Relations and Functions Test

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Relations and Functions Test - 76

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Weekly Quiz Competition

Question 1
1 / -0

Find $$g(x)$$, if $$f(x) = 5x^{2} + 4$$ and $$f(g(3)) = 84$$

A
$$3x - 10$$

B
$$4x - 7$$

C
$$6x - 17$$

D
$$x^{2} - 5$$

E
$$x^{2} - 3$$

Solution

Given, $$f(x)=5x^2+4$$, $$f(g(3))=84$$
$$\Rightarrow f(g(3)) = 5{g(3)}^{2} +4 = 84$$
$$\Rightarrow 5{g(3)}^{2}=80$$
$$\Rightarrow g(3) = \sqrt{16} = 4$$
$$\Rightarrow g(x) = {x}^{2}-5$$ satisfies the relation $$g(3) = 4$$
Therefore, the correct option is $$D$$.
Question 2
1 / -0

If the binary operation $$a$$ # $$b = a^{b} - \sqrt {b}$$, then $$(2$$ # $$4) - (4$$ # $$2) =$$

A
$$-32$$

B
$$\sqrt {2} - 2$$

C
$$0$$

D
$$\sqrt {2} - 2$$

E
$$32$$

Solution
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

If $$h(x) = x^2, g(x)= x^2 -3$$ and f(x)= x -2, what can you say about ho(gof) and (hog)of?

A
(hog)of $$\neq $$ ho(gof)

B
ho(gof) = (hog)of

C
(hog)of = 4 ho(gof)

D
ho(gof)=$$(x^2-4x-1)^2$$

Solution
Question 4
1 / -0

Which of the following functions are not identical?

A
$$f(x)=\frac{x}{x^2}$$ and $$g(x) = \frac{1}{x}$$

B
$$f(x)=\frac{x^2}{x}$$ and $$g(x) =$$

C
$$f(x)=In \,x^4$$ and g(x)= 4 In Xx

D
f(x) = In {(x-1)(x-2)} and g(x) = In (x-2)+In (x-3)
Question 5
1 / -0

The function $$f(x)={x}^{2}+bx+c$$, where $$b$$ and $$c$$ real constants, describes

A
one-to-one mapping

B
onto mapping

C
not one-to-one onto mapping

D
neither one-to-one nor onto mapping

Solution

$$f\left( x \right) =x^{2}+bx+c$$ when $$f\left( x_{1} \right) =f\left( x_2 \right)$$
Since this a quadratic function and quadratic function is neither one-one nor onto.
so option D is correct.
Question 6
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

If f(x)=x+2and $$g(x)=x^2-3$$, then which is true?

A
fog $$\neq $$ gog

B
2fog = gof

C
fog = gof

D
fog = 2 gof

Solution
Question 7
1 / -0

Given $$g(x)=9\log_8(x-3)-5, g^{-1}(13)=$$

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

B
$$6$$

C
$$61$$

D
$$67$$

E
$$259$$

Solution

Let $$y=9\log _{ 8 }{ (x-3) } -5$$
Solve for $$x$$ as shown below:
$$y=9\log _{ 8 }{ (x-3) } -5\\ y+5=9\log _{ 8 }{ (x-3) } \\ \log _{ 8 }{ (x-3) } =\frac { y+5 }{ 9 } \\ x-3={ 8 }^{ \left( \frac { y+5 }{ 9 } \right) }\\ x={ 8 }^{ \left( \frac { y+5 }{ 9 } \right) }+3$$
Therefore, $$g^{ -1 }(x)={ 8 }^{ \left( \frac { y+5 }{ 9 } \right) }+3$$
Now, substitute $$x=13$$ in $$g^{ -1 }(x)={ 8 }^{ \left( \frac { y+5 }{ 9 } \right) }+3$$ that is:
$$g^{ -1 }(13)={ 8 }^{ \left( \frac { 13+5 }{ 9 } \right) }+3$$
$$g^{ -1 }(13)={ 8 }^{ \left( \frac { 18 }{ 9 } \right) }+3$$
$$g^{ -1 }(13)={ 8 }^2+3$$
$$g^{ -1 }(13)=64+3$$
$$g^{ -1 }(13)=67$$
Question 8
1 / -0

If $$f(x) = 4x^{2} - 1$$ and $$g(x) = 8x + 7, g\circ f(2) =$$

A
$$15$$

B
$$23$$

C
$$127$$

D
$$345$$

E
$$2115$$

Solution
Question 9
1 / -0

Let $$f:\{x, y , z\} \rightarrow \{1, 2, 3\}$$ be a one-one mapping such that only one of the following three statements and remaining two are false : $$f(x) \neq 2, f(y) =2, f(z) \neq 1$$, then

A
$$f(x) > f(y) > f(z)$$

B
$$f(x) < f(y) < f(z)$$

C
$$f(y) < f(y) < f(z)$$

D
$$f(y) < f(z) < f(x)$$

Solution

$$f:\left\{ x,y,z \right\} \rightarrow \left\{ 1,2,3 \right\} $$
$$I-f\left( x \right) \neq 2$$
$$II-f\left( y \right) =2$$
$$III-f\left( z \right) \neq 1$$
1. Let statement I is true, $$\therefore $$ other two are false
$$\therefore f\left( x \right) \neq 2,f\left( y \right) \neq 2$$
$$\therefore f\left( z \right) =1$$
$$f\left( x \right) \& f\left( y \right) $$ can be $$2$$ or $$3$$
We don't know exact value of $$f\left( x \right) \& f\left( y \right) $$
2. Let statement II is true
$$f\left( y \right) =2\quad \therefore f\left( z \right) =1\quad $$ [$$\because f\left( z \right) \neq 1$$ is false]
$$f\left( x \right) =3$$
$$f\left( x \right) >f\left( y \right) >f\left( z \right) $$
Question 10
1 / -0

$$f(x)=x^2-x+5$$ and $$g(x) = f^{-1}(x)$$. Then $$g'(7)=$$

A
$$1$$

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

C
$$\dfrac{1}{13}$$

D
$$\dfrac{1}{4}$$

Solution

$$f\left( x \right) ={ x }^{ 2 }-x+5=y$$
$$g\left( x \right) =f^{ -1 }\left( x \right) $$
$$y={ x }^{ 2 }-x+5$$
$$y={ x }^{ 2 }-x+\cfrac { 1 }{ 4 } -\cfrac { 1 }{ 4 } +5$$
$$y={ \left( x-\cfrac { 1 }{ 2 } \right) }^{ 2 }+\cfrac { 19 }{ 4 } $$
$$\sqrt { y-\cfrac { 19 }{ 4 } } +\cfrac { 1 }{ 2 } =x$$
$$\therefore g\left( x \right) =\sqrt { y-\cfrac { 19 }{ 4 } } +\cfrac { 1 }{ 2 } $$
$$g'\left( x \right) =\cfrac { 1 }{ 2\sqrt { y-\cfrac { 19 }{ 4 } } } $$
when, $$x=7,y=47$$
$$g'\left( 7 \right) =\cfrac { 1 }{ 2\sqrt { 47-\cfrac { 19 }{ 4 } } } $$
$$=\cfrac { 1 }{ \sqrt { 188-19 } } $$
$$=\cfrac { 1 }{ \sqrt { 169 } } $$
$$g'\left( 7 \right) =\cfrac { 1 }{ 13 } $$

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Re-Attempt Weekly Quiz Competition

Relations and Functions Test - 76

Result

Relations and Functions Test - 76

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SHARING IS CARING

Question 1

$$3x - 10$$

$$4x - 7$$

$$6x - 17$$

$$x^{2} - 5$$

$$x^{2} - 3$$

Solution

Question 2

$$-32$$

$$\sqrt {2} - 2$$

$$0$$

$$\sqrt {2} - 2$$

$$32$$

Solution

Question 3

(hog)of $$\neq $$ ho(gof)

ho(gof) = (hog)of

(hog)of = 4 ho(gof)

ho(gof)=$$(x^2-4x-1)^2$$

Solution

Question 4

$$f(x)=\frac{x}{x^2}$$ and $$g(x) = \frac{1}{x}$$

$$f(x)=\frac{x^2}{x}$$ and $$g(x) =$$

$$f(x)=In \,x^4$$ and g(x)= 4 In Xx

f(x) = In {(x-1)(x-2)} and g(x) = In (x-2)+In (x-3)

Question 5

one-to-one mapping

onto mapping

not one-to-one onto mapping

neither one-to-one nor onto mapping

Solution

Question 6

fog $$\neq $$ gog

2fog = gof

fog = gof

fog = 2 gof

Solution

Question 7

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

$$6$$

$$61$$

$$67$$

$$259$$

Solution

Question 8

$$15$$

$$23$$

$$127$$

$$345$$

$$2115$$

Solution

Question 9

$$f(x) > f(y) > f(z)$$

$$f(x) < f(y) < f(z)$$

$$f(y) < f(y) < f(z)$$

$$f(y) < f(z) < f(x)$$

Solution

Question 10

$$1$$

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

$$\dfrac{1}{13}$$

$$\dfrac{1}{4}$$

Solution

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