Then the number of solutions of $| x^ 2 − [ x ] | = 1$ is?

Question

Let $[ x ]$ denote the greatest integer less than or equal to $x$ for any real number $x$ . Then the number of solutions of $$| x^ 2 − [ x ] | = 1$$ is?

A. 0

B. 1

C. 2

D. 3

By trial and error i got only $+\sqrt{2}$ as the solution.

However, is there any technical way to solve this from which i can be sure about my answer?

And is t

Answer 1

Hint: break the equation to $$\begin{align}&x^2-[x]=1 \\&\text{ Or }\quad x^2= 1+[x] \quad\text{and}\quad x^2-[x]=-1\\&\text{ Or }\quad x^2=[x]-1\end{align}$$

Draw the graphs of $x^2$ and the RHS side of the equation. The number of times these graphs cut would be the number of solutions.

Since the graphs cut once at $x=2^{1/2}$ it has only one solution.