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MATHS 3 Marks Question

Relations and Functions · Rajasthan Board (RBSE) STD 12 Science (Rajasthan - English Medium). 17 questions.

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Question 13 Marks
Let $A =\{-1, 0, 1, 2\}, B = \{-4, -2, 0, 2\}$ and $f, g : A \rightarrow B $ be the functions defined by $f(x) = x^2 - x, x \in A$ and $g(x) = 2\left| {x - \frac{1}{2}} \right| - 1,x \in A$. Are $f$ and $g$ equal? Justify your answer.
(Hint: One may note that two functions $f : A \rightarrow B$ and $g : A \rightarrow B$ such that $f(a) = g(a)$ $\forall$ a $\in A,$ are called equal functions).
Answer
When $\mathrm{x}=-1$ then $\mathrm{f}(\mathrm{x})=1^2+1=2$ and $g(x)=2\left|-1-\frac{1}{2}\right|-1=2$
At $\mathrm{x}=0, \mathrm{f}(0)=0$ and $g(0)=2\left|-\frac{1}{2}\right|-1=2 \times \frac{1}{2}-1=0$
At $\mathrm{x}=1, \mathrm{f}(1)=1^2-1=0$ and $g(1)=2\left|1-\frac{1}{2}\right|-1=2 \times \frac{1}{2}-1=0$
At $x=2, f(2)=2^2-2=2$ and $g(2)=2\left|2-\frac{1}{2}\right|-1=3-1=2$
Thus for each $a \in A, f(a)=g(a)$
Therefore, $f$ and $g$ are equal function.
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Question 23 Marks
If f: N $\to$ N is defined by f(n) = $\left\{ \begin{array} { l } { \frac { n + 1 } { 2 } , \text { if } n \text { is odd } } \\ { \frac { n } { 2 } , \text { if } n \text { is even } } \end{array} \right.$for all n $ \in$ N. State whether the function f is bijective. Justify your answer.
Answer
The given function is $f : N$ $ \to$ $N$ such that
$ f ( n ) = \left\{ \begin{array} { l l } { \frac { n + 1 } { 2 } , } & { \text { if } n \text { is odd } } \\ { \frac { n } { 2 } } & { \text { if } n \text { is even } } \end{array} \right.$
One-one function
Here,$f(1) =$ $\frac { 1 + 1 } { 2 } = \frac { 2 } { 2 } $$= 1$
and $f(2) =$ $\frac 22$$= 1$
$\therefore$ $f(n)$ is not a one-one function because at two distinct values of domain (N), f(n) has same image.

Onto function
If n is an odd natural number, then $ 2n- 1$ is also an odd natural number.
Now, $f(2n - 1) =$ $\frac { 2 n - 1 + 1 } { 2 }$$= n$
Again, if n is an even natural number, then $2n$ is also an even natural number.
$f(2n) =$ $\frac{2n}{2}$ $= n$

From Equations (i) and (ii), we observe that for each n (whether even or odd) there exists its pre-image in N,
i.e, Range = co-domain
$\therefore $ f is onto
f(x) is not one-one but onto
So, it is not a bijective function.

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Question 33 Marks
Let $A = R - \{3\}$ and $B = R - \{1\}$. Consider the function $f : A \rightarrow B$ defined by $f(x) = \left( {\frac{{x - 2}}{{x - 3}}} \right)$ Is f one-one and onto? Justify your answer.
Answer
$\mathrm{A}=\mathrm{R}-\{3\}$ and $\mathrm{B}=\mathrm{R}-\{1\}$ and $f(x)=\frac{x-2}{x-3}$
Let $\mathrm{x}_1, \mathrm{x}_2 \in \mathrm{~A}$, then $f\left(x_1\right)=\frac{x_1-2}{x_1-3}$ and $f\left(x_2\right)=\frac{x_2-2}{x_2-3}$
Now, for $f\left(x_1\right)=f\left(x_2\right)$
$\Rightarrow \frac{x_1-2}{x_1-3}=\frac{x_2-2}{x_2-3} $
$ \Rightarrow\left(\mathrm{x}_1-2\right)\left(\mathrm{x}_2-3\right)=\left(\mathrm{x}_2-2\right)\left(\mathrm{x}_1-3\right) $
$ \Rightarrow \mathrm{x}_1 \mathrm{x}_2-3 \mathrm{x}_1-2 \mathrm{x}_2+6=\mathrm{x}_1 \mathrm{x}_2-2 \mathrm{x}_1-3 \mathrm{x}_2+6 $
$ \Rightarrow-3 \mathrm{x}_1-2 \mathrm{x}_2=-2 \mathrm{x}_1-3 \mathrm{x}_2 $
$ \Rightarrow \mathrm{x}_1=\mathrm{x}_2 \therefore \text { fis one-one function. }$
Now, $y=\frac{x-2}{x-3}$
$ \Rightarrow \mathrm{y}(\mathrm{x}-3)=\mathrm{x}-2 $
$ \Rightarrow \mathrm{xy}-3 \mathrm{y}=\mathrm{x}-2 $
$ \Rightarrow \mathrm{x}(\mathrm{y}-1)=3 \mathrm{y}-2 $
$ \Rightarrow x=\frac{3 y-2}{y-1} $
$ \therefore f\left(\frac{3 y-2}{y-1}\right)=\frac{\frac{3 y-2}{y-1}-2}{\frac{3 y-2}{y-1}-3}=\frac{3 y-2-2 y+2}{3 y-2-3 y+3}=y $
$ \Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{y}$
Therefore, $f$ is an onto function.
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Question 43 Marks
Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R = {(a, b) : |a - b| is even}, is an equivalence relation. Show that all the elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to any element of {2, 4}.
Answer
A = {1, 2, 3, 4, 5} and R = {(a, b) : |a - b| is even}, then R = {(1, 3), (1, 5), (3, 5), (2, 4)}
  1. For (a, a), |a - a| = 0 which is even. $\therefore$ R is reflexive.
    If |a - b| is even, then |b - a| is also even. $\therefore$ R is symmetric.
    Now, if |a - b| and |b - c| is even then |a - b + b - c| is even
    $\Rightarrow$ |a - c| is also even. $\therefore$ R is transitive.
    Therefore, R is an equivalence relation.
  2. Elements of {1, 3, 5} are related to each other.
    Since |1 - 3| = 2, |3 - 5| = 2, |1 - 5| = 4 all are even numbers
    $\Rightarrow$ Elements of {1, 3, 5} are related to each other.
    Similarly elements of (2, 4) are related to each other.
    Since |2 - 4| = 2 an even number, then no element of the set {1, 3, 5} is related to any element of (2, 4).
    Hence no element of {1, 3, 5} is related to any element of {2, 4}.
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Question 53 Marks
Show that the relation R in the set A of all the books in a library of a college, given by R = {(x, y) : x and y have same number of pages} is an equivalence relation.
Answer
Books x and x have same number of pages $\Rightarrow$ (x, x) $\in$ R $\therefore$R is reflexive.
If (x, y) $\in$ R ,then x and y have same no. of pages
$\Rightarrow$ y and x have same no. of pages
$\Rightarrow$ (y, x) $\in$ R $\therefore$ R is symmetric.
Now if (x, y) $\in$ R, (y, z) $\in$ R. Then
x and y have same no. of pages and y and z have same no. of pages. This implies x and z have same no. of pages.
$\Rightarrow$ (x, z) $\in$ R $\therefore$ R is transitive.
Since R is reflexive, symmetric and transistive, therefore, R is an equivalence relation.
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Question 63 Marks
Check whether the relation $R$ in $R$ defined by $R = \{(a, b) : a  \leq  b^3\}$ is reflexive, symmetric or transitive.
Answer
  1. For $(a, a), a < a^3$ which is false. $\therefore R$ is not reflexive.
  2. For $(a, b), a < b^3$ and $(b, a), b > a^3$ which is false. $\therefore R$ is not symmetric.
  3. For $a < b^2 b < c^3$. Now $b < c^3$ implies $b^3 < c^9$
Thus, we get a $<c^9$, therefore $( a, c )$ does not belong to $\mathbb{R}$ and hence $R$ is not transitive. Therefore, $R$ is neither reflexive, nor symmetric and nor transitive.
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Question 73 Marks
Show that the relation R in R defined as R = {(a, b) : a $\leq$ b}, is reflexive and transitive but not symmetric.
Answer
  1. a $\leq$ a which is true, so (a, a) $\in$ R, $\therefore$ R is reflexive.
  2. a $\leq$ b but b $\leq$ a which is false. $\therefore$ R is not symmetric.
  3. a $\leq$ b and b $\leq$ c $\Rightarrow$ a $\leq$ c which is true. $\therefore$ R is transitive.

Therefore, R is reflexive and transitive but not symmetric.

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Question 83 Marks
Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b ): b = a +1} is reflexive, symmetric or transitive.
Answer
Let A = {1, 2, 3, 4, 5, 6}
Relation R is defined on set A as:
R = {(a, b): b = a + 1}
Therefore, R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}
We find (a, a) $\notin$ R, where a $\in$ A.
For instance (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6) $\notin$ R
Therefore, R is not reflexive.
It can be observed that (1, 2) $\in$ R, but (2, 1) $\notin$ R.
Therefore, R is not symmetric.
Now, (1, 2), (2, 3) $\in$ R
But, (1, 3) $\notin$ R
Therefore, R is not transitive
Hence, R is neither reflexive, nor symmetric, nor transitive.
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Question 93 Marks
Show that the relation $R$ in the set $R$ of real numbers, defined as $R = \{(a, b): a  \leq  b^2\}$ is neither reflexive nor symmetric nor transitive.
Answer
$R=\left\{(a, b): a \leq b^2\right\}$
It can be observed that
$\left(\frac{1}{2}, \frac{1}{2}\right) \in R \text {, since } \frac{1}{2}>\left(\frac{1}{2}\right)^2=\frac{1}{4}$
$\therefore \mathrm{R}$ is not reflexive.
Now, $(1,4) \in R$ as $1<4^2$.
But, $4$ is not less than $1^2$.
$\therefore(4,1) \notin R$
$\therefore \mathrm{R}$ is not symmetric.
Now,
$ (3,2),(2,1.5) \in R $
$ \left(\text { as } 3<2^2=4 \text { and } 2<(1.5)^2=2.25\right) $
$ \text { But, } 3>(1.5)^2=2.25 $
$ \therefore(3,1.5) \notin R $
$ \therefore R \text { is not transitive. }$
Hence, $R$ is neither reflexive, nor symmetric, nor transitive.
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Question 103 Marks
Let $L$ be the set of all lines in $xy$ plane and $R$ be the relation in $L$ define as $R = \{(L_1, L_2) : L_1 || L_2\}.$ Show that $R$ is an equivalence relation. Find the set of all lines related to the line $y = 2x + 4.$
Answer
$L_1 \| L_1$ i.e $\left(L_1, L_1\right) \in R$ Hence reflexive
Let $\left(L_1, L_2\right) \in R$, then
$\mathrm{L}_1 \mid \mid \mathrm{L}_2$ which implies $\mathrm{L}_2$ || $\mathrm{L_1}$
$\Rightarrow\left(L_2, L_1\right) \in R$ Hence symmetric
We know the
$\mathrm{L}_1 \| \mathrm{L}_2$ and $\mathrm{L}_2 \| \mathrm{L}_3$
Then $L_1| | L_3$
Therefore, $\left(L_1, L_2\right) \in R$ and $\left(L_2, L_3\right) \in R$ implies $\left(L_1, L_3\right) \in R$
Hence Transitive
Hence, $R$ is an equivalence relation.
Any line parallel to $y=2 x+4$ is of the form $y=2 x+K$, where $k$ is a real number.
Therefore, set of all lines parallel to $y=2 x+4$ is $\{y: y=2 x+k, k$ is a real number $\}$
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Question 113 Marks
Determine whether the below relation is reflexive, symmetric and transitive:
Relation R in the set Z of all integers defined as
R = {(x, y) : x – y is an integer}
Answer
It is given that Relation R in the set Z of all integers is defined as
R = {(x, y) : x – y is an integer}
Now, for every x $\in$ Z, (x, x) $\in$ R, as x - x = 0 is an integer.
$\Rightarrow$ R is reflexive.
Next, for every x, y $\in$ Z if (x, y) $\in$ R, then x - y is an integer.
$\Rightarrow$ -(x - y) is also an integer.
$\Rightarrow$ (y - x) is an integer.
$\Rightarrow$ (y - x) $\in$ R
$\Rightarrow$ R is symmetric.
Further, Take (x, y) ,(y, z) $\in$ R where x, y, z $\in$ R,
$\Rightarrow$ (x - y) and (y - z) are integers.
$\Rightarrow$ (x - z) = (x - y) + (y - x) is an integer.
$\Rightarrow$ (x, z) $\in$ R
$\Rightarrow$ R is transitive.
Therefore, R is reflexive, symmetric and transitive.
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Question 123 Marks
Show that the relation $R$ defined in the set A of all polygons as $R = \{(P_1, P_2) : P_1$ and $P_2$ have same number of sides$\},$ is an equivalence relation. What is the set of all elements in $A$ related to the right angle triangle T with sides $3, 4,$ and $5?$
Answer
Part I: $R = \{(P_1, P_2): P_1$ and $P_2$ have same number of sides $\}$
  1. Consider the element $(P_1, P_1),$ it shows $P_1$ and $P_1$ have same number of sides. Therefore, $R$ is reflexive.
  2. If $(P_1, P_2)$ $\in$ $R$ then $P_1$ and $P_2$ have same no.of pages .Then, $P_2$ and $P_1$ have same no. of pages so $(P_2, P_1)$ $\in$ R
    $\therefore$ R is symmetric.
  3. If $(P_1, P_2)$ $\in$ $R$ and $(P_2, P_3)$ $\in$ $R$ then $P_1$ and $P_2$ have same no. of pages and $P_2$ and $P_3$ have same no. of pages .This implies $P_1$ and $P_3$ have same no. of pages and hence $(P_1, P_3)$ $\in R$ ,therefore, $R$ is transitive.
Therefore, $R$ is an equivalent relation.
Part II: Since the relation considers only the number of sides, therefore, all the triangles are similar to the given triangle.
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Question 133 Marks
Determine whether the below relation is reflexive, symmetric and transitive:
Relation R in the set A = {1, 2, 3, 4, 5, 6} as
R = {(x, y) : y is divisible by x}
Answer
It is given that relation R on the set A = {1, 2, 3, 4, 5, 6} as
R = {(x, y) : y is divisible by x}
We know that any number 'x' is divisible by itself.
$\Rightarrow$ (x, x) $\in$ R $\forall$ $x\in A$
$\Rightarrow$ R is reflexive.
Now, (2, 4) $\in$ R but (4, 2) $\notin$ R.
$\Rightarrow$ R is not symmetric.
Let (x,y), (y,z) $\in$ R.
$\Rightarrow$y is divisible by x and z is divisible by y.
$\Rightarrow$ z is divisible by x.
$\Rightarrow$ (x,z) $\in$ R
$\Rightarrow$ R is transitive.
Therefore, R is reflexive and transitive but not symmetric.
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Question 143 Marks
Show that the relation $R$ defined in the set $A$ of all triangles as $R=\left\{\left(T_1, T_2\right): T_1\right.$ is similar to $\left.T_2\right\}$, is equivalence relation. Consider three right angle triangles $T_1$ with sides $3,4,5, T_2$ with sides $5,12,13$ and $T_3$ with sides $6,8,10$. Which triangles among $T_1, T_2$ and $T_3$ are related?
Answer
Part I: $R=\left\{\left(T_1, T_2\right): T_1\right.$ is similar to $\left.T_2\right\}$ and $T_1, T_2$ are triangles.
We know that each triangle similar to itself and thus $\left(T_1, T_1\right) \in R$
$\therefore \mathrm{R}$ is reflexive.
Also if two triangles are similar, then $T_1 \cong T_2 \Rightarrow T_1 \cong T_2$
$\therefore \mathrm{R}$ is symmetric.
Again, if $T_1 \cong T_2$ and $T_2 \cong T_3 \Rightarrow$ then $T_1 \cong T_3$
$\therefore \mathrm{R}$ is transitive.
Therefore, $R$ is an equivalent relation.
Part II: It is given that $T_1, T_2$ and $T_3$ are right angled triangles.
$\Rightarrow T_1$ with sides $3,4,5, T_2$ with sides $5,12,13$ and $T_3$ with sides $6,8,10$
Since, two triangles are similar if corresponding sides are proportional.
Therefore, $\frac{3}{6}=\frac{4}{8}=\frac{5}{10}=\frac{1}{2}$
Therefore, $T_1$ and $T_3$ are related.
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Question 153 Marks
Determine whether the below relation is reflexive, symmetric and transitive:
Relation R in the set N of natural numbers is defined as
R = {(x, y) : y = x + 5 and x < 4}
Answer
It is given that Relation R in the set N of natural numbers defined as
R = {(x, y) : y = x + 5 and x < 4}
Clearly,
R = {(1, 6), (2, 7), (3, 8)}
Reflexive
A relation is said to be reflexive if (x, x) $\in$ R, where x is from domain. we can see that (1,1) $\notin$ R.
$\Rightarrow$ R is not reflexive.
Symmetric
A relation is said to be symmetric if (y, x) $\in$ R whenever (x, y) $\in$ R.
Here, (1,6) $\in$ R, but (6,1) $\notin$ R
$\Rightarrow$ R is not symmetric.
Transitive
Now, since there is no pair in R such that (x, y) and (y, z) $\in$ R, then (x, z) cannot belong to R.
$\therefore$ R is not transitive.
Therefore, R is neither reflexive, nor symmetric, nor transitive.
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Question 163 Marks
Show that the relation $R$ in the set $A$ of points in a plane given by $R = \{(P, Q) :$ distance of the point $P$ from the origin is same as the distance of the point $Q$ from the origin$\}$, is an equivalence relation. Further, show that the set of all points related to a point $P \ne (0, 0)$ is the circle passing through $P$ with origin as centre.
Answer
Part I: $R = \{(P, Q):$ distance of the point $P$ from the origin is the same as the distance of the point $Q$ from the origin$\}$
Let $P\left(x_1, y_1\right)$ and $Q\left(x_2, y_2\right)$ and $O(0,0)$.
$ \therefore \mathrm{OP}=\mathrm{OQ} $
$ \Rightarrow \sqrt{x_1^2+y_1^2}=\sqrt{x_2^2+y_2^2} $
$ \Rightarrow x_1^2+y_1^2=x_2^2+y_2^2$
Now, For $(P, P), OP = OP$
$\therefore \mathrm{R}$ is reflexive.
Also $O P=O Q$ and $O Q=O P$
$\Rightarrow(P, Q)=(Q, P) \in R$
$\therefore \mathrm{R}$ is symmetric.
Also $O P=O Q$ and $O Q=O R$
$\Rightarrow O P=O Q \therefore R \text { is transitive. }$
Therefore, R is an equivalent relation.
Part II:$\text { As } x_1^2+y_1^2=x_2^2+y_2^2=r^2 \text { (let) } $
$ \Rightarrow x^2+\mathrm{y}^2=\mathrm{r}^2$
which represents a circle with centre $(0,0)$ and radius r .
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Question 173 Marks
Determine whether the below relations is reflexive, symmetric and transitive:
Relation R in the set A = {1, 2, 3, ..., 13, 14} defined as R = {(x, y) : 3x – y = 0}.
Answer
The Relation R on the set A = {1, 2, 3, ...., 13, 14}, is defined as
R = {(x, y) : 3x - y = 0}
Then, R = {(1, 3), (2, 6), (3, 9), (4, 12)}
Reflexive
A relation is said to be reflexive if (x, x) $\in$ R, for every x in the domain.
Clearly, R is not reflexive as (1,1), (2,2) …… (14,14) $\notin$ R
Symmetric
A relation is said to be symmetric if (y, x) $\in$ R whenever (x, y) $\in$ R.
But here, R is not symmetric as (1,3) $\in$ R, but (3,1) $\notin$ R
Transitive
A relation is said to be transitive if (x, z) $\in$ R whenever (x, y) $\in$ R and (y, z) $\in$ R
But here, R is not transitive as (1,3), (3,9) $\in$ R, but (1,9) $\notin$ R
Therefore, R is neither reflexive, nor symmetric, nor transitive.
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