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NCERT Solutions · Class 11 Mathematics Relations and Functions

36 exercises · 36 still being checked

EXERCISE 2.1 1–10 (part 1 of 4)

  1. Exercise 1

    If \(\displaystyle \left(\frac{x}{3}+1, y-\frac{2}{3}\right)=\left(\frac{5}{3}, \frac{1}{3}\right)\), find the values of \(\displaystyle x\) and \(\displaystyle y\).

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    NCERT’s answer
    $\displaystyle x=2$ and $\displaystyle y=1$
    Equality of ordered pairs. Two ordered pairs are equal exactly when their first components are equal and their second components are equal: \[(a,b)=(c,d) \iff a=c \text{ and } b=d\]Comparing first components: \[\frac{x}{3}+1=\frac{5}{3} \;\Rightarrow\; \frac{x}{3}=\frac{5}{3}-1=\frac{2}{3} \;\Rightarrow\; x=2\]Comparing second components: \[y-\frac{2}{3}=\frac{1}{3} \;\Rightarrow\; y=\frac{1}{3}+\frac{2}{3}=1\]Check: \(\displaystyle \left(\frac{2}{3}+1,\;1-\frac{2}{3}\right)=\left(\frac{5}{3},\frac{1}{3}\right)\). ✓\(\displaystyle x = 2\) and \(\displaystyle y = 1\).

  2. Exercise 2

    If the set A has $\displaystyle 3$ elements and the set \(\displaystyle \mathrm{B}=\{3,4,5\}\), then find the number of elements in (A×B).

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    NCERT’s answer
    The number of elements in $\displaystyle \mathrm{A} \times \mathrm{B}$ is $\displaystyle 9$ .
    Counting rule for a Cartesian product. If \(\displaystyle n(\mathrm{A})=p\) and \(\displaystyle n(\mathrm{B})=q\), then every element of \(\displaystyle \mathrm{A}\) can be paired with every element of \(\displaystyle \mathrm{B}\), so \[n(\mathrm{A}\times\mathrm{B})=n(\mathrm{A})\times n(\mathrm{B})=p\,q\]Here \(\displaystyle n(\mathrm{A})=3\) and \(\displaystyle \mathrm{B}=\{3,4,5\}\), so \(\displaystyle n(\mathrm{B})=3\). Hence \[n(\mathrm{A}\times\mathrm{B})=3\times 3=9\]\(\displaystyle \mathrm{A}\times\mathrm{B}\) has $\displaystyle 9$ elements.

  3. Exercise 3

    If \(\displaystyle \mathrm{G}=\{7,8\}\) and \(\displaystyle \mathrm{H}=\{5,4,2\}\), find \(\displaystyle \mathrm{G} \times \mathrm{H}\) and \(\displaystyle \mathrm{H} \times \mathrm{G}\).

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    NCERT’s answer
    $\displaystyle \mathrm{G} \times \mathrm{H}=\{(7,5),(7,4),(7,2),(8,5),(8,4),(8,2)\}$ $\displaystyle \mathrm{H} \times \mathrm{G}=\{(5,7),(5,8),(4,7),(4,8),(2,7),(2,8)\}$
    Cartesian product. \(\displaystyle \mathrm{G}\times\mathrm{H}=\{(g,h): g\in\mathrm{G},\, h\in\mathrm{H}\}\) — every element of the first set paired, in that order, with every element of the second.With \(\displaystyle \mathrm{G}=\{7,8\}\) and \(\displaystyle \mathrm{H}=\{5,4,2\}\), take each element of \(\displaystyle \mathrm{G}\) in turn: \[\mathrm{G}\times\mathrm{H}=\{(7,5),(7,4),(7,2),(8,5),(8,4),(8,2)\}\]Now reverse the roles, so the first component comes from \(\displaystyle \mathrm{H}\): \[\mathrm{H}\times\mathrm{G}=\{(5,7),(5,8),(4,7),(4,8),(2,7),(2,8)\}\]Both have \(\displaystyle 2\times 3=6\) elements, but the pairs are different — the Cartesian product is not commutative.\(\displaystyle \mathrm{G}\times\mathrm{H}=\{(7,5),(7,4),(7,2),(8,5),(8,4),(8,2)\}\) and \(\displaystyle \mathrm{H}\times\mathrm{G}=\{(5,7),(5,8),(4,7),(4,8),(2,7),(2,8)\}\).

  4. Exercise 4

    State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly.
    (i)
    If \(\displaystyle \mathrm{P}=\{m, n\}\) and \(\displaystyle \mathrm{Q}=\{n, m\}\), then \(\displaystyle \mathrm{P} \times \mathrm{Q}=\{(m, n),(n, m)\}\).
    (ii)
    If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs \(\displaystyle (x, y)\) such that \(\displaystyle x \in \mathrm{~A}\) and \(\displaystyle y \in \mathrm{~B}\).
    (iii)
    If \(\displaystyle \mathrm{A}=\{1,2\}, \mathrm{B}=\{3,4\}\), then \(\displaystyle \mathrm{A} \times(\mathrm{B} \cap \phi)=\phi\).

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    NCERT’s answer
    (i)
    False $\displaystyle \mathrm{P} \times \mathrm{Q}=\{(m, n),(m, m),(n, n),(n, m)\}$ (ii) True (iii) True
    Test each claim against the definition \(\displaystyle \mathrm{A}\times\mathrm{B}=\{(x,y): x\in\mathrm{A},\, y\in\mathrm{B}\}\).
    (i)
    \(\displaystyle \mathrm{P}=\{m,n\}\) and \(\displaystyle \mathrm{Q}=\{n,m\}\) are the same two-element set, since order of listing does not matter in a set. So \(\displaystyle n(\mathrm{P}\times\mathrm{Q})=2\times 2=4\), not \(\displaystyle 2\). Pairing every element of \(\displaystyle \mathrm{P}\) with every element of \(\displaystyle \mathrm{Q}\):
    \[\mathrm{P}\times\mathrm{Q}=\{(m,m),(m,n),(n,m),(n,n)\}\]
    The given set omits \(\displaystyle (m,m)\) and \(\displaystyle (n,n)\). False.
    Correct statement: if \(\displaystyle \mathrm{P}=\{m,n\}\) and \(\displaystyle \mathrm{Q}=\{n,m\}\), then \(\displaystyle \mathrm{P}\times\mathrm{Q}=\{(m,m),(m,n),(n,m),(n,n)\}\).
    (ii)
    This is exactly the definition. If \(\displaystyle \mathrm{A}\neq\phi\) and \(\displaystyle \mathrm{B}\neq\phi\), pick \(\displaystyle a\in\mathrm{A}\) and \(\displaystyle b\in\mathrm{B}\); then \(\displaystyle (a,b)\in\mathrm{A}\times\mathrm{B}\), so the product is non-empty, and every member of it is an ordered pair \(\displaystyle (x,y)\) with \(\displaystyle x\in\mathrm{A}\), \(\displaystyle y\in\mathrm{B}\). True.
    (iii)
    \(\displaystyle \mathrm{B}\cap\phi=\phi\) for every set \(\displaystyle \mathrm{B}\), and a Cartesian product with an empty factor has no pairs to form, so \(\displaystyle \mathrm{A}\times\phi=\phi\). Hence \(\displaystyle \mathrm{A}\times(\mathrm{B}\cap\phi)=\phi\). True.
    (i) False — \(\displaystyle \mathrm{P}\times\mathrm{Q}=\{(m,m),(m,n),(n,m),(n,n)\}\); (ii) True; (iii) True.

  5. Exercise 5

    If \(\displaystyle \mathrm{A}=\{-1,1\}\), find \(\displaystyle \mathrm{A} \times \mathrm{A} \times \mathrm{A}\).

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    NCERT’s answer
    $\displaystyle \mathrm{A} \times \mathrm{A}=\{(-1,-1),(-1,1),(1,-1),(1,1)\}$ $\displaystyle \mathrm{A} \times \mathrm{A} \times \mathrm{A}=\{(-1,-1,-1),(-1,-1,1),(-1,1,-1),(-1,1,1),(1,-1,-1),(1,-1,1)$, ($\displaystyle 1$, $\displaystyle 1$, -$\displaystyle 1$), ($\displaystyle 1$, $\displaystyle 1$, $\displaystyle 1$)\}
    Triple Cartesian product. \(\displaystyle \mathrm{A}\times\mathrm{A}\times\mathrm{A}=\{(a,b,c): a,b,c\in\mathrm{A}\}\) — every ordered triple whose three entries are drawn from \(\displaystyle \mathrm{A}\).With \(\displaystyle \mathrm{A}=\{-1,1\}\) each of the three slots can be filled in \(\displaystyle 2\) ways, so there are \(\displaystyle 2^{3}=8\) triples. Listing them systematically: \[\mathrm{A}\times\mathrm{A}\times\mathrm{A}=\{(-1,-1,-1),\,(-1,-1,1),\,(-1,1,-1),\,(-1,1,1),\,(1,-1,-1),\,(1,-1,1),\,(1,1,-1),\,(1,1,1)\}\]\(\displaystyle \mathrm{A}\times\mathrm{A}\times\mathrm{A}=\{(-1,-1,-1),(-1,-1,1),(-1,1,-1),(-1,1,1),(1,-1,-1),(1,-1,1),(1,1,-1),(1,1,1)\}\), which has $\displaystyle 8$ elements.

  6. Exercise 6

    If \(\displaystyle \mathrm{A} \times \mathrm{B}=\{(a, x),(a, y),(b, x),(b, y)\}\). Find A and B.

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    NCERT’s answer
    $\displaystyle \mathrm{A}=\{a, b\}, \mathrm{B}=\{x, y\}$
    Reading the factors off the product. In \(\displaystyle \mathrm{A}\times\mathrm{B}\) every first component comes from \(\displaystyle \mathrm{A}\) and every second component comes from \(\displaystyle \mathrm{B}\). So \(\displaystyle \mathrm{A}\) is the set of all first components and \(\displaystyle \mathrm{B}\) is the set of all second components.The given pairs are \(\displaystyle (a,x),(a,y),(b,x),(b,y)\).First components: \(\displaystyle a, a, b, b\), so \(\displaystyle \mathrm{A}=\{a,b\}\). Second components: \(\displaystyle x, y, x, y\), so \(\displaystyle \mathrm{B}=\{x,y\}\).Check: \(\displaystyle \{a,b\}\times\{x,y\}=\{(a,x),(a,y),(b,x),(b,y)\}\), which is the given set, and \(\displaystyle n=2\times 2=4\). ✓\(\displaystyle \mathrm{A}=\{a,b\}\) and \(\displaystyle \mathrm{B}=\{x,y\}\).

  7. Exercise 7

    Let \(\displaystyle \mathrm{A}=\{1,2\}, \mathrm{B}=\{1,2,3,4\}, \mathrm{C}=\{5,6\}\) and \(\displaystyle \mathrm{D}=\{5,6,7,8\}\). Verify that
    (i)
    \(\displaystyle \mathrm{A} \times(\mathrm{B} \cap \mathrm{C})=(\mathrm{A} \times \mathrm{B}) \cap(\mathrm{A} \times \mathrm{C})\).
    (ii)
    \(\displaystyle \mathrm{A} \times \mathrm{C}\) is a subset of \(\displaystyle \mathrm{B} \times \mathrm{D}\).

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    Verify by writing out both sides. Here \(\displaystyle \mathrm{A}=\{1,2\}\), \(\displaystyle \mathrm{B}=\{1,2,3,4\}\), \(\displaystyle \mathrm{C}=\{5,6\}\), \(\displaystyle \mathrm{D}=\{5,6,7,8\}\).(i) Left side: \(\displaystyle \mathrm{B}\) and \(\displaystyle \mathrm{C}\) have no element in common, so \(\displaystyle \mathrm{B}\cap\mathrm{C}=\phi\), and a product with an empty factor is empty: \[\mathrm{A}\times(\mathrm{B}\cap\mathrm{C})=\mathrm{A}\times\phi=\phi\]Right side: \[\mathrm{A}\times\mathrm{B}=\{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)\}\] \[\mathrm{A}\times\mathrm{C}=\{(1,5),(1,6),(2,5),(2,6)\}\] No pair appears in both lists (second components are \(\displaystyle 1,2,3,4\) on one side and \(\displaystyle 5,6\) on the other), so \[(\mathrm{A}\times\mathrm{B})\cap(\mathrm{A}\times\mathrm{C})=\phi\]Both sides equal \(\displaystyle \phi\), so the identity holds here.(ii) \(\displaystyle \mathrm{A}\times\mathrm{C}=\{(1,5),(1,6),(2,5),(2,6)\}\). Since \(\displaystyle \mathrm{A}=\{1,2\}\subset\mathrm{B}\) and \(\displaystyle \mathrm{C}=\{5,6\}\subset\mathrm{D}\), every pair \(\displaystyle (x,y)\) with \(\displaystyle x\in\mathrm{A}\), \(\displaystyle y\in\mathrm{C}\) also has \(\displaystyle x\in\mathrm{B}\), \(\displaystyle y\in\mathrm{D}\), hence lies in \(\displaystyle \mathrm{B}\times\mathrm{D}\). Indeed all four pairs \(\displaystyle (1,5),(1,6),(2,5),(2,6)\) occur among the \(\displaystyle 4\times 4=16\) elements of \(\displaystyle \mathrm{B}\times\mathrm{D}\).(i) Both sides equal \(\displaystyle \phi\), so \(\displaystyle \mathrm{A}\times(\mathrm{B}\cap\mathrm{C})=(\mathrm{A}\times\mathrm{B})\cap(\mathrm{A}\times\mathrm{C})\); (ii) \(\displaystyle \mathrm{A}\times\mathrm{C}\subseteq\mathrm{B}\times\mathrm{D}\) — verified.

  8. Exercise 8

    Let \(\displaystyle \mathrm{A}=\{1,2\}\) and \(\displaystyle \mathrm{B}=\{3,4\}\). Write \(\displaystyle \mathrm{A} \times \mathrm{B}\). How many subsets will \(\displaystyle \mathrm{A} \times \mathrm{B}\) have? List them.

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    NCERT’s answer
    $\displaystyle \mathrm{A} \times \mathrm{B}=\{(1,3),(1,4),(2,3),(2,4)\}$ $\displaystyle \mathrm{A} \times \mathrm{B}$ will have $\displaystyle 2^{4}=16$ subsets.
    Cartesian product, then the subset count. With \(\displaystyle \mathrm{A}=\{1,2\}\) and \(\displaystyle \mathrm{B}=\{3,4\}\): \[\mathrm{A}\times\mathrm{B}=\{(1,3),(1,4),(2,3),(2,4)\}\]A set with \(\displaystyle n\) elements has \(\displaystyle 2^{n}\) subsets, because each element is independently either in or out. Here \(\displaystyle n(\mathrm{A}\times\mathrm{B})=4\), so the number of subsets is \[2^{4}=16\]Listing them by size:
    Size $\displaystyle 0$: \(\displaystyle \phi\)
    Size $\displaystyle 1$: \(\displaystyle \{(1,3)\}\), \(\displaystyle \{(1,4)\}\), \(\displaystyle \{(2,3)\}\), \(\displaystyle \{(2,4)\}\)
    Size $\displaystyle 2$: \(\displaystyle \{(1,3),(1,4)\}\), \(\displaystyle \{(1,3),(2,3)\}\), \(\displaystyle \{(1,3),(2,4)\}\), \(\displaystyle \{(1,4),(2,3)\}\), \(\displaystyle \{(1,4),(2,4)\}\), \(\displaystyle \{(2,3),(2,4)\}\)
    Size $\displaystyle 3$: \(\displaystyle \{(1,3),(1,4),(2,3)\}\), \(\displaystyle \{(1,3),(1,4),(2,4)\}\), \(\displaystyle \{(1,3),(2,3),(2,4)\}\), \(\displaystyle \{(1,4),(2,3),(2,4)\}\)
    Size $\displaystyle 4$: \(\displaystyle \{(1,3),(1,4),(2,3),(2,4)\}\)
    That is \(\displaystyle 1+4+6+4+1=16\) subsets. ✓\(\displaystyle \mathrm{A}\times\mathrm{B}=\{(1,3),(1,4),(2,3),(2,4)\}\), and it has \(\displaystyle 2^{4}=16\) subsets, listed above.

  9. Exercise 9

    Let A and B be two sets such that \(\displaystyle n(\mathrm{~A})=3\) and \(\displaystyle n(\mathrm{~B})=2\). If ( \(\displaystyle x, 1\) ), ( \(\displaystyle y, 2\) ), ( \(\displaystyle z, 1\) ) are in \(\displaystyle \mathrm{A} \times \mathrm{B}\), find A and B , where \(\displaystyle x, y\) and \(\displaystyle z\) are distinct elements.

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    NCERT’s answer
    $\displaystyle \mathrm{A}=\{x, y, z\}$ and $\displaystyle \mathrm{B}=\{1,2\}$
    First components lie in A, second components lie in B. Since \(\displaystyle (x,1),(y,2),(z,1)\in\mathrm{A}\times\mathrm{B}\): \[x,\,y,\,z\in\mathrm{A} \qquad\text{and}\qquad 1,\,2\in\mathrm{B}\]We are told \(\displaystyle x, y, z\) are distinct, so \(\displaystyle \{x,y,z\}\subseteq\mathrm{A}\) already supplies three different elements. But \(\displaystyle n(\mathrm{A})=3\), so \(\displaystyle \mathrm{A}\) can hold nothing more: \[\mathrm{A}=\{x,y,z\}\]Similarly \(\displaystyle 1\) and \(\displaystyle 2\) are two distinct elements of \(\displaystyle \mathrm{B}\) and \(\displaystyle n(\mathrm{B})=2\), so \[\mathrm{B}=\{1,2\}\]\(\displaystyle \mathrm{A}=\{x,y,z\}\) and \(\displaystyle \mathrm{B}=\{1,2\}\).

  10. Exercise 10

    The Cartesian product A × A has $\displaystyle 9$ elements among which are found \(\displaystyle (-1,0)\) and $\displaystyle (0,1)$. Find the set A and the remaining elements of A × A.

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    NCERT’s answer
    $\displaystyle \mathrm{A}=\{-1,0,1\}$, remaining elements of $\displaystyle \mathrm{A} \times \mathrm{A}$ are $\displaystyle (-1, -1)$, $\displaystyle (-1, 1)$, $\displaystyle (0, -1)$, $\displaystyle (0, 0)$, $\displaystyle (1, -1)$, $\displaystyle (1,0),(1,1)$
    Use the count to fix \(\displaystyle n(\mathrm{A})\), then the given pairs to name the elements.Since \(\displaystyle n(\mathrm{A}\times\mathrm{A})=n(\mathrm{A})\times n(\mathrm{A})=[n(\mathrm{A})]^{2}=9\), we get \(\displaystyle n(\mathrm{A})=3\).In \(\displaystyle \mathrm{A}\times\mathrm{A}\) both components come from \(\displaystyle \mathrm{A}\). From \(\displaystyle (-1,0)\) we get \(\displaystyle -1\in\mathrm{A}\) and \(\displaystyle 0\in\mathrm{A}\); from \(\displaystyle (0,1)\) we get \(\displaystyle 0\in\mathrm{A}\) and \(\displaystyle 1\in\mathrm{A}\). That is three distinct elements \(\displaystyle -1, 0, 1\), and \(\displaystyle \mathrm{A}\) has room for exactly three, so \[\mathrm{A}=\{-1,0,1\}\]Now list all nine pairs: \[\mathrm{A}\times\mathrm{A}=\{(-1,-1),(-1,0),(-1,1),(0,-1),(0,0),(0,1),(1,-1),(1,0),(1,1)\}\]Deleting the two already given, \(\displaystyle (-1,0)\) and \(\displaystyle (0,1)\), leaves seven.\(\displaystyle \mathrm{A}=\{-1,0,1\}\); the remaining elements of \(\displaystyle \mathrm{A}\times\mathrm{A}\) are \(\displaystyle (-1,-1), (-1,1), (0,-1), (0,0), (1,-1), (1,0), (1,1)\).