Tag: introduction to set

Questions Related to introduction to set

A is a set containing $n$ elements. $A$ subset $P$ of $A$ is chosen. the set $A$ is reconstructed by replacing the elements of $P.A$ subset $Q$ of $A$ is again chosen. the number of ways of choosing $P$ and $Q$ so that $P \cap Q$

  1. $9. ^{n}C _{2}$

  2. $3^{n}- ^{n}C _{2}$

  3. $^{n}C _{2}.3^{n-2}$

  4. $4^{n}-3^{n}$


Correct Option: A

Let $A=\left{ a,b,c,d \right} ,B=\left{ b,c,d,e \right}$. Then $n\left[ \left( A\times B \right) \cap \left( B\times A \right)  \right]$ is equal to 

  1. $3$

  2. $6$

  3. $9$

  4. $none$


Correct Option: A

The set of all points where the function $f(x)=||x|$ is twice differentiable is

  1. $(-\infty, \infty)$

  2. $(-\infty, 0)\cup (0, \infty)$

  3. $(0, \infty)$

  4. $[0, \infty)$


Correct Option: A

Let $S=\left{ \left( x,y \right) :\dfrac { y\left( 3x-1 \right)  }{ x\left( 3x-2 \right)  } <0 \right}$ and $S'=\left{ \left( x,y \right) \in A\times B;\ -1\le A\le 1,-1\le B\le 1 \right} $ There area of $S\cap S'$ is

  1. $1$

  2. $3$

  3. $2$

  4. $4$


Correct Option: A

Let A={1, 2, 3, 4), B={2, 3, 4, 5}, then $n{ (A\times B)\cap (B\times A)} =$?

  1. 13

  2. 16

  3. 9

  4. 10


Correct Option: A

Let $P={ \theta :sin\theta -cos\theta =\sqrt { 2 } cos\theta } $ and $Q={ sin\theta + cos\theta =\sqrt { 2 } sin\theta } $ be two sets. Then:

  1. $P\subset Q\quad and\quad Q-P\neq \emptyset $

  2. $Q\subset P$

  3. $P\subset Q$

  4. $P=Q$


Correct Option: D
Explanation:

$P = \left{ {\theta :\sin \theta  - \cos \theta  = \sqrt 2 \cos \theta } \right}$

$Q = \left{ {\theta :\sin \theta  + \cos \theta  = \sqrt 2 \sin \theta } \right}$
From $P$
$\sin \theta  - \cos \theta  = \sqrt 2 \cos \theta $
$\sin \theta  = \left( {\sqrt 2  + 1} \right)\cos \theta $
$\frac{{\sin \theta }}{{\cos \theta }} = \left( {\sqrt 2  + 1} \right)$
$\tan \theta  = \left( {\sqrt 2  + 1} \right)$
from $Q$
$\sin \theta  + \cos \theta  = \sqrt 2 \sin \theta $
$\sin \theta \left( {\sqrt 2  - 1} \right) = \cos \theta $
$\frac{{\sin \theta }}{{\cos \theta }} = \left( {\sqrt 2  - 1} \right)$
$\tan \theta  = \left( {\sqrt 2  - 1} \right)$
$\tan \theta  = \frac{1}{{\sqrt 2  - 1}} \times \frac{{\sqrt 2  + 1}}{{\sqrt 2  + 1}} = \sqrt 2  + 1$
$\therefore P = Q$
Hence,
option $(D)$ is correct answer.

If $A = {1, 2, 3, 4, 5}, B = {2, 4, 6, 8}$ and C= ${3,4,5,6}$, 

then verify : $A - (B \cup C) = (A - B) \cap (A - C)$.

  1. True

  2. False


Correct Option: A
Explanation:
Given, $A = \{1, 2, 3, 4, 5\}, B = \{2, 4, 6, 8\}$ and $C=\{3,4,5,6\}$

For the LHS:

Union of two sets will have the elements of both sets.

So, $ B \cup C = \{2,3,4,5,6,8 \}$ 

$ A - (B \cup C) $ will have elements of $A$ which are not in $ (B \cup C) $

So, $ A - (B \cup C) = \{ 1 \}$ ..... $(1)$

For the RHS:

$ A - B $ will have elements of $A$ which are not in $B$.

So, $ A - B = \{ 1,3,5 \}$  

$ A - C $ will have elements of $A$ which are not in $C$.

So, $ A - C = \{ 1,2 \}$  

Intersection of two sets has the common elements of both the sets. 

$\Rightarrow (A - B) \cap (A - C) = \{1\}$ ..... $(2)$

From $(1)$ and $(2),$ we have

$ A - (B \cup C) =(A - B) \cap (A - C) $

Hence, the given expression is true.

Let $A = {$ multiples of $3$ less than $20 }$
      $B = {$ multiples of $5$ less than $20}$
Then  $A$ $\displaystyle\cap$ $B$ is

  1. ${3, 5}$

  2. ${5, 9}$

  3. ${15}$

  4. $\displaystyle\phi $


Correct Option: C
Explanation:

$A = {$ multiples of $ 3 $ less than $20}$

    $= {3,6,9,12,15,18}$
$B={ $ multiples of $5$ less than $20}$
    $= {5,10,15}$

$A \cap B = 15$

Let $P _1$ be the set of all prime numbers, i.e., $P _1=\left {2, 3, 5, 7, 11, ....\right }$, Let $Pn=\left {np|p\epsilon P _1|\right }$, i.e., the set of all prime multiples of n. Then which of the following sets is non empty?

  1. $P _1\cap P _{23}$

  2. $P _7\cap P _{21}$

  3. $P _{12}\cap P _{20}$

  4. $P _{20}\cap P _{24}$


Correct Option: C
Explanation:

Check by option
$P _{12}=\left {24, 36, 60, 84, ....\right }$
$P _{20}=\left {40, 60, 100, .....\right }$
$P _{12}\cap P _{20}$ has common element.

If $A={a,b,c,d,e,f}$ and $B={c,e,f,g,h}$, then the number of elements of $(A-B)\cap A$ is

  1. $0$

  2. $1$

  3. $2$

  4. $3$


Correct Option: D
Explanation:

$A-B={a,b,d}$
$(A-B)\cap A={a,b,d}$