Question

For $x\in R^{+}$, if $x,[x]$ and $\{x\}$ are in harmonic progression then the value of $x$ is
[Note: $[x]$ and $\{x\}$ denote largest integer less than or equal to $x$ and fractional part of $x$ respectively.]

• A. $\frac{1}{\sqrt{2}}\tan \frac{\pi }{8}$
• B. $\frac{1}{\sqrt{2}}\cot \frac{\pi }{8}$
• C. $\frac{1}{\sqrt{2}}\tan \frac{\pi }{12}$
• D. $\frac{1}{\sqrt{2}}\cot \frac{\pi }{12}$

Answer 1

Answer: (B)

As $x,[x],\{x\}$ are in H.P., so
$[x]=\frac{2\{x\}}{x+\{x\}}=\frac{2\{x\}(\{x\}+[x])}{[x]+2\{x\}}$
$[x{]}^2+2[x]\{x\}=2\{x{\}}^2+2\{x\}[x]$
$\Rightarrow [x{]}^2=2\{x{\}}^2$
$\Rightarrow [x]=\sqrt{2}\{x\}$
$\Rightarrow \{x\}=\frac{1}{\sqrt{2}},[x]=1$
So, $x=\{x\}+[x]=1+\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}\cot \frac{\pi }{8}$