Relations Test 27

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Relations Test 27

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Weekly Quiz Competition

Question 1
1 / -0

The relation P defined from R to R as a P b $$\Leftrightarrow$$ 1 + ab > 0, for all a, b $$\epsilon$$ R is

A
reflexive only

B
reflexive and symmetric only

C
transitive only

D
equivalence

Solution

$$Pb\Leftrightarrow 1+ab>0$$

i)Reflexive: $$a\times a$$ for any real value except 0 is positive.Hence $$0\times 0+1>1\Rightarrow a\times a+1>0$$

ii)Symmetric: if $$a\times b+1>0$$ then $$b\times a+1$$ will also be $$>0$$

iii)Not Transitive: a=-2,b=0\Rightarrow -2\times 0+1>0$$

$$b=0,c=4\Rightarrow 0\times 4+1>0$$

but a is not related to c $$\because a=-2,c=4\Rightarrow -2\times 4+1<0$$
Question 2
1 / -0

If R is a relation on a finite set having n elements, then the number of relations on A is :

A
$$2^n$$

B
$$2^{n^2}$$

C
$$n^2$$

D
$$n^n$$

Solution

Here, the required number of relations on A can be found out by the formula $$2^{n^2}$$.
Hence, option B is correct.
Question 3
1 / -0

If $$A=\left\{2,3,5\right\}, B=\left\{2,5,6\right\}$$, then $$\left( A-B \right) \times \left( A\cap B \right)$$ is

A
$$\left\{(3,2),(3,3),(3,5)\right\}$$

B
$$\left\{(3,2),(3,5),(3,6)\right\}$$

C
$$\left\{(3,2),(3,5)\right\}$$

D
$$none\ of\ these$$

Solution

Ans. $$(c)$$.
$$A-B=\{ 3\} ,A\cap B=(2,5)$$.
$$\therefore \left( A-B \right) \times \left( A\cap B \right) =\{ (3,2),(3,5)\}$$.
Question 4
1 / -0

If $$A=\left\{1,2,3\right\}$$ and $$B=\left\{3,8\right\},$$ then
$$\left( A\cup B \right) \times \left( A\cap B \right)$$ is

A
$$\left\{(3,1),(3,2),(3,3),(3,8)\right\}$$

B
$$\left\{(1,3),(2,3),(3,3),(8,3)\right\}$$

C
$$\left\{(1,2),(2,2),(3,3),(8,8)\right\}$$

D
$$\left\{(8,3),(8,2),(8,1),(8,8)\right\}$$

Solution

$$\left( A\cup B \right) =\{ 1,2,3,8\}$$
$$\left( A\cap B \right) =\{ 3\}$$

$$\therefore \left( A\cup B \right) \times \left( A\cap B \right)$$
$$=\left\{ (1,3),(2,3),(3,3),(8,3) \right\}$$

Ans.$$(b)$$.
Question 5
1 / -0

Let $$R = gS - 4$$. When $$S = 8, R = 16$$. When $$S = 10, R$$ is equal to

A
$$11$$

B
$$14$$

C
$$20$$

D
$$21$$

E
None of these

Solution

$$16 = 8g - 4; \therefore g = 2\dfrac {1}{2}; \therefore R = \left (2\dfrac {1}{2}\right ) 10 - 4 = 21$$.
Question 6
1 / -0

If the line $$x=\alpha$$ divides the area of the region $$R=\left\{(x,y)\in \mathbb{R}^2:x^3\le y\le x,0\le x\le 1\right\}$$ into two equal parts, then

A
$$0\lt \alpha\le \dfrac{1}{2}$$

B
$$\dfrac{1}{2}\lt \alpha\lt 1$$

C
$$2\alpha^4-4\alpha^2+1=0$$

D
$$\alpha^4+4\alpha^2-1=0$$

Solution

So, $$\int _{ 0 }^{ \alpha }{ (x-{ x }^{ 3 }) } dx=\int _{ \alpha }^{ 1 }{ (x-{ x }^{ 3 }) } dx\\ { [\cfrac { { x }^{ 2 } }{ 2 } -\cfrac { { x }^{ 4 } }{ 4 } ] }_{ 0 }^{ \alpha }={ [\cfrac { { x }^{ 2 } }{ 2 } -\cfrac { { x }^{ 4 } }{ 4 } ] }_{ \alpha }^{ 1 }\\ \cfrac { { \alpha }^{ 2 } }{ 2 } -\cfrac { { \alpha }^{ 4 } }{ 4 } =\cfrac { { 1 }^{ 2 } }{ 2 } -\cfrac { { 1 }^{ 4 } }{ 4 } -\cfrac { { \alpha }^{ 2 } }{ 2 } +\cfrac { { \alpha }^{ 4 } }{ 4 } \\ \cfrac { { 2\alpha }^{ 2 } }{ 2 } -\cfrac { 2{ \alpha }^{ 4 } }{ 4 } =\cfrac { { 1 } }{ 4 } ={ \alpha }^{ 2 }-\cfrac { { \alpha }^{ 4 } }{ 2 } \\ \cfrac { { 1 } }{ 2 } ={ 2\alpha }^{ 2 }-{ \alpha }^{ 4 }\\ { 2\alpha }^{ 4 }-4{ \alpha }^{ 2 }+1=0$$
Thus, $${ 2\alpha }^{ 4 }-4{ \alpha }^{ 2 }+1=0$$
Question 7
1 / -0

Find the domain of the function defined as $$f(x)=\dfrac{1}{1-x^2}$$.

A
$$x\epsilon R -\{1,-1\}$$

B
$$x\epsilon R -\{0\}$$

C
$$x\epsilon R -\{-1\}$$

D
$$x\epsilon R -\{1\}$$

Solution

$$f\left(x\right)=\dfrac{1}{1-x^{2}}$$
For existance $$1-x^{2}\neq 0$$
$$\Rightarrow x\neq+1,-1$$
$$\Rightarrow x\in R-\left\{1,-1\right\}$$
Question 8
1 / -0

If $$aN = (ax/x \ \epsilon \ N)$$ and $$bN \cap cN = dN,$$, where $$b, c \ \epsilon N$$ are relatively prime, then

A
d = bc

B
c = bd

C
b = cd

D
none

Solution
Question 9
1 / -0

Let $$f\left( x \right) :\begin{cases} x,\quad x\quad is\quad rational \\ 0,\quad x\quad is\quad irrational \end{cases}$$
and
$$g\left( x \right) :\begin{cases} 0,\quad x\quad is\quad rational \\ x,\quad x\quad is\quad irrational \end{cases}$$

If $$f:R\rightarrow R$$ and $$g:R\rightarrow R$$, then $$(f-g)$$ is

A
one-one and into

B
neither one-one nor onto

C
many-one and onto

D
one-one and onto

Solution

When x is rational $$f(x)=x,g(x)=0$$

$$\therefore (f-g)=x-0=x$$.So for each rational no. in $$(f-g)$$ there is the same rational number in co domain $$f(x)=0,g(x)=x$$ [when x is irrational]

$$\therefore (f-g)=0-x=-x$$

So for each irrational number x in domain, we have $$(-x)$$ in co domain.

Conversely for each negative irrational number, there is an element in domain.

But for positive irrational no. there is no corresponding element.

Thus $$(f-g)$$ is one -one and into.
Question 10
1 / -0

If $$f:R \to R$$ is a function satisfying $$f(x + y) = f(xy)$$ for all $$x, y \in R$$ and $$f\left(\dfrac{3}{4}\right) = \left( \dfrac{3}{4} \right)$$ , then $$f\left( \dfrac{9}{16} \right)$$ =

A
$$\dfrac{3}{4}$$

B
$$\dfrac{9}{16}$$

C
$$\dfrac{\sqrt{3}}{2}$$

D
$$0$$

Solution

$$f(x+y)=f(xy)\quad -(i)\\ f(\cfrac { 3 }{ 4 } )=(\cfrac { 3 }{ 4 } )\\ f(\cfrac { 9 }{ 16 } )=f(\cfrac { 3 }{ 4 } \times \cfrac { 3 }{ 4 } )\\ from(i)\\ f(\cfrac { 3 }{ 4 } \times \cfrac { 3 }{ 4 } )=f(\cfrac { 3 }{ 4 } +\cfrac { 3 }{ 4 } )\\ f(\cfrac { 3 }{ 4 } \times 2)=f(\cfrac { 3 }{ 2 } )\\ \because f(\cfrac { 3 }{ 4 } )=\cfrac { 3 }{ 4 } \quad constant\quad func.\\ \therefore f(\cfrac { 3 }{ 2 } )=\cfrac { 3 }{ 4 } $$

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Re-Attempt Weekly Quiz Competition

Relations Test 27

Result

Relations Test 27

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Rank

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SHARING IS CARING

Question 1

reflexive only

reflexive and symmetric only

transitive only

equivalence

Solution

Question 2

$$2^n$$

$$2^{n^2}$$

$$n^2$$

$$n^n$$

Solution

Question 3

$$\left\{(3,2),(3,3),(3,5)\right\}$$

$$\left\{(3,2),(3,5),(3,6)\right\}$$

$$\left\{(3,2),(3,5)\right\}$$

$$none\ of\ these$$

Solution

Question 4

$$\left\{(3,1),(3,2),(3,3),(3,8)\right\}$$

$$\left\{(1,3),(2,3),(3,3),(8,3)\right\}$$

$$\left\{(1,2),(2,2),(3,3),(8,8)\right\}$$

$$\left\{(8,3),(8,2),(8,1),(8,8)\right\}$$

Solution

Question 5

$$11$$

$$14$$

$$20$$

$$21$$

None of these

Solution

Question 6

$$0\lt \alpha\le \dfrac{1}{2}$$

$$\dfrac{1}{2}\lt \alpha\lt 1$$

$$2\alpha^4-4\alpha^2+1=0$$

$$\alpha^4+4\alpha^2-1=0$$

Solution

Question 7

$$x\epsilon R -\{1,-1\}$$

$$x\epsilon R -\{0\}$$

$$x\epsilon R -\{-1\}$$

$$x\epsilon R -\{1\}$$

Solution

Question 8

d = bc

c = bd

b = cd

none

Solution

Question 9

one-one and into

neither one-one nor onto

many-one and onto

one-one and onto

Solution

Question 10

$$\dfrac{3}{4}$$

$$\dfrac{9}{16}$$

$$\dfrac{\sqrt{3}}{2}$$

$$0$$

Solution

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