Question

If $$A = \left\{ {4n + 2|n{\text{ is a natural number}}} \right\}$$       and $$B = \left\{ {3n|n{\text{ is a natural number}}} \right\},$$       then what is $$\left( {A \cap B} \right)$$   equal to ?

A. $$\left\{ {12{n^2} + 6n|n{\text{ is a natural number}}} \right\}$$
B. $$\left\{ {24n - 12|n{\text{ is a natural number}}} \right\}$$
C. $$\left\{ {60n + 30|n{\text{ is a natural number}}} \right\}$$
D. $$\left\{ {12n - 6|n{\text{ is a natural number}}} \right\}$$  
Answer :   $$\left\{ {12n - 6|n{\text{ is a natural number}}} \right\}$$
Solution :
$$\eqalign{ & {\text{Let }}A = \left\{ {4n + 2:n\, \in N} \right\}\,{\text{and }}B = \left\{ {3n:n\, \in \,N} \right\} \cr & \Rightarrow \,A = \left\{ {6,\,10,\,14,\,18,\,22,\,26,\,30,\,34,\,38,\,42,.....} \right\} \cr & {\text{and }}B = \left\{ {3,\,6,\,9,\,12,\,15,\,18,\,21,\,24,\,27,\,30,.....} \right\} \cr & \therefore \,A \cap B = \left\{ {6,\,18,\,30,\,42,.....} \right\} \cr & = 6 + 12n - 12 \cr & = 12n - 6 \cr & {\text{Hence, }}A \cap B = \left\{ {12n - 6:n{\text{ is a natural number}}} \right\} \cr} $$

Releted MCQ Question on
Calculus >> Sets and Relations

Releted Question 1

If $$X$$ and $$Y$$ are two sets, then $$X \cap {\left( {X \cup Y} \right)^c}$$   equals.

A. $$X$$
B. $$Y$$
C. $$\phi $$
D. None of these
Releted Question 2

The expression $$\frac{{12}}{{3 + \sqrt 5 + 2\sqrt 2 }}$$    is equal to

A. $$1 - \sqrt 5 + \sqrt 2 + \sqrt {10} $$
B. $$1 + \sqrt 5 + \sqrt 2 - \sqrt {10} $$
C. $$1 + \sqrt 5 - \sqrt 2 + \sqrt {10} $$
D. $$1 - \sqrt 5 - \sqrt 2 + \sqrt {10} $$
Releted Question 3

If $${x_1},{x_2},.....,{x_n}$$    are any real numbers and $$n$$ is any positive integer, then

A. $$n\sum\limits_{i = 1}^n {{x_i}^2 < {{\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)}^2}} $$
B. $$\sum\limits_{i = 1}^n {{x_i}^2 \geqslant {{\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)}^2}} $$
C. $$\sum\limits_{i = 1}^n {{x_i}^2 \geqslant n{{\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)}^2}} $$
D. none of these
Releted Question 4

Let $$S$$ = {1, 2, 3, 4}. The total number of unordered pairs of disjoint subsets of $$S$$ is equal to

A. 25
B. 34
C. 42
D. 41

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