Relations and Functions Test

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Relations and Functions Test - 38

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Weekly Quiz Competition

Question 1
1 / -0

Find the domain of the function defined as $$f(x)=\dfrac{1}{1-x^2}$$.

A
$$x\epsilon R -\{1,-1\}$$

B
$$x\epsilon R -\{0\}$$

C
$$x\epsilon R -\{-1\}$$

D
$$x\epsilon R -\{1\}$$

Solution

$$f\left(x\right)=\dfrac{1}{1-x^{2}}$$
For existance $$1-x^{2}\neq 0$$
$$\Rightarrow x\neq+1,-1$$
$$\Rightarrow x\in R-\left\{1,-1\right\}$$
Question 2
1 / -0

If $$aN = (ax/x \ \epsilon \ N)$$ and $$bN \cap cN = dN,$$, where $$b, c \ \epsilon N$$ are relatively prime, then

A
d = bc

B
c = bd

C
b = cd

D
none

Solution
Question 3
1 / -0

Let $$f\left( x \right) :\begin{cases} x,\quad x\quad is\quad rational \\ 0,\quad x\quad is\quad irrational \end{cases}$$
and
$$g\left( x \right) :\begin{cases} 0,\quad x\quad is\quad rational \\ x,\quad x\quad is\quad irrational \end{cases}$$

If $$f:R\rightarrow R$$ and $$g:R\rightarrow R$$, then $$(f-g)$$ is

A
one-one and into

B
neither one-one nor onto

C
many-one and onto

D
one-one and onto

Solution

When x is rational $$f(x)=x,g(x)=0$$

$$\therefore (f-g)=x-0=x$$.So for each rational no. in $$(f-g)$$ there is the same rational number in co domain $$f(x)=0,g(x)=x$$ [when x is irrational]

$$\therefore (f-g)=0-x=-x$$

So for each irrational number x in domain, we have $$(-x)$$ in co domain.

Conversely for each negative irrational number, there is an element in domain.

But for positive irrational no. there is no corresponding element.

Thus $$(f-g)$$ is one -one and into.
Question 4
1 / -0

If $$f:R \to R$$ is a function satisfying $$f(x + y) = f(xy)$$ for all $$x, y \in R$$ and $$f\left(\dfrac{3}{4}\right) = \left( \dfrac{3}{4} \right)$$ , then $$f\left( \dfrac{9}{16} \right)$$ =

A
$$\dfrac{3}{4}$$

B
$$\dfrac{9}{16}$$

C
$$\dfrac{\sqrt{3}}{2}$$

D
$$0$$

Solution

$$f(x+y)=f(xy)\quad -(i)\\ f(\cfrac { 3 }{ 4 } )=(\cfrac { 3 }{ 4 } )\\ f(\cfrac { 9 }{ 16 } )=f(\cfrac { 3 }{ 4 } \times \cfrac { 3 }{ 4 } )\\ from(i)\\ f(\cfrac { 3 }{ 4 } \times \cfrac { 3 }{ 4 } )=f(\cfrac { 3 }{ 4 } +\cfrac { 3 }{ 4 } )\\ f(\cfrac { 3 }{ 4 } \times 2)=f(\cfrac { 3 }{ 2 } )\\ \because f(\cfrac { 3 }{ 4 } )=\cfrac { 3 }{ 4 } \quad constant\quad func.\\ \therefore f(\cfrac { 3 }{ 2 } )=\cfrac { 3 }{ 4 } $$
Question 5
1 / -0

Find the domain of the function defined as $$f(x)=\dfrac{x+1}{2x+3}$$.

A
$$x\epsilon$$ R$$-$${$$\dfrac{3}{2}$$}

B
$$x\epsilon$$ R$$-$${$$\dfrac{-3}{2}$$}

C
$$x\epsilon$$ R
D

$$x\epsilon$$ R$$-$${$$\dfrac{2}{3}$$}
Solution

Given:

$$ f\left ( x \right ) = \frac{x + 1}{2x + 3} $$

$$ 2x + 3 = 0 $$

$$ x = -\frac{3}{2} $$

The domain is $$ R - \left \{ -\frac{3}{2} \right \} $$.

Hence, the domain of the function is $$ R - \left \{ -\frac{3}{2} \right \} $$.
Question 6
1 / -0

Let R be a relation from$$ A=\left\{ 1,2,3,4 \right\} to B=\left\{ 1,3,5 \right\}$$ such that $$R=[(a,b):a<b,where\ a\epsilon A\ and\ b\epsilon B]$$. What is $$R$$ equal to?

A
$$(1,3),(1,5),(2,3),(2,5),(3,5),(4,5)$$

B
$$(3,1),(5,1),(3,2),(5,2),(5,3),(5,4)$$

C
$$(3,3),(3,5),(5,3),(5,5)$$

D
$$(3,3),(3,4),(4,5)$$

Solution

$$A=\{ 1,2,3,4\} \quad ,\quad B=\{ 1,3,5\} \\ R=[(a,b):a<b,where\quad a\epsilon A\quad \& \quad b\epsilon B]\\ R=[(1,3),(1,5),(2,3),(2,5),(3,5),(4,5)]$$
Question 7
1 / -0

$$x*y=\sqrt {\dfrac {(x+y)(y^{2}-12x)}{(x-2)(y-7)}}$$ then what will be the value of $$5*9$$?

A
$$7$$

B
$$8$$

C
$$10$$

D
$$9$$

Solution

$$x\ast y=\sqrt{\dfrac{(x+y)(y^{2}-12x)}{(x-2)(y-7)}}$$

$$5\ast 9=\sqrt{\dfrac{(5+9)(81-60)}{(3)(2)}}=\sqrt{\dfrac{(14)(21)}{(3)(2)}}=\sqrt{(7)(7)}$$

$$=\sqrt{49}$$

$$=7$$
Question 8
1 / -0

If $$f:R \rightarrow S$$ defined by $$f(x)=\sin x-\sqrt {3} \cos x+1$$ is onto, then the interval of $$S$$ is

A
$$[-1,3]$$

B
$$[1,1]$$

C
$$[0,1]$$

D
$$[0,-1]$$

Solution

$$f(x)=(\cfrac { \sin { x } -\sqrt { 3 } \cos { x } }{ 2 } )2+1$$
$$(\cfrac { 1 }{ 2 } \sin { x } -\cfrac { \sqrt { 3 } }{ 2 } \cos { x } )\times 2$$
$$2\cos { (\cfrac { \pi }{ 6 } +x)+1 } $$
$$-1\le \cos { (\cfrac { \pi }{ 6 } +x)+1 }\le 1$$
$$\Longrightarrow -2\le 2\cos { (\cfrac { \pi }{ 6 } +x)+1 }\le 2$$
$$\Longrightarrow -2+1\le 2\cos { (\cfrac { \pi }{ 6 } +x)+1 }+1\le 2+1$$
$$\Longrightarrow -1\le 2\cos { (\cfrac { \pi }{ 6 } +x)+1 }+1\le 3$$
$$\therefore$$ $$S\in[-1,3]$$.
Question 9
1 / -0

If $$f:R\rightarrow R$$ is defined by $$ f\left( x \right) =\left| x \right| $$, then:

A
$${ f }^{ -1 }(x)=-x$$

B
$${ f }^{ -1 }(x)=\dfrac { 1 }{ \left| x \right| } $$

C
$$f^{-1} (x)$$ does not exist

D
$${ f }^{ -1 }(x)=\dfrac { 1 }{ x } $$

Solution

$$f^{-1} (x)$$ does not exist.
For $${ f }^{ -1 }(x)={ x }$$ to exists ,$$f(x)$$ must be subjective.But in case when $$f\left( x \right) =\left| x \right| $$ ,$$f(x)$$ is not injective(one-one).
Therefore $$f^{-1}(x)$$ does not exist.
Question 10
1 / -0

If $$f:\,R\, \to R\,$$ is an even function which is differentiable on R and $${f^N}\left( \pi \right) = 1\,the\,{f^N}\left( { - \pi } \right)\,{\rm{is}}$$

A
-1

B
0

C
1

D
2

Solution

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Relations and Functions Test - 38

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Relations and Functions Test - 38

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Question 1

$$x\epsilon R -\{1,-1\}$$

$$x\epsilon R -\{0\}$$

$$x\epsilon R -\{-1\}$$

$$x\epsilon R -\{1\}$$

Solution

Question 2

d = bc

c = bd

b = cd

none

Solution

Question 3

one-one and into

neither one-one nor onto

many-one and onto

one-one and onto

Solution

Question 4

$$\dfrac{3}{4}$$

$$\dfrac{9}{16}$$

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

$$0$$

Solution

Question 5

$$x\epsilon$$ R$$-$${$$\dfrac{3}{2}$$}

$$x\epsilon$$ R$$-$${$$\dfrac{-3}{2}$$}

$$x\epsilon$$ R

$$x\epsilon$$ R$$-$${$$\dfrac{2}{3}$$}

Solution

Question 6

$$(1,3),(1,5),(2,3),(2,5),(3,5),(4,5)$$

$$(3,1),(5,1),(3,2),(5,2),(5,3),(5,4)$$

$$(3,3),(3,5),(5,3),(5,5)$$

$$(3,3),(3,4),(4,5)$$

Solution

Question 7

$$7$$

$$8$$

$$10$$

$$9$$

Solution

Question 8

$$[-1,3]$$

$$[1,1]$$

$$[0,1]$$

$$[0,-1]$$

Solution

Question 9

$${ f }^{ -1 }(x)=-x$$

$${ f }^{ -1 }(x)=\dfrac { 1 }{ \left| x \right| } $$

$$f^{-1} (x)$$ does not exist

$${ f }^{ -1 }(x)=\dfrac { 1 }{ x } $$

Solution

Question 10

-1

0

1

2

Solution

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