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NCERT Exemplar · Class 12 Mathematics Relations and Functions

62 questions · 62 still being checked

EXERCISE 1.3 11–20 (part 2 of 6)

  1. Exercise 11

    Let the function f:R→R\displaystyle f: \mathbf{R} \rightarrow \mathbf{R} be defined by f(x)=cos⁡x,∀x∈R\displaystyle f(x)=\cos x, \forall x \in \mathbf{R}. Show that f\displaystyle f is neither one-one nor onto.

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    Not one-one: \[f(0) = \cos 0 = 1, \qquad f(2\pi) = \cos 2\pi = 1 \] \[f(0) = f(2\pi) \quad\text{but}\quad 0 \ne 2\pi \] Not onto: \[-1 \le \cos x \le 1 \Rightarrow \text{range of } f = [-1, 1] \] \[2 \in \mathbf{R}, \quad 2 \notin [-1, 1] \Rightarrow \cos x = 2 \text{ has no solution} \] Answer: \(\displaystyle f\) is neither one-one nor onto.

  2. Exercise 12

    Let X={1,2,3}\displaystyle \mathrm{X}=\{1,2,3\} and Y={4,5}\displaystyle \mathrm{Y}=\{4,5\}. Find whether the following subsets of X×Y\displaystyle \mathrm{X} \times \mathrm{Y} are functions from X to Y or not.
    (i)
    f={(1,4),(1,5),(2,4),(3,5)}\displaystyle f=\{(1,4),(1,5),(2,4),(3,5)\}
    (ii)
    g={(1,4),(2,4),(3,4)}\displaystyle g=\{(1,4),(2,4),(3,4)\}
    (iii)
    h={(1,4),(2,5),(3,5)}\displaystyle h=\{(1,4),(2,5),(3,5)\}
    (iv)
    k={(1,4),(2,5)}\displaystyle k=\{(1,4),(2,5)\}.

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    NCERT’s answer
    (i)
    \(\displaystyle f\) is not function
    (ii)
    \(\displaystyle g\) is function
    (iii)
    \(\displaystyle h\) is function
    (iv)
    \(\displaystyle k\) is not function
    A function gives every element of X exactly one image. \[f:\ 1\mapsto 4 \text{ and } 1\mapsto 5 \quad\Rightarrow\quad \text{not a function} \] \[g:\ 1\mapsto 4,\ 2\mapsto 4,\ 3\mapsto 4 \quad\Rightarrow\quad \text{a function} \] \[h:\ 1\mapsto 4,\ 2\mapsto 5,\ 3\mapsto 5 \quad\Rightarrow\quad \text{a function} \] \[\text{dom}\,k = \{1, 2\} \ne \mathrm{X} \quad\Rightarrow\quad \text{not a function} \] Answer: (ii) \(\displaystyle g\) and (iii) \(\displaystyle h\) are functions; (i) \(\displaystyle f\) and (iv) \(\displaystyle k\) are not.

  3. Exercise 13

    If functions f:A→B\displaystyle f: \mathrm{A} \rightarrow \mathrm{B} and g:B→A\displaystyle g: \mathrm{B} \rightarrow \mathrm{A} satisfy gof=IA\displaystyle g o f=\mathrm{I}_{\mathrm{A}}, then show that f\displaystyle f is one-one and g\displaystyle g is onto.

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    \[g\circ f = \mathrm{I}_A \Rightarrow g(f(x)) = x \quad \forall x \in A \] \(\displaystyle f\) is one-one: take \(\displaystyle x_1, x_2 \in A\) with \(\displaystyle f(x_1) = f(x_2)\). \[g(f(x_1)) = g(f(x_2)) \] \[(g\circ f)(x_1) = (g\circ f)(x_2) \] \[x_1 = x_2 \quad (g\circ f = \mathrm{I}_A) \] \(\displaystyle g\) is onto: take any \(\displaystyle y \in A\). Then \(\displaystyle f(y) \in \mathrm{B}\) and \[g(f(y)) = (g\circ f)(y) = y \] So every \(\displaystyle y \in A\) is the image under \(\displaystyle g\) of \(\displaystyle f(y) \in \mathrm{B}\). Answer: \(\displaystyle f\) is one-one and \(\displaystyle g\) is onto.

  4. Exercise 14

    Let f:R→R\displaystyle f: \mathbf{R} \rightarrow \mathbf{R} be the function defined by f(x)=12−cos⁡x∀x∈R\displaystyle f(x)=\frac{1}{2-\cos x} \forall x \in \mathbf{R}.Then, find the range of f\displaystyle f.

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    NCERT’s answer
    \(\displaystyle \left[\frac{1}{3}, 1\right]\)
    \[-1 \le \cos x \le 1 \] \[1 \le 2 - \cos x \le 3 \] \[\frac{1}{3} \le \frac{1}{2 - \cos x} \le 1 \] \[\cos x = 1 \Rightarrow f(x) = 1, \qquad \cos x = -1 \Rightarrow f(x) = \frac{1}{3} \] \(\displaystyle f\) is continuous, so it takes every value between \(\displaystyle \frac13\) and \(\displaystyle 1\). Answer: range of \(\displaystyle f\) is \(\displaystyle \left[\frac{1}{3},\, 1\right]\)

  5. Exercise 15

    Let n\displaystyle n be a fixed positive integer. Define a relation R in Z\displaystyle \mathbf{Z} as follows: ∀a,b∈Z\displaystyle \forall a, b \in \mathbf{Z}, aRb\displaystyle a \mathrm{R} b if and only if a−b\displaystyle a-b is divisible by n\displaystyle n. Show that R is an equivalance relation.

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    Reflexive: for any \(\displaystyle a \in \mathbf{Z}\) \[a - a = 0 = n \cdot 0 \Rightarrow a \mathrm{R} a \] Symmetric: let \(\displaystyle a \mathrm{R} b\), so \(\displaystyle a - b = nk\) for some \(\displaystyle k \in \mathbf{Z}\) \[b - a = n(-k), \quad -k \in \mathbf{Z} \Rightarrow b \mathrm{R} a \] Transitive: let \(\displaystyle a \mathrm{R} b\) and \(\displaystyle b \mathrm{R} c\), so \(\displaystyle a - b = nk\) and \(\displaystyle b - c = nm\) \[a - c = (a - b) + (b - c) = n(k + m), \quad k + m \in \mathbf{Z} \Rightarrow a \mathrm{R} c \] Answer: R is reflexive, symmetric and transitive, so it is an equivalence relation.

  6. Exercise 16

    If A={1,2,3,4}\displaystyle \mathrm{A}=\{1,2,3,4\}, define relations on A which have properties of being:
    (a)
    reflexive, transitive but not symmetric
    (b)
    symmetric but neither reflexive nor transitive
    (c)
    reflexive, symmetric and transitive.

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    (a) Take \[R_1 = \{(1,1),(2,2),(3,3),(4,4),(1,2)\} \] Reflexive: \(\displaystyle (a,a) \in R_1\) for every \(\displaystyle a \in A\). \[(1,2) \in R_1 \quad\text{but}\quad (2,1) \notin R_1 \Rightarrow \text{not symmetric} \] \[(1,1),(1,2) \Rightarrow (1,2) \in R_1; \qquad (1,2),(2,2) \Rightarrow (1,2) \in R_1 \Rightarrow \text{transitive} \] (b) Take \[R_2 = \{(1,2),(2,1)\} \] \[(1,2) \in R_2 \Rightarrow (2,1) \in R_2 \Rightarrow \text{symmetric} \] \[(1,1) \notin R_2 \Rightarrow \text{not reflexive} \] \[(1,2),(2,1) \in R_2 \quad\text{but}\quad (1,1) \notin R_2 \Rightarrow \text{not transitive} \] (c) Take \[R_3 = \{(1,1),(2,2),(3,3),(4,4)\} \] \[(a,a) \in R_3 \ \forall a \Rightarrow \text{reflexive and symmetric} \] \[(a,a),(a,a) \Rightarrow (a,a) \in R_3 \Rightarrow \text{transitive} \] Answer: \(\displaystyle R_1\), \(\displaystyle R_2\), \(\displaystyle R_3\) above.

  7. Exercise 17

    Let R be relation defined on the set of natural number N\displaystyle \mathbf{N} as follows: R={(x,y):x∈N,y∈N,2x+y=41}\displaystyle \mathrm{R}=\{(x, y): x \in \mathbf{N}, y \in \mathbf{N}, 2 x+y=41\}. Find the domain and range of the relation R. Also verify whether R is reflexive, symmetric and transitive.

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    NCERT’s answer
    Domain of \(\displaystyle \mathrm{R}=\{1,2,3,4, \ldots . .20\}\) and Range of \(\displaystyle \mathrm{R}=\{1,3,5,7,9, \ldots . .39\}\). R is neither reflective, nor symmetric and nor transitive.
    Since \(\displaystyle y\ge 1\), \(\displaystyle x\) is bounded: \[y = 41-2x \ge 1 \Rightarrow x \le 20 \] \[\text{Domain} = \{1,2,3,\ldots,20\} \] \[\text{Range} = \{41-2x : x=1,\ldots,20\} = \{39,37,\ldots,3,1\} \] Reflexive test: \[(1,1):\ 2(1)+1 = 3 \ne 41 \Rightarrow (1,1)\notin R \] Symmetric test: \[(20,1)\in R \ \text{since}\ 2(20)+1 = 41 \] \[(1,20):\ 2(1)+20 = 22 \ne 41 \Rightarrow (1,20)\notin R \] Transitive test: \[(20,1)\in R,\quad (1,39)\in R \ \text{since}\ 2(1)+39 = 41 \] \[(20,39):\ 2(20)+39 = 79 \ne 41 \Rightarrow (20,39)\notin R \] Answer: Domain \(\displaystyle =\{1,2,\ldots,20\}\), Range \(\displaystyle =\{1,3,5,\ldots,39\}\); \(\displaystyle R\) is not reflexive, not symmetric and not transitive.

  8. Exercise 18

    Given A={2,3,4},B={2,5,6,7}\displaystyle \mathrm{A}=\{2,3,4\}, \mathrm{B}=\{2,5,6,7\}. Construct an example of each of the following:
    (a)
    an injective mapping from A to B
    (b)
    a mapping from A to B which is not injective
    (c)
    a mapping from B to A.

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    (a) Every element of A gets a different image: \[f = \{(2,2),\,(3,5),\,(4,6)\} \] \[f(2),\,f(3),\,f(4) = 2,\,5,\,6 \ \text{(distinct)} \Rightarrow f \ \text{is one-one} \] (b) Two elements of A share an image: \[g = \{(2,5),\,(3,5),\,(4,6)\} \] \[g(2)=g(3)=5 \ \text{but}\ 2\ne 3 \Rightarrow g \ \text{is not injective} \] (c) Every element of B gets exactly one image in A: \[h = \{(2,2),\,(5,3),\,(6,4),\,(7,4)\} \] Answer: (a) \(\displaystyle f=\{(2,2),(3,5),(4,6)\}\); (b) \(\displaystyle g=\{(2,5),(3,5),(4,6)\}\); (c) \(\displaystyle h=\{(2,2),(5,3),(6,4),(7,4)\}\).

  9. Exercise 19

    Give an example of a map
    (i)
    which is one-one but not onto
    (ii)
    which is not one-one but onto
    (iii)
    which is neither one-one nor onto.

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    (i) \(\displaystyle f:\mathbf{N}\to\mathbf{N},\ f(x)=2x\) \[f(x_1)=f(x_2)\Rightarrow 2x_1=2x_2\Rightarrow x_1=x_2 \quad\text{(one-one)} \] \[f(x)=1\Rightarrow x=\tfrac12\notin\mathbf{N} \quad\text{(not onto)} \] (ii) \(\displaystyle g:\mathbf{R}\to[0,\infty),\ g(x)=x^2\) \[g(-1)=g(1)=1,\ \ -1\ne 1 \quad\text{(not one-one)} \] For every \(\displaystyle y\ge 0\): \[g(\sqrt{y})=y \quad\text{(onto)} \] (iii) \(\displaystyle h:\mathbf{R}\to\mathbf{R},\ h(x)=x^2\) \[h(-1)=h(1)=1,\ \ -1\ne 1 \quad\text{(not one-one)} \] \[x^2=-1 \ \text{has no real solution} \quad\text{(not onto)} \] Answer: (i) \(\displaystyle f(x)=2x\) on \(\displaystyle \mathbf{N}\); (ii) \(\displaystyle g(x)=x^2\) from \(\displaystyle \mathbf{R}\) to \(\displaystyle [0,\infty)\); (iii) \(\displaystyle h(x)=x^2\) from \(\displaystyle \mathbf{R}\) to \(\displaystyle \mathbf{R}\).

  10. Exercise 20

    Let A=R−{3},B=R−{1}\displaystyle \mathrm{A}=\mathbf{R}-\{3\}, \mathrm{B}=\mathbf{R}-\{1\}. Let f:A→B\displaystyle f: \mathrm{A} \rightarrow \mathrm{B} be defined by f(x)=x−2x−3\displaystyle f(x)=\frac{x-2}{x-3} ∀x∈ A\displaystyle \forall x \in \mathrm{~A}. Then show that f\displaystyle f is bijective.

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    One-one: let \(\displaystyle x_1,x_2\in A\) with \(\displaystyle f(x_1)=f(x_2)\). \[\frac{x_1-2}{x_1-3}=\frac{x_2-2}{x_2-3} \] \[(x_1-2)(x_2-3)=(x_2-2)(x_1-3) \] \[x_1x_2-3x_1-2x_2+6=x_1x_2-3x_2-2x_1+6 \] \[-x_1=-x_2 \Rightarrow x_1=x_2 \] Onto: let \(\displaystyle y\in B\), so \(\displaystyle y\ne 1\). Solve \(\displaystyle y=f(x)\) for \(\displaystyle x\). \[y(x-3)=x-2 \Rightarrow x(y-1)=3y-2 \Rightarrow x=\frac{3y-2}{y-1} \] This \(\displaystyle x\) exists since \(\displaystyle y\ne 1\), and it is not \(\displaystyle 3\): \[\frac{3y-2}{y-1}=3 \Rightarrow 3y-2=3y-3 \quad\text{(impossible)} \] So \(\displaystyle x\in A\), and \[f(x)=\frac{(3y-2)-2(y-1)}{(3y-2)-3(y-1)}=\frac{y}{1}=y \] Answer: \(\displaystyle f\) is one-one and onto, hence bijective.