Question

Let $\left[x\right]$ denote the greatest integer in $x$. Then in the interval $[0,3]$ the number of solutions of the equation $x^2-3x+\left[x\right]=0$ are

• A. $6$
• B. $4$
• C. $2$
• D. $0$

Answer 1

The correct option is C $2$

$x^2-3x+\left[x\right]=0$
Case I : If $x\in [0,1)$
$\Rightarrow \left[x\right]=0$
So, the equation will be $x^2-3x=0$
$\Rightarrow x=0,3$
$x=0$ lies in the assumed domain and is a solution to the equation.
Case II : If $x\in [1,2)$
$\Rightarrow \left[x\right]=1$
So, the equation will be $x^2-3x+1=0$

$\Rightarrow x=\frac{-(-3)\pm \sqrt{(-3{)}^2-4(1\left)\right(1})}{2\left(1\right)}=\frac{3\pm \sqrt{5}}{2}$Both values lie outside the assumed domain.

Case III : If $x\in [2,3)$
$\Rightarrow \left[x\right]=2$
So, the equation will be $x^2-3x+2=0\Rightarrow x^2-2x-x+2=0\Rightarrow (x-2)(x-1)=0$
$\Rightarrow x=1,2$
So, $2$ lies in the assumed domain and is a solution of the equation.
At $x=3,\left[3\right]=3$
The equation will be $x^2-3x+3=0$
Here, $D=9-4\left(1\right)\left(3\right)=-3<0$ and has no real solution in the assumed domain.

Hence, there are two roots $0$ and $2$.