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

If $A=\left{1, 2, 3\right}$, then the numbers of subsets of set $A$ containing element $3$, is 

  1. $24$
  2. $28$
  3. $8$
  4. $16$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
The set $\left\{1, 2, 3\right\}$ has $8$ subsets. The first subset would be the null or empty subset, which contains none of the numbers: $\left\{\right\}$.
 The null set is a subset of every set. The other subsets would include some of the numbers in the set, but not all of them: $\left\{1\right\}$,$\left\{2\right\}$,$\left\{3\right\}$,$\left\{1,2\right\}$,$\left\{1,3\right\}$,$\left\{2,3\right\},\{1,2,3\}$
Multiple choice maths set concepts finite and infinite sets types of sets set language

Let ${ a } _{ 1 },{ a } _{ 2 },{ a } _{ 3 },............{ a } _{ 10 }$ be in G.P. with ${ a } _{ i }>0$ for $i=1,2,....,10$ and $S$ be the set of pairs $(r,k),r\quad k\in N$ ( the set of natural numbers) for which
$\left| { log } _{ e }{ a } _{ 1 }^{ r }{ a } _{ 2 }^{ k }\quad { log } _{ e }{ a } _{ 2 }^{ r }{ a } _{ 3 }^{ k }\quad { log } _{ e }{ a } _{ 3 }^{ r }{ a } _{ 4 }^{ k }\ { log } _{ e }{ a } _{ 4 }^{ r }{ a } _{ 5 }^{ k }\quad { log } _{ e }{ a } _{ 5 }^{ r }{ a } _{ 6 }^{ k }\quad { log } _{ e }{ a } _{ 6 }^{ r }{ a } _{ 7 }^{ k }\ { log } _{ e }{ a } _{ 7 }^{ r }a _{ 8 }^{ k }\quad { log } _{ e }{ a } _{ 8 }^{ r }{ a } _{ 9 }^{ k }\quad { log } _{ e }{ a } _{ 9 }^{ r }{ a } _{ 10 }^{ k } \right| =0$
Then the number of elements in S, is :

  1. Infinitely many

  2. 4

  3. 10

  4. 2
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The determinant of the matrix involves logarithmic terms of a geometric progression. Since the rows/columns are linearly dependent based on the properties of logarithms and geometric sequences, the determinant is zero for any r and k in N, leading to infinitely many solutions.

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

If $A$ is finite set. Let $n(A)$ denote the number of elements in $A$ and $B$ are finite sets, $A\neq B$ and $n(A) = n(B)$. Then $n(A\cap B)$ is

  1. $ > n(A)$
  2. $ < n(A)$
  3. $ \neq n(A)$
  4. $ \leq n(A)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If A and B are finite sets with the same number of elements (n(A) = n(B)) but A is not equal to B, then the intersection A intersect B must contain fewer elements than A. If it contained n(A) elements, then A would have to be a subset of B, and since they have the same size, A would equal B.

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

Identify the type of Set
$A= { x| x \epsilon N, 2 \leq x \leq 3}$

  1. Finite Set

  2. Infinite Set

  3. Null Set

  4. Singleton Set
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We have to identify the type of set.

Given $A={x|x\in N, 2 \leq x \leq 3 }$

               $={2,3}$ which is a finite set.

Therefore $A$ is a finite set.

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

A finite set $S$ is given by $S={x:x\in N: x\le15}.$ Find the cardinality of its power set.

  1. 32952

  2. 16384

  3. 32768

  4. 16476
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$S$ is given by $S={x:x\in N: x\le15}$
$S={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}$
$\therefore n(S)=15$
The power set is set of all the possible subsets of $S.$
Number of elements in power set $=$ Total number of subsets
So, cardinality of power set $S=2^{15}=32768$