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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Relations and Functions Test - 45

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Question 1

1x−1\dfrac {1}{x-1}x−11​

xx−1\dfrac {x}{x-1}x−1x​

1x+1\dfrac {1}{x+1}x+11​

1x\dfrac {1}{x}x1​

Solution

Question 2

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

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

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

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

Question 3

111

2n2n2n

2n−12n-12n−1

(2n−1)a2\dfrac{(2n-1)a}{2}2(2n−1)a​

Solution

Question 4

000

111

222

333

Solution

Question 5

2(n2)2^{ \left( { n }^{ 2 } \right) }2(n2)

2(n2)−12^{ \left( { n }^{ 2 } \right) }-12(n2)−1

2n2^{ n }2n

None of these.

Solution

Question 6

222

444

161616

none of these

Solution

Question 7

288

1

2

12

Solution

Question 8

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

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

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

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

Solution

Question 9

{2,3,5}

{3,5}

{2,5}

{2,3,4,5}

Solution

Question 10

115\dfrac{1}{15}151​

215\dfrac{2}{15}152​

15\dfrac{1}{5}51​

NoneoftheseNone of theseNoneofthese

Solution

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

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

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

$\dfrac {1}{x}$

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

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

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

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

$1$

$2n$

$2n-1$

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

$0$

$1$

$2$

$3$

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

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

$2^{ n }$

$2$

$4$

$16$

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

$\dfrac{1}{15}$

$\dfrac{2}{15}$

$\dfrac{1}{5}$

$None of these$