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JEE Advanced 2024 · Complex Numbers and Quadratic Equations

Let S={a+b2:a,b∈Z}S = \{a + b\sqrt{2} : a, b \in \mathbb{Z}\}, T1={(−1+2)n:n∈N}T_1 = \{(-1+\sqrt{2})^{n} : n \in \mathbb{N}\} and T2={(1+2)n:n∈N}T_2 = \{(1+\sqrt{2})^{n} : n \in \mathbb{N}\}. Then which of the following statements is (are) TRUE?

Answers: A. Z∪T1∪T2⊂S\mathbb{Z} \cup T_1 \cup T_2 \subset S · C. T2∩(2024,∞)≠ϕT_2 \cap (2024, \infty) \neq \phi · D. For any given a,b∈Za, b \in \mathbb{Z}, cos⁡(π(a+b2))+isin⁡(π(a+b2))∈Z\cos(\pi(a + b\sqrt{2})) + i \sin(\pi(a + b\sqrt{2})) \in \mathbb{Z} if and only if b=0b = 0, where i=−1i = \sqrt{-1}.

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Let S = {a + b√2 : a, b ∈ Z}, T₁ = {(-1+√2)ⁿ : n ∈ N} and T₂ = {(1+√2)ⁿ : n ∈ N}. Then which of the following statements is (are) TRUE? S={a+b2:a,b∈Z}S = \{a + b\sqrt{2} : a, b \in \mathbb{Z}\} T1={(−1+2)n:n∈N},T2={(1+2)n:n∈N}T_1 = \{(-1+\sqrt{2})^{n} : n \in \mathbb{N}\},\quad T_2 = \{(1+\sqrt{2})^{n} : n \in \mathbb{N}\} To find: which of (1), (2), (3), (4) are TRUE.\text{To find: which of (1), (2), (3), (4) are TRUE.} Plan: verify (1), (2), (3), (4) one by one.\text{Plan: verify (1), (2), (3), (4) one by one.} Key: (−1+2)(1+2)=1\text{Key: } (-1+\sqrt{2})(1+\sqrt{2}) = 1 (−1+2)(1+2)=−1+2=1(-1+\sqrt{2})(1+\sqrt{2}) = -1 + 2 = 1 ⇒ (−1+2)n=(1+2)−n\Rightarrow\ (-1+\sqrt{2})^{n} = (1+\sqrt{2})^{-n} (−1+2), (1+2)∈S(-1+\sqrt{2}),\ (1+\sqrt{2}) \in S S closed under ×⇒(−1+2)n, (1+2)n∈SS \text{ closed under } \times \Rightarrow (-1+\sqrt{2})^{n},\ (1+\sqrt{2})^{n} \in S (1) Z⊂S, T1⊂S, T2⊂S(1)\ \mathbb{Z} \subset S,\ T_1 \subset S,\ T_2 \subset S ⇒Z∪T1∪T2⊂S  ✓\Rightarrow \mathbb{Z} \cup T_1 \cup T_2 \subset S\ \ \checkmark tn=(2−1)n, 0<2−1<1t_n = (\sqrt{2}-1)^{n},\ 0 < \sqrt{2}-1 < 1 tn↓0 as n→∞t_n \downarrow 0 \text{ as } n \to \infty (2−1)8≈8.66×10−4>12024(\sqrt{2}-1)^8 \approx 8.66\times 10^{-4} > \dfrac{1}{2024} (2−1)9≈3.59×10−4<12024(\sqrt{2}-1)^9 \approx 3.59\times 10^{-4} < \dfrac{1}{2024} (2−1)9∈T1∩(0,12024)(\sqrt{2}-1)^9 \in T_1 \cap \left(0,\dfrac{1}{2024}\right) (2) FALSE ×(2)\ \text{FALSE}\ \times un=(1+2)n, 1+2>1u_n = (1+\sqrt{2})^{n},\ 1+\sqrt{2} > 1 un↑∞ as n→∞u_n \uparrow \infty \text{ as } n \to \infty (1+2)8≈1154<2024(1+\sqrt{2})^8 \approx 1154 < 2024 (1+2)9≈2786>2024(1+\sqrt{2})^9 \approx 2786 > 2024 (1+2)9∈T2∩(2024,∞)≠ϕ(1+\sqrt{2})^9 \in T_2 \cap (2024,\infty) \neq \phi (3) TRUE ✓(3)\ \text{TRUE}\ \checkmark cos⁡(π(a+b2))+isin⁡(π(a+b2))=eiπ(a+b2)\cos(\pi(a+b\sqrt{2})) + i\sin(\pi(a+b\sqrt{2})) = e^{i\pi(a+b\sqrt{2})} =eiπa⋅eiπb2=(−1)a(cos⁡(πb2)+isin⁡(πb2))= e^{i\pi a}\cdot e^{i\pi b\sqrt{2}} = (-1)^{a}\left(\cos(\pi b\sqrt{2}) + i\sin(\pi b\sqrt{2})\right) Real integer requires sin⁡(πb2)=0\text{Real integer requires } \sin(\pi b\sqrt{2}) = 0 ⇒b2=k∈Z⇒b=0 (2 irrational)\Rightarrow b\sqrt{2} = k \in \mathbb{Z} \Rightarrow b = 0\ (\sqrt{2}\ \text{irrational}) b=0⇒(−1)a∈Zb = 0 \Rightarrow (-1)^{a} \in \mathbb{Z} (4) TRUE ✓(4)\ \text{TRUE}\ \checkmark (1), (3), (4) are TRUE\boxed{(1),\ (3),\ (4)\ \text{are TRUE}}

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