JEE Advanced 2024 Paper 1 Q05 Mathematics Sets, Relations & Functions Sets & Set Operations Hard

JEE Advanced 2024 Paper 1 · Q05 · Sets & Set Operations

Let $S = \{a + b\sqrt{2} : a, b \in \mathbb{Z}\}$, $T_1 = \{(-1+\sqrt{2})^n : n \in \mathbb{N}\}$, and $T_2 = \{(1+\sqrt{2})^n : n \in \mathbb{N}\}$. Then which of the statements is (are) TRUE?

  1. A. $\mathbb{Z} \cup T_1 \cup T_2 \subset S$
  2. B. $T_1 \cap \left(0, \dfrac{1}{2024}\right) = \phi$, where $\phi$ denotes the empty set
  3. C. $T_2 \cap (2024, \infty) \neq \phi$
  4. D. For any given $a, b \in \mathbb{Z}$, $\cos\!\left(\pi(a+b\sqrt{2})\right) + i\sin\!\left(\pi(a+b\sqrt{2})\right) \in \mathbb{Z}$ if and only if $b = 0$, where $i = \sqrt{-1}$.
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:A, C, D

Solution

(A) Any integer $n=n+0\sqrt{2}\in S$. Powers of $\pm 1+\sqrt{2}$ expand by binomial into form $a+b\sqrt{2}$ with $a,b\in\mathbb{Z}$, so $T_1, T_2\subset S$. TRUE. (B) $-1+\sqrt{2}\approx 0.4142<1$, and powers shrink: $(-1+\sqrt{2})^n\to 0^+$. For large $n$, $(-1+\sqrt{2})^n<\frac{1}{2024}$, so $T_1\cap(0,1/2024)\neq\phi$. FALSE. (C) $(1+\sqrt{2})^n\to\infty$, so for large $n$ it exceeds $2024$. TRUE. (D) $\cos\theta+i\sin\theta\in\mathbb{Z}$ iff $\sin\theta=0$ and $\cos\theta=\pm1$, i.e. $\theta=k\pi$. Need $\pi(a+b\sqrt{2})=k\pi\Rightarrow a+b\sqrt{2}\in\mathbb{Z}$. Since $\sqrt{2}$ irrational and $a,b\in\mathbb{Z}$, this forces $b=0$. TRUE. Correct options: A, C, D.

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