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, $x=\{x\}+[x]=1+\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}} \cot \frac{\pi}{8}$