Functions Test 2

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Functions Test 2

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Weekly Quiz Competition

Question 1
1 / -0

$$f:R\rightarrow R , g:R\rightarrow R$$ and $$f(x)= \sin x$$, $$g(x)=x^{2}$$ then $$fog(x)=$$

A
$$x^{2}+\sin x$$

B
$$x^{2}\sin x$$

C
$$\sin^{2}x$$

D
$$\sin x^{2}$$

Solution

$$fog\left ( x \right )=f\left(g ( x \right ))=f (x^{2})$$
$$=\sin x^{2}$$
Question 2
1 / -0

Find the value of $$\displaystyle \left( g\circ f \right) \left( 6 \right) $$ if $$\displaystyle g\left( x \right) ={ x }^{ 2 }+\frac { 5 }{ 2 } $$ and $$\displaystyle f\left( x \right) =\frac { x }{ 4 } -1$$.

A
2.75

B
3

C
3.5

D
8.625

Solution

$$f(x)=\dfrac {x}{4}-1$$ and $$g(x)=x^2+\dfrac {5}{2}$$
$$\therefore f(6)=\dfrac{6}{4}-1=\dfrac{3}{2}-1=\dfrac{1}{2}$$
$$\therefore (gof)(6)=g(\dfrac{1}{2})=\left(\dfrac{1}{2}\right)^2+\dfrac{5}{2}=\dfrac{11}{4}=2.75$$
Question 3
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 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 $$f(x) = \sqrt {x^{2} - 3x + 6}$$ and $$g(x) = \dfrac {156}{x +17}$$, find the value of the composite function $$g(f(4))$$.

A
$$5.8$$

B
$$7.4$$

C
$$7.7$$

D
$$8.2$$

E
$$10.3$$

Solution

$$f(x)=\sqrt { x^{ 2 }-3x+6 } ,\quad g(x)=\frac{ 156 }{ x+17 }\\ g(f(4))=\quad ?\\ f(4)=\sqrt { 10 } \\ g(f(4))=g(\sqrt { 10 } )=\quad 3.16\\ g(f(4))=7.7$$
Question 6
1 / -0

Find the correct expression for $$\displaystyle f\left( g\left( x \right) \right) $$ given that $$\displaystyle f\left( x \right) =4x+1$$ and $$\displaystyle g\left( x \right) ={ x }^{ 2 }-2$$

A
$$\displaystyle -{ x }^{ 2 }+4x+1$$

B
$$\displaystyle { x }^{ 2 }+4x-1$$

C
$$\displaystyle 4{ x }^{ 2 }-7$$

D
$$\displaystyle 4{ x }^{ 2 }-1$$

E
$$\displaystyle 16{ x }^{ 2 }+8x-1$$

Solution

Given, $$g(x)=x^2-2$$
$$f(g(x))=f(x^2-2)$$
Also given, $$f(x)=4x+1$$
$$\therefore f(x^2-2)=4(x^2-2)+1$$
$$\Rightarrow 4x^2-8+1$$
$$\Rightarrow 4x^2-7$$
Question 7
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 8
1 / -0

Which of the following functions is/are constant ?

A
$$f(x)=x^{2}+2$$

B
$$f(x)=x+\dfrac{1}{x}$$

C
$$f(x)=7$$

D
$$f(x)=6+x$$

Solution

$$f(x)=7$$ is constant function as its values do not depend on the variable $$x$$.
Its value is $$7$$ for any value of $$x$$.
Option $$C$$ is correct.
Question 9
1 / -0

Let $$f:\mathbb{R}\rightarrow \mathbb{R}$$ be a function such that for any irrational number $$r,$$ and any real number $$x$$ we have $$f(x)=f(x+r)$$. Then, $$f$$ is

A
an identity function

B
a constant function

C
a zero function

D
onto function

Solution

A constant function is a function that has the same output value no matter what your input value is. Because of this, a constant function has the form y=k where k is a constant(a single value that doesn't change).
So in here our input is $$x$$ i f we change our input to say $$x+r$$
and still output does'nt change i.e.
$$f(x)=f(x+r)$$
it means it is a constant function,
example:
$$ f(x)=5=5(x)^{0} $$
$$ f(x+r)=5(x+r)^{0}$$
$$ f(x+r)=5\times 1 $$
$$ f(x+r)=5=f(x) $$
Question 10
1 / -0

If $$f: A \rightarrow B$$ is a bijective function and if n(A) = 5, then n(B) is equal to

A
10

B
4

C
5

D
25

Solution

We are given $$f$$ is a bijective function.

It means $$f$$ is one-one as well as onto function.

Therefore every element of $$B$$ is a image in $$f$$.

$$f$$ is one-one therefore image of every element is different.

$$\Rightarrow$$ number of elements in $$B$$ should be equal to number of elements in $$A$$.

$$\therefore n(B)=n(A)=5$$

Option $$(C)$$ is correct.

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Re-Attempt Weekly Quiz Competition

Functions Test 2

Result

Functions Test 2

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SHARING IS CARING

Question 1

$$x^{2}+\sin x$$

$$x^{2}\sin x$$

$$\sin^{2}x$$

$$\sin x^{2}$$

Solution

Question 2

2.75

3

3.5

8.625

Solution

Question 3

$$1$$

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

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

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

Solution

Question 4

$$1, 2$$

$$-1, 2$$

$$-1, -2$$

$$1, -2$$

Solution

Question 5

$$5.8$$

$$7.4$$

$$7.7$$

$$8.2$$

$$10.3$$

Solution

Question 6

$$\displaystyle -{ x }^{ 2 }+4x+1$$

$$\displaystyle { x }^{ 2 }+4x-1$$

$$\displaystyle 4{ x }^{ 2 }-7$$

$$\displaystyle 4{ x }^{ 2 }-1$$

$$\displaystyle 16{ x }^{ 2 }+8x-1$$

Solution

Question 7

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

$$-1$$

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

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

Solution

Question 8

$$f(x)=x^{2}+2$$

$$f(x)=x+\dfrac{1}{x}$$

$$f(x)=7$$

$$f(x)=6+x$$

Solution

Question 9

an identity function

a constant function

a zero function

onto function

Solution

Question 10

10

4

5

25

Solution

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