2 Marks Questions

Question 12 Marks

Given a non empty set X, consider P (X) which is the set of all subsets of X.
Define the relation R in P (X) as follows:
For subsets A, B in P (X), ARB if and only if A $\subset$ B. Is R an equivalence relation on P (X)? Justify your answer.

Answer

A $\subset$ A $\therefore$ R is reflexive.
If A $\subset$ B then B $\subset$ A is not true $\therefore$ R is not symmetric.
If A $\subset$ B, B $\subset$ C, then A $\subset$ C $\therefore$ R is transitive.

Therefore, R is not equivalent relation.

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Question 22 Marks

Show that the function f : R $\rightarrow$ {x $\in$ R : -1 < x < 1} defined by $f(x) = \frac{x}{{1 + |x|}}$, x $\in$ R is one-one and onto function.

Answer

f is one-one: For any x, y $\in$ R, we have f(x) : f(y)
$\Rightarrow \frac{x}{{1 + |x|}} = \frac{y}{{|y| + 1}}$
$\Rightarrow$ xy + x = xy + y
$\Rightarrow$ x = y
Therefore, f is one-one function.
If f is one-one, let y = R – {1}, then f(x) = y
$\Rightarrow \frac{x}{{x + 1}} = y$
$\Rightarrow x = \frac{y}{{1 - y}}$
It is clear that x $\in$ R for all y = R – {1}, also x $\ne$=-1
Because x = -1
$\Rightarrow \frac{y}{{1 - y}} = - 1$
$\Rightarrow$ y = -1 + y which is not possible.
Thus for each R – {1} there exists $x = \frac{y}{{1 - y}} \in$ R – {1} such that
$f(x) = \frac{x}{{x + 1}} = \frac{{\frac{y}{{1 - y}}}}{{\frac{y}{{1 - y}} + 1}} = y$
Therefore f is onto function.

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Question 32 Marks

Let $A$ and $B$ be sets. Show that $f : A \times B \rightarrow B \times A$ such that $f(a, b) = (b, a)$ is a bijective function.

Answer

Injectivity: Let $\left(a_1, b_1\right)$ and $\left(a_2, b_2\right) \in A \times B$ such that $f\left(a_1, b_1\right)=f\left(a_2, b_2\right)$
$ \Rightarrow\left(b_1, a_1\right)=\left(b_2, a_2\right) $
$ \Rightarrow b_1=b_2 \text { and } a_1=a_2 $
$ \Rightarrow\left(a_1, b_1\right)=\left(a_2, b_2\right) \text { for all }\left(a_1, b_1\right),\left(a_2, b_2\right) \in A \times B$
So, $f$ is injective.
Surjectivity: Let $(b, a)$ be an arbitrary element of $B \times A$. Then $b \in B$ and $a \in A$.
$\Rightarrow(a, b) \in A \times B$
Thus, for all $(b, a) \in B \times A$, there exists $(a, b) \in A \times B$ such that $f(a, b)=(b, a)$
So, $f: A \times B \rightarrow B \times A$ is an onto function, therefore $f$ is bijective.

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Question 42 Marks

State whether the function is one-one, onto or bijective. Justify your answer. $f: R \rightarrow R$ defined by $f(x) = 1+ x^2$

Answer

Let $x_1, x_2 \in R$
If $f(x_1) = f(x_2)$
$1 + {x_1}^2 = 1 + {x_1}^2$
${x_1}^2 = {x_1}^2$
${x_1} = \pm\ {x_2}$
Hence not one - one
$y = 1 + x^2$
$x = \pm \left( {\sqrt {1 - y} } \right)$
$f\left( {\sqrt {1 - y} } \right) = 1 + (1 - y) = 2 - y \ne y$
Therefore, $f$ is not onto.

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Question 52 Marks

State whether the function is one-one, onto or bijective. Justify your answer. $f: R \rightarrow R$ defined by $f(x) = 3 - 4x.$

Answer

Let $\left(\mathrm{x}_1, \mathrm{x}_2\right) \in R$ such that
$f\left(x_1\right)=f\left(x_2\right) $
$ 3-4 x_1=3-4 x_2 $
$ x_1=x_2$
Hence one-one
$ \mathrm{Y}=3-4 \mathrm{x} $
$ x=\left(\frac{3-y}{4}\right) $
$ f\left(\frac{3-y}{4}\right)=3-4\left(\frac{3-y}{4}\right) $
$ \mathrm{f}(\mathrm{x})=\mathrm{y} $
$ =\mathrm{y}$
Hence onto also.

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Question 62 Marks

Show that the Signum Function f : R $\rightarrow$ R, given by $f(x) = \left\{ {\begin{array}{*{20}{c}} {1,\;if\;x > 0} \\ {0,\;if\;x = 0} \\ { 1,\;if\;x < 0} \end{array}} \right.$ is neither one-one nor onto.

Answer

Signum Function f : R $\rightarrow$ R, given by $f(x) = \left\{ {\begin{array}{*{20}{c}} {1,\;if\;x > 0} \\ {0,\;if\;x = 0} \\ { - 1,\;if\;x < 0} \end{array}} \right.$
f(1) = f(2) = 1
Two distinct elements have same image.
$\therefore$ f is not one-one.
Except -1, 0, 1 no other members of co-domain of f has any pre-image its domain.
$\therefore$ f is not onto.
Therefore, f is neither one-one nor onto.

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Question 72 Marks

Show that the Modulus Function f : R $\rightarrow$ R, given by f(x) = |x|, is neither one-one nor onto, where |x| is x, if x is positive or 0 and |x| is -x, if x is negative.

Answer

Modulus Function f : R $\rightarrow$ R, given by f(x) = |x|
One-one: f(1) = |1| = 1 and f(2) = |2| = 2,so distinct elements have same image.So, f is not one-one.
onto: f takes only positive values, so range(f) = set of positive real numbers $\neq$ R, codomain. So, f is not onto.

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Question 82 Marks

Prove that the Greatest integer Function f : R $\rightarrow$ R, given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.

Answer

Function f : R $\rightarrow$ R, given by f(x) = [x]
$\because \;1 \leqslant x \leqslant 2$, f(x) = 1
$\therefore$ f(1) = 1 and f(1.1) = 1
$\therefore$ f is not one-one.
f takes only integer values, therefore range(f) = set of integers,which is not equal to R, codomain.
Therefore, f is not onto.

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Question 92 Marks

Check the injectivity and surjectivity of the below function:
$f : Z \rightarrow Z$ given by $f(x) = x^3$

Answer

It is given that $f: Z \rightarrow Z$ given by $f(x)=x^3$
We can see that for $x, y \in N$,
$f(x)=f(y) $
$ \Rightarrow x^3=y^3 $
$ \Rightarrow x=y $
$ \Rightarrow f \text { is injective. }$
Now, let $2 \in Z$. But, we can see that there does not exists any $x$ in $Z$ such that
$f(x)=x^3=2$
$\Rightarrow f$ is not surjective.
Therefore, function f is injective but not surjective.

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Question 102 Marks

Check the injectivity and surjectivity of the below function:
$f : N \rightarrow N$ given by $f(x) = x^3$

Answer

The function $f: N \rightarrow N$ is given by $f(x)=x^3$
Clearly for $x, y \in N$
$ f(x)=f(y) $
$ \Rightarrow x^3=y^3 $
$ \Rightarrow x=y$
$\Rightarrow \mathrm{f}$ is injective.
Now, let $2 \in N$. But, we can see that there does not exists any $x$ in $N$ such that
$f(x)=x^3=2$
$\Rightarrow \mathrm{f}$ is not surjective.
Therefore, function f is injective but not surjective.

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Question 112 Marks

Check the injectivity and surjectivity of the below function:
$f : R \rightarrow R$ given by $f(x) = x^2$

Answer

$f : R \rightarrow R$ given by $f(x) = x^2$
As $f(-1) = f(1) = 1$
$\Rightarrow -1$ and $1$ have same image.
$\therefore$ f is not injective.
e.g. $-2 \in$ co-domain, but $\sqrt { - 2} \notin R=$ domain of $f.$
$\therefore$ f is not surjective.

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Question 122 Marks

Check the injectivity and surjectivity of the below function:
$f : Z \rightarrow Z$ given by $f(x) = x^2$

Answer

$f : Z \rightarrow Z$ given by $f(x) = x^2$
Since, $z = \left\{ {0,\; \pm 1,\; \pm 2,\; \pm 3,\;....} \right\}$ therefore, $f(-1) = f(1) = 1$
$\Rightarrow -$1 and $1$ have same image.
$\therefore$ f is not injective.
There are such numbers of co-domain which have no image in domain $Z.$
e.g. 3 $\in$ co-domain, but $\sqrt 3 \notin$ domain of f.
$\therefore$ f is not surjective.

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Question 132 Marks

Check the injectivity and surjectivity of the below function:
$f : N \rightarrow N$ given by $f(x) = x^2$

Answer

$f : N \rightarrow N$ given by $f(x) = x^2$
If $f(x_1) = f(x_2)$ then $x_1^2 = x_2^2$
$\Rightarrow$ $x_1 = x_2$
$\therefore$ $f$ is injective.
There are such numbers of co-domain which have no image in domain $N.$
e.g. $ 3\in$ co-domain $N$, but there is no pre-image in domain of $f.$
therefore $f $ is not onto.
$\therefore$ $f$ is not surjective.

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Question 142 Marks

Show that the function $ f : R_+ \rightarrow R_+$ defined by $f(x) = \frac{1}{x}$ is one-one and onto, where $R_+$ is the set of all non-zero real numbers. Is the result true, if the domain $R_+$ is replaced by N with co-domain being same as $R_+?$

Answer

$f(x)=\frac{1}{x}, f: R_* \rightarrow R_*$
Part I: $f\left(x_1\right)=\frac{1}{x_1}$ and $f\left(x_2\right)=\frac{1}{x_2}$
If $f\left(x_1\right)=f\left(x_2\right)$ then $\frac{1}{x_1}=\frac{1}{x_2}$
$\Rightarrow x_1=x_2$
$\therefore \mathrm{f}$ is one-one.
$ f(x)=\frac{1}{x} $
$ \Rightarrow y=\frac{1}{x} $
$ \Rightarrow x=\frac{1}{y} $
$ \Rightarrow f\left(\frac{1}{y}\right)=y $
$\therefore \mathrm{f} \text { is onto. }$
Part II: When domain $R$ is replaced by $N$, co-domain remaining the same, then, $f: N \rightarrow R$
$ \text { If } \mathrm{f}\left(\mathrm{x}_1\right)=\mathrm{f}\left(\mathrm{x}_2\right) $
$ \Rightarrow \frac{1}{n_1}=\frac{1}{n_2} $
$ \Rightarrow \mathrm{n}_1=\mathrm{n}_2 \text { where } \mathrm{n}_1, \mathrm{n}_2 \in \mathrm{~N}$
$\therefore \mathrm{f}$ is one-one.
But, every real number belonging to co-domain may not have a pre-image in $N$.
e.g. $2$ in codomain $R_*$ does not have pre-image in $N$ as if
$x$ in $N$ be pre-image of $2$ ,then $\mathrm{f}(\mathrm{x})=2 \Rightarrow \frac{1}{x}=2 \Rightarrow \mathrm{x}=\frac{1}{2} \notin N$.
$\therefore \mathrm{f}$ is not onto.

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Question 152 Marks

Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.

Answer

R = {(1, 2), (2, 1)}, so for (a, a), (1, 1) $\notin$ R. $\therefore$R is not reflexive.
Also if (a, b) $\in$R then (b, a) $\in$ R $\therefore$ R is symmetric.
Now (a, b) $\in$ R and (b, c) $\in$ R ,then does not imply (a, c) $\notin$ R as (1,2) $\in$ R and (2,1) $\in$ R but (1,1) $\notin$ R $\therefore$ R is not transitive.
Therefore, R is symmetric but neither reflexive nor transitive.

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Question 162 Marks

Determine whether the relation is reflexive, symmetric and transitive:
Relation R in the set A of human beings in a town at a particular time given by
R = {(x, y) : x is father of y}

Answer

It is given that R = {(x, y) : x is father of y}
$\Rightarrow$ (x, x) $\notin$ R as x cannot be the father of himself.
$\Rightarrow$ R is not reflexive.
Now, if (x, y) $\in$ R, then x is the father of y.
$\Rightarrow$ But y is not father of x.
$\Rightarrow$ (y, x) $\notin$ R
$\Rightarrow$ R is not symmetric.
Now, let (x, y), (y, z) $\in$ R
$\Rightarrow$ x is the father of y and y is the father of z.
$\Rightarrow$ x is not the father of z.
$\Rightarrow$ Indeed x is the grandfather of z.
$\Rightarrow$ (x, z) $\notin$ R
$\Rightarrow$ R is not transitive.
Therefore, R is neither reflexive, nor symmetric, nor transitive.

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Question 172 Marks

Determine whether the relation is reflexive, symmetric and transitive:
Relation R in the set A of human beings in a town at a particular time given by
R = {(x, y) : x is wife of y}

Answer

It is given that R = {(x, y) : x is wife of y}
Clearly, (x,x) $\notin$ R as x cannot be the wife of herself.
⇒ R is not reflexive.
Now, if (x,y) $\in$ R, then x is the wife of y.
$\Rightarrow$ But y is not wife of x.
$\Rightarrow$ (y,x) $\notin$ R
$\Rightarrow$ R is not symmetric.
Further, let (x,y), (y,z) $\in$ R
$\Rightarrow$ x is the wife of y and y is the wife of z.
$\Rightarrow$ This is not possible.
$\Rightarrow$ (x,z) $\notin$ R
$\Rightarrow$ R is not transitive.
Therefore, R is neither reflexive, nor symmetric, nor transitive.

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Question 182 Marks

Determine whether the relation is reflexive, symmetric and transitive:
Relation R in the set A of human beings in a town at a particular time given by
R = {(x, y) : x is exactly 7 cm taller than y}

Answer

It is given that R = {(x, y) : x is exactly 7 cm taller than y}
Clearly, (x,x) $\notin$ R as a human being x cannot be taller than himself.
$\Rightarrow$ R is not reflexive.
Now, if (x,y) $\in$ R, then x is exactly 7 cm taller than y.
$\Rightarrow$ But y is not taller than x.
$\Rightarrow$ (y,x) $\notin$ R
$\Rightarrow$ R is not symmetric.
Further, let (x,y), (y,z) $\in$ R
$\Rightarrow$ x is exactly 7 cm taller than y and y is exactly 7 cm taller than z.
$\Rightarrow$ x is exactly 14 cm taller than z.
$\Rightarrow$ (x,z) $\notin$ R
$\Rightarrow$ R is not transitive.
Therefore, R is neither reflexive, nor symmetric, nor transitive.

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Question 192 Marks

Determine whether the below relation is reflexive, symmetric and transitive:
Relation R in the set A of human beings in a town at a particular time given by
R = {(x, y) : x and y live in the same locality}

Answer

Given that R = {(x, y) : x and y live in the same locality}
Clearly, (x, x) $\in$ R as x and x live in the same locality.
$\Rightarrow$ R is reflexive.
Now, if (x, y) $\in$ R, then x and y live in the same locality.
$\Rightarrow$ y and x live in the same locality.
$\Rightarrow$ (y, x) $\in$ R
$\Rightarrow$ R is symmetric.
Further, let (x, y), (y, z) $\in$ R
$\Rightarrow$ x and y live in the same locality and y and z live in the same locality.
$\Rightarrow$ x and z live in the same locality
$\Rightarrow$ (x, z) $\in$ R
$\Rightarrow$ R is transitive.
Therefore, R is reflexive, symmetric and transitive.

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Question 202 Marks

Determine whether the below relation is reflexive, symmetric and transitive:
Relation R in the set A of human beings in a town at a particular time given by
R = {(x, y) : x and y work at the same place}

Answer

It is given that R = {(x, y) : x and y work at the same place}
Clearly, (x, x) $\in$ R, as we can say x and x work at the same place.
$\Rightarrow$ R is reflexive.
Now, if (x, y) $\in$ R, then x and y work on the same place.
$\Rightarrow$ y and x work at the same place.
$\Rightarrow$ (y, x) $\in$ R
$\Rightarrow$ R is symmetric.
Further, let (x, y), (y, z) $\in$ R
$\Rightarrow$ x and y work at the same place and y and z work at the same place.
$\Rightarrow$ x and z work at the same place
$\Rightarrow$ (x, z) $\in$ R
$\Rightarrow$ R is transitive.
Therefore, R is reflexive, symmetric and transitive.

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Question 212 Marks

Give an example of a relation which is symmetric and transitive but not reflexive

Answer

Let A = {-7, -8}
Define a relation R on A as:
R = { (-7, -7)}
Relation R is not reflexive as (a, a) $\notin$ R
Relation R is symmetric as (-7, -7) $\in$ R and (-7, -7) $\in$ R
Clearly R is transitive.
Therefore, relation R is symmetric and transitive but not reflexive.

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Question 222 Marks

Give an example of a relation which is reflexive and transitive but not symmetric.

Answer

Let us define a relation $R$ in $R$ as
$R=\left\{(a, b): a^3 \geq b^3\right\}$
It is clear that $(a, a) \in R$ as $a^3=a^3$
$\Rightarrow R$ is reflexive.
Now, $(2,1) \in R$, but $(1,2) \notin R$
$\Rightarrow R$ is not symmetric.
Now, let $(a, b)(b, c) \in R$
$ \Rightarrow a^3 \geq b^3 \text { and } b^3 \geq c^3 $
$ \Rightarrow a^3 \geq c^3 $
$ \Rightarrow(a, c) \in R $
$ \Rightarrow R \text { is transitive. }$
Therefore, relation R is reflexive and transitive but not symmetric.

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Question 232 Marks

Give an example of a relation which is reflexive and symmetric but not transitive.

Answer

Let us take A = {2,4,6}
Define a relation R on A as:
A = {(2,2), (4,4), (6,6), (2,4), (4,2), (4,6), (6,4)}
Relation of R is reflexive as for every a $\in$ A,
(a,a) $\in$ R
$\Rightarrow$ (2,2), (4,4), (6,6) $\in$ R,
Relation R is symmetric as (a,b) $\in$ R
$\Rightarrow$ (b,a) $\in$ R for all a ,b $\in$ R
And Relation R is not transitive as (2,4), (4,6) $\in$ R,
but (2,6) $\notin$ R
Therefore, relation R is reflexive and symmetric but not transitive.

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Question 242 Marks

Give an example of a relation which is transitive but neither reflexive nor symmetric.

Answer

Let a relation R is defined as:
R = {(a,b): a < b}
For any a $\in$ R, we have (a,a) $\notin$ R as a cannot be strictly less than itself.
In fact, a = a,
Therefore, R is not reflexive.
Now, (1,2) $\in$ R but 2 > 1
$\Rightarrow$ (2,1)) $\notin$ R.
$\Rightarrow$ R is not symmetric.
Now, let (a,b), (b,c) $\in$ R
$\Rightarrow$ a < b and b < c
$\Rightarrow$ a < c
$\Rightarrow$ (a,c) $\in$ R
$\Rightarrow$ R is transitive.
Therefore, relation R is transitive but not reflexive and symmetric.

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Question 252 Marks

Give an example of a relation which is symmetric but neither reflexive nor transitive.

Answer

Let A = {3,4,5}
Define a relation R on A as R = {(3,4), (4,3)}
Relation R is not reflexive as (3,3), (4,4) ,(5,5) $\notin$ R.
Now, as (3,4) $\in$ R and (4,3) $\in$ R,
R is symmetric.
Further, (3,4),(4,3) $\in$ R, but (3,3) $\notin$ R
$\Rightarrow$ R is not transitive.
Therefore, relation R is symmetric but not reflexive or transitive.

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