Question

Let f (x) be a function such that f (x) = x - [x], where [x] is the greatest integer less than or equal to x. Then the number of solutions of the equation $f\left(x\right)+f\left(\frac{1}{x}\right)=1$ is (are)

• A. 0
• B. 1
• C. 2
• D. infinite

Answer 1

The correct option is D

infinite

Given, $f\left(x\right)=x-\left[x\right],xϵR-\left\{0\right\}$

$Nowf\left(x\right)+f\left(\frac{1}{x}\right)=1\therefore x-\left[x\right]+\frac{1}{x}-\left[\frac{1}{x}\right]=1\Rightarrow (x+\frac{1}{x})-\left(\right[x]+\left[\frac{1}{x}\right])=1\Rightarrow (x+\frac{1}{x})=\left[x\right]+\left[\frac{1}{x}\right]+1....\left(i\right)$

Clearly ,R.H.S is an integer $\therefore$ L. H. S. is also an integer

$Letx+\frac{1}{x}=kaninteger\Rightarrow x^2-kx+1=0\therefore x=\frac{k\pm \sqrt{k^2-4}}{2}$

For real values of x, $k^2-4\ge 0\Rightarrow k\ge 2ork\le -2$

We also observe that k = 2 and -2 does not satisfy equation (i)

$\therefore$ The equation (i) will have solutions if k > 2 or k < -2, where $kϵz$.

Hence equation (i) has infinite number of solutions.