SolveItClass 11 · NCERT

NCERT Solutions · Class 11 Mathematics Relations and Functions

36 questions · 36 still being checked

EXERCISE 2.2 1–9 (part 2 of 4)

  1. Exercise 1

    Let A={1,2,3,…,14}\displaystyle \mathrm{A}=\{1,2,3, \ldots, 14\}. Define a relation R from A to A by R={(x,y):3x−y=0\displaystyle \mathrm{R}=\{(x, y): 3 x-y=0, where x,y∈ A}\displaystyle x, y \in \mathrm{~A}\}. Write down its domain, codomain and range.

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    NCERT’s answer
    $\displaystyle \mathrm{R}=\{(1,3),(2,6),(3,9),(4,12)\}$ Domain of $\displaystyle \mathrm{R}=\{1,2,3,4\}$ Range of $\displaystyle \mathrm{R}=\{3,6,9,12\}$ Co domain of $\displaystyle \mathrm{R}=\{1,2, \ldots, 14\}$
    Roster form, then read off the three sets. For a relation \(\displaystyle \mathrm{R}\) from \(\displaystyle \mathrm{A}\) to \(\displaystyle \mathrm{A}\): the domain is the set of first components, the codomain is the whole second set as declared (here \(\displaystyle \mathrm{A}\)), and the range is the set of second components that actually occur.The condition \(\displaystyle 3x-y=0\) means \(\displaystyle y=3x\). Run \(\displaystyle x\) through \(\displaystyle \mathrm{A}=\{1,2,\ldots,14\}\) and keep only those \(\displaystyle x\) for which \(\displaystyle 3x\) is also in \(\displaystyle \mathrm{A}\), i.e. \(\displaystyle 3x\le 14\), i.e. \(\displaystyle x\le 4\):\[x=1\Rightarrow y=3,\quad x=2\Rightarrow y=6,\quad x=3\Rightarrow y=9,\quad x=4\Rightarrow y=12\](\(\displaystyle x=5\) would give \(\displaystyle y=15\notin\mathrm{A}\).) Hence \[\mathrm{R}=\{(1,3),(2,6),(3,9),(4,12)\}\]Domain \(\displaystyle =\{1,2,3,4\}\); codomain \(\displaystyle =\{1,2,3,\ldots,14\}\); range \(\displaystyle =\{3,6,9,12\}\).

  2. Exercise 2

    Define a relation R on the set N of natural numbers by R={(x,y):y=x+5\displaystyle \mathrm{R}=\{(x, y): y=x+5, x\displaystyle x is a natural number less than 4;x,y∈N}\displaystyle 4 ; x, y \in \mathbf{N}\}. Depict this relationship using roster form. Write down the domain and the range.

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    NCERT’s answer
    $\displaystyle \mathrm{R}=\{(1,6),(2,7),(3,8)\}$ Domain of $\displaystyle \mathrm{R}=\{1,2,3\}$ Range of $\displaystyle \mathrm{R}=\{6,7,8\}$
    Roster form. List every ordered pair the rule produces.The rule is \(\displaystyle y=x+5\) with \(\displaystyle x\) a natural number less than \(\displaystyle 4\), so \(\displaystyle x\in\{1,2,3\}\):\[x=1\Rightarrow y=6,\qquad x=2\Rightarrow y=7,\qquad x=3\Rightarrow y=8\]\[\mathrm{R}=\{(1,6),(2,7),(3,8)\}\]The domain is the set of first components and the range is the set of second components.\(\displaystyle \mathrm{R}=\{(1,6),(2,7),(3,8)\}\); domain \(\displaystyle =\{1,2,3\}\); range \(\displaystyle =\{6,7,8\}\).

  3. Exercise 3

    A={1,2,3,5}\displaystyle \mathrm{A}=\{1,2,3,5\} and B={4,6,9}\displaystyle \mathrm{B}=\{4,6,9\}. Define a relation R from A to B by R={(x,y)\displaystyle \mathrm{R}=\{(x, y) : the difference between x\displaystyle x and y\displaystyle y is odd; x∈ A,y∈ B}\displaystyle x \in \mathrm{~A}, y \in \mathrm{~B}\}. Write R in roster form.

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    NCERT’s answer
    $\displaystyle \mathrm{R}=\{(1,4),(1,6),(2,9),(3,4),(3,6),(5,4),(5,6)\}$
    Roster form. Test every pair \(\displaystyle (x,y)\) with \(\displaystyle x\in\mathrm{A}=\{1,2,3,5\}\) and \(\displaystyle y\in\mathrm{B}=\{4,6,9\}\), keeping those for which the difference is odd. A difference is odd exactly when one of the two numbers is even and the other is odd.In \(\displaystyle \mathrm{B}\), the numbers \(\displaystyle 4\) and \(\displaystyle 6\) are even and \(\displaystyle 9\) is odd. So:
    \(\displaystyle x=1\) (odd): pairs with the even \(\displaystyle 4\) and \(\displaystyle 6\) — \(\displaystyle (1,4),(1,6)\); \(\displaystyle 1-9=-8\) is even, rejected.
    \(\displaystyle x=2\) (even): pairs with the odd \(\displaystyle 9\) — \(\displaystyle (2,9)\); \(\displaystyle 2-4\) and \(\displaystyle 2-6\) are even, rejected.
    \(\displaystyle x=3\) (odd): \(\displaystyle (3,4),(3,6)\).
    \(\displaystyle x=5\) (odd): \(\displaystyle (5,4),(5,6)\).
    \[\mathrm{R}=\{(1,4),(1,6),(2,9),(3,4),(3,6),(5,4),(5,6)\}\]\(\displaystyle \mathrm{R}=\{(1,4),(1,6),(2,9),(3,4),(3,6),(5,4),(5,6)\}\) — $\displaystyle 7$ ordered pairs.

  4. Exercise 4

    The Fig2.7 shows a relationship between the sets P and Q. Write this relation
    (i)
    in set-builder form
    (ii)
    roster form. What is its domain and range?

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    NCERT’s answer
    (i)
    $\displaystyle \mathrm{R}=\{(x, y): y=x-2$ for $\displaystyle x=5,6,7\}$ (ii) $\displaystyle \mathrm{R}=\{(5,3),(6,4),(7,5)\}$. Domain of $\displaystyle \mathrm{R}=\{5,6,7\}$, Range of $\displaystyle \mathrm{R}=\{3,4,5\}$
    Read the arrow diagram, then describe the rule. In Fig $\displaystyle 2.7$ the left oval is \(\displaystyle \mathrm{P}=\{5,6,7\}\) and the right oval is \(\displaystyle \mathrm{Q}=\{3,4,5\}\), with one arrow from each element of \(\displaystyle \mathrm{P}\): \(\displaystyle 5\to 3\), \(\displaystyle 6\to 4\), \(\displaystyle 7\to 5\). Each arrow \(\displaystyle x\to y\) contributes the ordered pair \(\displaystyle (x,y)\).(i) Set-builder form. In every pair the second entry is \(\displaystyle 2\) less than the first, since \(\displaystyle 5-3=6-4=7-5=2\): \[\mathrm{R}=\{(x,y): y=x-2,\; x\in\mathrm{P},\; y\in\mathrm{Q}\}\](ii) Roster form. \[\mathrm{R}=\{(5,3),(6,4),(7,5)\}\]The domain is the set of first components, the range the set of second components.\(\displaystyle \mathrm{R}=\{(x,y): x-y=2,\, x\in\mathrm{P},\, y\in\mathrm{Q}\}=\{(5,3),(6,4),(7,5)\}\); domain \(\displaystyle =\{5,6,7\}\); range \(\displaystyle =\{3,4,5\}\).

  5. Exercise 5

    Let A={1,2,3,4,6}\displaystyle \mathrm{A}=\{1,2,3,4,6\}. Let R be the relation on A defined by {(a,b):a,b∈ A,b\displaystyle \{(a, b): a, b \in \mathrm{~A}, b is exactly divisible by a}\displaystyle a\}.
    (i)
    Write R in roster form
    (ii)
    Find the domain of R
    (iii)
    Find the range of R.

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    NCERT’s answer
    (i)
    $\displaystyle \mathrm{R}=\{(1,1),(1,2),(1,3),(1,4),(1,6),(24),(2,6),(2,2),(4,4),(6,6)$, $\displaystyle (3, 3)$, $\displaystyle (3,6)\}$ (ii) Domain of $\displaystyle \mathrm{R}=\{1,2,3,4,6\}$ (iii) Range of $\displaystyle \mathrm{R}=\{1,2,3,4,6\}$
    Roster form. For each \(\displaystyle a\in\mathrm{A}=\{1,2,3,4,6\}\), keep those \(\displaystyle b\in\mathrm{A}\) that \(\displaystyle a\) divides exactly (i.e. \(\displaystyle b\) is a multiple of \(\displaystyle a\)).
    \(\displaystyle a=1\): \(\displaystyle 1\) divides everything — \(\displaystyle (1,1),(1,2),(1,3),(1,4),(1,6)\)
    \(\displaystyle a=2\): multiples of \(\displaystyle 2\) in \(\displaystyle \mathrm{A}\) are \(\displaystyle 2,4,6\) — \(\displaystyle (2,2),(2,4),(2,6)\)
    \(\displaystyle a=3\): multiples of \(\displaystyle 3\) in \(\displaystyle \mathrm{A}\) are \(\displaystyle 3,6\) — \(\displaystyle (3,3),(3,6)\)
    \(\displaystyle a=4\): only \(\displaystyle 4\) (since \(\displaystyle 8\notin\mathrm{A}\)) — \(\displaystyle (4,4)\)
    \(\displaystyle a=6\): only \(\displaystyle 6\) — \(\displaystyle (6,6)\)
    (i) \[\mathrm{R}=\{(1,1),(1,2),(1,3),(1,4),(1,6),(2,2),(2,4),(2,6),(3,3),(3,6),(4,4),(6,6)\}\] That is \(\displaystyle 5+3+2+1+1=12\) pairs.(ii) Every element of \(\displaystyle \mathrm{A}\) appears as a first component (each divides itself), so the domain is all of \(\displaystyle \mathrm{A}\).(iii) Every element of \(\displaystyle \mathrm{A}\) appears as a second component (again from the pairs \(\displaystyle (a,a)\)), so the range is all of \(\displaystyle \mathrm{A}\).\(\displaystyle \mathrm{R}=\{(1,1),(1,2),(1,3),(1,4),(1,6),(2,2),(2,4),(2,6),(3,3),(3,6),(4,4),(6,6)\}\); domain \(\displaystyle =\{1,2,3,4,6\}\); range \(\displaystyle =\{1,2,3,4,6\}\).

  6. Exercise 6

    Determine the domain and range of the relation R defined by R={(x,x+5):x∈{0,1,2,3,4,5}}\displaystyle \mathrm{R}=\{(x, x+5): x \in\{0,1,2,3,4,5\}\}.

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    NCERT’s answer
    Domain of $\displaystyle \mathrm{R}=\{0,1,2,3,4,5$, Range of $\displaystyle \mathrm{R}=\{5,6,7,8,9,10\}$ Range of $\displaystyle \mathrm{R}=\{5,6,7,8,9,10\}$
    Domain and range from the pairs. The domain of a relation is the set of all first components; the range is the set of all second components.Here the first component is \(\displaystyle x\) and the second is \(\displaystyle x+5\), with \(\displaystyle x\) running over \(\displaystyle \{0,1,2,3,4,5\}\), so \[\mathrm{R}=\{(0,5),(1,6),(2,7),(3,8),(4,9),(5,10)\}\]Reading the two columns:Domain \(\displaystyle =\{0,1,2,3,4,5\}\); range \(\displaystyle =\{5,6,7,8,9,10\}\).

  7. Exercise 7

    Write the relation R={(x,x3):x\displaystyle \mathrm{R}=\left\{\left(x, x^{3}\right): x\right. is a prime number less than 10\displaystyle 10}\displaystyle \} in roster form.

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    NCERT’s answer
    $\displaystyle \mathrm{R}=\{(2,8),(3,27),(5,125),(7,343)\}$
    Roster form. First identify the values of \(\displaystyle x\): the prime numbers less than \(\displaystyle 10\) are \(\displaystyle 2, 3, 5, 7\). (\(\displaystyle 1\) is not prime; \(\displaystyle 4, 6, 8, 9\) are composite.)Now cube each one: \[2^{3}=8,\qquad 3^{3}=27,\qquad 5^{3}=125,\qquad 7^{3}=343\]\(\displaystyle \mathrm{R}=\{(2,8),(3,27),(5,125),(7,343)\}\).

  8. Exercise 8

    Let A={x,y,z}\displaystyle \mathrm{A}=\{x, y, z\} and B={1,2}\displaystyle \mathrm{B}=\{1,2\}. Find the number of relations from A to B .

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    NCERT’s answer
    No. of relations from A into $\displaystyle \mathrm{B}=2^{6}$
    Counting relations. A relation from \(\displaystyle \mathrm{A}\) to \(\displaystyle \mathrm{B}\) is any subset of \(\displaystyle \mathrm{A}\times\mathrm{B}\). So the number of relations is the number of subsets of \(\displaystyle \mathrm{A}\times\mathrm{B}\), namely \(\displaystyle 2^{\,n(\mathrm{A}\times\mathrm{B})}\).Here \(\displaystyle n(\mathrm{A})=3\) and \(\displaystyle n(\mathrm{B})=2\), so \[n(\mathrm{A}\times\mathrm{B})=3\times 2=6\] \[\text{number of relations}=2^{6}=64\](This includes the empty relation \(\displaystyle \phi\) and the whole product \(\displaystyle \mathrm{A}\times\mathrm{B}\).)There are \(\displaystyle 2^{6}=64\) relations from \(\displaystyle \mathrm{A}\) to \(\displaystyle \mathrm{B}\).

  9. Exercise 9

    Let R be the relation on Z\displaystyle \mathbf{Z} defined by R={(a,b):a,b∈Z,a−b\displaystyle \mathrm{R}=\{(a, b): a, b \in \mathbf{Z}, a-b is an integer }\displaystyle \}. Find the domain and range of R.

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    NCERT’s answer
    Domain of $\displaystyle \mathrm{R}=\mathbf{Z}$ Range of $\displaystyle \mathrm{R}=\mathbf{Z}$
    Read the condition against the set it is stated over.Here \(\displaystyle a \) and \(\displaystyle b \) are already restricted to \(\displaystyle \mathbf{Z} \), and the difference of two integers is always an integer. So the condition "\(\displaystyle a-b \) is an integer" is satisfied by every pair of integers — it rules nothing out.Hence \[\mathrm{R}=\{(a,b): a,b \in \mathbf{Z}\}=\mathbf{Z}\times\mathbf{Z}. \]Domain. The set of all first components. Every integer \(\displaystyle a \) appears, since \(\displaystyle (a,a) \in \mathrm{R} \) for each \(\displaystyle a \in \mathbf{Z} \).Range. The set of all second components. Every integer \(\displaystyle b \) appears, since \(\displaystyle (b,b) \in \mathrm{R} \) for each \(\displaystyle b \in \mathbf{Z} \).Domain of R \(\displaystyle = \mathbf{Z} \) and range of R \(\displaystyle = \mathbf{Z} \).