Let $S=\{a+b \sqrt{2}: a, b \in \mathbb{Z}\}, T_1=\left\{(-1+\sqrt{2})^n: n \in \mathbb{N}\right\}$, and…

Let $S=\{a+b \sqrt{2}: a, b \in \mathbb{Z}\}, T_1=\left\{(-1+\sqrt{2})^n: n \in \mathbb{N}\right\}$, and $T_2=\left\{(1+\sqrt{2})^n: n \in \mathbb{N}\right\}$. Then which of the following statements is (are) TRUE?
  1. $\mathbb{Z} \cup T_1 \cup T_2 \subset S$
  2. $T_1 \cap\left(0, \frac{1}{2024}\right)=\phi$, where $\phi$ denotes the empty set.
  3. $T_2 \cap(2024, \infty) \neq \phi$
  4. For any given $a, b \in \mathbb{Z}, \cos (\pi(a+b \sqrt{2}))+i \sin (\pi(a+b \sqrt{2})) \in \mathbb{Z}$ if and only if $b=0$, where $i=\sqrt{-1}$.

Solution

(1) $\begin{aligned}& (-1+\sqrt{2})^{\mathrm{n}}=\mathrm{m}+\sqrt{2} \mathrm{n}, \mathrm{m}, \text{ where, } \mathrm{n} \in \mathbb{Z} \\ & (1+\sqrt{2})^{\mathrm{n}}=\mathrm{m}_1+\sqrt{2} \mathrm{n}_1, \text{ where, } \mathrm{~m}_1, \mathrm{n}_1 \in \mathbb{Z} \\& \Rightarrow \mathbb{Z} \cup \mathrm{T}_1 \cup \mathrm{T}_2 \subseteq \mathrm{S}\end{aligned}$ but $b \sqrt{2} \in S$ for negative $b \in \mathbb{Z}$. So $\mathbb{Z} \cup \mathrm{T}_1 \cup \mathrm{T}_2 \subset \mathrm{S}$ (2) $\begin{aligned} & (\sqrt{2}-1)^{\mathrm{n}}=\frac{1}{(\sqrt{2}+1)^{\mathrm{n}}} < \frac{1}{2024} \\ & \Rightarrow 2024 < (\sqrt{2}+1)^{\mathrm{n}}, \text{ where, } \exists \mathrm{n} \in \mathbb{N} \\ & \Rightarrow \mathrm{T}_1 \cap\left(0, \frac{1}{2024}\right) \neq \phi\end{aligned}$ (3) $\begin{aligned} & (1+\sqrt{2})^{\mathrm{n}}>2024, \text{ where, } \exists \mathrm{n} \in \mathbb{N} \\ & \Rightarrow \mathrm{T}_2 \cap(2024, \infty) \neq \phi\end{aligned}$ (4) $\sin (\pi(a+b \sqrt{2})=0) \Rightarrow b=0, \text{ where, } a \in \mathbb{Z}$ $\Rightarrow$ Options (1), (3), (4) are Correct.

Asked in: JEE Advanced 2024 (Paper 1)

Practice more Complex Number questions on Aicharya