- Home
- Standard 11
- Mathematics
Let $S=\{a+b \sqrt{2}: a, b \in Z \}, T_1=\left\{(-1+\sqrt{2})^n: n \in N \right\}$ and $T_2=\left\{(1+\sqrt{2})^n: n \in N \right\}$. Then which of the following statements is (are) $TRUE$?
$(A)$ $Z \cup T_1 \cup T_2 \subset S$
$(B)$ $T_1 \cap\left(0, \frac{1}{2024}\right)=\phi$, where $\phi$ denotes the empty set
$(C)$ $T_2 \cap(2024, \infty) \neq \phi$
$(D)$ For any given $a, b \in Z , \cos (\pi(a+b \sqrt{2}))+i \sin (\pi(a+b \sqrt{2})) \in Z$ if and only if $b=0$, where $i=\sqrt{-1}$
$A,B,C$
$A,B$
$A,C$
$A,B,D$
Solution
$(A)(-1+\sqrt{2})^{ n }= m +\sqrt{2} n , m , n \in Z$
$(1+\sqrt{2})^{ n }= m _1+\sqrt{2} n _1, m _1, n _1 \in Z$
$\Rightarrow Z \cup T _1 \cup T _2 \subseteq S$
but $b \sqrt{2} \in S$ for negative $b \in Z$.
So $\quad Z \cup T _1 \cup T _2 \subset S$
$(B)(\sqrt{2}-1)^{ n }=\frac{1}{(\sqrt{2}+1)^2}<\frac{1}{2024}$
$\Rightarrow 2024<(\sqrt{2}+1)^{ n }, \exists n \in N$
$\Rightarrow T _1 \cap\left(0, \frac{1}{2024}\right) \neq \phi$
$(C)(1+\sqrt{2})^2>2024, \exists n \in N$
$\Rightarrow T _2 \cap(2024, \infty) \neq \phi$
$(D)\sin (\pi(a+b \sqrt{2})=0) \Rightarrow b=0, a \in Z$.
$\Rightarrow$ Options $(A), (C), (D)$ are Correct.
