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JEE Advance - Mathematics (2024 - Paper 1 Online - No. 5)

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?
$\mathbb{Z} \cup T_1 \cup T_2 \subset S$
$T_1 \cap\left(0, \frac{1}{2024}\right)=\phi$, where $\phi$ denotes the empty set.
$T_2 \cap(2024, \infty) \neq \phi$
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 $i=\sqrt{-1}$.

Explanation

$$\begin{aligned} \text{(A)} \quad & (-1+\sqrt{2})^{\mathrm{n}}=\mathrm{m}+\sqrt{2} \mathrm{n}, \mathrm{m}, \mathrm{n} \in \mathbb{Z} \\ & (1+\sqrt{2})^{\mathrm{n}}=\mathrm{m}_1+\sqrt{2} \mathrm{n}_1, \mathrm{~m}_1, \mathrm{n}_1 \in \mathbb{Z} \\ & \Rightarrow \mathbb{Z} \cup \mathrm{T}_1 \cup \mathrm{T}_2 \subseteq \mathrm{S} \end{aligned}$$

but $$b \sqrt{2} \in S$$ for negative $$b \in \mathbb{Z}$$.

So $$\quad \mathbb{Z} \cup T_1 \cup T_2 \subset \mathrm{S}$$

$$\begin{aligned} \text{(B)}\quad& (\sqrt{2}-1)^{\mathrm{n}}=\frac{1}{(\sqrt{2}+1)^{\mathrm{n}}}<\frac{1}{2024} \\ & \Rightarrow 2024<(\sqrt{2}+1)^{\mathrm{n}}, \exists \mathrm{n} \in \mathbb{N} \\ & \Rightarrow \mathrm{T}_1 \cap\left(0, \frac{1}{2024}\right) \neq \phi \end{aligned}$$

$$\begin{aligned} \text{(C)}\quad& (1+\sqrt{2})^{\mathrm{n}}>2024, \exists \mathrm{n} \in \mathbb{N} \\ & \Rightarrow \mathrm{T}_2 \cap(2024, \infty) \neq \phi \end{aligned}$$

$$\mathrm{(D)}$$ $$\quad \sin (\pi(a+b \sqrt{2})=0) \Rightarrow b=0, a \in \mathbb{Z}$$.

$$\Rightarrow$$ Options (A), (C), (D) are Correct.

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