Question

Let $\left[x\right]$ denote the greatest integer function. What is the number of solutions of the equation $x^2-4x+\left[x\right]=0$ in the interval $[0,2]$?

• A. Zero (no solution)
• B. One
• C. Two
• D. Three

Answer 1

Answer: (B)

One

$x^2-4x+\left[x\right]=0$

Now $x^2+4x=(-[x\left]\right)$

(1) When $xϵ[0,1)\Rightarrow \left[x\right]=0$

$x^2-4x=0\Rightarrow x=0,4$ but we have assumed

$xϵ[0,1)$

So only ${}^{\prime }{0}^{\prime }$ is solution

(2) When $xϵ[1,2)\Rightarrow \left[x\right]=1$

$4x^2-4x=-1\Rightarrow x=(2+\sqrt{3}),(2-\sqrt{3})$

Both are not in assumed range so rejected

(3) When $x=2\Rightarrow \left[x\right]=2$

$x^2-4x=-2\Rightarrow x=(2+\sqrt{2}),(2-\sqrt{2})$, both rejected as we have assumed $x=2$

$\therefore x^2=4x+\left[x\right]=0$ has only one solution

$\ln [0,2]$ i.e. ${}^{\prime }{0}^{\prime }$.