Question

If the equation $x^3-3x+1=0$ 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)

$f(x)=x^3-3x+1$
$f^{\prime }(x)=3\left(x^2-1\right)=(x+1)(x-1)$
$\therefore f(x)\text{ is increasing in }(-\infty ,-1)\cup (1,\infty )\text{ and decreasing in }(-1,1)$
$\text{ Now, }f(-2)=-1$f (1)=-1 $<br/>$f(-1)=3$f(2)=3$
$f(0)=1$
$f(-2)\cdot f(-1)<0\Rightarrow \text{ Hence one root lies in }(-2,-1)$
$\therefore \left[x_1\right]=-2$
$\text{ Now }f(0)\cdot f(1)<0\Rightarrow \text{ Hence one root lies in }(0,1)$
$\therefore \left[x_2\right]=0$
$\text{ Also }f(1)\cdot f(2)<0\Rightarrow \text{ Hence one root lies in }(1,2)$
$\therefore \left[x_3\right]=1$

$\Rightarrow x_1+x_2+x_3=0$
$\therefore \left\{x_1\right\}+\left\{x_2\right\}+\left\{x_3\right\}=x_1-\left[x_1\right]+x_2-\left[x_2\right]+x_3-\left[x_3\right]$
$=\left(x_1+x_2+x_3\right)-\left(\left[x_1\right]+\left[x_2\right]+\left[x_3\right]\right)$
$=0-(-2+0+1)=1$