Relations Test 9

By the definition for a relation with $$n$$ elements to be  reflexive relation  it must have at least $$n$$ ordered pairs.
it has $$m$$ ordered pairs therefore $$\mathrm{m}\geq \mathrm{n}$$.

 Domain $$=$$ value of $$\mathrm{x}$$ in $$f=2\mathrm{x}^{2}+3\mathrm{y}^{2}\leq 6$$  then
$$2\mathrm{x}^{2}+3\mathrm{y}^{2}\leq 6$$ = $$\displaystyle \frac{\mathrm{x}^{2}}{3}+\frac{\mathrm{y}^{2}}{2}\leq 1$$ 
Since it is an equation of ellipse then value of $$x$$ $$ \Rightarrow  \mathrm{x}\in[-\sqrt{3},\sqrt{3}]$$

The possible values of $$x$$ which satisfies the given relation is the domain of the relation

Given $$A\times A$$ has $$9$$ elements $$\Rightarrow n(A)=3$$
Elements of $$A\times A  $$ are  $$(-1,0)$$  &  $$(0,1)$$.
$$\therefore$$ $$A$$ needs to contain $$-1, 0$$ and $$1$$.
$$\therefore$$ Since $$n(A)=3 \Rightarrow  A=\left \{ -1,0,1 \right \}$$

$$R$$ is reflexive if it contains $$(1, 1), (2, 2), (3, 3)$$.
$$\because (1, 2)\in R$$ and $$ (2,3)\in R$$
$$\therefore R$$ is symmetric if $$(2, 1), (3, 2) \in R.$$

A relation is equivalence if it is symmetric, transitive and reflexive
Consider, $$xR_1y\Leftrightarrow \left | x \right |=\left | y \right |$$

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.

$$n\left ( B \right )=4$$
$$n\left ( A \right )=3$$
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 $$\sin n\pi =0$$
$$\therefore f(x)=0$$