JEE-Mains 2014 (1/30)

Probability Level 2

If $X=\{ 4^n-3n-1 : n \in \mathbb{N}\}$ and $Y=\{9(n-1): n \in \mathbb{N} \}$ , where $\mathbb{N}$ is the set of natural numbers, then $X \cup Y$ is equal to :

\mathbb{N}

X

Y

Y-X

1 solution

Rohit Ner
Jun 14, 2015

Let $P\left( n \right) :{ 4 }^{ n }-3n-1$ . Let $P(k)$ be divisible by $9$ .
$P\left( k \right) :{ 4 }^{ k }-3k-1 = 9m \\ \Rightarrow {4}^{k} = 9m+3k+1$ $\begin{aligned} P\left( k+1 \right) & ={ 4 }^{ k+1 }-3(k+1)-1 \\ & =4.{4}^{k}-3k-3-1 \\ & =4(9m+3k+1)-3k-4 \\ & =9(4m+k)\end{aligned}$
$\Rightarrow$ $X$ is the set of natural numbers divisible by $9$ . However, it doesnt contain all the multiples of $9$ . Since $Y$ contains all the multiples of $9$ , therefore $X\subset Y$ which means $X\cup Y$ $\Huge\color{#3D99F6}{=\boxed{Y}}$

Moderator note:

Yes this is correct. For clarity, you should explain why you started with the intention to prove that all the elements in set $X$ is divisible by 9. For example, you noticed for small $n$ , its elements are $0,9,54,243$ which suggests that all of them are divisible by 9.

JEE-Mains 2014 (1/30)

1 solution

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