Relations and Functions Test

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Relations and Functions Test - 75

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Weekly Quiz Competition

Question 1
1 / -0

If $$f(x)=\displaystyle\frac{1}{3}x+3$$, then find $$f^{-1}(x)$$.

A
$$-\displaystyle\frac{1}{3}x-3$$

B
$$\displaystyle -3x+\frac{1}{3}$$

C
$$3x+\displaystyle\frac{1}{3}$$

D
$$3x-3$$

E
$$3x-9$$

Solution

$$f(x) = \dfrac { 1 }{ 3 } x + 3$$
Let the equation be $$y$$, then we get
$$f(x) = \dfrac { 1 }{ 3 } x+3 = y$$
Therefore $${ f }^{ -1 }y= x$$
$$\therefore \dfrac { 1 }{ 3 } x+3=y$$
$$\therefore x = 3(y-3)$$
$$\therefore { f }^{ -1 }y=x=3(y-3)$$
Now replace $$y$$ with $$x$$.
So, the answer is $$3x-9$$.
Question 2
1 / -0

If $$f(x)=2x^3$$ and $$g(x)=3x$$, calculate the value of $$g(f(-2))-f(g(2))$$.

A
$$-480$$

B
$$-384$$

C
$$0$$

D
$$384$$

E
$$480$$
Solution
- $$f(-2) = 2 \times { (-2) }^{ 3 } = -16$$
- $$g(2) = 3 \times 2 =6$$
- $$g(f(-2)) = g(-16) = 3 \times -16 = -48$$
- $$f(g(2)) = f(6) = 2 \times { (6) }^{ 3 } = 432$$
- $$g(f(-2)) - f(g(2)) = -48-432 = -480$$
Question 3
1 / -0

Find the inverse of the quadratic function $$f$$.
$$f(x)=-{x}^{2}+2, x>=0$$

A
$$\sqrt{(2+x)}$$

B
$$(2-x)$$

C
$$\sqrt{(2+x^2)}$$

D
$$\sqrt{(2-x)}$$

Solution

Let $$y=f(x)=-x^2+2$$ and now solve for $$x$$ to find the inverse of the given function.
$$\Rightarrow y=x^2+2$$
$$\Rightarrow -x^2=y-2$$
$$\Rightarrow x^2=2-y$$
$$\Rightarrow x=\pm \sqrt { 2-y }$$
Since it is given that $$x>=0$$, therefore,
$$x=\sqrt { 2-y }$$
Therefore, the inverse of the function in terms of $$x$$ is $$\sqrt { 2-x }$$.
Question 4
1 / -0

Find the inverse of the cube root function $$f$$.
$$f(x)={(x+1)}^{1/3}$$

A
$${x}^{3}+1$$

B
$${x}^{3}-1$$

C
$${x}-1$$

D
$${x}+1$$

Solution

Let $$y=f(x)={(x+1)}^{1/3}$$ and now solve for $$x$$ to find the inverse of the given function.
$$y={(x+1)}^{1/3}$$, cubing both sides we get,
$$y^3=x+1$$
$$x=y^3-1$$
Therefore, the inverse of the function in terms of $$x$$ is $$x^3-1$$.
Question 5
1 / -0

Find the inverse of the logarithmic function $$f$$.
$$f(x)=\ln(x)$$

A
$${e}^{x}$$

B
$$-{e}^{x}$$

C
$${e}^{2x}$$

D
$$-{e}^{2x}$$

Solution

Let $$y=f(x)=\ln { (x) }$$ and now solve for $$x$$ to find the inverse of the given function.
$$y=\ln { (x) }$$
$$x=e^y$$
Therefore, the inverse of the function in terms of $$x$$ is $$e^x$$.
Question 6
1 / -0

Find the inverse of the square root function $$f$$.
$$f(x)=\sqrt{x-1}$$

A
$$x^2-1$$

B
$$x^2+1$$

C
$$x+1$$

D
$$x-1$$

Solution

Let $$y=f(x)=\sqrt { x-1 }$$ and now solve for $$x$$ to find the inverse of the given function.
$$y=\sqrt { x-1 }$$, squaring both sides we get,
$$y^2=x-1$$
$$x=y^2+1$$
Therefore, the inverse of the function in terms of $$x$$ is $$x^2+1$$.
Question 7
1 / -0

Given two functions $$f(x)$$ and $$g(x)$$ such that $$f(x) = \sin (arctan x), g(x) =\tan (arc\sin x)$$, and $$0\leq x < \dfrac {\pi}{2}$$. The value of the composite function $$f\left (g\left (\dfrac {\pi}{10}\right )\right ) $$ is:

A
$$0.314$$

B
$$0.354$$

C
$$0.577$$

D
$$0.707$$

E
$$0.866$$

Solution

$$g(x)=\tan { (\sin ^{ -1 }{ x } ) } =\tan { \theta } $$
$$=\dfrac { x }{ \sqrt { 1-{ x }^{ 2 } } } $$
$$f(x)=\sin { (\tan ^{ -1 }{ x } ) } =\dfrac { x }{ \sqrt { 1+{ x }^{ 2 } } } $$
$$f(g(x))=\dfrac { g(x) }{ \sqrt { 1+{ g(x) }^{ 2 } } } $$
$$=\dfrac { \dfrac { x }{ \sqrt { 1-{ x }^{ 2 } } } }{ \sqrt { 1+\dfrac { x }{ \sqrt { 1-{ x }^{ 2 } } } } } =x$$
$$f(g(\dfrac { \pi }{ 10 } ))=0.314$$
Question 8
1 / -0

Find the maximum value of $$g(f(x))$$ if:
$$f(x) = x + 4$$ and
$$g(x) = 6 - x^{2}$$

A
$$-6$$

B
$$-4$$

C
$$2$$

D
$$4$$

E
$$6$$

Solution

Given, $$f(x)=x+4, g(x)=6-x^2$$
$$\therefore g(f(x)) = g(x+4) = 6-{(x+4)}^{2}$$
The minimum value of $${(x+4)}^{2}$$ is $$0$$ which occurs at $$x=-4$$.
So, the maximum value of $$g(f(x))$$ is $$6-0 = 6$$.
Question 9
1 / -0

If $$f(x) = 1 - 4x$$, and $$f^{-1}(x)$$ is the inverse of $$f(x)$$, then the value of $$f(-3) f{-1}(-3)=$$ is:

A
$$1$$

B
$$3$$

C
$$4$$

D
$$10$$

E
$$13$$

Solution

Let $$f(x) = 1-4x = y$$
$$\therefore x =\dfrac { (1-y)}{4}$$ and $$x = f^{ -1 }\left( y \right) $$
Therefore $$x = f^{ -1 }\left( y \right) =\dfrac {(1-y)}{4}$$
So, $$ f(-3)f^{ -1 }\left( -3 \right) = \dfrac {(1+12)(1+3)}{4} = 13$$
Question 10
1 / -0

Given that $$f(x) = \dfrac {2x}{5} + \dfrac {7}{3}$$, find $$ f^{-1} (x) $$.

A
$$\dfrac {15}{6x + 35}$$

B
$$\dfrac {6}{15x + 35}$$

C
$$\dfrac {15x - 6}{35}$$

D
$$\dfrac {15x - 35}{6}$$

E
$$\dfrac {15 - 6x}{35}$$

Solution

Given, $$f(x) =\dfrac { 2x}{5} + \dfrac {7}{3} $$
Let $$f(x)$$ be $$y$$ , then $$f^{ -1 }\left( y \right) $$ is $$x$$
$$\therefore f(x) = \dfrac {2x}{5} + \dfrac {7}{3} = y$$
$$\therefore \dfrac {2x}{5} = y-\dfrac {7}{3} =\dfrac { (3y-7)}{3}$$
$$\therefore x =\dfrac { (15y-35)}{6} = f^{ -1 }\left( y \right) $$
$$\therefore f^{ -1 }\left( x \right) =\dfrac { (15x-35)}{6}$$

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Re-Attempt Weekly Quiz Competition

Relations and Functions Test - 75

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Relations and Functions Test - 75

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SHARING IS CARING

Question 1

$$-\displaystyle\frac{1}{3}x-3$$

$$\displaystyle -3x+\frac{1}{3}$$

$$3x+\displaystyle\frac{1}{3}$$

$$3x-3$$

$$3x-9$$

Solution

Question 2

$$-480$$

$$-384$$

$$0$$

$$384$$

$$480$$

Solution

Question 3

$$\sqrt{(2+x)}$$

$$(2-x)$$

$$\sqrt{(2+x^2)}$$

$$\sqrt{(2-x)}$$

Solution

Question 4

$${x}^{3}+1$$

$${x}^{3}-1$$

$${x}-1$$

$${x}+1$$

Solution

Question 5

$${e}^{x}$$

$$-{e}^{x}$$

$${e}^{2x}$$

$$-{e}^{2x}$$

Solution

Question 6

$$x^2-1$$

$$x^2+1$$

$$x+1$$

$$x-1$$

Solution

Question 7

$$0.314$$

$$0.354$$

$$0.577$$

$$0.707$$

$$0.866$$

Solution

Question 8

$$-6$$

$$-4$$

$$2$$

$$4$$

$$6$$

Solution

Question 9

$$1$$

$$3$$

$$4$$

$$10$$

$$13$$

Solution

Question 10

$$\dfrac {15}{6x + 35}$$

$$\dfrac {6}{15x + 35}$$

$$\dfrac {15x - 6}{35}$$

$$\dfrac {15x - 35}{6}$$

$$\dfrac {15 - 6x}{35}$$

Solution

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