Sets Test

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Sets Test - 25

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Weekly Quiz Competition

Question 1
1 / -0

If universal set $$\xi = \{a, b, c, d, e, f, g, h\}, A = \{b, c, d, e, f\}, B =\{a, b, c, g, h\}$$ and $$C = \{c, d, e, f, g\}$$, then find $$B - A$$

A
$$\{b,c,e,f\}$$

B
$$\{a, b, f, h\}$$

C
$$\{a, g, h\}$$

D
$$\{a,c,e,g\}$$

Solution

Given : $$\xi = \{a, b, c, d, e, f, g, h\}, A = \{b, c, d, e, f\}, B =\{a, b, c, g, h\}$$ and $$C = \{c, d, e, f, g\}$$
$$ B - A $$ will have elements of $$ B $$ which are not in $$ A $$

So, $$ B - A =\{a,g,h \}$$.
Question 2
1 / -0

If universal set $$\xi = \{a, b, c, d, e, f, g, h\}, A = \{b, c, d, e, f\}, B =\{a, b, c, g, h\}$$ and $$C = \{c, d, e, f, g\}$$ find $$(B - C)'$$

A
$$\{d, e, f, g\}$$

B
$$\{c, d, e, f, g\}$$

C
$$\{c, d, f, g\}$$

D
$$\{a, c, d, e, f, g\}$$

Solution

Given : $$\xi = \{a, b, c, d, e, f, g, h\}, A = \{b, c, d, e, f\}, B =\{a, b, c, g, h\}$$ and $$C = \{c, d, e, f, g\}$$
$$ B - C $$ will have elements of $$ B $$ which are not in $$ C $$

So, $$ B - C = \{ a, b, h\} $$

And $$ {(B-C)}'= \xi - (B-C) = $$ { $$ c,d,e,f,g $$ }
Question 3
1 / -0

If $$A = (6, 7, 8, 9), B = (4, 6, 8, 10)$$ and $$C = \{x : x \,\,\epsilon\,\,N : 2 < x \leq 7\}$$ ; find : $$B - B$$

A
$$\phi$$

B
$$\{0\}$$

C
$$\{6,7\}$$

D
$$\{4\}$$

Solution

Given : $$A = (6, 7, 8, 9), B = (4, 6, 8, 10)$$ and $$C = \{x : x \,\,\epsilon\,\,N : 2 < x \leq 7\}$$
$$ B - B $$ will always be a null set it will contain elements of $$B$$ which are not in $$B$$ i.e. no elements.

So, $$ B - B = \phi $$
Question 4
1 / -0

Find the set of all solutions of the equation $$2^{\left | y \right |}-\left | 2^{y-1}-1 \right |=2^{y-1}+1$$, the solution includes

A
$$y =-1$$

B
$$y>1$$

C
$$y=1$$

D
$$y<1$$

Solution

Here, $$2^{\left | y \right |}-\left | 2^{y-1}-1 \right |=2^{y-1}+1$$
We know to define modulus, we have three cases as
Case I: y< 0
$$\Rightarrow $$   $$2^{-y}+\left ( 2^{y-1}-1 \right )=2^{y-1}+1$$
$$\Rightarrow $$   $$2^{-y}=2^{1}$$          (as when y< 0 |y|=-y and $$\left | 2^{y-1}-1 \right |=-\left ( 2^{y-1}-1 \right )$$)
Hence, y=-1, which is true when y< 0          (i)
Case II: $$0\leq y< 1$$
$$\Rightarrow $$   $$2^{y}+\left ( 2^{y-1}-1 \right )=2^{y-1}+1$$
$$\Rightarrow $$   $$2^{y}=2$$          (as when $$0\leq y< 1$$ |y|=-y and $$\left | 2^{y-1}-1 \right |=-\left ( 2^{y-1}-1 \right )$$)
$$\Rightarrow $$   $$y=1$$, which shows no solution as,
   $$0\leq y< 1$$          (ii)
Case III: $$y\geq 1$$
$$\Rightarrow $$   $$2^{y}-\left ( 2^{y-1}-1 \right )=2^{y-1}+1$$
$$\Rightarrow $$   $$2^{y}=2^{y-1}+2^{y-1}$$
$$\Rightarrow $$   $$2^{y}=2.2^{y-1}$$          (as when $$y\geq 0$$ |y|=y and $$\left | 2^{y-1}-1 \right |=\left ( 2^{y-1}-1 \right )$$)
$$\Rightarrow $$   $$2^{y}=2^{y}$$, which is an identity therefor, it is true $$\forall y\geq 1$$          (iii)
Hence, from Eqs. (i), (ii), (iii) the solution of set is $$\left \{ y: y\geq 1\cup y=-1 \right \}$$.
Question 5
1 / -0

If $$Q=\{ x:x=\cfrac { 1 }{ y },$$ where $$ y\in N \} $$, then

A
$$0\in Q$$

B
$$1\in Q$$

C
$$2\in Q$$

D
$$\cfrac{2}{3} \in Q$$

Solution

Since $$\cfrac{1}{y}\ne 0$$, $$\cfrac{1}{y}\ne 2$$, $$\cfrac{1}{y}\ne \cfrac{-2}{3}$$ [$$\because y\in N$$]
$$\therefore \cfrac{1}{y}$$ can be $$1$$

Ans: B
Question 6
1 / -0

$$\displaystyle A=\left\{ x:x\neq x \right\} $$ represents:

A
$${0}$$

B
$$\phi$$

C
$${1}$$

D
$${x}$$

Solution

$$x=x$$ for all $$x$$
Hence, there is no element in A.
A is an empty set .
$$A=\{ \}$$
Question 7
1 / -0

$$A=\left\{ x:x\neq x \right\} $$ represents-

A
$$\left\{ 0 \right\} $$

B
$$\left\{ \right\} $$

C
$$\left\{ 1 \right\} $$

D
$$\left\{ x \right\} $$

Solution

It is fundamental concept that a void set is defined as, $$A =\{x : x\neq x\}=\{\}$$
Question 8
1 / -0

Which of the following statements is false?

A
$$2$$ $$\in$$ {first five counting numbers.}

B
$$f$$ $${\notin } $$ {consonants}

C
Rose $${\notin }$$ {the set of all fruits.}

D
Asia $${\in }$$ {Continents}

Solution

$$\textbf{Step-1: Observing the Option 1}$$
$$\text{2 belongs to the sets of first five counting numbers}$$
$$\text{so option (A) is True.}$$
$$\textbf{Step-2: Observing other option}$$
$$\text{f belongs to the set of consonants so option B is false}$$
$$\text{Rose belongs to the set of all fruits so option C is True}$$
$$\text{Asia belongs to the sets of continents so option D is also True}$$
$$\textbf{Hence, the false statement is option B}$$
Question 9
1 / -0

The solution set of $$x+2<9$$ over a set of positive even integers is

A
$$\displaystyle \left \{ 8,10,12,... \right \}$$

B
$$\displaystyle \left \{ 2,4,6 \right \}$$

C
$$\displaystyle \left \{ 1,2,3,4,5,6 \right \}$$

D
$$\displaystyle \left \{ 2,4,6,8 \right \}$$

Solution

$$x+2<9$$
$$x<7$$
Since $$x$$ is even positive integer then $$x\in [2,7)$$.
Hence the set of positive even integers which fall in the above set is
$$\{2,4,6\}$$.
Question 10
1 / -0

The solution set of $$3 x - 4 < 8$$ over the set of non-negative square numbers is

A
$$\displaystyle \left \{ 1,2,3 \right \}$$

B
$$\displaystyle \left \{ 1,4 \right \}$$

C
$$\displaystyle \left \{ 1 \right \}$$

D
$$\displaystyle \left \{ 16 \right \}$$

Solution

$$3x-4<8$$
$$3x<12$$
$$x<4$$
Hence set of non-negative square numbers belonging to the above set is
$$\{1\}$$.

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Re-Attempt Weekly Quiz Competition

Sets Test - 25

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Sets Test - 25

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SHARING IS CARING

Question 1

$$\{b,c,e,f\}$$

$$\{a, b, f, h\}$$

$$\{a, g, h\}$$

$$\{a,c,e,g\}$$

Solution

Question 2

$$\{d, e, f, g\}$$

$$\{c, d, e, f, g\}$$

$$\{c, d, f, g\}$$

$$\{a, c, d, e, f, g\}$$

Solution

Question 3

$$\phi$$

$$\{0\}$$

$$\{6,7\}$$

$$\{4\}$$

Solution

Question 4

$$y =-1$$

$$y>1$$

$$y=1$$

$$y<1$$

Solution

Question 5

$$0\in Q$$

$$1\in Q$$

$$2\in Q$$

$$\cfrac{2}{3} \in Q$$

Solution

Question 6

$${0}$$

$$\phi$$

$${1}$$

$${x}$$

Solution

Question 7

$$\left\{ 0 \right\} $$

$$\left\{ \right\} $$

$$\left\{ 1 \right\} $$

$$\left\{ x \right\} $$

Solution

Question 8

$$2$$ $$\in$$ {first five counting numbers.}

$$f$$ $${\notin } $$ {consonants}

Rose $${\notin }$$ {the set of all fruits.}

Asia $${\in }$$ {Continents}

Solution

Question 9

$$\displaystyle \left \{ 8,10,12,... \right \}$$

$$\displaystyle \left \{ 2,4,6 \right \}$$

$$\displaystyle \left \{ 1,2,3,4,5,6 \right \}$$

$$\displaystyle \left \{ 2,4,6,8 \right \}$$

Solution

Question 10

$$\displaystyle \left \{ 1,2,3 \right \}$$

$$\displaystyle \left \{ 1,4 \right \}$$

$$\displaystyle \left \{ 1 \right \}$$

$$\displaystyle \left \{ 16 \right \}$$

Solution

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