Relations and Functions Test

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Relations and Functions Test - 31

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Weekly Quiz Competition

Question 1
1 / -0

If $$A \times B = \{(3, a), (3, -1), (3, 0), (5, a), (5, -1), (5, 0)\}$$, find $$A$$.

A
$$\{a, 5\}$$

B
$$\{a, -1\}$$

C
$$\{0, 5\}$$

D
$$\{3, 5\}$$

Solution
Question 2
1 / -0

Given $$(a - 2, b + 3) = (6, 8)$$, are equal ordered pair. Find the value of $$a$$ and $$b$$.

A
$$a = 8$$ and $$b = 5$$

B
$$a = 8$$ and $$b = 3$$

C
$$a = 5$$ and $$b = 5$$

D
$$a = 8$$ and $$b = 6$$

Solution

By equality of ordered pairs, we have
$$(a - 2, b + 3) = (6, 8)$$
On equating we get
$$a - 2 = 6$$
$$a = 8$$
$$b + 3 = 8$$
$$b = 5$$
So, the value of$$ a = 8 $$ and $$ b = 5.$$
Question 3
1 / -0

What is the second component of an ordered pair $$(3, -0.2)$$?

A
$$3$$

B
$$0.2$$

C
$$1$$

D
$$-0.2$$

Solution

In an ordered pair $$(x,y)$$, the first component is $$x$$ and the second component is $$y$$.
Therefore, in an ordered pair $$(3,-0.2)$$, the second component is $$-0.2$$.
Question 4
1 / -0

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

A
$$\{(2, 1), (2, 2), (3, 1), (3, 2)\}$$

B
$$\{(2, 1), (2, 1), (3, 1), (3, 2)\}$$

C
$$\{(2, 1), (2, 2), (2, 1), (3, 2)\}$$

D
$$\{2, 1), (2, 2), (3, 1), (2, 2)\}$$

Solution
Question 5
1 / -0

If $$A \times B =$$ $${(2, 4), (2, a), (2, 5), (1, 4), (1, a), (1, 5)}$$, find $$B$$.

A
$$\{4, 2, 5\}$$

B
$$\{4, a, 5\}$$

C
$$\{4, 1, 5\}$$

D
$$\{2, a, 5\}$$

Solution
Question 6
1 / -0

Ordered pairs $$(x, y)$$ and $$(3, 6)$$ are equal if $$x = 3$$ and $$y = ?$$

A
$$3$$

B
$$6$$

C
$$-6$$

D
$$-3$$

Solution

Given
$$(x,y)= (3,6)$$
$$x=3$$
$$y=6$$
The value of $$x=3$$ and $$y=6$$.
Question 7
1 / -0

If the function $$f$$ is defined by $$f(x)=3x+4$$, then $$2 f(x)+ 4=$$____ .

A
$$5x+4$$

B
$$5x+8$$

C
$$6x+4$$

D
$$6x+8$$

E
$$6x+12$$

Solution

Given
$$f(x)$$ $$=$$ $$3x$$ $$+$$ $$4$$

$$2f(x)$$ $$+$$ $$4$$ $$=$$ $$(2$$ $$\times$$ $$f(x))$$ $$+$$ $$4$$
From above,
$$=$$ $$(2$$ $$\times$$ $$(3x$$ $$+$$ $$4))$$ $$+$$ $$4$$
$$=$$ $$(2$$ $$\times$$ $$3x)$$ $$+$$ $$(2$$ $$\times$$ $$4)$$ $$+$$ $$4$$
$$=$$ $$6x$$ $$+$$ $$8$$ $$+$$ $$4$$
$$=$$ $$6x$$ $$+$$ $$12$$

Therefore, the value of $$2f(x)$$ $$+$$ $$4$$ is $$'$$$$6x$$ $$+$$ $$12$$$$'$$.
Question 8
1 / -0

Let $$S=\{(a, b, c) \in N \times N\times N:a+b+c=21, a \le b \le c\}$$ and $$T=\{(a, b, c)\in N\times N\times N: a, b, c\, are\, in\,AP\}$$, where $$N$$ is the set of all natural numbers. Then, the number of elements in the set $$S\cap T$$ is

A
$$6$$

B
$$7$$

C
$$13$$

D
$$14$$

Solution

We have, $$a+b+c=21$$ and $$\dfrac{a+c}{2}=b$$
$$\Rightarrow a+c+\dfrac{a+c}{2}=21$$
$$\Rightarrow a+c=14\Rightarrow \dfrac{a+c}{2}=7$$
$$\Rightarrow b=7$$
So, a can take values from $$1$$ to $$6$$ and $$c$$ can have values from $$8$$ to $$13$$
$$a=b=c=7$$
$$[\because a\le b \le c]$$
So, there are $$7$$ such triplets.
Question 9
1 / -0

Let $$f\left( x \right)=x+\dfrac { 1 }{ 2 } $$. Then, the number of real values of $$x$$ for which the three unequal terms $$f\left( x \right),f\left( 2x \right),f\left( 4x \right)$$ are in HP is

A
$$1$$

B
$$0$$

C
$$3$$

D
$$2$$

Solution

Given,
$$f\left( x \right) =x+\dfrac { 1 }{ 2 } =\dfrac { 2x+1 }{ 2 } $$
$$\therefore f\left( 2x \right) =2x+\dfrac { 1 }{ 2 } $$
$$\Rightarrow f\left( 2x \right) =\dfrac { 4x+1 }{ 2 } $$
and
$$f\left( 4x \right) =4x+\dfrac { 1 }{ 2 } \Rightarrow f\left( 4x \right) =\dfrac { 8x+1 }{ 2 } $$

Since, $$f\left( x \right) ,f\left( 2x \right) $$ and $$f\left( 4x \right) $$ are in HP.

$$\therefore \dfrac { 1 }{ f\left( x \right) } ,\dfrac { 1 }{ f\left( 2x \right) } $$ and $$\dfrac { 1 }{ f\left( 4x \right) } $$ are in AP.

$$\Rightarrow \dfrac { 1 }{ f\left( 2x \right) } =\dfrac { \dfrac { 1 }{ f\left( x \right) } +\dfrac { 1 }{ f\left( 4x \right) } }{ 2 } $$

$$\Rightarrow \dfrac { 2 }{ 4x+1 } =\dfrac { \dfrac { 2 }{ 2x+1 } +\dfrac { 2 }{ 8x+1 } }{ 2 } $$

$$\Rightarrow \dfrac { 2 }{ 4x+1 } =\dfrac { 10x+2 }{ \left( 2x+1 \right) \left( 8x+1 \right) } $$

$$\Rightarrow \left( 2x+1 \right) \left( 8x+1 \right) =\left( 5x+1 \right) \left( 4x+1 \right) $$
$$\Rightarrow 16{ x }^{ 2 }+10x+1=20{ x }^{ 2 }+9x+1$$
$$\Rightarrow 4{ x }^{ 2 }-x=0$$
$$\Rightarrow x\left( 4x-1 \right) =0$$
$$\Rightarrow x=\dfrac { 1 }{ 4 } $$ $$\left[ \because x\neq 0 \right] $$
Hence, one real value of $$x$$ for which the three unequal terms are in HP.
Question 10
1 / -0

Find the maximum value of $$f(x) = 1 - x^{2}$$ if $$-2\leq x\leq 2$$

A
$$2$$

B
$$1$$

C
$$0$$

D
$$-1$$

E
$$-2$$

Solution

Given, $$f(x) = 1-{ x }^{ 2 }$$
Here $$f(x)$$ is maximum, if $$1-{ x }^{ 2 }$$ is maximum, which implies $${ x }^{ 2 }$$ should be minimum.
The minimum value of $${ x }^{ 2 }$$ is $$0$$, which occurs at $$x=0$$.
Therefore, the maximum value of $$f(x)$$ is $$1-0 = 1$$.

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Relations and Functions Test - 31

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Relations and Functions Test - 31

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Question 1

$$\{a, 5\}$$

$$\{a, -1\}$$

$$\{0, 5\}$$

$$\{3, 5\}$$

Solution

Question 2

$$a = 8$$ and $$b = 5$$

$$a = 8$$ and $$b = 3$$

$$a = 5$$ and $$b = 5$$

$$a = 8$$ and $$b = 6$$

Solution

Question 3

$$3$$

$$0.2$$

$$1$$

$$-0.2$$

Solution

Question 4

$$\{(2, 1), (2, 2), (3, 1), (3, 2)\}$$

$$\{(2, 1), (2, 1), (3, 1), (3, 2)\}$$

$$\{(2, 1), (2, 2), (2, 1), (3, 2)\}$$

$$\{2, 1), (2, 2), (3, 1), (2, 2)\}$$

Solution

Question 5

$$\{4, 2, 5\}$$

$$\{4, a, 5\}$$

$$\{4, 1, 5\}$$

$$\{2, a, 5\}$$

Solution

Question 6

$$3$$

$$6$$

$$-6$$

$$-3$$

Solution

Question 7

$$5x+4$$

$$5x+8$$

$$6x+4$$

$$6x+8$$

$$6x+12$$

Solution

Question 8

$$6$$

$$7$$

$$13$$

$$14$$

Solution

Question 9

$$1$$

$$0$$

$$3$$

$$2$$

Solution

Question 10

$$2$$

$$1$$

$$0$$

$$-1$$

$$-2$$

Solution

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