CBSE Questions for Class 11 Commerce Applied Mathematics Relations Quiz 4 - MCQExams.com

$$n\left ( B \right )=4$$
$$n\left ( A \right )=3$$
Therefore there exists $$2^{3 \times 4}$$ relations from $$A$$ to $$B$$
i.e $$2^{12}$$ relations

$$f(x)=\dfrac {\sin(\pi[x])}{x^2+1}$$
$$\because$$ $$[x]$$ always gives integer value and $$\sin n\pi =0$$
$$\therefore f(x)=0$$

Option A, $$aRb$$ if $$a+b$$ is an even integer.

Let $$a, b,c \in I$$
Clearly, $$a+a =2a$$ is an even integer.
Hence, $$aRa$$ for all $$a\in I$$

Now, let $$aRb$$ 
$$\Rightarrow a+b$$ is an even integer.
or $$b+a$$ is an even integer.
$$\Rightarrow bRa$$

Next, let $$aRb, bRc$$
$$\Rightarrow a+b $$ in an even integer and $$ b+c$$ is an even integer
$$\Rightarrow  a$$ and $$b$$ both are even or both are odd . Also, $$b$$ and $$c$$ both are even or both are odd.
Hence, if $$a$$ and $$b$$ are even .So $$c$$ is also even . Hence, $$a+c$$ is even.
And if $$a$$ and $$b$$ are odd.So, $$c$$ is also odd . Again $$a+c$$ is even.
Hence, $$aRc$$

Hence, option A is equivalence relation.

Option B, $$aRb$$ if $$a-b$$ is an even integer.

Let $$a, b,c \in I$$
Clearly, $$a-a =0$$ is an even integer.
Hence, $$aRa$$ for all $$a\in I$$

Now, let $$aRb$$ 
$$\Rightarrow a-b$$ is an even integer.
or $$-(a-b)$$ is also an even integer.
$$\Rightarrow \ bRa$$

Next, let $$aRb, bRc$$
$$\Rightarrow a-b $$ in an even integer and $$ b-c$$ is an even integer
$$a+c=a-b+b-c=$$ even+even 
Hence, $$a+c$$ is even.
Hence, $$aRc$$

Hence, option $$B$$ is equivalence relation.

Option C $$aRb$$ if $$a<b$$

Let $$a, b,c \in I$$
Clearly,symmetry does not hold here 
Now, let $$aRb$$ 
$$\Rightarrow a<b$$ 
But it does not implies $$b<a$$

Hence, option C is not an equivalence relation.

Option D, $$aRb$$ if $$a=b$$ 

Let $$a, b,c \in I$$
Clearly, $$a=a$$ 
Hence, $$aRa$$ for all $$a\in I$$

Now, let $$aRb$$ 
$$\Rightarrow a=b$$ 
or $$b=a$$ is an even integer.
$$\Rightarrow bRa$$

Next, let $$aRb, bRc$$
$$\Rightarrow a=b $$and $$ b=c$$ 
$$\Rightarrow a=b=c$$
Hence, $$aRc$$

Hence, option D is equivalence relation.

Numbers of elements of is $$\{x_{1},x_{2},x_{3},x_{4}\}$$.

Since,$$8\mathrm{x}\equiv 6(mod \ 14)$$
i.e., $$8\mathrm{x}-6=14\mathrm{k}$$ for $$\mathrm{k}\in \mathrm{I}$$.
$$ \Rightarrow 8x = 14k+6$$
$$ \Rightarrow 4x = 7k+3$$
The values $$6$$ and $$13$$ satisfy this equation (when $$k =3$$ and $$k=7$$), 

We have $$x\in N, 1\le x<4$$
$$\Rightarrow x =1,2,3$$
Hence $$R =\{(1,3),(2,4),(3,5)\}$$

for equivalence relation R
$$(a,a)\in R$$
if $$(a,b)\in R$$
then $$(b,a)\in R$$
if $$(a,b)\in R$$
$$(b,c)\in R$$
then $$(a,c)\in R$$
$$R={(1,2), (2,3), (1,1), (2,2), (3,3), (2,1), (3,2), (1,3), (3,1)}$$
No of added ordered pairs $$=9-2=7$$

$$\because  R$$ is the relation 'less than' from $$A$$ to $$B$$.

$$A = \{2, 4, 6\}, B = \{2,3,5\}$$
$$n (A\times B) = 3\times 3 = 9$$
Number of relations from $$A$$ to $$B $$ is $$2^9$$

Here $$R$$ is a relation $$A$$ to $$B$$ defined by '$$x$$ is greater then $$y$$'
$$R=\left\{ (2,1);(3,1) \right\} $$.Hence, range of $$R=\left\{ 1 \right\} $$

$$\implies \mathrm{R}_{1}=\{(1,3),(2, 4),(3,5),(4,6),(5,7)\}$$

Consider, $$x^2-3x+2=0$$
$$\implies x^2-2x-x+2=0$$
$$\implies (x-2)(x-1)=0$$
$$\implies x=1,2$$

$$A\times A \Rightarrow$$ the first element will be from $$A$$ and the second element will also be from $$A$$.

C is not even logical while in the first 2 cases:

Given $$A=\{b,c,d\}$$ and  $$B=\{x,y\}$$, then

For two sets, A and B the set difference of set B from set A is the set of all element in A but not in B.
Example:
$$A=\{a,b,c,d\}$$
$$B=\{c,d,e\}$$
$$A-B=\{a,b\}$$

For two non-empty sets $$A$$ and $$B$$, the Cartesian product $$A\times B$$ is the set of all ordered pair of elements from $$A$$ and $$B$$.
$$A \times B = \{(x, y) : x \in A, y \in  B\}$$ 

If $$A= \{a,b\}$$ and $$B= \{x, y\},$$
then $$A\times B = \{(a, x); (a, y); (b, x); (b, y)\}$$

If $$n(A)=m$$,and $$n(B)=n$$,then $$n(A\times B)=mn$$

$$R$$ is a relation in $$A$$ and $$(a,b)$$.

Since $$R=\left \{ (1, 2),(2, 3) \right \}$$
i.e.,  $$A=\left \{ (1, 2) \right \} $$ and $$B=\left \{ 2, 3 \right \}$$
Now, if $$R_{r}$$ is the reflexive relation, such that 
$$R_{1}=\left \{ (1, 2),(2, 3),(1, 1),(2, 2),(3, 3) \right \}$$ has $$5$$ elements.
Now, if  R' is both symmetric and reflexive relation, then 
$$R_{2}=\left \{ (1, 2),(2, 3),(1, 1),(2, 2),(2, 1),(3, 2),(3, 3) \right \}$$ has $$7$$ elements
Again, if $$R_{3}$$ is reflexive, symmetric and transitive all together, than 
$$R_{3}=\begin{Bmatrix} (1, 2) & (2, 3) & (1, 1) \\ (2, 2) & (2, 1) & (3, 2) \\ (3, 3) & (1, 3) & (3, 1)\end{Bmatrix} $$
has $$9$$ elements. Starting from $$2$$ elements, therefore, the minimum number of elements to be added is $$7$$.

As monotonic and range $$=$$ Codomain $$\Rightarrow$$ Bijective

$${\textbf{Step -1: Define the Cartesian product.}}$$

               $${\text{Cartesian product: If }}A{\text{ and }}B{\text{ are two non empty sets, then }}$$ 

               $${\text{Cartesian product }}A \times B{\text{ is set of all ordered pairs }}\left( {a,b} \right)$$ $${\text{such that }}a \in A{\text{ and }}b \in B.$$

$${\textbf{Step -2: Find the Cartesian product of given sets.}}$$

               $${\text{We have given,}}$$ 

               $$A = \left\{ {1,2} \right\}$$ $${\text{and}}$$ $$B = \left\{ {3,4} \right\}$$

               $${\text{So,}}$$ $$A \times B = \left\{ {\left( {1,3} \right),\left( {1,4} \right),\left( {2,3} \right),\left( {2,4} \right)} \right\}$$

$${\textbf{Hence, option A. }}\left\{ \mathbf{\left( {1,3} \right),\left( {1,4} \right),\left( {2,3} \right),\left( {2,4} \right)} \right\}$$ $${\textbf{is correct answer.}}$$