3,4,5,6,9,12,15,\ldots\}$. In (v), union with $\phi$ leaves the set unchanged.

Every element of $A$ is in $B$, so $A\subset B$. Since $A$ adds no new element beyond those already in $B$, the union is $B$.

Since $A\subset B$, every element of $A$ is already in $B$. Therefore the elements that are in $A$ or $B$ are exactly the elements of $B$.

For each part, combine all elements occurring in the listed sets and write each distinct element once.

The intersection contains only common elements. In Q1(iii), $A=\{3,6,9,\ldots\}$ and $B=\{1,2,3,4,5\}$, so the only common element is 3. In Q1(iv), $A=\{2,3,4,5,6\}$ and $B=\{7,8,9\}$, so there is no common element. Any set intersected with $\phi$ gives $\phi$.

First compute the needed unions: $B\cup C=\{7,9,11,13,15\}$ and $B\cup D=\{7,9,11,13,15,17\}$. Also $A\cap B=\{7,9,11\}$ and $A\cup D=\{3,5,7,9,11,15,17\}$. Taking common elements gives the listed intersections.

Since $B,C,D$ are subsets of the natural numbers, intersecting them with $A$ gives the same sets. No natural number is both even and odd, so $B\cap C=\phi$. The only even prime number is 2, so $B\cap D=\{2\}$. The odd primes are exactly the primes other than 2, so $C\cap D$ is the set of odd prime numbers.

Two sets are disjoint if their intersection is empty. In (i), the second set is $\{4,5,6\}$, and 4 is common. In (ii), $e$ is common. In (iii), no integer can be both even and odd, so the intersection is empty.

For $P-Q$, keep the elements of $P$ that are not in $Q$. Applying this rule to each ordered pair gives the listed differences; note that set difference is not commutative.

The common elements are $b$ and $d$. Removing these from $X$ leaves $a,c$; removing them from $Y$ leaves $f,g$.

The set $R-Q$ contains all real numbers that are not rational. By definition, these are the irrational numbers.

(i) The common element is 3. (ii) The common element is $a$. (iii) The first set contains numbers congruent to 2 modulo 4, while the second contains odd numbers listed; there is no common element. (iv) There is no common element between $\{2,6,10\}$ and $\{3,7,11\}$.

Complements are taken with respect to $U$. $A\cup C=\{1,2,3,4,5,6\}$, so its complement is $\{7,8,9\}$. $A\cup B=\{1,2,3,4,6,8\}$, so its complement is $\{5,7,9\}$. Also $B-C=\{2,8\}$, so $(B-C)'=U-\{2,8\}=\{1,3,4,5,6,7,9\}$.

For each set, remove its elements from $U$. The remaining elements form the complement.

Complements are taken in $N$. For (viii), $x+5=8$ gives $x=3$, so the complement is all natural numbers except 3. For (ix), $2x+5=9$ gives $x=2$, so the complement is all natural numbers except 2. For (x), the complement of $\{x:x\ge7\}$ in $N$ is $\{1,2,3,4,5,6\}$. For (xi), $2x+1\gt10$ gives $x\gt\frac92$, so in $N$ the set is $\{5,6,7,\ldots\}$ and its complement is $\{1,2,3,4\}$.

$A'=\{1,3,5,7,9\}$ and $B'=\{1,4,6,8,9\}$. For (i), $A\cup B=\{2,3,4,5,6,7,8\}$, so $(A\cup B)'=\{1,9\}$. Also $A'\cap B'=\{1,9\}$. Hence $(A\cup B)'=A'\cap B'$. For (ii), $A\cap B=\{2\}$, so $(A\cap B)'=\{1,3,4,5,6,7,8,9\}$. Also $A'\cup B'=\{1,3,4,5,6,7,8,9\}$. Hence $(A\cap B)'=A'\cup B'$.

By De Morgan's laws, $(A\cup B)'=A'\cap B'$, so (i) and (ii) have the same Venn diagram: only the part of $U$ outside both circles is shaded. Also $(A\cap B)'=A'\cup B'$, so (iii) and (iv) have the same Venn diagram: all of $U$ except the overlap $A\cap B$ is shaded.

In a triangle, if no angle is different from $60^\circ$, then all three angles are $60^\circ$. Thus the complement of triangles with at least one angle different from $60^\circ$ is the set of triangles with all angles equal to $60^\circ$, i.e. equilateral triangles.

A set together with its complement gives the universal set, so $A\cup A'=U$. Since $\phi'=U$, $\phi'\cap A=U\cap A=A$. A set and its complement have no common element, so $A\cap A'=\phi$. Since $U'=\phi$, $U'\cap A=\phi\cap A=\phi$.

Solve $x^2-8x+12=0$: $(x-2)(x-6)=0$, so $A=\{2,6\}$. Now $B=\{2,4,6\}$, $C=\{2,4,6,8,\ldots\}$ and $D=\{6\}$. Every element of $D$ is in $A$, every element of $A$ is in $B$, and every element of $B$ is in $C$.

(i) False: take $A=\{1\}$, $B=\{\{1\}\}$ and $x=1$. Then $x\in A$ and $A\in B$, but $x\notin B$. (ii) False: take $A=\{1\}$, $B=\{1,2\}$ and $C=\{B\}$. Then $A\subset B$ and $B\in C$, but $A\notin C$. (iii) True: if $x\in A$, then $x\in B$ because $A\subset B$, and then $x\in C$ because $B\subset C$; hence $A\subset C$. (iv) False: take $A=\{1,2\}$, $B=\{2,3\}$ and $C=\{1,2,4\}$. Then $A\not\subset B$ and $B\not\subset C$, but $A\subset C$. (v) False: take $A=\{1,2\}$, $B=\{1\}$ and $x=2$. Then $x\in A$, $A\not\subset B$, but $x\notin B$. (vi) True: if $x\in A$, then $x\in B$ since $A\subset B$. This contradicts $x\notin B$, so $x\notin A$.

Let $x\in B$. If $x\in A$, then $x\in A\cap B=A\cap C$, so $x\in C$. If $x\notin A$, then $x\in A\cup B=A\cup C$; since $x\notin A$, it follows that $x\in C$. Hence $B\subset C$. Similarly, by interchanging $B$ and $C$, every element of $C$ belongs to $B$, so $C\subset B$. Therefore $B=C$.

If $A\subset B$, then no element of $A$ lies outside $B$, so $A-B=\phi$; also $A\cup B=B$ and $A\cap B=A$. Conversely, if $A-B=\phi$, every element of $A$ is in $B$, so $A\subset B$. If $A\cup B=B$, every element of $A$ is already in $B$, so $A\subset B$. If $A\cap B=A$, every element of $A$ is common to $A$ and $B$, so $A\subset B$. Thus each condition is true exactly when $A\subset B$.

Let $x\in C-B$. Then $x\in C$ and $x\notin B$. Since $A\subset B$, any element of $A$ must be in $B$; therefore $x\notin B$ implies $x\notin A$. Hence $x\in C$ and $x\notin A$, so $x\in C-A$. Thus $C-B\subset C-A$.

Every element of $A$ either belongs to $B$ or does not belong to $B$. The elements of $A$ that belong to $B$ form $A\cap B$, and the elements of $A$ that do not belong to $B$ form $A-B$. Therefore $A=(A\cap B)\cup(A-B)$. For the second identity, $B-A$ contains exactly the elements of $B$ not already in $A$. Adding these to $A$ gives every element that is in $A$ or in $B$, so $A\cup(B-A)=A\cup B$.

(i) Since $A\cap B\subset A$, the union of $A$ with its subset $A\cap B$ is $A$. (ii) Since $A\subset A\cup B$, the intersection of $A$ with its superset $A\cup B$ is $A$. These are the absorption laws.

Compute the intersections: $\{1\}\cap\{1,2\}=\{1\}$ and $\{1\}\cap\{1,3\}=\{1\}$. However, $2\in B$ but $2\notin C$, and $3\in C$ but $3\notin B$. Hence equal intersections with $A$ do not force $B=C$.

Using the hint, $A=A\cap(A\cup X)$. Since $A\cup X=B\cup X$, $A=A\cap(B\cup X)=(A\cap B)\cup(A\cap X)$. But $A\cap X=\phi$, so $A=A\cap B$. Hence $A\subset B$. Similarly, $B=B\cap(B\cup X)=B\cap(A\cup X)=(A\cap B)\cup(B\cap X)=A\cap B$, since $B\cap X=\phi$. Hence $B\subset A$. Therefore $A=B$.

Here $A\cap B=\{2\}$, $B\cap C=\{3\}$ and $A\cap C=\{1\}$, so all pairwise intersections are non-empty. But there is no element common to all three sets, so $A\cap B\cap C=\phi$.