Relations and Functions Test

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Relations and Functions Test - 37

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Weekly Quiz Competition

Question 1
1 / -0

Let $$\rho$$ be a relation defined on$$N$$, the set of natural numbers, as
$$\rho =\left\{ \left( x,y \right) \in N\times N:2x+y=41 \right\} $$ then

A
$$\rho$$ is an equivalence relation

B
$$\rho$$ is only reflexive relation

C
$$\rho$$ is only symmetric relation

D
$$\rho$$ is not transitive

Solution

$$\rho =\left\{ \left( x,y \right) \in N\times N:2x+y=41 \right\} $$
for reflexive relation $$xRx\Rightarrow 2x+x=41\Rightarrow x=\cfrac { 41 }{ 3 } \notin N$$
for symmetric $$\Rightarrow xRy\Rightarrow 2x+y=41\neq yRx\quad $$ (Not symmetric)
for transitive $$xRy\Rightarrow 2x+y=41$$ and $$\quad yRz\Rightarrow 2y+z=41,x$$ (Not transitive)
Question 2
1 / -0

For real values of x ,the range of $$\frac {x^2+2x+1}{X^2+2x-1}$$ is

A
$$(-\infty,0) \cup (1,\infty)$$

B
$$\begin{bmatrix}\frac{1}{2},2\end{bmatrix}$$

C
$$\begin{bmatrix} -\infty,\frac {-2}{9} \end{bmatrix}\cup (1,\infty)$$

D
$$(-\infty,-6)\cup(-2,\infty)$$

Solution

Let $$y=\frac{x^2+2x+1} {x^2+2x-1}$$
$$\Rightarrow yx^2+2xy-y=x^2+2x+1$$
$$\Rightarrow (y-1)x^2+2x(y-1)-(y+1)=0$$
$$B^2-4AC\ge 0$$ & $$y\neq 1,y\neq 0$$
Question 3
1 / -0

Let $$R = gS - 4$$. When $$S = 8, R = 16$$. When $$S = 10, R$$ is equal to

A
$$11$$

B
$$14$$

C
$$20$$

D
$$21$$

E
None of these

Solution

$$16 = 8g - 4; \therefore g = 2\dfrac {1}{2}; \therefore R = \left (2\dfrac {1}{2}\right ) 10 - 4 = 21$$.
Question 4
1 / -0

If $$f:R\rightarrow R,g\quad :R\rightarrow R$$ are defined by $$f\left( x \right) =5x-3,g(x)={ x }^{ 2 }+3,$$ then $$\left( { gof }^{ -1 } \right) \left( 3 \right) =$$

A
$$\frac { 25 }{ 3 }$$

B
$$\frac { 111 }{ 25 } $$

C
$$\frac { 9 }{ 25 } $$

D
$$\frac { 25 }{ 111 } $$

Solution

$$f\left( x \right) =5x-3$$
$${ f }^{ -1 }\left( x \right) =\frac { x+3 }{ 5 } $$
$$g\left( x \right) ={ x }^{ 2 }+3$$
$${ gof }^{ -1 }\left( 3 \right) =g\left( { f }^{ -1 }\left( 3 \right) \right) =g\left( \frac { 6 }{ 5 } \right) =\frac { 111 }{ 25 } $$
Question 5
1 / -0

If $$f(g(x))=g(f(x))=x$$ for all real numbers $$x$$ and $$f(2)=5$$ and $$f(5)=3$$, then the value of $$g(3)+g(f(2))$$ is

A
$$7$$

B
$$5$$

C
$$3$$

D
$$2$$

Solution

Since $$g$$ and $$f$$ inverse, $$g(3)=5$$ and os
$$g(3)+g(f(2))=5+2=7$$
$${ g }^{ -1 }(x)=f(x);{ f }^{ -1 }(x)=g(x)$$
$$f(2)=5$$
$${ f }^{ -1 }(5)=2;g(5)=2$$
$${ f }^{ -1 }(3)=5;g(3)=5\quad $$
$$ g(3) + g(f(2)) = g(3) + g(5) = 7$$
Question 6
1 / -0

The relation P defined from R to R as a P b $$\Leftrightarrow$$ 1 + ab > 0, for all a, b $$\epsilon$$ R is

A
reflexive only

B
reflexive and symmetric only

C
transitive only

D
equivalence

Solution

$$Pb\Leftrightarrow 1+ab>0$$

i)Reflexive: $$a\times a$$ for any real value except 0 is positive.Hence $$0\times 0+1>1\Rightarrow a\times a+1>0$$

ii)Symmetric: if $$a\times b+1>0$$ then $$b\times a+1$$ will also be $$>0$$

iii)Not Transitive: a=-2,b=0\Rightarrow -2\times 0+1>0$$

$$b=0,c=4\Rightarrow 0\times 4+1>0$$

but a is not related to c $$\because a=-2,c=4\Rightarrow -2\times 4+1<0$$
Question 7
1 / -0

If R is a relation on a finite set having n elements, then the number of relations on A is :

A
$$2^n$$

B
$$2^{n^2}$$

C
$$n^2$$

D
$$n^n$$

Solution

Here, the required number of relations on A can be found out by the formula $$2^{n^2}$$.
Hence, option B is correct.
Question 8
1 / -0

If $$f(x) = \dfrac {4x + x^{4}}{1 + 4x^{3}}$$ and $$g(x) = \ln \left (\dfrac {1 + x}{1 - x}\right )$$, then what is the value of $$f\cdot g \left (\dfrac {e - 1}{e + 1}\right )$$ equal to?

A
$$2$$

B
$$1$$

C
$$0$$

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

Solution

$$g\left(\dfrac{e-1}{e+1}\right)=\log_e\left\lbrace\dfrac{1+\left(\dfrac{e-1}{e+1}\right)}{1-\left(\dfrac{e-1}{e+1}\right)}\right\rbrace=\log_e\left(\dfrac{2e}{2}\right)=\log_ee=1$$

$$\therefore fog\left(\dfrac{e-1}{e+1}\right)=f\left(g\left(\dfrac{e-1}{e+1}\right)\right)=f(1)=\dfrac{4+1}{1+4}=1$$
Question 9
1 / -0

If the line $$x=\alpha$$ divides the area of the region $$R=\left\{(x,y)\in \mathbb{R}^2:x^3\le y\le x,0\le x\le 1\right\}$$ into two equal parts, then

A
$$0\lt \alpha\le \dfrac{1}{2}$$

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

C
$$2\alpha^4-4\alpha^2+1=0$$

D
$$\alpha^4+4\alpha^2-1=0$$

Solution

So, $$\int _{ 0 }^{ \alpha }{ (x-{ x }^{ 3 }) } dx=\int _{ \alpha }^{ 1 }{ (x-{ x }^{ 3 }) } dx\\ { [\cfrac { { x }^{ 2 } }{ 2 } -\cfrac { { x }^{ 4 } }{ 4 } ] }_{ 0 }^{ \alpha }={ [\cfrac { { x }^{ 2 } }{ 2 } -\cfrac { { x }^{ 4 } }{ 4 } ] }_{ \alpha }^{ 1 }\\ \cfrac { { \alpha }^{ 2 } }{ 2 } -\cfrac { { \alpha }^{ 4 } }{ 4 } =\cfrac { { 1 }^{ 2 } }{ 2 } -\cfrac { { 1 }^{ 4 } }{ 4 } -\cfrac { { \alpha }^{ 2 } }{ 2 } +\cfrac { { \alpha }^{ 4 } }{ 4 } \\ \cfrac { { 2\alpha }^{ 2 } }{ 2 } -\cfrac { 2{ \alpha }^{ 4 } }{ 4 } =\cfrac { { 1 } }{ 4 } ={ \alpha }^{ 2 }-\cfrac { { \alpha }^{ 4 } }{ 2 } \\ \cfrac { { 1 } }{ 2 } ={ 2\alpha }^{ 2 }-{ \alpha }^{ 4 }\\ { 2\alpha }^{ 4 }-4{ \alpha }^{ 2 }+1=0$$
Thus, $${ 2\alpha }^{ 4 }-4{ \alpha }^{ 2 }+1=0$$
Question 10
1 / -0

The period of the function
$$f(x)=2^{\sqrt{\dfrac{2}{1-\cos 2x}}}+4^{\sqrt{\dfrac{1+\tan^2x}{4}}}$$

A
$$2\pi$$

B
$$\pi$$

C
$$\dfrac{\pi}{2}$$

D
$$\dfrac{\pi}{4}$$

Solution

$$f(x)=2^{\sqrt{\dfrac{2}{1-\cos 2x}}}+4^{\sqrt{\dfrac{1+\tan^2 x}{4}}}$$

$$=2^{\sqrt{\dfrac{2}{1+2\sin^2 x-1}}}+4^{\sqrt{\dfrac{\sec^2 x}{4}}}$$

$$=2^{\sqrt{\dfrac{1}{\sin^2 x}}}+2^ {2 \times {\sqrt{\dfrac{\sec^2 x}{4}}}}$$

$$=2^{cosec x}+ 2^{sec x}$$

In case of $${cosec x} $$ and $${sec x}$$
Period $$ = 2π$$
Domain: $$-∞ < x <+∞ , x≠ (2n+1) π )/2 , n ε Z$$

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Relations and Functions Test - 37

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Relations and Functions Test - 37

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SHARING IS CARING

Question 1

$$\rho$$ is an equivalence relation

$$\rho$$ is only reflexive relation

$$\rho$$ is only symmetric relation

$$\rho$$ is not transitive

Solution

Question 2

$$(-\infty,0) \cup (1,\infty)$$

$$\begin{bmatrix}\frac{1}{2},2\end{bmatrix}$$

$$\begin{bmatrix} -\infty,\frac {-2}{9} \end{bmatrix}\cup (1,\infty)$$

$$(-\infty,-6)\cup(-2,\infty)$$

Solution

Question 3

$$11$$

$$14$$

$$20$$

$$21$$

None of these

Solution

Question 4

$$\frac { 25 }{ 3 }$$

$$\frac { 111 }{ 25 } $$

$$\frac { 9 }{ 25 } $$

$$\frac { 25 }{ 111 } $$

Solution

Question 5

$$7$$

$$5$$

$$3$$

$$2$$

Solution

Question 6

reflexive only

reflexive and symmetric only

transitive only

equivalence

Solution

Question 7

$$2^n$$

$$2^{n^2}$$

$$n^2$$

$$n^n$$

Solution

Question 8

$$2$$

$$1$$

$$0$$

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

Solution

Question 9

$$0\lt \alpha\le \dfrac{1}{2}$$

$$\dfrac{1}{2}\lt \alpha\lt 1$$

$$2\alpha^4-4\alpha^2+1=0$$

$$\alpha^4+4\alpha^2-1=0$$

Solution

Question 10

$$2\pi$$

$$\pi$$

$$\dfrac{\pi}{2}$$

$$\dfrac{\pi}{4}$$

Solution

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