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← JEE Advanced 2024 · Paper 1
MathematicsMultiple correct+4 / -2Question 5 of 51

JEE Advanced 2024 · Paper 1, Mathematics question 5

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

Choose one or more options
Answer and worked solution

Official answer: A, C, D

(A) Integers are a+02a + 0\sqrt2. Expanding (±1+2)n(\pm1+\sqrt2)^n binomially gives integer coefficients of 11 and 2\sqrt2, so T1,T2⊂ST_1, T_2 \subset S. True.

(B) 0<2−1<10 < \sqrt2 - 1 < 1, so (2−1)n→0+(\sqrt2-1)^n \to 0^+; for large nn it lies in (0,12024)\left(0, \frac{1}{2024}\right). The intersection is non-empty. False.

(C) 1+2>11+\sqrt2 > 1, so (1+2)n→∞(1+\sqrt2)^n \to \infty and eventually exceeds 2024. True.

(D) cos⁡(πt)+isin⁡(πt)=eiπt\cos(\pi t) + i\sin(\pi t) = e^{i\pi t} has modulus 1; it is an integer only if it equals ±1\pm1, i.e. t=a+b2t = a + b\sqrt2 is an integer. Since 2\sqrt2 is irrational, that happens exactly when b=0b = 0. True.