Tag: set language

Quizzes (7) Questions (71)

Questions Related to set language

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

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

  1. $(i)$ only
  2. $(ii)$ only
  3. $(i)$ and $(ii)$ both
  4. Neither $(i)$ nor $(ii)$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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 $(0,0).$

  1. $(i),(ii)$ and $(iv)$
  2. $(ii)$ only
  3. $(i),(iii)$ and $(iv)$
  4. $(i),(ii)$ and $(iii)$$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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 $100.$ 

  1. $(i)$ and $(iii)$
  2. $(i)$ only
  3. $(ii),(iii)$ and $(iv)$
  4. $(ii)$ and $(iv)$
Reveal answer Fill a bubble to check yourself
A Correct answer
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. 

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

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

  1. $(i)$ only
  2. $(ii)$ only
  3. $(i)$ and $(ii)$ both
  4. Neither $(i)$ nor $(ii)$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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.

  1. $-2$
  2. $-\displaystyle\frac{1}{2}$
  3. $0$
  4. $2$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $1/2$
Multiple choice maths set language different sets de morgan's law de morgan's law for set theory

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

  1. $2$
  2. $7$
  3. $5$

  4. can not be determined
Reveal answer Fill a bubble to check yourself
D Correct answer
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. 

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

$A\cup B=A\cap B$ if and only if

  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
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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

  1. 30

  2. 16

  3. 26

  4. 40
Reveal answer Fill a bubble to check yourself
B Correct answer
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