Question

For each positive real number $\lambda$, let $A_{\lambda }$ be the set of all natural numbers n such that $|\sin \left(\sqrt{n+1}\right)-\sin \left(\sqrt{n}\right)|<\lambda$. Let $A_{\lambda }^c$ be the complement of $A_{\lambda }$ in the set of all natural numbers. Then,

• A. $A_{\frac{1}{2}},A_{\frac{1}{3}},A_{\frac{2}{5}}$ are all finite sets
• B. $A_{\frac{1}{2}}$ is a finite set but $A_{\frac{1}{3}},A_{\frac{2}{5}}$ are infinite sets
• C. $A_{\frac{1}{2}}^c,A_{\frac{1}{3}}^c,A_{\frac{2}{5}}^c$ are all finite sets
• D. $A_{\frac{1}{3}},A_{\frac{2}{5}}$ are finite sets and $A_{\frac{1}{2}}$ is an infinite set

Answer 1

Answer: (C)

$A_{\frac{1}{2}}^c,A_{\frac{1}{3}}^c,A_{\frac{2}{5}}^c$ are all finite sets

$A_{\lambda }^c$ is the set of natural numbers such that :

$\left|sin\right(\sqrt{n+1})-sin(\sqrt{n}\left)\right|\ge \lambda$

$\Rightarrow \left|cos\left(\frac{\sqrt{n+1}+\sqrt{n}}{2}\right)sin\left(\frac{\sqrt{n+1}-\sqrt{n}}{2}\right)\right|\ge \frac{\lambda }{2}$

$\Rightarrow \left|cos\left(\frac{\sqrt{n+1}+\sqrt{n}}{2}\right)sin\left(\frac{1}{2(\sqrt{n+1}+\sqrt{n})}\right)\right|\ge \frac{\lambda }{2}$

Let $\theta =\sqrt{n+1}+\sqrt{n}$. $\therefore \theta \ge \sqrt{2}+1$.

For $\lambda =\frac{1}{2}$,

$\left|cos\left(\frac{\theta }{2}\right)\right|\left|sin\left(\frac{1}{2\theta }\right)\right|\ge \frac{1}{4}$ ....equation (1)

$sin\left(\frac{1}{2\theta }\right)$ is a decreasing function of $\theta$. $sin\left(\frac{1}{2\theta }\right)\in (0,sin\left(\frac{1}{2(\sqrt{2}+1)}\right)]$

If $sin\left(\frac{1}{2\theta }\right)<\frac{1}{4}$ then no solution exists to equation (1) and the set is finite.

Otherwise let $\theta ={\theta }_0$ such that $sin\left(\frac{1}{2{\theta }_0}\right)=\frac{1}{4}$. Thus, if solutions to equation (1) exist, they must satisfy $\sqrt{2}+1\le \theta \le {\theta }_0$.

Thus the set is finite.

Similar arguments are made for other values of $\lambda$.