- Home
- Standard 11
- Mathematics
For each positive real number $\lambda$. Let $A_\lambda$ be the set of all natural numbers $n$ such that $|\sin (\sqrt{n+1})-\sin (\sqrt{n})|<\lambda$. Let $A_\lambda^c$ be the complement of $A_\lambda$ in the set of all natural numbers. Then,
$A_{1 / 2}, A_{1 / 3}, A_{25}$ are all finite sets
$A_{1 / 3}$ is a finite set but $A_{1 / 2}, A_{25}$ are infi,nite sets
$A_{12}^c, A_{13}^c, A_{25}^c$ are all finites sets
$A_{1 / 3}, A_{2 / 5}$ are finite sets and $A_{1 / 2}$ is an infinite set
Solution
(c)
We have,
$|\sin \sqrt{n+1}-\sin \sqrt{n}| < \lambda, \lambda \in R$
When $n \rightarrow$
$|\sin \sqrt{n+1}-\sin \sqrt{n}| \rightarrow 0$
$\therefore$ There exist infinite natural number for which $|\sin \sqrt{n+1}-\sin \sqrt{n}| < \lambda, \forall \gamma \lambda \in 0$
Hence, $A_{1 / 2}, A_{1 / 3}, A_{2 / 5}$ are infinite sets. $\therefore A_{1 / 2}^e, A_{1 / 3}^e, A_{2 / 5}^e$ are finite sets.
