Question

If the equation $x ^3-3 x +1=0$ has three real roots $x _1, x _2, x _3$, where $x _1< x _2< x _3$, then the value of $\left(\left\{x_1\right\}+\left\{x_2\right\}+\left\{x_3\right\}\right)$ is equal to
[Note: $\{x\}$ denotes the fractional part of $x$.]

• A. $\frac{3}{2}$
• B. 1
• C. 2
• D. $\frac{5}{2}$

Answer 1

Answer: (B)

$f(x)=x^3-3 x+1 $
$f ^{\prime}( x )=3\left( x ^2-1\right)=( x +1)( x -1)$
$\therefore f ( x ) \text { is increasing in }(-\infty,-1) \cup(1, \infty) \text { and decreasing in }(-1,1)$
$\text { Now, } f (-2)=-1 $f (1)=-1 $
$f(-1)=3$f(2)=3 $
$f (0)=1$
$f (-2) \cdot f (-1)<0 \Rightarrow \text { Hence one root lies in }(-2,-1) $
$\therefore\left[ x _1\right]=-2 $
$\text { Now } f (0) \cdot f (1)<0 \Rightarrow \text { Hence one root lies in }(0,1)$
$\therefore\left[ x _2\right]=0$
$\text { Also } f (1) \cdot f (2)<0 \Rightarrow \text { Hence one root lies in }(1,2) $
$\therefore\left[ x _3\right]=1$

$\Rightarrow x _1+ x _2+ x _3=0 $
$\therefore\left\{ x _1\right\}+\left\{ x _2\right\}+\left\{ x _3\right\} = x _1-\left[ x _1\right]+ x _2-\left[ x _2\right]+ x _3-\left[ x _3\right]$
$ =\left( x _1+ x _2+ x _3\right)-\left(\left[ x _1\right]+\left[ x _2\right]+\left[ x _3\right]\right) $
$ =0-(-2+0+1)=1$