Relations Test 26

Result

Relations Test 26

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The number of ordered pairs $$\left( m,n \right)$$, where $$ m, n \in \left\{ 1,2,3,\dots ,50 \right\}$$, such that $$ { 6 }^{ m }+{ 9 }^{ n }$$ is a multiple of $$5$$ is

A
$$1250$$

B
$$2500$$

C
$$625$$

D
$$500$$

Solution

For a number to be divisible by $$5$$ the last digit shall be a multiple of $$5$$
Any power of 6 has its last digit(or unit's digit) equal to 6.
The power of $$9$$ of the form
$$4p+1$$ has it's last digit as 9
$$4p+2$$ has it's last digit as 1
$$4p+3$$ has it's last digit as 9
$$4p$$ has it's last digit as 1
Thus $$6^{m}+9^{n}$$ will be a multiple of $$5$$ only if $$m$$ is any number from the given set and $$n$$ is a number of the form $$4p+1$$ or $$4p+3$$.
Therefore the number of ways to select $$m$$ is $$\binom{50}{1}$$ and the number of ways to select $$n$$ from the given set is $$\binom{25}{1}$$
hence the total number of ordered pairs $$(m,n)$$ is $$\binom{50}{1} \times \binom{25}{1}=50 \times 25=1250$$
Question 2
1 / -0

Let $$n$$ be a fixed positive integer. Define a relation $$R$$ in the set $$Z$$ of integers by $$aRb$$ if and only if $$\dfrac {n}{a - b}$$. The relation $$R$$ is

A
Reflexive

B
Symmetric

C
Transitive

D
An equivalence relation

Solution

Given $$aRb$$ such that $$R:\dfrac{n}{a-b}\epsilon Z$$
$$(A) aRa:$$ $$\dfrac{n}{a-a}=\infty$$ and $$\infty$$ does not belongs to $$Z$$
So not a reflexive relation and hence (A) and (D) options are ruled out.
$$(B)aRb\epsilon Z $$ we need to check whether $$bRa\epsilon Z$$ or not.
$$\dfrac{n}{a-b}\epsilon Z$$ Now, $$\dfrac{n}{b-a}\rightarrow -\dfrac{n}{a-b}$$
As $$\dfrac{n}{a-b}$$ is integer so negative of it also integer
$$\therefore bRa\epsilon Z$$
Hence relation $$R$$ is Symmetric.
$$(C)$$ Now $$aRb\epsilon Z$$ and $$bRc\epsilon Z $$ but we can't say anything about $$aRc\epsilon Z.$$ Hence not a transitive relation.
Question 3
1 / -0

If two sets $$A$$ and $$B$$ are having $$39$$ elements in common, then the number of elements common to each of the sets $$A\times B$$ and $$B\times A$$ are

A
$${ 2 }^{ 39 }$$

B
$${ 39 }^{ 2 }$$

C
$$78$$

D
$$351$$

Solution

If set $$A$$ and set $$B$$ have $$39$$ common elements, then the number of common elements in set $$A\times B$$ and set $$B\times A\,=39^2$$
Question 4
1 / -0

Let $$S$$ be a relation on $$\mathbb{R}^{+}$$ defined by $$xSy\Leftrightarrow { x }^{ 2 }-{ y }^{ 2 }=2\left( y-x \right)$$, then $$S$$ is

A
Only reflecxive

B
Only symmetric

C
Only Trasitive

D
Equivalence

Solution

Let $$S$$ be the relation on $$\mathbb{R}^{+}$$ defined by $$xSy\Leftrightarrow x^2-y^2=2(y-x)$$.
Now, $$xSx$$ holds as $$x^2-x^2=2(x-x)$$ as $$0=0$$. So, $$S$$ is reflexive.
Also, $$xSy\Leftrightarrow ySx$$ holds as $$x^2-y^2=2(y-x)\Leftrightarrow y^2-x^2=2(x-y)$$. So, $$S$$ is symmetric.
Let, $$xSy,\ ySz$$ holds.
Then, $$x^2-y^2=2(y-x)$$.........(1) and $$y^2-z^2=2(z-y)$$.......(2).
Now, adding (1) and (2) we get, $$x^2-z^2=2(z-x)\Rightarrow xSz$$ holds. So, $$S$$ is also transitive.
Moreover $$S$$ is an equivalence relation.
Question 5
1 / -0

If $${C}_{r}$$ stands for $${ _{ }^{ n }{ C } }_{ r }$$ then $$\left( { C }_{ 0 }+{ C }_{ 1 } \right) +\left( { C }_{ 1 }+{ C }_{ 2 } \right) +....\left( { C }_{ n-1 }+{ C }_{ n } \right) \quad $$ is equal to

A
$${ 2 }^{ n }-1$$

B
$${ 2 }^{ n+1 }+1$$

C
$${ 2 }^{ n+1 }-1$$

D
$${ 2 }^{ n+1 }-2$$

Solution

$$(1+x)^{n}=^{}C_0+^{}C_{1}x+^{}C_2x^2+C_{3}x^{3}...........+^{}C_{n}x^{n}$$

Put x =1

Using the relation of sum of coefficent
$$(C_0+C_1+$$ $$C_2+C_3....C_n)=2^n$$
According to the question ,

$$(C_0+C_1)+$$ $$(C_1+C_2)+$$ $$(C_2+C_3)+$$ $$(C_{n-1}+C_n)\rightarrow$$

on simplifying
$$(C_0+C_1+$$ $$C_2+C_3....C_n)+$$ $$(C_1+C_2+C_3+....$$ $$C_{n-1})\rightarrow$$

Adding and subtracting $$C_0and C_n$$ to above equation

$$(C_0+C_1+$$ $$C_2+C_3....C_n)+$$ $$(C_0+C_1+C_2+C_3+....$$ $$C_{n})-C_0-C_n\rightarrow$$

$$=2^n+2^n-1-1=2^{n+1}-2$$
Question 6
1 / -0

Let $$\rho$$ be a relation defined on$$N$$, the set of natural numbers, as
$$\rho =\left\{ \left( x,y \right) \in N\times N:2x+y=41 \right\} $$ then

A
$$\rho$$ is an equivalence relation

B
$$\rho$$ is only reflexive relation

C
$$\rho$$ is only symmetric relation

D
$$\rho$$ is not transitive

Solution

$$\rho =\left\{ \left( x,y \right) \in N\times N:2x+y=41 \right\} $$
for reflexive relation $$xRx\Rightarrow 2x+x=41\Rightarrow x=\cfrac { 41 }{ 3 } \notin N$$
for symmetric $$\Rightarrow xRy\Rightarrow 2x+y=41\neq yRx\quad $$ (Not symmetric)
for transitive $$xRy\Rightarrow 2x+y=41$$ and $$\quad yRz\Rightarrow 2y+z=41,x$$ (Not transitive)
Question 7
1 / -0

For real values of x ,the range of $$\frac {x^2+2x+1}{X^2+2x-1}$$ is

A
$$(-\infty,0) \cup (1,\infty)$$

B
$$\begin{bmatrix}\frac{1}{2},2\end{bmatrix}$$

C
$$\begin{bmatrix} -\infty,\frac {-2}{9} \end{bmatrix}\cup (1,\infty)$$

D
$$(-\infty,-6)\cup(-2,\infty)$$

Solution

Let $$y=\frac{x^2+2x+1} {x^2+2x-1}$$
$$\Rightarrow yx^2+2xy-y=x^2+2x+1$$
$$\Rightarrow (y-1)x^2+2x(y-1)-(y+1)=0$$
$$B^2-4AC\ge 0$$ & $$y\neq 1,y\neq 0$$
Question 8
1 / -0

Let $$R$$ be a relation defined as $$aRb$$ if $$1 + ab > 0$$, then the relation $$R$$ is

A
reflexive and symmetric

B
symmetric but not reflexive

C
transitive

D
equivalence

Solution

Given relation is $$aRb$$ is $$1+ab>0$$,
Considering both $$a$$ and $$b$$ are real numbers,
We know that $$ab=ba$$,
$$\Longrightarrow aRb=1+ab>0=1+ba>0=bRa$$,
$$\therefore$$ $$R$$ is a symmetric relation.

Now, $$aRa=1+{ a }^{ 2 }$$ as $$a^2$$ is always a positive real number
$$\therefore 1+a^2 >0$$
$$\therefore$$ $$R$$ is a reflexive relation.

Now consider $$aRb$$ which implies $$1+ab>0$$ and also $$bRc$$ which implies $$1+bc>0$$
If we take $$a=0.5,$$ $$b=-0.5$$ and $$c=-4$$, then
$$1+(0.5)(-0.5) = 0.75>0$$ and $$1+(-0.5)(-4) = 3>$$
$$\Rightarrow $$ Both $$aRb$$ and $$bRc$$ are satisfied
But, $$aRc=1+(0.5)(-4) =-2<0$$
$$\therefore$$ $$aRc$$ is not a relation
Hence $$R$$ is not a equivalence relation, but is a reflexive and symmetric relation.
Question 9
1 / -0

Let $$A=\left\{ a,b,c \right\} $$ and $$B=\left\{ 1,2 \right\} $$. Consider a relation $$R$$ defined from set $$A$$ to set $$B$$. Then $$R$$ is equal to set

A
$$A$$

B
$$B$$

C
$$A\times B$$

D
$$B\times A$$

Solution

The relation defined from a set $$A$$ to set $$B$$ is defined as a subset of $$A\times B$$.

$$R:A\rightarrow B=\{(x,y)|x\in A, y\in B , (x,y)\in A\times B\}$$

Here, $$A\times B=\{(a,1),(a,2),(b,1),(b,2),(c,1),(c,2)\}$$

Relation from A to B will be a subset of this $$A\times B$$.
Question 10
1 / -0

If $$f:R\rightarrow R,g\quad :R\rightarrow R$$ are defined by $$f\left( x \right) =5x-3,g(x)={ x }^{ 2 }+3,$$ then $$\left( { gof }^{ -1 } \right) \left( 3 \right) =$$

A
$$\frac { 25 }{ 3 }$$

B
$$\frac { 111 }{ 25 } $$

C
$$\frac { 9 }{ 25 } $$

D
$$\frac { 25 }{ 111 } $$

Solution

$$f\left( x \right) =5x-3$$
$${ f }^{ -1 }\left( x \right) =\frac { x+3 }{ 5 } $$
$$g\left( x \right) ={ x }^{ 2 }+3$$
$${ gof }^{ -1 }\left( 3 \right) =g\left( { f }^{ -1 }\left( 3 \right) \right) =g\left( \frac { 6 }{ 5 } \right) =\frac { 111 }{ 25 } $$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Relations Test 26

Result

Relations Test 26

Score

-

Rank

-

SHARING IS CARING

Question 1

$$1250$$

$$2500$$

$$625$$

$$500$$

Solution

Question 2

Reflexive

Symmetric

Transitive

An equivalence relation

Solution

Question 3

$${ 2 }^{ 39 }$$

$${ 39 }^{ 2 }$$

$$78$$

$$351$$

Solution

Question 4

Only reflecxive

Only symmetric

Only Trasitive

Equivalence

Solution

Question 5

$${ 2 }^{ n }-1$$

$${ 2 }^{ n+1 }+1$$

$${ 2 }^{ n+1 }-1$$

$${ 2 }^{ n+1 }-2$$

Solution

Question 6

$$\rho$$ is an equivalence relation

$$\rho$$ is only reflexive relation

$$\rho$$ is only symmetric relation

$$\rho$$ is not transitive

Solution

Question 7

$$(-\infty,0) \cup (1,\infty)$$

$$\begin{bmatrix}\frac{1}{2},2\end{bmatrix}$$

$$\begin{bmatrix} -\infty,\frac {-2}{9} \end{bmatrix}\cup (1,\infty)$$

$$(-\infty,-6)\cup(-2,\infty)$$

Solution

Question 8

reflexive and symmetric

symmetric but not reflexive

transitive

equivalence

Solution

Question 9

$$A$$

$$B$$

$$A\times B$$

$$B\times A$$

Solution

Question 10

$$\frac { 25 }{ 3 }$$

$$\frac { 111 }{ 25 } $$

$$\frac { 9 }{ 25 } $$

$$\frac { 25 }{ 111 } $$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.