Question

Let f(x) be a function defined by f(x) = x - [x], 0 $\ne$ x $ϵ$ R where [x] 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. Infinite
• C. 1
• D. 2

Answer 1

The correct option is B

Infinite

Given f(x) = x - [x]. 0 $\ne$ x $ϵ$ R.

f(x) + f$\left(\frac{1}{x}\right)$ = x - [x] + $\frac{1}{x}-\left[\frac{1}{x}\right]=1$

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

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

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

$\because$ x is real so $(k+1{)}^2$ - 4 $\ge$ 0 [ $b^2-4ac\ge 0$ ]

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

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