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Let $[\mathrm{x}]$ denote the greatest integer function. What is the number of solutions of the equation $x^{2}-4 x+[x]=0$ in the interval [0,2]?
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The correct answer is:
One
$\mathrm{x}^{2}-4 \mathrm{x}+[\mathrm{x}]=0$
Given internal, $[0,2]$
Case $1:$ Let $0 \leq \mathrm{x} < 1$
$[\mathrm{x}]=0$
$\therefore \mathrm{x}^{2}-4 \mathrm{x}+0=0 \Rightarrow \mathrm{x}(\mathrm{x}-4)=0 \Rightarrow \mathrm{x}=0, \mathrm{x}=4$
$\mathrm{x}=4$ can't be taken in $0 \leq \mathrm{x} < 1$
$x=0$
Case $2:$ Let $1 \leq x < 2$ $[\mathrm{x}]=1$
$\therefore \mathrm{x}^{2}-4 \mathrm{x}+1=0$
roots are $\frac{4 \pm \sqrt{16-4}}{2}=\frac{4 \pm \sqrt{12}}{2}=2 \pm \sqrt{3}$
In internal $1 \leq \mathrm{x} < 2,2 \pm \sqrt{3}$ are not the roots. Case $3:$ Let $x=2$ $[\mathrm{x}]=2$
$\therefore x^{2}-4 x+2=0$
roots are $\frac{4 \pm \sqrt{16-8}}{2}=\frac{4 \pm \sqrt{8}}{2}=2 \pm \sqrt{2}$
Since, $x=2$, roots can't be $2 \pm \sqrt{2}$
$\therefore$ There is only one solution, $x=0$
Given internal, $[0,2]$
Case $1:$ Let $0 \leq \mathrm{x} < 1$
$[\mathrm{x}]=0$
$\therefore \mathrm{x}^{2}-4 \mathrm{x}+0=0 \Rightarrow \mathrm{x}(\mathrm{x}-4)=0 \Rightarrow \mathrm{x}=0, \mathrm{x}=4$
$\mathrm{x}=4$ can't be taken in $0 \leq \mathrm{x} < 1$
$x=0$
Case $2:$ Let $1 \leq x < 2$ $[\mathrm{x}]=1$
$\therefore \mathrm{x}^{2}-4 \mathrm{x}+1=0$
roots are $\frac{4 \pm \sqrt{16-4}}{2}=\frac{4 \pm \sqrt{12}}{2}=2 \pm \sqrt{3}$
In internal $1 \leq \mathrm{x} < 2,2 \pm \sqrt{3}$ are not the roots. Case $3:$ Let $x=2$ $[\mathrm{x}]=2$
$\therefore x^{2}-4 x+2=0$
roots are $\frac{4 \pm \sqrt{16-8}}{2}=\frac{4 \pm \sqrt{8}}{2}=2 \pm \sqrt{2}$
Since, $x=2$, roots can't be $2 \pm \sqrt{2}$
$\therefore$ There is only one solution, $x=0$
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