Relations and Functions Test

Result

Relations and Functions Test - 30

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

For the pair of functions,
$$f\left( x \right) =9x-4, g\left( x \right) =7x-9$$
Find $$\left( f-g \right) \left( x \right) $$.

A
$$16x-13$$

B
$$2x+5$$

C
$$2x-13$$

D
$$-2x-5$$

Solution

$$(f-g)(x)=f(x)-g(x)= 9x-4-7x+9=2x+5$$
Question 2
1 / -0

Let a relation $$R$$ be defined by $$R=\left \{(4,5), (1,4), (4,6), (7,6), (3,7)\right \}$$. The relation $$R^{-1}\circ R$$ is given by

A
$$\left \{(1,1), (4,4), (7,4), (4,7), (7,7)\right \}$$

B
$$\left \{(1,1), (4,4), (4,7), (7,4), (7,7),(3,3)\right \}$$

C
$$\left \{(1,5), (1,6), (3,6)\right \}$$

D
None of these

Solution

We have $$R=\left \{(4,5), (1,4), (4,6), (7,6), (3,7)\right \}$$.
$$\therefore R^{-1}=\left \{(5,4), (4,1), (6,4), (6,7), (7,3)\right \}$$
$$(4,4)\in R^{-1}\circ R$$ because $$(4,5)\in R$$ and $$(5,4)\in R^{-1}$$
$$(1,1)\in R^{-1}\circ R$$ because $$(1,4)\in R$$ and $$(4,1)\in R^{-1}$$
$$(4,4)\in R^{-1}\circ R$$ because $$(4,6)\in R$$ and $$(6,4)\in R^{-1}$$
$$(4,7)\in R^{-1}\circ R$$ because $$(4,6)\in R$$ and $$(6,7)\in R^{-1}$$
$$(7,4)\in R^{-1}\circ R$$ because $$(7,6)\in R$$ and $$(6,4)\in R^{-1}$$
$$(7,7)\in R^{-1}\circ R$$ because $$(7,6)\in R$$ and $$(6,7)\in R^{-1}$$
$$(3,3)\in R^{-1}\circ R$$ because $$(3,7)\in R$$ and $$(7,3)\in R^{-1}$$
$$\therefore R^{-1}\circ R=\left \{(4,4), (1,1), (4,7), (7,4), (7,7), (3,3)\right \}$$.
$$\therefore$$ The correct answer is $$B$$.
Question 3
1 / -0

If $$A=\left \{1, 2,3\right \}$$ and $$B=\left \{3,8\right \}$$, then $$(A\cup B)\times (A\cap B)$$ is equal to

A
$$\left \{(8,3), (8,2), (8,1), (8,8)\right \}$$

B
$$\left \{(1,2), (2,2), (3,3), (8,8)\right \}$$

C
$$\left \{(3,1), (3,2), (3,3), (3,8)\right \}$$

D
$$\left \{(1,3), (2,3), (3,3), (8,3)\right \}$$

Solution

Given, $$A=\left \{1, 2,3\right \}$$ and $$B=\left \{3,8\right \}$$,
Therefore, $$A\cup B=\left \{1,2,3\right \}\cup \left \{3,8\right \}=\left \{1,2,3,8\right \}$$
and $$A\cap B=\left \{1,2,3\right \}\cap \left \{3,8\right \}=\left \{3\right \}$$
$$\therefore (A\cup B)\times (A\cap B)=\left \{1,2,3,8\right \}\times \left \{3\right \}$$
$$=\left \{(1,3), (2,3), (3,3), (8,3)\right \}$$
Question 4
1 / -0

The relation $$R$$ defined on the set $$A=\left \{1,2,3,4,5\right \}$$ by $$R=\left \{(x,y):|x^2-y^2| < 16\right \}$$ is given by

A
$$\left \{(1,1), (2,1), (3,1), (4,1), (2,3)\right \}$$

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

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

D
None of these

Solution

We have $$R=\left \{(x,y):|x^2-y^2| < 16\right \}$$.

Take $$x=1,$$ we have
$$ |x^2-y^2| < 16\Rightarrow |1-y^2| < 16$$
$$\Rightarrow |y^2-1| < 16\Rightarrow y=1,2,3,4$$.

Take $$x=2,$$ we have
$$ |x^2-y^2| < 16 \Rightarrow |4-y^2| < 16$$
$$\Rightarrow |y^2-4| < 16\Rightarrow y=1,2,3,4$$.

Take $$x=3,$$ we have
$$ |x^2-y^2| < 16\Rightarrow |9-y^2| < 16$$
$$\Rightarrow |y^2-9| < 16\Rightarrow y=1,2,3,4$$.

Take $$x=4,$$ we have
$$ |x^2-y^2| < 16\Rightarrow |16-y^2| < 16$$
$$\Rightarrow |y^2-16| < 16\Rightarrow y=1,2,3,4,5$$.

Take $$x=5,$$
$$|x^2-y^2| < 16 \Rightarrow |25-y^2| < 16$$
$$\Rightarrow |y^2-25|<16 \Rightarrow y = 4,5$$

$$\therefore R=\left \{(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), \\(3,3), (3,4), (4,1), (4,2), (4,3), (4,4), (4,5), (5,4), (5,5)\right \}$$
$$\therefore$$ The correct answer is $$D$$.
Question 5
1 / -0

What is the Cartesian product of $$A = \left \{1, 2\right \}$$ and $$B = \left \{a, b\right \}$$?

A
$$\left \{(1, a), (1, b), (2, a), (b, b)\right \}$$

B
$$\left \{(1, 1), (2, 2), (a, a), (b, b)\right \}$$

C
$$\left \{(1, a), (2, a), (1, b), (2, b)\right \}$$

D
$$\left \{(1, 1), (a, a), (2, a), (1, b)\right \}$$

Solution

If $$A $$ and $$B$$ are two non empty sets, then the Cartesian product $$A \times B$$ is set of all ordered pairs $$(a,b)$$ such that $$a\in A$$ and $$b\in B$$.

Given $$A =\{1,2\}$$ and $$B = \{a,b\}$$

Hence $$A\times B = \{(1,a),(1,b),(2,a),(2,b)\}$$
Question 6
1 / -0

Given that $$f\left( x \right) =5{ x }^{ 2 }-8x,g\left( x \right) ={ x }^{ 2 }-5x-24$$. Find $$\left( \displaystyle\frac { f }{ g } \right) \left( x \right) $$

A
$$\displaystyle\frac { 5x }{ x+1 } $$

B
$$\displaystyle\frac { 5{ x }^{ 2 }-8x }{ { x }^{ 2 }-5x-24 } $$

C
$$\displaystyle\frac { 5x-8 }{ -5 } $$

D
$$\displaystyle\frac { 5-x }{ 24 } $$

Solution

$$\left (\dfrac{f}{g}\right)(x)$$ means $$\dfrac{f(x)}{g(x)}$$
Given, $$f(x)=5x^2-8x$$ and $$g(x)=x^2-5x-24$$
So, $$\left (\dfrac{f}{g}\right)(x)=\dfrac{5x^2-8x}{x^2-5x-24}$$
Hence, option B is correct.
Question 7
1 / -0

Find $$\left( f.g \right) \left( 2 \right) $$ when $$f\left( x \right) =x-1$$ and $$g\left( x \right) =-5{ x }^{ 2 }+14x+7$$.

A
$$45$$

B
$$-39$$

C
$$39$$

D
$$15$$

Solution

$$(f.g)(2)$$ means $$f(2).g(2)$$
Given, $$f(x)=x-1, g(x)=-5x^2+14x+7$$
So, $$f(2)=2-1=1$$
and $$g(2)=-5(2)^2+14(2)+7$$
$$=-20+28+7$$
$$=15$$
Thus $$(f.g)(2)=f(2).g(2)=15\times 1=15$$
Question 8
1 / -0

$$(x, y)$$ and $$(p, q)$$ are two ordered pairs. Find the values of $$p$$ and $$y$$, if $$(4y + 5, 3p - 1) = (25, p + 1)$$

A
$$p = 0, y = 5$$

B
$$p = 1, y = 5$$

C
$$p = 0, y = 1$$

D
$$p = 1, y = 1$$

Solution

Given $$(x,y)=(p,q)$$
$$(4y + 5, 3p - 1) = (25, p + 1)$$
On equating we get
$$4y + 5 = 25$$
$$4y = 25 - 5$$
$$4y = 20$$
$$y = 5$$
$$3p - 1 = p + 1$$
$$3p - p = 1 + 1$$
$$2p = 2$$
$$p = 1$$
So, the value of$$ p = 1, y = 5$$
Question 9
1 / -0

What is the first component of an ordered pair $$(1, -1)$$?

A
$$1$$

B
$$-1$$

C
$$2$$

D
$$0$$

Solution

In an ordered pair $$(x,y)$$, the first component is $$x$$ and the second component is $$y$$.
Therefore, in an ordered pair $$(1,-1)$$, the first component is $$1$$.
Question 10
1 / -0

$$(x, y)$$ and $$(p, q)$$ are two ordered pairs. Find the values of $$x$$ and $$p$$, if $$(3x - 1, 9) = (11, p + 2)$$

A
$$x = 4, p = 9$$

B
$$x = 6, p = 7$$

C
$$x = 4, p = 5$$

D
$$x = 4, p = 7$$

Solution

Given$$(x,y)=(p,q)$$
$$(3x - 1, 9) = (11, p + 2)$$
By equating
$$3x - 1 = 11$$
$$3x = 12$$
$$x = 4$$
$$9 = p + 2$$
$$p = 7$$
So, the value of $$x = 4, p = 7.$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Relations and Functions Test - 30

Result

Relations and Functions Test - 30

Score

-

Rank

-

SHARING IS CARING

Question 1

$$16x-13$$

$$2x+5$$

$$2x-13$$

$$-2x-5$$

Solution

Question 2

$$\left \{(1,1), (4,4), (7,4), (4,7), (7,7)\right \}$$

$$\left \{(1,1), (4,4), (4,7), (7,4), (7,7),(3,3)\right \}$$

$$\left \{(1,5), (1,6), (3,6)\right \}$$

None of these

Solution

Question 3

$$\left \{(8,3), (8,2), (8,1), (8,8)\right \}$$

$$\left \{(1,2), (2,2), (3,3), (8,8)\right \}$$

$$\left \{(3,1), (3,2), (3,3), (3,8)\right \}$$

$$\left \{(1,3), (2,3), (3,3), (8,3)\right \}$$

Solution

Question 4

$$\left \{(1,1), (2,1), (3,1), (4,1), (2,3)\right \}$$

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

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

None of these

Solution

Question 5

$$\left \{(1, a), (1, b), (2, a), (b, b)\right \}$$

$$\left \{(1, 1), (2, 2), (a, a), (b, b)\right \}$$

$$\left \{(1, a), (2, a), (1, b), (2, b)\right \}$$

$$\left \{(1, 1), (a, a), (2, a), (1, b)\right \}$$

Solution

Question 6

$$\displaystyle\frac { 5x }{ x+1 } $$

$$\displaystyle\frac { 5{ x }^{ 2 }-8x }{ { x }^{ 2 }-5x-24 } $$

$$\displaystyle\frac { 5x-8 }{ -5 } $$

$$\displaystyle\frac { 5-x }{ 24 } $$

Solution

Question 7

$$45$$

$$-39$$

$$39$$

$$15$$

Solution

Question 8

$$p = 0, y = 5$$

$$p = 1, y = 5$$

$$p = 0, y = 1$$

$$p = 1, y = 1$$

Solution

Question 9

$$1$$

$$-1$$

$$2$$

$$0$$

Solution

Question 10

$$x = 4, p = 9$$

$$x = 6, p = 7$$

$$x = 4, p = 5$$

$$x = 4, p = 7$$

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.