Tag: set language

Questions Related to set language

State which of the following are infinite sets.
$(i)A={x:x\in Z: x^2 $ is even $}$
$(ii)B={x:x\in R:-4<x<-2}$

maths set concepts finite and infinite sets types of sets set language

  1. $(i)$ only

  2. $(ii)$ only

  3. $(i)$ and $(ii)$ both

  4. Neither $(i)$ nor $(ii)$

Show answer
Correct Option: C
Explanation:

$(i)A={x:x\in Z: x^2 $ is even $}$
$A={...,-6,-4,-2,0,2,4,6,...}$ which is an infinite set.
$(ii)B={x:x\in R:-4<x<-2}$
There will be infinite real numbers any two numbers so, it is an infinite set.

Which of the following are infinite set?
$(i)$The set of lines which are parallel to x-axis.
$(ii)$The set of animals living on the earth.
$(iii)$ The set of numbers which are multiple of $5.$
$(iv)$ The set of the circles passing through the origin $(0,0).$

maths set concepts finite and infinite sets types of sets set language

  1. $(i),(ii)$ and $(iv)$

  2. $(ii)$ only

  3. $(i),(iii)$ and $(iv)$

  4. $(i),(ii)$ and $(iii)$$

Show answer
Correct Option: C
Explanation:

$(i)$The set of lines which are parallel to x-axis is an infinite set because line parallel to x-axis are infinite in number.
$(ii)$The set of animals living on the earth is a finite set because the number of animals living on the earth is finite (although it is quite a big number)
$(iii)$ The set of numbers which are multiple of $5$ is an infinite numbers multiples of $5$ are infinite in number.
$(iv)$ The set of the circles passing through the origin $(0,0)$ is an infinite set because infinite number of circles can pass through the origin.

Which of the following sets are finite sets.
$(i)$ The sets of months in a year.
$(ii){1,2,3,....}$
$(iii){1,2,3,...,99,100}$
$(iv)$ The set of positive integers greater than $100.$ 

maths set concepts finite and infinite sets types of sets set language

  1. $(i)$ and $(iii)$

  2. $(i)$ only

  3. $(ii),(iii)$ and $(iv)$

  4. $(ii)$ and $(iv)$

Show answer
Correct Option: A
Explanation:

$(i)$ The sets of months in a year is a finite set because it has $12$ elements.
$(ii){1,2,3,....}$ is an infinite set as it has infinite number of elements.
$(iii){1,2,3,...,99,100}$ is a finite set as it has number from $1$ to $100$ which is finite in number.
$(iv)$ The set of positive integers greater than $100$ is an infinite set because positive integers greater than $100$ are infinite in number. 

If $A={a,{b}},$ find $P(A).$

maths set concepts finite and infinite sets types of sets set language

  1. $P(A)={\phi,a,{b},{a,{b}}}$

  2. $P(A)={a,{b},{a,{b}}}$

  3. $P(A)={\phi,{a,{b}}}$

  4. $P(A)={{a,{b}}}$

Show answer
Correct Option: A
Explanation:

Given that:
$A={a,{b}},$
 $P(A)={\phi,a,{b},{a,{b}}}$

State which of the following are infinite sets.
$(i)A={x:x\in Z: x $ is odd$}$
$(ii)B={x:x\in R:<-10}$

maths set concepts finite and infinite sets types of sets set language

  1. $(i)$ only

  2. $(ii)$ only

  3. $(i)$ and $(ii)$ both

  4. Neither $(i)$ nor $(ii)$

Show answer
Correct Option: C
Explanation:

$(i)A={x:x\in Z: x^2 $ is even $}$
$A={...,-3,-1-1,3,...}$ which is an infinite set.
$(ii)B={x:x\in R:-2<x<-4}$
$B={...,-14,-13,-12,-11}$ so it is an infinite set.

State whether the following statement is True or False
$A= { x| x\ is\ a\ negative\ integer\ ;x>-5 }$ is a finite set.

maths set concepts finite and infinite sets types of sets set language

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

We have to state whether the statement "$A={x|x:is:a:negative:integer:;x>-5} : is:a:finite : set$" is true or false.

Consider $A={x|x:is:a:negative:integer:;x>-5} $
                     $={-4,-3,-2,-1} $ which has finite number of elements.

So $A$ is a finite set.

Hence the given statement is true.

If the system of equation $x+2y-3z=1$, $(p+2)z=3$, $(2p+1)y+z=2$ has infinite number of solutions, then the value of p is not equal to.

maths set concepts finite and infinite sets types of sets set language

  1. $-2$

  2. $-\displaystyle\frac{1}{2}$

  3. $0$

  4. $2$

Show answer
Correct Option: A
Explanation:
$x+2y-3z=1$
$(p+2)z=3$
$(2p+1)y+z=2$
let $p=5$, $s\in R/ \left\{ -2, 1/2\right\}$
$\therefore z=\dfrac{3}{s+2}$
$\Rightarrow y=\left(2-\dfrac{3}{s+2}\right)\dfrac{1}{(2s+1)}\Rightarrow \dfrac{2s+1}{(s+2)(2s+1)}-\dfrac{1}{s+2}$
$[As\ 2s+1\neq 0]$
$\therefore x=3z+1-2y$
$=\dfrac{9}{s+2}+1-\dfrac{2}{s+2}=\dfrac{7}{s+2}+1$
$\therefore$ solutions $(x, y, z)=\left(\dfrac{7}{s+2}+1, \dfrac{1}{s+2}, \dfrac{3}{s+2}\right)$
is an infinite set,
$\therefore p$ cannot be equal to $-2$ or $1/2$

Let $A$ and $B$ are two finite sets such that $n(A)=3$ and $n(B)=4$ then  the number of elements in $A\Delta B$.

maths set language different sets de morgan's law de morgan's law for set theory

  1. $2$

  2. $7$

  3. $5$

  4. can not be determined

Show answer
Correct Option: D
Explanation:

Now, we have,

$A\Delta B=(A-B)\cup(B-A)$.
But it is impossible to find the number of elements in the set $A\Delta B$ as the sets $A$ and $B$ are not given explicitly. 

$A\cup B=A\cap B$ if and only if

maths set language different sets de morgan's law de morgan's law for set theory

  1. A is an empty set

  2. B is an empty set

  3. Both A and B are empty sets

  4. Both A and B are non-empty sets

Show answer
Correct Option: C
Explanation:

Solution:- Lets assume A is an empty set and B=$\left{ a,b \right}$

Now $A\cup B=\left{ a,b \right}$  and $A\cap B=\oslash $, so in all cases other than C , the condition is not satisfied. So C is the correct answer.

If A and B be two sets such that n(A) = 15, n(B) =25, then number of possible values of $n(A\Delta B)$(symmetric difference of  A and B) is

maths set language different sets de morgan's law de morgan's law for set theory

  1. 30

  2. 16

  3. 26

  4. 40

Show answer
Correct Option: B
Explanation:
$n(A \triangle B)= n (A \cup B)- n (A \cap B)$
for $n$ (A \triangle B)$ to be max. $n (A \cap B)=0$
We know, that 
$n(A \cup B)= n (A)+ n (B)- n (A \cap B) = 15+25-0=40$
$\Rightarrow n (A \triangle B)_{max} = 40-0 =40$
For minimum value of $n (A \triangle B)$
$n (A \cup B)$ should be min, $n (A \cap B)$ should be max.
$n (A \triangle B)$ min $=25-15= 10$
So. value of 
$n (A \triangle  B)= n (A \cup B)- n(A \cap B)$ lies om the set
${10,11,12,......, 3,9,40}$
Now, when $n (A \triangle B)$ is max. i.e. when 
$n( A \cup B )=40$ & $n (A \cap B)=0$
If we decrease $n (A \cup B)$ by $1$ then $n (A \cap B)$
Will increase by $1$
$n (A \triangle B)=39-1= 38$
Similarly on for the decrease of $1$ you will get in $(A \triangle B)$ as $36$ and $30$ so on.
Hence 
Range of $n (A \triangle B)= {10,12,14,16,18,20,......,38,40}$ 
$=16$ values