Question

If the sum of roots of the equation $\{x+1\}+2x=4[x+1]-6$ is $p$, then the value of $3p$ is equal to
(where $\left\{x\right\}$ represents fractional part of $x$ and $\left[x\right]$ represents greatest integer function of $x$)

Answer 1

Given : $\{x+1\}+2x=4[x+1]-6$
$\Rightarrow \left\{x\right\}+2\left(\right[x]+\{x\left\}\right)=4\left[x\right]+4-6\Rightarrow 3\left\{x\right\}=2\left[x\right]-2\Rightarrow \left\{x\right\}=\frac{2\left[x\right]-2}{3}\cdots \left(1\right)$
We know,
$0\le \left\{x\right\}<1\Rightarrow 0\le \frac{2\left[x\right]-2}{3}<1\Rightarrow 0\le 2\left[x\right]-2<3\Rightarrow 1\le \left[x\right]\le \frac{5}{2}\therefore \left[x\right]=1,2$
From equation $\left(1\right)$,
If $\left[x\right]=1\Rightarrow \left\{x\right\}=0$
If $\left[x\right]=2\Rightarrow \left\{x\right\}=\frac{2}{3}$
$\because x=\left[x\right]+\left\{x\right\}\therefore x=1+0=1\text{and}x=2+\frac{2}{3}=\frac{8}{3}$
So, the sum of roots is
$p=1+\frac{8}{3}=\frac{11}{3}\therefore 3p=3\times \frac{11}{3}=11$