Relations and Functions Test

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Relations and Functions Test - 39

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Weekly Quiz Competition

Question 1
1 / -0

If $$f\left( x \right) = \cfrac{{{4^x}}}{{{4^x} + 2}}$$, then $$f\left( {\cfrac{1}{{97}}} \right) + f\left( {\cfrac{2}{{97}}} \right) + .... + f\left( {\cfrac{{96}}{{97}}} \right)$$ is equal to :

A
$$1$$

B
$$48$$

C
$$- 48$$

D
None of these

Solution

$$f\left( x \right) = \cfrac{{{4^x}}}{{{4^x} + 2}}$$

$$f\left( {1 - x} \right) = \cfrac{{{4^{1 - x}}}}{{{4^{1 - x}} + 2}} = \cfrac{{\cfrac{4}{{{4^x}}}}}{{\cfrac{4}{{{4^x}}} + 2}} = \cfrac{4}{{4 + {{2.4}^x}}} = \cfrac{2}{{2 + {4^x}}}$$

$$f\left( x \right) + f\left( {1 - x} \right) = \cfrac{{{4^x} + 2}}{{{4^x} + 2}} = 1$$

$$ \Rightarrow f\left( {\cfrac{1}{{97}}} \right) + f\left( {\cfrac{2}{{97}}} \right) + ...... + f\left( {\cfrac{{96}}{{97}}} \right)$$

$$ = \left[ {f\left( {\cfrac{1}{{97}}} \right) + f\left( {\cfrac{{96}}{{97}}} \right)} \right] + \left[ {f\left( {\cfrac{2}{{97}}} \right) + f\left( {\cfrac{{95}}{{97}}} \right)} \right] + ..... + \left[ {f\left( {\cfrac{{48}}{{97}}} \right) + f\left( {\cfrac{{49}}{{97}}} \right)} \right]$$

$$ = 1 + 1 + 1 + 1 + ....... + 1(\therefore 48\,ones)$$

$$ = 48$$
Question 2
1 / -0

Let R be a relation from N to N defined by
$$R = \left\{ {\left( {a,\,b} \right):a,\,b\, \in \,N\,\,and\,\,a = {b^2}} \right\}$$

A
$$\left( {a,a} \right) \in R\,$$, for all $$\,a \in N$$

B
$$\left( {a,b} \right) \in R$$, implies $$\left( {b,a} \right) \in R$$

C
$$\left( {a,b} \right) \in R,\,\left( {b,c} \right) \in \,R$$ implies $$\left( {a,c} \right) \in R$$

D
None of these

Solution
Question 3
1 / -0

If $$A = \{2, 3, 5\}$$ and $$B = \{5, 7\}$$, find the set with highest number of elements:

A
$$A \times B$$

B
$$ B \times A$$

C
$$A \times A$$

D
$$B \times B$$

Solution

$$A=\left \{ 2,3,5 \right \}$$

$$B=\left \{ 5,7 \right \}$$

$$A\times B=\left \{ (2,5),(2,7),(3,5),(3,7),(5,5),(5,7) \right \}$$

$$B\times A=\left \{ (5,2),(5,3),(5,5),(7,2),(7,3),(7,5) \right \}$$

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

$$B\times B=\left \{ (5,5),(5,7),(7,5),(7,7) \right \}$$

$$\therefore A\times A$$ has the highest number of elements
Question 4
1 / -0

Let $$A = \left\{ {a,\,b,\,c} \right\}$$ and $$B = \left\{ {4,\,5} \right\}$$. Consider a relation defined from set A to set B, then R is equal to

A
A

B
B

C
$$A \times B$$

D
$$B \times A$$

Solution

A relation from set $$A$$ to set $$B$$ is called
$$A\times B$$ and have following elements
$$A\times B=\left\{(a, 4),\ (a, 5),\ (b, 4),\ (b, 5),\ (c, 4),\ (c, 5)\right\}$$
Total $$6$$ elements
$$C$$ is correct
Question 5
1 / -0

Let $$A$$ and $$B$$ be two sets containing four and two elements respectively.Then the number of subset of the set $$A \times B$$, each having at least three elements is

A
$$256$$

B
$$275$$

C
$$510$$

D
$$219$$

Solution

$$A$$ and $$B$$ containing $$4$$ and $$2$$ respectively
set A has 4 elements
set B has 2 elements
total number of element in $$(A\times B)=4\times 2=8$$
total number of subsets of $$(A\times B)={2}^{8}=256$$
Number of subsets having 0 element $$={ _{ }^{ 8 }{ C } }_{ 0 }=1$$
Number of subsets having 1 element $$={ _{ }^{ 8 }{ C } }_{ 1 }=8$$
number of subsets having 2 element $$={ _{ }^{ 8 }{ C } }_{ 2 }=\cfrac{8\times 7}{2}=28$$
Number of subsets having atleast 3 elements
$$=256-28-8-1=219$$
Question 6
1 / -0

If $$f:R \to R$$ is defined by $$f\left( x \right) = {x^2} - 3x + 2$$ and $$f\left( {{x^2} - 3x - 2} \right) = a{x^4} + b{x^3} + c{x^2} + dx + e$$ then $$a + b + c + d + e = $$

A
$$1$$

B
$$2$$

C
$$30$$

D
$$20$$

Solution
Question 7
1 / -0

Let $$A = \left\{ {x,\,y,\,z} \right\}$$ and $$B = \left\{ {1,\,2} \right\}$$. The number relations from A to B is

A
64

B
32

C
16

D
28

Solution

$$\Rightarrow$$ $$A=\{X,\,Y,\,Z\}$$ and $$B=\{1,\,2\}$$ [ Given ]
$$\Rightarrow$$ Number of relation from $$A$$ to $$B$$ $$=2^{Number\,of\,elements\,in\,A\times B}$$
$$=2^{n(A)\times n(B)}$$
$$\Rightarrow$$ Number of elements in set $$A=3$$
$$\Rightarrow$$ Number of elements in set $$B=2$$
$$\Rightarrow$$ Number of relation from $$A$$ to $$B$$ $$=2^{n(A)\times n(B)}$$
$$=2^{3\times 2}$$
$$=2^6$$
$$=2\times 2\times 2\times 2\times 2\times 2$$
$$=64$$
Question 8
1 / -0

If $$f:R\rightarrow R$$ and $$g:R\rightarrow R$$ are defined by $$f\left( x \right) =x-\left[ x \right]$$ and $$ g\left( x \right) =\left[ x \right]$$ for $$x\in R$$,where $$[x]$$ is the greatest integer not exceeding $$x$$,then for every $$x\in R$$,$$f(g(x))$$ is equal to

A
$$x$$

B
$$0$$

C
$$f(x)$$

D
$$g(x)$$

Solution

$$f(g(x))=[x]-[x]$$
As $$[[x]]=[x]$$
$$fog(x)=[x]-[x]=0$$
Question 9
1 / -0

If $$A = \left \{1, 2, 3, 4\right \}$$ and $$I_{A}$$ be the identify relation on $$A$$, then

A
$$(1, 2) \epsilon I_{A}$$

B
$$(2, 2) \epsilon I_{A}$$

C
$$(2, 1) \epsilon I_{A}$$

D
$$(3, 4) \epsilon I_{A}$$

Solution
Question 10
1 / -0

The number of pairs (a, b) of positive real numbers satisfying $$a^4+b^4 < 1$$ and $$a^2+b^2 > 1$$ is

A
$$0$$

B
$$1$$

C
$$2$$

D
More than $$2$$

Solution

$$a^4+b^4<1$$
$$a^2+b^2>1$$
$$-a^2-b^2<-1$$
$$a^4-a^2+b^4-b^2<0$$
$$(a^2-\cfrac{1}{2})^2 + (b^2 - \cfrac{1}{2})^2 < \cfrac{1}{2}$$
Hence, it is a circle of radius $$\cfrac{1}{\sqrt2}$$ and all the points within it satisfies this equation. Hence, more than 2

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Relations and Functions Test - 39

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Relations and Functions Test - 39

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Question 1

$$1$$

$$48$$

$$- 48$$

None of these

Solution

Question 2

$$\left( {a,a} \right) \in R\,$$, for all $$\,a \in N$$

$$\left( {a,b} \right) \in R$$, implies $$\left( {b,a} \right) \in R$$

$$\left( {a,b} \right) \in R,\,\left( {b,c} \right) \in \,R$$ implies $$\left( {a,c} \right) \in R$$

None of these

Solution

Question 3

$$A \times B$$

$$ B \times A$$

$$A \times A$$

$$B \times B$$

Solution

Question 4

A

B

$$A \times B$$

$$B \times A$$

Solution

Question 5

$$256$$

$$275$$

$$510$$

$$219$$

Solution

Question 6

$$1$$

$$2$$

$$30$$

$$20$$

Solution

Question 7

64

32

16

28

Solution

Question 8

$$x$$

$$0$$

$$f(x)$$

$$g(x)$$

Solution

Question 9

$$(1, 2) \epsilon I_{A}$$

$$(2, 2) \epsilon I_{A}$$

$$(2, 1) \epsilon I_{A}$$

$$(3, 4) \epsilon I_{A}$$

Solution

Question 10

$$0$$

$$1$$

$$2$$

More than $$2$$

Solution

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