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JEE Main 20146 April
Exam
JEE Main
Year
2014
Date
6 April
Subject
Mathematics
Chapter
Sets and Relations

JEE Main 2014 · 6 April · Mathematics · Sets and Relations

If X = \ 4 ^ n - 3n - 1: n ∈ N\ and Y = \ 9(n - 1): n ∈ N\ , where...

If X={4n−3n−1:n∈N}X = \left\{{4}^{n} - 3n - 1: n \in N\right\} and Y={9(n−1):n∈N}Y = \left\{9\left(n - 1\right): n \in N\right\}, where NN is the set of natural numbers, then X∪YX \cup Y is equal to

Options

  1. AXX
  2. BYY
  3. CNN
  4. DY−XY - X

Correct answer

B. YY

Step-by-step solution

X={4n−3n−1:n∈N}X = \left\{{4}^{n} - 3n - 1: n \in N\right\}

4n−3n−1=(3+1)n−3n−1{4}^{n} - 3n - 1 = {\left(3 + 1\right)}^{n} - 3n - 1

=nC0(3)o1n+nC1(3)1(1)n−1+.....+nCn(3)n−3n−1={}^{n}{C}_{0}{\left(3\right)}^{o}{1}^{n} + {}^{n}{C}_{1}{\left(3\right)}^{1}{\left(1\right)}^{n - 1} + . .... + {}^{n}{C}_{n}{\left(3\right)}^{n} - 3n - 1

=1+n(3)+nC232+.....+3n−3n−1= 1 + n\left(3\right) + {}^{n}{C}_{2}{3}^{2} + . .... + {3}^{n} - 3n - 1

=nC232+nC333+.....+3n={}^{n}{C}_{2}{3}^{2} + {}^{n}{C}_{3}{3}^{3} + . .... + {3}^{n}

=32(nC2+nC331+nC432+.....+3n−2)= {3}^{2}\left({}^{n}{C}_{2} + {}^{n}{C}_{3}{3}^{1} + {}^{n}{C}_{4}{3}^{2} + . .... + {3}^{n - 2}\right)

=9k= 9k, where kk is a natural number.

Y={9(n−1):n∈N}Y = \left\{9\left(n - 1\right): n \in N\right\}

∴X⊆Y∴ X ⊆ Y

∴X∪Y=Y∴ X \cup Y = Y
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