Mathematics - Quadratic Equation Question with Solution | TestHub

Given, $f(x)=x-[x],x\in R-\{0\}$

Now $f(x)+f\left(\frac{1}{x}\right)=1$

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

$\begin{array}{rlr}& & \Rightarrow \left(x+\frac{1}{x}\right)-[x]+\left[\frac{1}{x}\right]=1\end{array}$

$\begin{array}{rlr}& & \Rightarrow \left(x+\frac{1}{x}\right)=[x]+\left[\frac{1}{x}\right]+1\end{array}..\dots \dots \left(1\right)$

Clearly ,R.H.S is an integer ∴ L. H. S. is also an integer

Let $x+\frac{1}{x}=k$ an integer x²-kx+1 = 0

$\therefore x=\frac{k\pm \sqrt{k^2-4}}{2}$

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

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

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

Hence equation (i) has infinite number of solutions.