Relations and Functions Test

Result

Relations and Functions Test - 28

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

$$A$$ and $$B$$ are two sets having $$3$$ and $$4$$ elements respectively and having $$2$$ elements in common. The number of relations which can be defined from $$A$$ to $$B$$ is

A
$$2^5$$

B
$$2^{10} - 1$$

C
$$2^{12} - 1$$

D
$$none\ of\ these$$

Solution

`if $$A$$ and $$B$$ are two sets which contains 3 and 4 elements respectively.
Then no of elements in relation from $$A$$ to $$B$$ is $$2^(3\times 4)$$
means $$2^{12}$$
Question 2
1 / -0

A function $$f : R \rightarrow R$$ satisfies the equation $$f(x)f(y) - f(xy) = x + y$$ for all $$x, y \in R $$ and $$f(1) > 0$$ then

A
$$f(x) = x + \dfrac{1}{2}$$

B
$$f(x) = \dfrac{x}{2} + 1$$

C
$$f(x) = -\dfrac{x}{2} + 1$$

D
$$f(x) = x + 1$$

Solution

Taking $$x=y=1$$, we get
$$f\left( 1 \right) f\left( 1 \right) -f\left( 1 \right) =1+1\Rightarrow { \left( f\left( 1 \right) \right) }^{ 2 }-f\left( 1 \right) -2=0\\ \Rightarrow \left( f\left( 1 \right) -2 \right) \left( f\left( 1 \right) +1 \right) =0\Rightarrow f\left( 1 \right) =2\quad \left( \because f\left( 1 \right) >0 \right) $$
Taking $$y=1$$, we get
$$f\left( x \right) f\left( 1 \right) -f\left( x \right) =x+1\Rightarrow 2f\left( x \right) -f\left( x \right) =x+1\\ \Rightarrow f\left( x \right) =x+1$$
Question 3
1 / -0

If $$R$$ is a relation from a finite set $$A$$ having $$m$$ elements to a finite set $$B$$ having $$n$$ elements, then the number of relations from $$A$$ to $$B$$ is:

A
$$2^{mn}$$

B
$$2^{mn} - 1$$

C
$$2mn$$

D
$$m^n$$

Solution

$$R$$ is a relation from a finite set $$A$$ having $$m$$ elements to a finite set $$B$$ having $$n$$ elements.
Number of elements in $$A\times B$$ $$=mn$$
No of relations $$=$$ number of subsets of $$A\times B$$ $$=2^{mn}$$
Hence, option A is correct.
Question 4
1 / -0

The relation $$R$$ is defined in $$A = \left\{1,2,3\right\}$$ by $$a\ R$$ $$b$$ if $$|a^2 - b^2| \le 5$$. Which of the following is false?

A
$$R = \left\{(1,1), (2,2), (3,3), (2,1), (1,2), (2,3), (3,2)\right\}$$

B
$$R^{-1} = R$$

C
Domain of $$R = \left\{1,2,3\right\}$$

D
Range of $$R = \left\{5\right\}$$

Solution

$$R=\left\{ (1,1),(2,2),(3,3),(2,1),(1,2),(2,3),(3,2) \right\} $$
Clearly $$R^{-1} = R$$
Domain $$=\{1,2,3\}$$ Range $$=\{1,2,3\}$$
Hence option 'D' is correct choice.
Question 5
1 / -0

If $$A = \{1, 2 \}$$ and $$B = \{3, 4\}$$ then find $$A \times B$$

A
$$A \times B = \{(1,3),(1,2),(2,3),(2,4)\}$$

B
$$A \times B = \{(1,3),(1,4),(2,3),(2,4)\}$$

C
$$A \times B = \{(1,3),(1,4),(2,1),(2,4)\}$$

D
$$A \times B = \{(1,3),(1,4),(2,3),(2,1)\}$$

Solution
Question 6
1 / -0

If $$R = \{(x, y):3x + 2y = 15 \text{ and }x\,, y\,\in \, N\}$$, the range of the relation R is .........

A
$$\{1, 2\}$$

B
$$\{1, 2,..., 5\}$$

C
$$\{1, 2,..., 7\}$$

D
$$\{3, 6\}$$

Solution

Since x and y belong to natural numbers we need to find positive integral solutions for $$3x+2y=15$$

$$2y=15-3x$$

$$\Rightarrow y =\dfrac{15-3x}{2}$$

For $$x=0,2,4$$ $$y$$ will become fractional which is not possible

For $$x\geq 5$$ y becomes 0 or negative which again is not possible.

Hence for $$x=1,3$$ y= 6,3$$ respectively.

Hence the range is $$\{3,6\}$$
Question 7
1 / -0

If $$h(x)=x^2-2$$, then the value of $$\displaystyle\frac{1}{2}h\left(\frac{1}{2}\right)$$ is

A
$$0$$

B
$$1$$

C
$$\displaystyle-\frac{7}{8}$$

D
$$\displaystyle\frac{7}{8}$$

Solution

$$h(x)=x^2-2$$

$$\displaystyle\therefore\frac{1}{2}h\left(\frac{1}{2}\right)=\frac{1}{2}\left(\frac{1}{4}-2\right)=-\frac{7}{8}$$
Question 8
1 / -0

If f is an identity function then f(-5) = __________

A
-5

B
1

C
-1

D
None of these

Solution

In an identity function, we have $$ f(x) = x $$

So, $$ f(-5) = -5 $$
Question 9
1 / -0

The domain of the relation R = $$\displaystyle \left \{ \left ( x,y \right ):x,y\epsilon N \ and\ x+y\leq 3 \right \}$$ is____

A
{1,2,3}

B
{1,2}

C
{...-1,0,1,2,3}

D
None of these

Solution

The only natrual numbers, less than equal to $$ 3 $$ are $$ 1, 2, 3 $$

From these, only the sum of $$ 1 ,2 $$ gives $$ 3 $$

So the domain of he relation is {$$ 1,2$$}
Question 10
1 / -0

If $$(3p+q,p-q)=(p-q,3p+q)$$, then:

A
$$p=q=0$$

B
$$p=q$$

C
$$p=2q$$

D
$$p+q=0$$

Solution

As the ordered pairs are equal,
$$ 3p + q = p - q $$
$$ => 2p = -2q $$
$$ => 2p + 2q = 0 $$
$$ => p + q = 0 $$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Relations and Functions Test - 28

Result

Relations and Functions Test - 28

Score

-

Rank

-

SHARING IS CARING

Question 1

$$2^5$$

$$2^{10} - 1$$

$$2^{12} - 1$$

$$none\ of\ these$$

Solution

Question 2

$$f(x) = x + \dfrac{1}{2}$$

$$f(x) = \dfrac{x}{2} + 1$$

$$f(x) = -\dfrac{x}{2} + 1$$

$$f(x) = x + 1$$

Solution

Question 3

$$2^{mn}$$

$$2^{mn} - 1$$

$$2mn$$

$$m^n$$

Solution

Question 4

$$R = \left\{(1,1), (2,2), (3,3), (2,1), (1,2), (2,3), (3,2)\right\}$$

$$R^{-1} = R$$

Domain of $$R = \left\{1,2,3\right\}$$

Range of $$R = \left\{5\right\}$$

Solution

Question 5

$$A \times B = \{(1,3),(1,2),(2,3),(2,4)\}$$

$$A \times B = \{(1,3),(1,4),(2,3),(2,4)\}$$

$$A \times B = \{(1,3),(1,4),(2,1),(2,4)\}$$

$$A \times B = \{(1,3),(1,4),(2,3),(2,1)\}$$

Solution

Question 6

$$\{1, 2\}$$

$$\{1, 2,..., 5\}$$

$$\{1, 2,..., 7\}$$

$$\{3, 6\}$$

Solution

Question 7

$$0$$

$$1$$

$$\displaystyle-\frac{7}{8}$$

$$\displaystyle\frac{7}{8}$$

Solution

Question 8

-5

1

-1

None of these

Solution

Question 9

{1,2,3}

{1,2}

{...-1,0,1,2,3}

None of these

Solution

Question 10

$$p=q=0$$

$$p=q$$

$$p=2q$$

$$p+q=0$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.