JEE Advanced · 2024 · 26 May 2024 · Paper 1 · Complex Numbers

(A) By binomial expansion, $(\pm 1+\sqrt{2}{)}^n=m+n\sqrt{2}$ for some integers $m,n$. Thus $T_1,T_2\subset S$. Also $\mathbb{Z}\subset S$ (with $b=0$). However, $S$ contains elements like $-\sqrt{2}$ (with $a=0,b=-1$) that are not in $\mathbb{Z}\cup T_1\cup T_2$, so the inclusion is proper.

(B) $(\sqrt{2}-1{)}^n=\frac{1}{(\sqrt{2}+1{)}^n}$. For large enough $n$, $(\sqrt{2}+1{)}^n>2024$, so $(\sqrt{2}-1{)}^n<\frac{1}{2024}$. Thus $T_1\cap (0,\frac{1}{2024})\ne \varphi$. Statement (B) is FALSE.

(C) Since $(1+\sqrt{2}{)}^n$ grows without bound, there exists $n$ such that $(1+\sqrt{2}{)}^n>2024$. Thus $T_2\cap (2024,\infty )\ne \varphi$. TRUE.

(D) $\cos (\pi (a+b\sqrt{2}))+i\sin (\pi (a+b\sqrt{2}))=e^{i\pi (a+b\sqrt{2})}$. This is an integer only when the imaginary part is zero, which requires $\sin (\pi (a+b\sqrt{2}))=0$. This happens when $\pi (a+b\sqrt{2})=k\pi$ for integer $k$, i.e., $a+b\sqrt{2}=k$. Since $\sqrt{2}$ is irrational, this requires $b=0$. TRUE.