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

$fog\left ( x \right )=f\left(g ( x \right ))=f (x^{2})$
$=\sin x^{2}$

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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 $g(x)=x^2+\dfrac {5}{2}$
$\therefore f(6)=\dfrac{6}{4}-1=\dfrac{3}{2}-1=\dfrac{1}{2}$
$\therefore (gof)(6)=g(\dfrac{1}{2})=\left(\dfrac{1}{2}\right)^2+\dfrac{5}{2}=\dfrac{11}{4}=2.75$

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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.

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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.

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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

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

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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, $f(x)=4x+1$
$\therefore f(x^2-2)=4(x^2-2)+1$
$\Rightarrow 4x^2-8+1$
$\Rightarrow 4x^2-7$

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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

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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.

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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:
$f(x)=5=5(x)^{0}$
$f(x+r)=5(x+r)^{0}$
$f(x+r)=5\times 1$
$f(x+r)=5=f(x)$

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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

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Functions Test 2

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Question 1

x2+sin⁡xx^{2}+\sin xx2+sinx

x2sin⁡xx^{2}\sin xx2sinx

sin⁡2x\sin^{2}xsin2x

sin⁡x2\sin x^{2}sinx2

Solution

Question 2

2.75

3

3.5

8.625

Solution

Question 3

111

12\dfrac {1}{2}21​

13\dfrac {1}{3}31​

15\dfrac {1}{5}51​

Solution

Question 4

1,21, 21,2

−1,2-1, 2−1,2

−1,−2-1, -2−1,−2

1,−21, -21,−2

Solution

Question 5

5.85.85.8

7.47.47.4

7.77.77.7

8.28.28.2

10.310.310.3

Solution

Question 6

−x2+4x+1\displaystyle -{ x }^{ 2 }+4x+1−x2+4x+1

x2+4x−1\displaystyle { x }^{ 2 }+4x-1x2+4x−1

4x2−7\displaystyle 4{ x }^{ 2 }-74x2−7

4x2−1\displaystyle 4{ x }^{ 2 }-14x2−1

16x2+8x−1\displaystyle 16{ x }^{ 2 }+8x-116x2+8x−1

Solution

Question 7

−23\dfrac{-2}{3}3−2​

−1-1−1

32\dfrac{3}{2}23​

−32\dfrac{-3}{2}2−3​

Solution

Question 8

f(x)=x2+2f(x)=x^{2}+2f(x)=x2+2

f(x)=x+1xf(x)=x+\dfrac{1}{x}f(x)=x+x1​

f(x)=7f(x)=7f(x)=7

f(x)=6+xf(x)=6+xf(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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$x^{2}+\sin x$

$x^{2}\sin x$

$\sin^{2}x$

$\sin x^{2}$

$1$

$\dfrac {1}{2}$

$\dfrac {1}{3}$

$\dfrac {1}{5}$

$1, 2$

$-1, 2$

$-1, -2$

$1, -2$

$5.8$

$7.4$

$7.7$

$8.2$

$10.3$

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

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

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

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

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

$\dfrac{-2}{3}$

$-1$

$\dfrac{3}{2}$

$\dfrac{-3}{2}$

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

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

$f(x)=7$

$f(x)=6+x$