Relations and Functions Test

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Relations and Functions Test - 45

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Weekly Quiz Competition

Question 1
1 / -0

If $$f(x)=1-\dfrac {1}{x}$$, then $$ f\left[ f\left( \dfrac { 1 }{ x } \right) \right] $$ is

A
$$\dfrac {1}{x-1}$$

B
$$\dfrac {x}{x-1}$$

C
$$\dfrac {1}{x+1}$$

D
$$\dfrac {1}{x}$$

Solution
Question 2
1 / -0

If $$A=\left\{ 2,3,5 \right\} ,B=\left\{ 4,6,8 \right\} $$, then the relation from $$A$$ into $$B$$ is

A
$$\left\{(2,4),(3,5),(5,6)\right\}$$

B
$$\left\{(2,4),(5,8),(6,5)\right\}$$

C
$$\left\{(2,4),(3,6),(5,6)\right\}$$

D
$$\left\{(2,5),(3,6)\right\}$$
Question 3
1 / -0

If $$f(x)=\dfrac{a^{x}}{a^{x}+\sqrt{a}}(a>0)$$, then $$\sum_{r=1}^{2n-1}{2f\left(\dfrac{r}{2n}\right)}$$ is equal to

A
$$1$$

B
$$2n$$

C
$$2n-1$$

D
$$\dfrac{(2n-1)a}{2}$$

Solution

$$ \begin{aligned} \text { Solution } &-f(x)=\frac{a^{x}}{a^{x}+\sqrt{a}}(a>0) \\ & \Rightarrow \sum_{x=1}^{2 n-1} 2 f\left(\frac{r}{2 n}\right) \\ & \Rightarrow 2 f\left(\frac{1}{2 n}\right)+2 f\left(\frac{2}{2 n}\right)+\cdots+2 f\left(\frac{2 n-2}{2 n}\right)+2 f\left(\frac{2 n-1}{2 n}\right) \\ & \Rightarrow 2\left[f\left(\frac{1}{2 n}\right)+f\left(\frac{2}{2 n}\right)+\cdots+f\left(\frac{n}{2 n}\right)+f\left(\frac{n+1}{2 n}\right)+\cdots+f\left(\frac{2 n-1}{2 n}\right)\right] \end{aligned} $$ (1)

Now, $$\begin{aligned} f(x)+f(1-x) &=\frac{a^{x}}{a^{x}+\sqrt{a}}+\frac{\sqrt{a}}{\sqrt{a}+a^{x}} \\ &=\frac{a^{x}+\sqrt{a}}{a^{x}+\sqrt{a}}=1 \quad-(2) \end{aligned}$$

$$ \begin{array}{c} \text { Using (2) we get } \\ \Rightarrow 2\left[f\left(\frac{1}{2 n}\right)+f\left(\frac{2 n-1}{2 n}\right)+\ldots+f\left(\frac{n-1}{2 n}\right)+f\left(\frac{n+1}{2 n}\right)\right]+2 f\left(\frac{n}{2 n}\right) \\ \Rightarrow 2(n-1)+2 f\left(\frac{1}{2}\right)-(3) \\ \text { Now put } x=\frac{1}{2} \text { in } f(x)+f(1-x) \\ \Rightarrow f\left(\frac{1}{2}\right)+f\left(\frac{1}{2}\right)=1 \\ \Rightarrow f\left(\frac{1}{2}\right)=\frac{1}{2} \end{array} $$

$$ \begin{array}{l} \text { from equation } 3 \\ \Rightarrow \sum_{n=1}^{2 n-1} 2 f\left(\frac{n}{2 n}\right)=2(n-1)+\frac{1}{2} \\ =2\left(n-1+\frac{1}{2}\right) \\ =2 n-1 \\ \text { then },(c) \text { is the correct option. } \end{array} $$
Question 4
1 / -0

Directions For Questions

Let $$3f(x)+2f(-x)=x^{2}+x$$.

...view full instructions

Number real roots of $$f(f(x))=0$$ is-

A
$$0$$

B
$$1$$

C
$$2$$

D
$$3$$

Solution

$$3 f(x)+2 f(-x)=x^{2}+x$$
replace $$x$$ by $$-x$$ in equation
(1) $$\therefore \quad 3 f(-x)+2 f(x)=x^{2}-x$$
Multiply equation (1) by 2 and (2) by 3 and sulutract
$$\begin{aligned} & 2\{3 f(x)+2 f(-x)\}-3\{3 f(-x)+2 f(x)\}=2 x^{2}+x-3 x^{2}+3 x \\ \Rightarrow & 6 f(x)+4 f(-x)-9 f(-x)-6 f(x)=4 x-x^{2} \\ \Rightarrow &-5 f(-x)=4 x-x^{2} \end{aligned}$$
$$\Rightarrow \quad 5 f(-x)=x^{2}-4 x$$(3)
replace $$(-x)$$ by $$x$$ in equation (3)
$$\begin{aligned} & 5 f(x)=x^{2}+4 x \\ \therefore & f(x)=\frac{1}{5}\left(x^{2}+4 x\right) \end{aligned}$$
$$f(f(x))=0$$ if $$f(x)=0$$ and $$f(x)=-4$$
For, $$f(x)=0$$ at $$x=-4$$ and $$x=0$$
For, $$f(x)=-4$$
$$\Rightarrow \frac{x^{2}}{5}+\frac{4}{5} x=-4 \Rightarrow x^{2}+4 x+20=0$
$x^{2}+4 x+20$$ has no real roots.
Real roots of $$f(f(x))=0$$ are
$$x=-4$$ and $$x=0$$
so Answer: option (c).
Question 5
1 / -0

If a set $$A$$ has $$n$$ elements then the number of relations defined on $$A$$ is

A
$$2^{ \left( { n }^{ 2 } \right) }$$

B
$$2^{ \left( { n }^{ 2 } \right) }-1$$

C
$$2^{ n }$$

D
None of these.

Solution
Question 6
1 / -0

If A= {a, b} then possible number of relation on the set A.

A
$$2$$

B
$$4$$

C
$$16$$

D
none of these

Solution
Question 7
1 / -0

If n(A)=4, n(B)=3, $$n(A\times B\times C)=24,then\quad n(C)$$ is equal to

A
288

B
1

C
2

D
12

Solution

$$n\left(A\right)=4\,\,n\left(B\right)=3\,\,n\left(A\times B\times C\right)=24$$
We have $$n\left(A\times B\times C\right)=n\left(A\right)\times n\left(B\right)\times n\left(C\right)$$
$$\Rightarrow 24=4\times 3\times n\left(C\right)$$
$$\therefore n\left(C\right)=\dfrac{24}{12}=2$$
Question 8
1 / -0

Let A and B be two sets such that $$A\times B=\left\{ \left( a,1 \right) ,\left( b,3 \right) ,\left( a,3 \right) ,\left( b,1 \right) ,\left( a,2 \right) ,\left( b,2 \right) \right\} ,$$ then

A
$$A=\left\{ 1,2,3 \right\} $$ and $$B=\left\{ a,b \right\} $$

B
$$A=\left\{ a,b \right\} $$ and$$ B=\left\{ 1,2,3 \right\} $$

C
$$A=\left\{ 1,2,3 \right\} $$ and $$B\subset \left\{ a,b \right\} $$

D
$$A\subset \left\{ a,b \right\} $$ and $$B\subset \left\{ 1,2,3 \right\} $$

Solution

$$A$$ is the first element in the cartesian product $$A\times B=\left\{a\,,b\,,a\,,b\,,a\,,b\right\}$$
and $$B$$ is the second element in the cartesian product $$A\times B=\left\{1,\,3\,,3\,,1\,,2\,,2\right\}$$
$$\therefore$$ elements of $$A=\left\{a,b\right\}$$ and $$B=\left\{1\,,2\,,3\right\}$$
Question 9
1 / -0

A relation R is defined from {2,3,4,5} to {3,6,7,10} by XRY $$\Leftrightarrow $$ X is relatively prime to Y, then domain of R is

A
{2,3,5}

B
{3,5}

C
{2,5}

D
{2,3,4,5}

Solution

Given:
From $$\left\{2,3,4,5\right\}$$ to $$\left\{3,6,7,10\right\}$$

$$x$$ is related to $$y$$ iff $$x$$ is relatively prime to $$y$$

$$2$$ is relatively prime to $$3,7$$

$$3$$ is relatively prime to $$7,10$$

$$4$$ is relatively prime to $$3,7$$

$$5$$ is relatively prime to $$3,6,7$$

So, domain of $$R$$ is $$\left\{2,3,4,5\right\}$$
Question 10
1 / -0

If $$p(x)=\dfrac{x}{15}, x=1,2,3,4,5=0$$ otherwise, then $$p(x=1\ or\ 2)$$ is

A
$$\dfrac{1}{15}$$

B
$$\dfrac{2}{15}$$

C
$$\dfrac{1}{5}$$

D
$$None of these$$

Solution

$$ \begin{array}{l} P(x)=\frac{x}{15}, x=1,2,3,4,5 \\ =0 \text { other wise } \\ \qquad \begin{aligned} p(x=1 \operatorname{or} 2) &=p(x=1)+p(x=2) \\ &=\frac{1}{15}+\frac{2}{15}=\frac{1}{5} \end{aligned} \end{array} $$

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Re-Attempt Weekly Quiz Competition

Relations and Functions Test - 45

Result

Relations and Functions Test - 45

Score

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Rank

-

SHARING IS CARING

Question 1

$$\dfrac {1}{x-1}$$

$$\dfrac {x}{x-1}$$

$$\dfrac {1}{x+1}$$

$$\dfrac {1}{x}$$

Solution

Question 2

$$\left\{(2,4),(3,5),(5,6)\right\}$$

$$\left\{(2,4),(5,8),(6,5)\right\}$$

$$\left\{(2,4),(3,6),(5,6)\right\}$$

$$\left\{(2,5),(3,6)\right\}$$

Question 3

$$1$$

$$2n$$

$$2n-1$$

$$\dfrac{(2n-1)a}{2}$$

Solution

Question 4

$$0$$

$$1$$

$$2$$

$$3$$

Solution

Question 5

$$2^{ \left( { n }^{ 2 } \right) }$$

$$2^{ \left( { n }^{ 2 } \right) }-1$$

$$2^{ n }$$

None of these.

Solution

Question 6

$$2$$

$$4$$

$$16$$

none of these

Solution

Question 7

288

1

2

12

Solution

Question 8

$$A=\left\{ 1,2,3 \right\} $$ and $$B=\left\{ a,b \right\} $$

$$A=\left\{ a,b \right\} $$ and$$ B=\left\{ 1,2,3 \right\} $$

$$A=\left\{ 1,2,3 \right\} $$ and $$B\subset \left\{ a,b \right\} $$

$$A\subset \left\{ a,b \right\} $$ and $$B\subset \left\{ 1,2,3 \right\} $$

Solution

Question 9

{2,3,5}

{3,5}

{2,5}

{2,3,4,5}

Solution

Question 10

$$\dfrac{1}{15}$$

$$\dfrac{2}{15}$$

$$\dfrac{1}{5}$$

$$None of these$$

Solution

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