JEE Advanced 2024 — Paper 1 — Question 5

(A) $S$ contains all numbers $a+b\sqrt2$ with $a,b\in\mathbb Z$.

$\mathbb Z\subset S$ (by taking $b=0$).

For $T_1$, any element is either $(-1+\sqrt2)^n$ for $n\in\mathbb Z$.

Since $(-1+\sqrt2)^n = \alpha_n+\beta_n\sqrt2$ with integers $\alpha_n,\beta_n$ (by binomial expansion), it belongs to $S$.

Similarly for $T_2$. Hence $\mathbb Z\cup T_1\cup T_2\subset S$. (B) $T_1$ contains numbers like $(-1+\sqrt2)^1 = -1+\sqrt2 \approx 0.414$.

Since $0.414\in(0,12024)$, the intersection is nonempty, so statement (B) is false. (C) For $T_2$, $(1+\sqrt2)^n$ grows without bound as $n$ increases. For instance, $(1+\sqrt2)^20 > 2024$.

Thus $T_2\cap(2024,\infty) eq\phi$, so statement (C) is true. (D) The expression is $e^{i\pi(a+b\sqrt2)}$.

It is an integer (real) iff its imaginary part is zero, i.e., $\sin(\pi(a+b\sqrt2))=0$, which requires $a+b\sqrt2\in\mathbb Z$.

Since $\sqrt2$ is irrational, this holds iff $b=0$.

Then $e^{i\pi a}=\cos(\pi a)=\pm1$, an integer.

Hence statement (D) is true.