Relations and Functions Test

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Relations and Functions Test - 42

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Weekly Quiz Competition

Question 1
1 / -0

$$f : R \rightarrow R$$ is a function defined by $$f(x)=10x-7$$. If $$g=f^{-1}$$ then $$g(x) = $$

A
$$\displaystyle \frac{1}{10x-7}$$

B
$$\displaystyle \frac{1}{10x+7}$$

C
$$\displaystyle \frac{x+7}{10}$$

D
$$\displaystyle \frac{x-7}{10}$$

Solution

Here, $$f(x)=10x-7$$
which is clearly bijective as linear in $$x$$.
Let $$y=f(x)$$. So,
$$y=10x-7$$
$$\Rightarrow y+7=10x$$
$$\Rightarrow \displaystyle \frac{y+7}{10} = x =g(y)$$ as $$(f(x)=y\Rightarrow x=f^{-1}(y))$$
$$\Rightarrow g(x) = \dfrac{x+7}{10}$$.
Question 2
1 / -0

If $$f(x) = \dfrac{2x+5}{x^{2} + x + 5}$$, then $$f\left [ f(- 1 ) \right ]$$ is equal to

A
$$\dfrac{149}{155}$$

B
$$\dfrac{155}{147}$$

C
$$\dfrac{155}{149}$$

D
$$\dfrac{147}{155}$$

Solution

Given expression is $$ f(x)=\dfrac { 2x+5 }{ { x }^{ 2 }+x+5 } $$
$$ \therefore f(-1)=\dfrac { 2\times (-1)+5 }{ (-1)^{ 2 }+(-1)+5 } =\dfrac { 3 }{ 5 } $$
$$ \therefore f\left( f\left( -1 \right) \right) =\dfrac { 2\times \dfrac { 3 }{ 5 } +5 }{ \left( \dfrac { 3 }{ 5 } \right) ^{ 2 }+\dfrac { 3 }{ 5 } +5 } =\dfrac { 155 }{ 149 }$$
Question 3
1 / -0

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

A
$$1$$

B
$$\dfrac12$$

C
$$\dfrac13$$

D
$$\dfrac15$$

Solution

$$ f(x)=3x-4$$
$$ \Rightarrow { f }^{ -1 }(x)=\dfrac { x+4 }{ 3 } $$;
$$ g(x)=2+3x$$
$$ \Rightarrow g^{ -1 }(x)=\dfrac { x-2 }{ 3 } $$
$$ { f }^{ -1 }(5)=\dfrac { 5+4 }{ 3 } =3$$
$$ \left( { g }^{ -1 }o{ f }^{ -1 } \right) (5)=g^{ -1 }\left[ { f }^{ -1 }(5) \right] =g^{ -1 }(3)$$
$$ =\dfrac { 3-2 }{ 3 } =\dfrac { 1 }{ 3 } $$
Question 4
1 / -0

If $$ABC\sim PQR$$, then AB:PQ =

A
$$AC:PB$$

B
$$AC:PR$$

C
$$AB:PR$$

D
$$AC:RQ$$

Solution
Question 5
1 / -0

If $$f$$ and $$g$$ are one-one functions from $$R\to R$$, then

A
$$f+g$$ is one-one

B
$$fg$$ is one-one

C
$$fog$$ is one-one

D
none of these

Solution

Let $${x_1},{x_2} \in R$$ be two distinct elements, then $$g\left( {{x_1}} \right) \ne g\left( {{x_2}} \right)$$, as $$g$$ is one-one function. Similarly $$f\left( {g\left( {{x_1}} \right)} \right) \ne f\left( {g\left( {{x_2}} \right)} \right)$$ as $$f$$ is also one-one function.
Hence, $$f\circ g$$ is one-one function.
Note that $$f+g$$ and $$f\cdot g$$ may not one-one functions even if $$f$$ and $$g$$ are one one. For example consider $$f\left( x \right) = x$$ and $$g\left( x \right) =- x$$, then $$f+g$$ and $$f\cdot g$$ are not one-one.
Question 6
1 / -0

If $$f : R^+ \rightarrow R$$ such that $$f(x) = \log_3 x$$ then $$f^{-1}(x) = $$

A
$$\log x^3$$

B
$$3^x$$

C
$$3^{-x}$$

D
$$3^{\tfrac1x}$$

Solution

Let $$f^{-1}(x)=y$$
$$\Rightarrow x=f(y)=\log_3y$$
$$\Rightarrow y = 3^x$$
Therefore,
$$y= f^{-1}(x)=3^x$$
Question 7
1 / -0

Which of the following functions are one-one?

A
$$f:R\rightarrow R$$ given by $$ f(x)={ 2x }^{ 2 }+1$$ for all $$\quad x\in R$$

B
$$g:Z\rightarrow Z$$ given by $$ g(x)={ x }^{ 4 }$$ for all $$\quad x\in R$$

C
$$h:R\rightarrow R$$ given by $$ h(x)={ x }^{ 3 }+4$$ for all $$\quad x\in R$$

D
$$\phi :C\rightarrow C$$ given $$ \phi (z)={ 2z }^{ 6 }+4$$ for all $$\quad x\in R$$

Solution

For all even powered function,
$$f(x_{1})=f(x_{2})$$ where $$x_{1}=\pm(x_{2})$$
Hence, the only possible one-one function is $$f(x)=x^3+4$$
$$f(x)$$ is a strictly increasing function. So, it is one-one.
Question 8
1 / -0

A mapping function $$f:X\rightarrow Y$$ is one-one, if

A
$$f({ x }_{ 1 })\neq f({ x }_{ 2 })\ $$for all $$ { x }_{ 1 },{ x }_{ 2 }\in X$$

B
$$f({ x }_{ 1 })=f({ x }_{ 2 })\Rightarrow { x }_{ 1 }={ x }_{ 2 }$$ for all $${ x }_{ 1 },{ x }_{ 2 }\in X$$

C
$${ x }_{ 1 }={ x }_{ 2 }\Rightarrow f({ x }_{ 1 })=f({ x }_{ 2 })$$ for all $${ x }_{ 1 },{ x }_{ 2 }\in X$$

D
none of these

Solution

For a function to be one one
$$f(x_{1})=f(x_{2})$$
Implies that
$$x_{1}=x_{2}$$
Hence the answer is option B
Question 9
1 / -0

Let $$\displaystyle f:R\rightarrow A=\left \{ y: 0\leq y< \dfrac{\pi}{2} \right \}$$ be a function such that $$\displaystyle f(x)=\tan^{-1}(x^{2}+x+k),$$ where $$k$$ is a constant. The value of $$k$$ for which $$f$$ is an onto function is

A
$$1$$

B
$$0$$

C
$$\displaystyle \frac{1}{4}$$

D
none of these

Solution

For $$f$$ to be an onto function, Range and co-domain of $$f$$ should be equal

$$\Rightarrow f(x)\ge 0 \forall x\in R$$

$$\Rightarrow \tan^{-1}(x^2+x+k)\ge 0$$

For the above equation to be valid for all $$x$$

We must have, the discriminant of $$x^2+x+k=0$$, is zero

$$ b^2-4ac=0$$

$$\Rightarrow 1-4k=0\Rightarrow k =\dfrac{1}{4}$$
Question 10
1 / -0

If $$f:R\rightarrow R$$ given by $$f(x)={ x }^{ 3 }+({ a+2)x }^{ 2 }+3ax+5$$ is one-one, then $$a$$ belongs to the interval

A
$$(-\infty ,1)$$

B
$$(1 ,\infty)$$

C
$$(1 ,4)$$

D
$$(4 ,\infty)$$

Solution

If $$ f(x) $$ is one one then $$ f'\left( x \right) $$ should be either positive or negative for all $$ x $$.
$$f'\left( x \right) = 3{ x }^{ 2 }+2(a+2)x+3a$$
$$\Rightarrow D = 4{ (a+2) }^{ 2 }-4\times3\times3a<0$$
$$\Rightarrow$$ $${ a }^{ 2 }-5a+4<0$$
$$\Rightarrow$$ $$(a-1)(a-4)<0$$
Hence, $$a\in \left( 1,4 \right) $$

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Re-Attempt Weekly Quiz Competition

Relations and Functions Test - 42

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Relations and Functions Test - 42

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SHARING IS CARING

Question 1

$$\displaystyle \frac{1}{10x-7}$$

$$\displaystyle \frac{1}{10x+7}$$

$$\displaystyle \frac{x+7}{10}$$

$$\displaystyle \frac{x-7}{10}$$

Solution

Question 2

$$\dfrac{149}{155}$$

$$\dfrac{155}{147}$$

$$\dfrac{155}{149}$$

$$\dfrac{147}{155}$$

Solution

Question 3

$$1$$

$$\dfrac12$$

$$\dfrac13$$

$$\dfrac15$$

Solution

Question 4

$$AC:PB$$

$$AC:PR$$

$$AB:PR$$

$$AC:RQ$$

Solution

Question 5

$$f+g$$ is one-one

$$fg$$ is one-one

$$fog$$ is one-one

none of these

Solution

Question 6

$$\log x^3$$

$$3^x$$

$$3^{-x}$$

$$3^{\tfrac1x}$$

Solution

Question 7

$$f:R\rightarrow R$$ given by $$ f(x)={ 2x }^{ 2 }+1$$ for all $$\quad x\in R$$

$$g:Z\rightarrow Z$$ given by $$ g(x)={ x }^{ 4 }$$ for all $$\quad x\in R$$

$$h:R\rightarrow R$$ given by $$ h(x)={ x }^{ 3 }+4$$ for all $$\quad x\in R$$

$$\phi :C\rightarrow C$$ given $$ \phi (z)={ 2z }^{ 6 }+4$$ for all $$\quad x\in R$$

Solution

Question 8

$$f({ x }_{ 1 })\neq f({ x }_{ 2 })\ $$for all $$ { x }_{ 1 },{ x }_{ 2 }\in X$$

$$f({ x }_{ 1 })=f({ x }_{ 2 })\Rightarrow { x }_{ 1 }={ x }_{ 2 }$$ for all $${ x }_{ 1 },{ x }_{ 2 }\in X$$

$${ x }_{ 1 }={ x }_{ 2 }\Rightarrow f({ x }_{ 1 })=f({ x }_{ 2 })$$ for all $${ x }_{ 1 },{ x }_{ 2 }\in X$$

none of these

Solution

Question 9

$$1$$

$$0$$

$$\displaystyle \frac{1}{4}$$

none of these

Solution

Question 10

$$(-\infty ,1)$$

$$(1 ,\infty)$$

$$(1 ,4)$$

$$(4 ,\infty)$$

Solution

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