JEE Main
Exam JEE Main
Subject Mathematics
Chapter Sets and Relations JEE Main · Mathematics · Sets and Relations
If X = ( 4 ^ n -3n - 1: n ∈ N) and Y = \ \ 9(n - 1): n ∈ N\ , then...
If
X = ( 4 n − 3 n − 1 : n ∈ N ) X = \left({4}^{n} -3n - 1: n \in N\right) X = ( 4 n − 3 n − 1 : n ∈ N ) and
Y = \left\$\{9\left(n - 1\right): n \in N\right\} , then
X ∩ Y = X \cap Y = X ∩ Y = Options A X X X B Y Y Y C ϕ ϕ ϕ D \left\{0\right\}
Step-by-step solution
X = 4 n − 3 n − 1 : n ∈ N X = { {4}^{n} - 3n - 1: n \in N } X = 4 n − 3 n − 1 : n ∈ N X = ( 1 + 3 ) n − 3 n − 1 : n ∈ N X = { {( 1 + 3 )}^{n} - 3n - 1: n \in N } X = ( 1 + 3 ) n − 3 n − 1 : n ∈ N X = 1 + 3 n + 3 2 n ( n − 1 ) 2 ! + 3 3 n ( n − 1 ) ( 2 n − 1 ) 6 . . . . 3 n − 1 X = { 1 + 3n + \dfrac{{3}^{2} n( n - 1 )}{2 !} + \dfrac{{3}^{3} n( n - 1 )( 2n - 1 )}{6} ....3n - 1 } X = 1 + 3 n + 2 ! 3 2 n ( n − 1 ) + 6 3 3 n ( n − 1 ) ( 2 n − 1 ) ....3 n − 1 X = 3 2 ( n − 1 ) n ( 1 2 ! + 3 ( 2 n − 1 ) 6 + . . . . . ) X = { {3}^{2} ( n - 1 )n( \dfrac{1}{2 !} + \dfrac{3( 2n - 1 )}{6} + ..... ) } X = 3 2 ( n − 1 ) n ( 2 ! 1 + 6 3 ( 2 n − 1 ) + ..... ) X = 9 ( n − 1 ) n ( 1 2 ! + . . . . . . ) X = { 9( n - 1 )n( \dfrac{1}{2 !} + ...... ) } X = 9 ( n − 1 ) n ( 2 ! 1 + ...... ) Y = 9 ( n − 1 ) : n ∈ N Y = { 9( n - 1 ): n \in N } Y = 9 ( n − 1 ) : n ∈ N ∴ X ⊆ Y ∴ X ⊆ Y ∴ X ⊆ Y ∴ X ∩ Y = X ∴ X \cap Y = X ∴ X ∩ Y = X (or)
Put
n = 1 , 2 , 3 , 4 , 5 , 6 , . . . . . n = 1, 2, 3, 4, 5, 6, ..... n = 1 , 2 , 3 , 4 , 5 , 6 , ..... in X and Y
X = 0 , 9 , 54 , . . . . X = { 0, 9, 54, .... } X = 0 , 9 , 54 , .... Y = 0 , 9 , 18 , 27 , 36 , 45 , 54 , 63...... Y = { 0, 9, 18, 27, 36, 45, 54, 63...... } Y = 0 , 9 , 18 , 27 , 36 , 45 , 54 , 63...... X ⊆ Y X ⊆ Y X ⊆ Y ∴ X ∩ Y = X ∴ X \cap Y = X ∴ X ∩ Y = X
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