Question

If $f\left(x\right)=x-\left[x\right],x(\ne 0)\in R,$ where $\left[x\right]$ is the greatest integer less than or equal to $x,$ then the number of solutions of $f\left(x\right)+f\left(\frac{1}{x}\right)=1$ are

• A. $0$
• B. $1$
• C. infinte
• D. $2$

Answer 1

Answer: (C)

infinte

We have,$f\left(x\right)+f\left(\frac{1}{x}\right)=1$

$\Rightarrow x-\left[x\right]+\frac{1}{x}-\left[\frac{1}{x}\right]=1$

$\Rightarrow x+\frac{1}{x}-1=\left[x\right]+\left[\frac{1}{x}\right]$

$\Rightarrow \frac{x^2+1-x}{x}=$ $($integer$)$ $k$ $($say$)$

$\Rightarrow x^2-(k+1)x+1=0$

Since $x$ is real, so${(k+1)}^2-4\ge 0$

$\Rightarrow k^2+2k-3\ge 0\Rightarrow (k+3)(k-1)\ge 0$

$\Rightarrow k\le -3\Rightarrow k\ge 1.$

Therefore, number of solutions is infinite.