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NCERT Solutions · Class 11 Mathematics Sets

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EXERCISE 1.4 1–12 (part 4 of 6)

  1. Exercise 1

    Find the union of each of the following pairs of sets:
    (i)
    X={1,3,5}Y={1,2,3}\displaystyle \mathrm{X}=\{1,3,5\} \quad \mathrm{Y}=\{1,2,3\}
    (ii)
    A=[a,e,i,o,u}B={a,b,c}\displaystyle \mathrm{A}=[a, e, i, o, u\} \quad \mathrm{B}=\{a, b, c\}
    (iii)
    A={x:x\displaystyle \mathrm{A}=\{x: x is a natural number and multiple of 3\displaystyle 3 }\displaystyle \} B={x:x\displaystyle \mathrm{B}=\{x: x is a natural number less than 6\displaystyle 6}\displaystyle \}
    (iv)
    A={x:x\displaystyle \mathrm{A}=\{x: x is a natural number and 1<x≤6}\displaystyle 1<x \leq 6\} B={x:x\displaystyle \mathrm{B}=\{x: x is a natural number and 6<x<10}\displaystyle 6<x<10\}
    (v)
    A={1,2,3},B=ϕ\displaystyle \mathrm{A}=\{1,2,3\}, \mathrm{B}=\phi

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    NCERT’s answer
    (i)
    $\displaystyle \mathrm{X} \cup \mathrm{Y}=\{1,2,3,5\}$ (ii) $\displaystyle \mathrm{A} \cup \mathrm{B}=\{a, b, c, e, i, o, u\}$ (iii) $\displaystyle \mathrm{A} \cup \mathrm{B}=\{x: x=1,2,4,5$ or a multiple of $\displaystyle 3 \}$ (iv) $\displaystyle \mathrm{A} \cup \mathrm{B}=\{x: 1<x<10, x \in \mathrm{~N}\}$ (v) $\displaystyle \mathrm{A} \cup \mathrm{B}=\{1,2,3\}$
    Union of two sets. \(\displaystyle \mathrm{A} \cup \mathrm{B} \) is the set of elements belonging to \(\displaystyle \mathrm{A} \), or to \(\displaystyle \mathrm{B} \), or to both; an element common to the two is listed only once.
    (i)
    \(\displaystyle \mathrm{X} = \{1,3,5\},\ \mathrm{Y} = \{1,2,3\} \). Collecting all elements and keeping \(\displaystyle 1 \) and \(\displaystyle 3 \) once each:
    \[\mathrm{X} \cup \mathrm{Y} = \{1,2,3,5\} \]
    (ii)
    \(\displaystyle \mathrm{A} = \{a,e,i,o,u\},\ \mathrm{B} = \{a,b,c\} \). The letter \(\displaystyle a \) is common, listed once:
    \[\mathrm{A} \cup \mathrm{B} = \{a,b,c,e,i,o,u\} \]
    (iii)
    \(\displaystyle \mathrm{A} = \{x : x \text{ is a natural number and a multiple of } 3\} = \{3,6,9,12,\ldots\} \) and \(\displaystyle \mathrm{B} = \{x : x \text{ is a natural number less than } 6\} = \{1,2,3,4,5\} \). Their union adds \(\displaystyle 1,2,4,5 \) to the multiples of \(\displaystyle 3 \):
    \[\mathrm{A} \cup \mathrm{B} = \{1,2,3,4,5,6,9,12,15,\ldots\} = \{x : x = 1,\,2,\,4,\,5 \text{ or } x \text{ is a multiple of } 3\} \]
    (iv)
    \(\displaystyle \mathrm{A} = \{x : x \in \mathbf{N},\ 1 < x \leq 6\} = \{2,3,4,5,6\} \) and \(\displaystyle \mathrm{B} = \{x : x \in \mathbf{N},\ 6 < x < 10\} = \{7,8,9\} \):
    \[\mathrm{A} \cup \mathrm{B} = \{2,3,4,5,6,7,8,9\} = \{x : x \in \mathbf{N},\ 1 < x < 10\} \]
    (v)
    \(\displaystyle \mathrm{A} = \{1,2,3\},\ \mathrm{B} = \phi \). The empty set contributes nothing, so
    \[\mathrm{A} \cup \mathrm{B} = \{1,2,3\} \]
    (i) {$\displaystyle 1$, $\displaystyle 2$, $\displaystyle 3$, $\displaystyle 5$} (ii) {a, b, c, e, i, o, u} (iii) {$\displaystyle 1$, $\displaystyle 2$, $\displaystyle 3$, $\displaystyle 4$, $\displaystyle 5$, $\displaystyle 6$, $\displaystyle 9$, $\displaystyle 12$, $\displaystyle 15$, …} (iv) {$\displaystyle 2$, $\displaystyle 3$, $\displaystyle 4$, $\displaystyle 5$, $\displaystyle 6$, $\displaystyle 7$, $\displaystyle 8$, $\displaystyle 9$} (v) {$\displaystyle 1$, $\displaystyle 2$, $\displaystyle 3$}

  2. Exercise 2

    Let A={a,b},B={a,b,c}\displaystyle \mathrm{A}=\{a, b\}, \mathrm{B}=\{a, b, c\}. Is A⊂B\displaystyle \mathrm{A} \subset \mathrm{B} ? What is A∪B\displaystyle \mathrm{A} \cup \mathrm{B} ?

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    NCERT’s answer
    Yes, $\displaystyle \mathrm{A} \cup \mathrm{B}=\{a, b, c\}$
    Subset test, then union. \(\displaystyle \mathrm{A} \subset \mathrm{B} \) means every element of \(\displaystyle \mathrm{A} \) is also an element of \(\displaystyle \mathrm{B} \).Here \(\displaystyle \mathrm{A} = \{a,b\} \) and \(\displaystyle \mathrm{B} = \{a,b,c\} \). Checking each element of \(\displaystyle \mathrm{A} \): \(\displaystyle a \in \mathrm{B} \) and \(\displaystyle b \in \mathrm{B} \). So every element of \(\displaystyle \mathrm{A} \) lies in \(\displaystyle \mathrm{B} \), i.e. \(\displaystyle \mathrm{A} \subset \mathrm{B} \).For the union, collect the elements of both sets, writing repeated ones once: \[\mathrm{A} \cup \mathrm{B} = \{a,b\} \cup \{a,b,c\} = \{a,b,c\} = \mathrm{B} \]Yes, A ⊂ B; and A ∪ B = {a, b, c} = B

  3. Exercise 3

    If A and B are two sets such that A C B , then what is A∪B\displaystyle \mathrm{A} \cup \mathrm{B} ?

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    NCERT’s answer
    B
    Union when one set is contained in the other. Suppose \(\displaystyle \mathrm{A} \subset \mathrm{B} \).Take any \(\displaystyle x \in \mathrm{A} \cup \mathrm{B} \). Then \(\displaystyle x \in \mathrm{A} \) or \(\displaystyle x \in \mathrm{B} \). If \(\displaystyle x \in \mathrm{A} \), then \(\displaystyle x \in \mathrm{B} \) as well, because \(\displaystyle \mathrm{A} \subset \mathrm{B} \). Either way \(\displaystyle x \in \mathrm{B} \), so \[\mathrm{A} \cup \mathrm{B} \subset \mathrm{B}. \]Conversely \(\displaystyle \mathrm{B} \subset \mathrm{A} \cup \mathrm{B} \) always holds, since every element of \(\displaystyle \mathrm{B} \) is in the union.Both inclusions together give equality.If A ⊂ B, then A ∪ B = B

  4. Exercise 4

    If A={1,2,3,4},B={3,4,5,6},C={5,6,7,8}\displaystyle \mathrm{A}=\{1,2,3,4\}, \mathrm{B}=\{3,4,5,6\}, \mathrm{C}=\{5,6,7,8\} and D={7,8,9,10}\displaystyle \mathrm{D}=\{7,8,9,10\}; find
    (i)
    A∪B\displaystyle \mathrm{A} \cup \mathrm{B}
    (ii)
    A∪C\displaystyle \mathrm{A} \cup \mathrm{C}
    (iii)
    B∪C\displaystyle \mathrm{B} \cup \mathrm{C}
    (iv)
    B∪D\displaystyle \mathrm{B} \cup \mathrm{D}
    (v)
    A∪B∪C\displaystyle \mathrm{A} \cup \mathrm{B} \cup \mathrm{C}
    (vi)
    A∪B∪D\displaystyle \mathrm{A} \cup \mathrm{B} \cup \mathrm{D}
    (vii)
    B∪C∪D\displaystyle \mathrm{B} \cup \mathrm{C} \cup \mathrm{D}

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    Union of sets. List all elements that occur in at least one of the sets, each written once. Here
    \[\mathrm{A} = \{1,2,3,4\},\quad \mathrm{B} = \{3,4,5,6\},\quad \mathrm{C} = \{5,6,7,8\},\quad \mathrm{D} = \{7,8,9,10\}. \]
    (i)
    \(\displaystyle \mathrm{A} \cup \mathrm{B} = \{1,2,3,4,5,6\} \) ($\displaystyle 3$ and $\displaystyle 4$ are common, kept once)
    (ii)
    \(\displaystyle \mathrm{A} \cup \mathrm{C} = \{1,2,3,4,5,6,7,8\} \)
    (iii)
    \(\displaystyle \mathrm{B} \cup \mathrm{C} = \{3,4,5,6,7,8\} \) ($\displaystyle 5$ and $\displaystyle 6$ are common)
    (iv)
    \(\displaystyle \mathrm{B} \cup \mathrm{D} = \{3,4,5,6,7,8,9,10\} \)
    (v)
    \(\displaystyle \mathrm{A} \cup \mathrm{B} \cup \mathrm{C} = (\mathrm{A} \cup \mathrm{B}) \cup \mathrm{C} = \{1,2,3,4,5,6\} \cup \{5,6,7,8\} = \{1,2,3,4,5,6,7,8\} \)
    (vi)
    \(\displaystyle \mathrm{A} \cup \mathrm{B} \cup \mathrm{D} = \{1,2,3,4,5,6\} \cup \{7,8,9,10\} = \{1,2,3,4,5,6,7,8,9,10\} \)
    (vii)
    \(\displaystyle \mathrm{B} \cup \mathrm{C} \cup \mathrm{D} = \{3,4,5,6,7,8\} \cup \{7,8,9,10\} = \{3,4,5,6,7,8,9,10\} \)
    (i) {$\displaystyle 1,2,3,4,5,6$} (ii) {$\displaystyle 1,2,3,4,5,6,7,8$} (iii) {$\displaystyle 3,4,5,6,7,8$} (iv) {$\displaystyle 3,4,5,6,7,8,9,10$} (v) {$\displaystyle 1,2,3,4,5,6,7,8$} (vi) {$\displaystyle 1,2,3,4,5,6,7,8,9,10$} (vii) {$\displaystyle 3,4,5,6,7,8,9,10$}

  5. Exercise 5

    Find the intersection of each pair of sets of question 1\displaystyle 1 above.

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    NCERT’s answer
    (i)
    $\displaystyle \mathrm{X} \cap \mathrm{Y}=\{1,3\}$ (ii) $\displaystyle \mathrm{A} \cap \mathrm{B}=\{a\}$ (iii) \{ $\displaystyle 3$ \} (iv) $\displaystyle \phi$ (v) $\displaystyle \phi$
    Intersection of two sets. \(\displaystyle \mathrm{A} \cap \mathrm{B} \) is the set of elements that belong to \(\displaystyle \mathrm{A} \) and to \(\displaystyle \mathrm{B} \). Taking the same five pairs as in Question $\displaystyle 1$:
    (i)
    \(\displaystyle \mathrm{X} = \{1,3,5\},\ \mathrm{Y} = \{1,2,3\} \). The common elements are \(\displaystyle 1 \) and \(\displaystyle 3 \):
    \[\mathrm{X} \cap \mathrm{Y} = \{1,3\} \]
    (ii)
    \(\displaystyle \mathrm{A} = \{a,e,i,o,u\},\ \mathrm{B} = \{a,b,c\} \). Only \(\displaystyle a \) is common:
    \[\mathrm{A} \cap \mathrm{B} = \{a\} \]
    (iii)
    \(\displaystyle \mathrm{A} = \{3,6,9,12,\ldots\},\ \mathrm{B} = \{1,2,3,4,5\} \). The only multiple of \(\displaystyle 3 \) below \(\displaystyle 6 \) is \(\displaystyle 3 \):
    \[\mathrm{A} \cap \mathrm{B} = \{3\} \]
    (iv)
    \(\displaystyle \mathrm{A} = \{2,3,4,5,6\},\ \mathrm{B} = \{7,8,9\} \). No element is common:
    \[\mathrm{A} \cap \mathrm{B} = \phi \]
    (v)
    \(\displaystyle \mathrm{A} = \{1,2,3\},\ \mathrm{B} = \phi \). The empty set has no element to share:
    \[\mathrm{A} \cap \mathrm{B} = \phi \]
    (i) {$\displaystyle 1$, $\displaystyle 3$} (ii) {a} (iii) {$\displaystyle 3$} (iv) φ (v) φ

  6. Exercise 6

    If A={3,5,7,9,11},B={7,9,11,13},C={11,13,15}\displaystyle \mathrm{A}=\{3,5,7,9,11\}, \mathrm{B}=\{7,9,11,13\}, \mathrm{C}=\{11,13,15\} and D={15,17}\displaystyle \mathrm{D}=\{15,17\}; find
    (i)
    A∩B\displaystyle \mathrm{A} \cap \mathrm{B}
    (ii)
    B∩C\displaystyle \mathrm{B} \cap \mathrm{C}
    (iii)
    A∩C∩D\displaystyle \mathrm{A} \cap \mathrm{C} \cap \mathrm{D}
    (iv)
    A∩C\displaystyle \mathrm{A} \cap \mathrm{C}
    (v)
    B∩D\displaystyle \mathrm{B} \cap \mathrm{D}
    (vi)
    A∩(B∪C)\displaystyle \mathrm{A} \cap(\mathrm{B} \cup \mathrm{C})
    (vii)
    A∩D\displaystyle \mathrm{A} \cap \mathrm{D}
    (viii)
    A∩(B∪D)\displaystyle \mathrm{A} \cap(\mathrm{B} \cup \mathrm{D})
    (ix)
    (A∩B)∩(B∪C)\displaystyle (\mathrm{A} \cap \mathrm{B}) \cap(\mathrm{B} \cup \mathrm{C})
    (x)
    (A∪D)∩(B∪C)\displaystyle (\mathrm{A} \cup \mathrm{D}) \cap(\mathrm{B} \cup \mathrm{C})

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    Intersection, and brackets first. \(\displaystyle \mathrm{A} \cap \mathrm{B} \) keeps only the elements common to both sets; where a bracket appears, evaluate it before intersecting. Here
    \[\mathrm{A} = \{3,5,7,9,11\},\ \mathrm{B} = \{7,9,11,13\},\ \mathrm{C} = \{11,13,15\},\ \mathrm{D} = \{15,17\}. \]
    (i)
    \(\displaystyle \mathrm{A} \cap \mathrm{B} = \{7,9,11\} \)
    (ii)
    \(\displaystyle \mathrm{B} \cap \mathrm{C} = \{11,13\} \)
    (iii)
    \(\displaystyle \mathrm{A} \cap \mathrm{C} \cap \mathrm{D} = (\mathrm{A} \cap \mathrm{C}) \cap \mathrm{D} = \{11\} \cap \{15,17\} = \phi \)
    (iv)
    \(\displaystyle \mathrm{A} \cap \mathrm{C} = \{11\} \)
    (v)
    \(\displaystyle \mathrm{B} \cap \mathrm{D} = \phi \) (B has no element in common with \(\displaystyle \{15,17\} \))
    (vi)
    \(\displaystyle \mathrm{B} \cup \mathrm{C} = \{7,9,11,13,15\} \), so \(\displaystyle \mathrm{A} \cap (\mathrm{B} \cup \mathrm{C}) = \{7,9,11\} \)
    (vii)
    \(\displaystyle \mathrm{A} \cap \mathrm{D} = \phi \)
    (viii)
    \(\displaystyle \mathrm{B} \cup \mathrm{D} = \{7,9,11,13,15,17\} \), so \(\displaystyle \mathrm{A} \cap (\mathrm{B} \cup \mathrm{D}) = \{7,9,11\} \)
    (ix)
    \(\displaystyle (\mathrm{A} \cap \mathrm{B}) \cap (\mathrm{B} \cup \mathrm{C}) = \{7,9,11\} \cap \{7,9,11,13,15\} = \{7,9,11\} \)
    (x)
    \(\displaystyle \mathrm{A} \cup \mathrm{D} = \{3,5,7,9,11,15,17\} \) and \(\displaystyle \mathrm{B} \cup \mathrm{C} = \{7,9,11,13,15\} \), so
    \[(\mathrm{A} \cup \mathrm{D}) \cap (\mathrm{B} \cup \mathrm{C}) = \{7,9,11,15\} \]
    (i) {$\displaystyle 7,9,11$} (ii) {$\displaystyle 11,13$} (iii) φ (iv) {$\displaystyle 11$} (v) φ (vi) {$\displaystyle 7,9,11$} (vii) φ (viii) {$\displaystyle 7,9,11$} (ix) {$\displaystyle 7,9,11$} (x) {$\displaystyle 7,9,11,15$}

  7. Exercise 7

    If A={x:x\displaystyle \mathrm{A}=\{x: x is a natural number },B={x:x\displaystyle \}, \mathrm{B}=\{x: x is an even natural number }\displaystyle \} C={x:x\displaystyle \mathrm{C}=\{x: x is an odd natural number }\displaystyle \} andD ={x:x\displaystyle =\{x: x is a prime number \}, find
    (i)
    A∩B\displaystyle \mathrm{A} \cap \mathrm{B}
    (ii)
    A∩C\displaystyle \mathrm{A} \cap \mathrm{C}
    (iii)
    A∩D\displaystyle \mathrm{A} \cap \mathrm{D}
    (iv)
    B∩C\displaystyle \mathrm{B} \cap \mathrm{C}
    (v)
    B∩D\displaystyle \mathrm{B} \cap \mathrm{D}
    (vi)
    C∩D\displaystyle \mathrm{C} \cap \mathrm{D}

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    Intersection of described sets. Read each intersection as a condition: an element must satisfy the defining property of both sets. Here
    \[\mathrm{A} = \mathbf{N},\quad \mathrm{B} = \{2,4,6,\ldots\},\quad \mathrm{C} = \{1,3,5,\ldots\},\quad \mathrm{D} = \{2,3,5,7,11,\ldots\}. \]
    (i)
    Every even natural number is a natural number, so \(\displaystyle \mathrm{B} \subset \mathrm{A} \) and
    \[\mathrm{A} \cap \mathrm{B} = \mathrm{B} = \{x : x \text{ is an even natural number}\} \]
    (ii)
    Similarly \(\displaystyle \mathrm{C} \subset \mathrm{A} \), so \(\displaystyle \mathrm{A} \cap \mathrm{C} = \mathrm{C} = \{x : x \text{ is an odd natural number}\} \)
    (iii)
    Every prime number is a natural number, so \(\displaystyle \mathrm{A} \cap \mathrm{D} = \mathrm{D} = \{x : x \text{ is a prime number}\} \)
    (iv)
    No natural number is both even and odd, so
    \[\mathrm{B} \cap \mathrm{C} = \phi \]
    (v)
    A number in \(\displaystyle \mathrm{B} \cap \mathrm{D} \) is an even prime. An even number greater than \(\displaystyle 2 \) has \(\displaystyle 2 \) as a divisor besides \(\displaystyle 1 \) and itself, so it is not prime. Hence
    \[\mathrm{B} \cap \mathrm{D} = \{2\} \]
    (vi)
    A number in \(\displaystyle \mathrm{C} \cap \mathrm{D} \) is an odd prime, i.e. every prime except \(\displaystyle 2 \):
    \[\mathrm{C} \cap \mathrm{D} = \{x : x \text{ is an odd prime number}\} = \mathrm{D} - \{2\} \]
    (i) B (ii) C (iii) D (iv) φ (v) {$\displaystyle 2$} (vi) the set of odd prime numbers

  8. Exercise 8

    Which of the following pairs of sets are disjoint
    (i)
    {1,2,3,4}\displaystyle \{1,2,3,4\} and {x:x\displaystyle \{x: x is a natural number and 4≤x≤6}\displaystyle 4 \leq x \leq 6\}
    (ii)
    {a,e,i,o,u}\displaystyle \{a, e, i, o, u\} and {c,d,e,f}\displaystyle \{c, d, e, f\}
    (iii)
    {x:x\displaystyle \{x: x is an even integer }\displaystyle \} and {x:x\displaystyle \{x: x is an odd integer }\displaystyle \}

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    NCERT’s answer
    (iii)
    Disjoint test. Two sets are disjoint when they have no element in common, i.e. when \(\displaystyle \mathrm{A} \cap \mathrm{B} = \phi \). So form the intersection in each case.
    (i)
    \(\displaystyle \{1,2,3,4\} \) and \(\displaystyle \{x : x \in \mathbf{N},\ 4 \leq x \leq 6\} = \{4,5,6\} \). The element \(\displaystyle 4 \) lies in both, so the intersection is \(\displaystyle \{4\} \neq \phi \) — not disjoint.
    (ii)
    \(\displaystyle \{a,e,i,o,u\} \) and \(\displaystyle \{c,d,e,f\} \). The letter \(\displaystyle e \) lies in both, so the intersection is \(\displaystyle \{e\} \neq \phi \) — not disjoint.
    (iii)
    \(\displaystyle \{x : x \text{ is an even integer}\} \) and \(\displaystyle \{x : x \text{ is an odd integer}\} \). An integer is either even or odd and never both, so the intersection is \(\displaystyle \phi \) — disjoint.
    Only the pair in (iii) is disjoint; (i) and (ii) are not.

  9. Exercise 9

    If A={3,6,9,12,15,18,21},B={4,8,12,16,20}\displaystyle \mathrm{A}=\{3,6,9,12,15,18,21\}, \mathrm{B}=\{4,8,12,16,20\}, C={2,4,6,8,10,12,14,16},D={5,10,15,20}\displaystyle \mathrm{C}=\{2,4,6,8,10,12,14,16\}, \mathrm{D}=\{5,10,15,20\}; find
    (i)
    A−B\displaystyle \mathrm{A}-\mathrm{B}
    (ii)
    A - C
    (iii)
    A - D
    (iv)
    B - A
    (v)
    C - A
    (vi)
    D - A
    (vii)
    B - C
    (viii)
    B−D\displaystyle \mathrm{B}-\mathrm{D}
    (ix)
    C−B\displaystyle \mathrm{C}-\mathrm{B}
    (x)
    D−B\displaystyle \mathrm{D}-\mathrm{B}
    (xi)
    C - D
    (xii)
    D - C

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    Difference of sets. \(\displaystyle \mathrm{A} - \mathrm{B} \) consists of those elements of \(\displaystyle \mathrm{A} \) that do not belong to \(\displaystyle \mathrm{B} \); start from the first set and strike out whatever the second set also holds. Here
    \[\mathrm{A} = \{3,6,9,12,15,18,21\},\ \mathrm{B} = \{4,8,12,16,20\},\ \mathrm{C} = \{2,4,6,8,10,12,14,16\},\ \mathrm{D} = \{5,10,15,20\}. \]
    (i)
    \(\displaystyle \mathrm{A} - \mathrm{B} \): strike \(\displaystyle 12 \) from A \(\displaystyle \Rightarrow \{3,6,9,15,18,21\} \)
    (ii)
    \(\displaystyle \mathrm{A} - \mathrm{C} \): strike \(\displaystyle 6, 12 \) \(\displaystyle \Rightarrow \{3,9,15,18,21\} \)
    (iii)
    \(\displaystyle \mathrm{A} - \mathrm{D} \): strike \(\displaystyle 15 \) \(\displaystyle \Rightarrow \{3,6,9,12,18,21\} \)
    (iv)
    \(\displaystyle \mathrm{B} - \mathrm{A} \): strike \(\displaystyle 12 \) from B \(\displaystyle \Rightarrow \{4,8,16,20\} \)
    (v)
    \(\displaystyle \mathrm{C} - \mathrm{A} \): strike \(\displaystyle 6, 12 \) from C \(\displaystyle \Rightarrow \{2,4,8,10,14,16\} \)
    (vi)
    \(\displaystyle \mathrm{D} - \mathrm{A} \): strike \(\displaystyle 15 \) from D \(\displaystyle \Rightarrow \{5,10,20\} \)
    (vii)
    \(\displaystyle \mathrm{B} - \mathrm{C} \): C holds \(\displaystyle 4,8,12,16 \) \(\displaystyle \Rightarrow \{20\} \)
    (viii)
    \(\displaystyle \mathrm{B} - \mathrm{D} \): strike \(\displaystyle 20 \) \(\displaystyle \Rightarrow \{4,8,12,16\} \)
    (ix)
    \(\displaystyle \mathrm{C} - \mathrm{B} \): strike \(\displaystyle 4,8,12,16 \) from C \(\displaystyle \Rightarrow \{2,6,10,14\} \)
    (x)
    \(\displaystyle \mathrm{D} - \mathrm{B} \): strike \(\displaystyle 20 \) \(\displaystyle \Rightarrow \{5,10,15\} \)
    (xi)
    \(\displaystyle \mathrm{C} - \mathrm{D} \): strike \(\displaystyle 10 \) \(\displaystyle \Rightarrow \{2,4,6,8,12,14,16\} \)
    (xii)
    \(\displaystyle \mathrm{D} - \mathrm{C} \): strike \(\displaystyle 10 \) from D \(\displaystyle \Rightarrow \{5,15,20\} \)
    (i) {$\displaystyle 3,6,9,15,18,21$} (ii) {$\displaystyle 3,9,15,18,21$} (iii) {$\displaystyle 3,6,9,12,18,21$} (iv) {$\displaystyle 4,8,16,20$} (v) {$\displaystyle 2,4,8,10,14,16$} (vi) {$\displaystyle 5,10,20$} (vii) {$\displaystyle 20$} (viii) {$\displaystyle 4,8,12,16$} (ix) {$\displaystyle 2,6,10,14$} (x) {$\displaystyle 5,10,15$} (xi) {$\displaystyle 2,4,6,8,12,14,16$} (xii) {$\displaystyle 5,15,20$}

  10. Exercise 10

    If X={a,b,c,d}\displaystyle \mathrm{X}=\{a, b, c, d\} and Y={f,b,d,g}\displaystyle \mathrm{Y}=\{f, b, d, g\}, find
    (i)
    X−Y\displaystyle \mathrm{X}-\mathrm{Y}
    (ii)
    Y−X\displaystyle \mathrm{Y}-\mathrm{X}
    (iii)
    X∩Y\displaystyle \mathrm{X} \cap \mathrm{Y}

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    Difference and intersection. \(\displaystyle \mathrm{X} - \mathrm{Y} \) keeps the elements of \(\displaystyle \mathrm{X} \) that are not in \(\displaystyle \mathrm{Y} \); \(\displaystyle \mathrm{X} \cap \mathrm{Y} \) keeps the ones that are in both. With \(\displaystyle \mathrm{X} = \{a,b,c,d\} \) and \(\displaystyle \mathrm{Y} = \{f,b,d,g\} \), the common letters are \(\displaystyle b \) and \(\displaystyle d \).
    (i)
    Remove \(\displaystyle b, d \) from \(\displaystyle \mathrm{X} \):
    \[\mathrm{X} - \mathrm{Y} = \{a,c\} \]
    (ii)
    Remove \(\displaystyle b, d \) from \(\displaystyle \mathrm{Y} \):
    \[\mathrm{Y} - \mathrm{X} = \{f,g\} \]
    (iii)
    The letters lying in both sets:
    \[\mathrm{X} \cap \mathrm{Y} = \{b,d\} \]
    (i) X − Y = {a, c} (ii) Y − X = {f, g} (iii) X ∩ Y = {b, d}

  11. Exercise 11

    If R\displaystyle \mathbf{R} is the set of real numbers and Q\displaystyle \mathbf{Q} is the set of rational numbers, then what is R - Q?

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    Difference of sets. \(\displaystyle \mathbf{R} - \mathbf{Q} \) is the set of real numbers that are not rational.By definition a real number is called irrational exactly when it cannot be written as \(\displaystyle \dfrac{p}{q} \) with \(\displaystyle p, q \) integers and \(\displaystyle q \neq 0 \) — that is, exactly when it is real but not rational. So removing \(\displaystyle \mathbf{Q} \) from \(\displaystyle \mathbf{R} \) leaves precisely the irrational numbers, such as \(\displaystyle \sqrt{2},\ \sqrt{3},\ \pi \).R − Q = the set of all irrational numbers

  12. Exercise 12

    State whether each of the following statement is true or false. Justify your answer.
    (i)
    {2,3,4,5}\displaystyle \{2,3,4,5\} and {3,6}\displaystyle \{3,6\} are disjoint sets.
    (ii)
    {a,e,i,o,u}\displaystyle \{a, e, i, o, u\} and {a,b,c,d}\displaystyle \{a, b, c, d\} are disjoint sets.
    (iii)
    {2,6,10,14}\displaystyle \{2,6,10,14\} and {3,7,11,15}\displaystyle \{3,7,11,15\} are disjoint sets.
    (iv)
    {2,6,10}\displaystyle \{2,6,10\} and {3,7,11}\displaystyle \{3,7,11\} are disjoint sets.

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    Disjoint test. Two sets are disjoint when they share no element, i.e. when their intersection is \(\displaystyle \phi \). Test each statement by forming the intersection.
    (i)
    \(\displaystyle \{2,3,4,5\} \cap \{3,6\} = \{3\} \neq \phi \), since \(\displaystyle 3 \) belongs to both. False.
    (ii)
    \(\displaystyle \{a,e,i,o,u\} \cap \{a,b,c,d\} = \{a\} \neq \phi \), since \(\displaystyle a \) belongs to both. False.
    (iii)
    \(\displaystyle \{2,6,10,14\} \cap \{3,7,11,15\} = \phi \) — the first set has only even numbers, the second only odd ones, so nothing is common. True.
    (iv)
    \(\displaystyle \{2,6,10\} \cap \{3,7,11\} = \phi \), again even against odd. True.
    (i) False (ii) False (iii) True (iv) True