Relations and Functions Test

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Relations and Functions Test - 32

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Weekly Quiz Competition

Question 1
1 / -0

Let A be a finite set containing n distinct elements. The number of relations that can be defined on A is.

A
$$2^n$$

B
$$n^2$$

C
$$2n^2$$

D
$$2n$$
Question 2
1 / -0

If $$f: R \rightarrow R$$ and $$g: R \rightarrow R$$ are defined by $$f(x) =3x -4$$, and $$g(x)=2 + 3x$$, find $$(g^{-1}\, of^{-1})(5)$$.

A
$$1$$

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

C
$$\dfrac {1}{3}$$

D
$$\dfrac {1}{5}$$

Solution

Given, $$f(x)=3x-4$$
Let $$y=3x-4$$
$$ \Rightarrow x=\dfrac { y+4 }{ 3 } \\ \Rightarrow f^{ -1 }\left( x \right) =\dfrac { x+4 }{ 3 } $$
Also given $$ g(x)=2+3x$$
$$\Rightarrow y=2+3x\\ \Rightarrow x=\dfrac { y-2 }{ 3 } \\ \Rightarrow g^{ -1 }\left( x \right) =\dfrac { x-2 }{ 3 } $$
Thus $$g^{ -1 }\left( \text{of}^{ -1 } \right) \left( x \right) =\dfrac { \left( \dfrac { x+4 }{ 3 } \right) -2 }{ 3 } $$
$$ \Rightarrow g^{ -1 }\left( \text{of}^{ -1 } \right) \left( x \right) =\dfrac { x-2 }{ 9 } \\ \Rightarrow g^{ -1 }\left( \text{of}^{ -1 } \right) \left( 5 \right) =\dfrac { 5-2 }{ 9 } =\dfrac { 1 }{ 3 } $$
So, option C is correct.
Question 3
1 / -0

Directions For Questions

Let $$f: A\rightarrow B, g: B\rightarrow C$$ and $$h:C\rightarrow D$$ be three functions, then the function $$gof:A \rightarrow C$$ defined by gof(x) g[f(x)] for all $$x \epsilon A$$ is called the composition of f and g.

...view full instructions

For what value of x is $$fog = gof$$ if $$f(x)=x - 2$$ and $$g(x)=x^3+3$$?

A
$$\dfrac{-2}{3}$$

B
$$-1$$

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

D
$$\dfrac{-3}{2}$$

Solution

$$f(x)=x-2\\ g(x)={ x }^{ 3 }+3\\ f[g(x)]=g(x)-2\\ \Rightarrow fog={ x }^{ 3 }+3-2={ x }^{ 3 }+1\\ g[f(x)]={ \{ f(x)\} }^{ 3 }+3\\ \Rightarrow gof={ (x-2) }^{ 3 }+3\\ fog=gof\\ { x }^{ 3 }+1={ (x-2) }^{ 3 }+3\\ { x }^{ 3 }-2={ (x-2) }^{ 3 }\\ { x }^{ 3 }-2={ x }^{ 3 }-8-6x(x-2)\\ { x }^{ 3 }-2={ x }^{ 3 }-8-6{ x }^{ 2 }+12x\\ \Rightarrow { x }^{ 2 }+1+2x=0\\ \Rightarrow { (x+1) }^{ 2 }=0\\ \Rightarrow x=-1$$
So none of the above options are correct
Question 4
1 / -0

If $$f: R \rightarrow R$$ and $$g: R \rightarrow R$$ are defined by $$f(x) =2x +3, g(x)=x^2 + 7$$, what are the values of $$x$$ such that $$g(f(x))=8$$?

A
$$1, 2$$

B
$$-1, 2$$

C
$$-1, -2$$

D
$$1, -2$$

Solution

Given, $$f(x)=2x+3$$
Also given $$ g(x)={ x }^{ 2 }+7$$
Therefore, $$ g[f(x)]={ \left( f(x) \right) }^{ 2 }+7$$
$$ \Rightarrow g[f(x)]={ \left( 2x+3 \right) }^{ 2 }+7$$
Thus $$ { \left( 2x+3 \right) }^{ 2 }+7=8\\ \Rightarrow { \left( 2x+3 \right) }^{ 2 }=1\\ \Rightarrow 2x+3=\pm 1\\ \Rightarrow 2x=\pm 1-3\\ \Rightarrow 2x=-2,-4\\ \Rightarrow x=-1,-2$$
Option C is correct.
Question 5
1 / -0

If $$*$$ is a binary operation in $$A$$ then

A
$$A$$ is closed under $$*$$

B
$$A$$ is not closed under $$*$$

C
$$A$$ is closed under $$+$$

D
None of these

Solution

Given that $$*$$ is a binary operation in $$A$$
Since $$*$$ is a binary operation in $$A$$ , $$A$$ must be closed under $$*$$
Therefore correct option is $$A$$
Question 6
1 / -0

Directions For Questions

Given a function $$f : A \rightarrow B$$; where $$A = \left \{1, 2, 3, 4, 5\right \}$$ and $$B = \left \{6, 7, 8\right \}$$.

...view full instructions

Find number of all such functions $$y = f(x)$$ which are one-one?

A
$$0$$

B
$$3^{5}$$

C
$$^{5}P_{3}$$

D
$$5^{3}$$

Solution

One-one functions are those in which each element in the domain has a unique image in the range of the function.

Here $$B$$ is the co-domain of the function which has lesser number of elements than the domain itself. This shows that it is not possible for the function to be one-one.
Question 7
1 / -0

If $$U=\left \{ 1,2,3,4,5,6,7,8,9,10 \right \}$$ and $$A=\left \{ 2,5,6,9,10 \right \}$$ then $${A}'$$ is

A
$$\left \{ 2,5,6,9,10 \right \}$$

B
$$\phi$$

C
$$\left \{ 1,3,5,10 \right \}$$

D
$$\left \{ 1,3,4,7,8 \right \}$$

Solution

$$A'=U-A=\{1,3,4,7,8\}$$
Question 8
1 / -0

If $$f : [ 1, \infty) \rightarrow [2, \infty) $$ is given by $$f(x) = x + \dfrac{1}{x},$$ then $$f^{-1}(x)$$ is equal to

A
$$\dfrac{x + \sqrt{x^2 - 4}}{2}$$

B
$$\dfrac{x}{1 +x^2}$$

C
$$\dfrac{x- \sqrt{x^2 - 4}}{2}$$

D
$$1 + \sqrt{x - 4}$$

Solution

$$f(x)=x+\dfrac { 1 }{ x } \\ y=x+\dfrac { 1 }{ x } \\ xy={ x }^{ 2 }+1\\ { x }^{ 2 }-xy+1=0$$
Using quadratic formula
$$x=\dfrac { y\pm \sqrt { { y }^{ 2 }-4(1)(1) } }{ 2(1) } \\ x=\dfrac { y\pm \sqrt { { y }^{ 2 }-4 } }{ 2 } $$

$$\Rightarrow x=\dfrac { y+\sqrt { { y }^{ 2 }-4 } }{ 2 } \\ \Rightarrow f^{ -1 }\left( x \right) =\dfrac { x+\sqrt { { x }^{ 2 }-4 } }{ 2 } $$
Question 9
1 / -0

The inverse of the function $$y=\cfrac { { 2 }^{ x } }{ 1+{ 2 }^{ x } } $$ is

A
$$x=\log _{ 2 }{ \cfrac { 1 }{ 1-{ 2 }^{ y } } } $$

B
$$x=\log _{ 2 }{ \left( 1-\cfrac { 1 }{ y } \right) } $$

C
$$x=\log _{ 2 }{ \left( \cfrac { 1 }{ 1-y } \right) } $$

D
$$x=\log _{ 2 }{ \left( \cfrac { y }{ 1-y } \right) } $$

Solution

Given,
$$y=\cfrac { { 2 }^{ x } }{ 1+{ 2 }^{ x } } \Rightarrow y+{ 2 }^{ x }y={ 2 }^{ x }$$
$$\Rightarrow y={ 2 }^{ x }(1-y)\Rightarrow { 2 }^{ x }=\cfrac { y }{ 1-y } $$
Taking log on both sides at base $$2$$, we get
$$\log _{ 2 }{ \left( { 2 }^{ x } \right) } =\log _{ 2 }{ \cfrac { y }{ 1-y } } $$
$$\Rightarrow x=\log _{ 2 }{ \cfrac { y }{ 1-y } } $$
$$\therefore$$ Inverse is $$\log _{ 2 }{ \cfrac { y }{ 1-y } } $$
Question 10
1 / -0

The number of real linear functions $$f(x)$$ satisfying $$f\left\{ f(x) \right\} =x+f(x)$$

A
$$0$$

B
$$4$$

C
$$5$$

D
$$2$$

Solution

Let $$f(x) = ax+b $$
$$f(f(x))=a(ax+b)+b = a^2x+ab+b$$

Given, $$f(f(x))=x+f(x)$$
$$\implies a^2x+ab+b = x+ ax+b$$
$$\implies (a^2-a-1)x+ab=0$$
Since above equation is valid for all values of x, we have
$$a^2-a-1 =0$$ and $$ ab=0$$

$$a^2-a-1 =0$$
$$a=\dfrac{ 1\pm \sqrt{1+4}}{2}$$
$$\implies a=\dfrac{ 1\pm \sqrt{5}}{2}$$

$$ab=0$$
Since $$a$$ is non-zero,
$$\implies b=0$$

$$f(x) =\dfrac{ 1+ \sqrt{5}}{2}x$$ and $$f(x) =\dfrac{ 1- \sqrt{5}}{2}x$$ satisfy the condition.
Hence the answer is option (D)

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Re-Attempt Weekly Quiz Competition

Relations and Functions Test - 32

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Relations and Functions Test - 32

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SHARING IS CARING

Question 1

$$2^n$$

$$n^2$$

$$2n^2$$

$$2n$$

Question 2

$$1$$

$$\dfrac {1}{2}$$

$$\dfrac {1}{3}$$

$$\dfrac {1}{5}$$

Solution

Question 3

$$\dfrac{-2}{3}$$

$$-1$$

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

$$\dfrac{-3}{2}$$

Solution

Question 4

$$1, 2$$

$$-1, 2$$

$$-1, -2$$

$$1, -2$$

Solution

Question 5

$$A$$ is closed under $$*$$

$$A$$ is not closed under $$*$$

$$A$$ is closed under $$+$$

None of these

Solution

Question 6

$$0$$

$$3^{5}$$

$$^{5}P_{3}$$

$$5^{3}$$

Solution

Question 7

$$\left \{ 2,5,6,9,10 \right \}$$

$$\phi$$

$$\left \{ 1,3,5,10 \right \}$$

$$\left \{ 1,3,4,7,8 \right \}$$

Solution

Question 8

$$\dfrac{x + \sqrt{x^2 - 4}}{2}$$

$$\dfrac{x}{1 +x^2}$$

$$\dfrac{x- \sqrt{x^2 - 4}}{2}$$

$$1 + \sqrt{x - 4}$$

Solution

Question 9

$$x=\log _{ 2 }{ \cfrac { 1 }{ 1-{ 2 }^{ y } } } $$

$$x=\log _{ 2 }{ \left( 1-\cfrac { 1 }{ y } \right) } $$

$$x=\log _{ 2 }{ \left( \cfrac { 1 }{ 1-y } \right) } $$

$$x=\log _{ 2 }{ \left( \cfrac { y }{ 1-y } \right) } $$

Solution

Question 10

$$0$$

$$4$$

$$5$$

$$2$$

Solution

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