Question

Find all posible values of $x$ satisfying $\frac{\left[x\right]}{[x-2]}-\frac{[x-2]}{\left[x\right]}=\frac{8\left\{x\right\}+12}{[x-2]\left[x\right]}$ (where $[\cdot ]$ denotes the greatest integer function and $\{\cdot \}$ is fractional part).

• A. $x\in \{4,\frac{11}{2}\}$
• B. $x\in \{3,\frac{11}{2}\}$
• C. $x\in \{2,\frac{7}{2}\}$
• D. $x\in \{1,\frac{9}{2}\}$

Answer 1

The correct option is A $x\in \{4,\frac{11}{2}\}$

Here,

$\frac{\left[x\right]}{[x-2]}-\frac{[x-2]}{\left[x\right]}=\frac{8\left\{x\right\}+12}{[x-2]\left[x\right]}$

$\Rightarrow \frac{{\left[x\right]}^2-{[x-2]}^2}{[x-2]\left[x\right]}=\frac{8\left\{x\right\}+12}{\left[x\right][x-2]}$

$\Rightarrow$ $\frac{(\left[x\right]-[x-2])(\left[x\right]+[x-2])}{[x-2]\left[x\right]}=\frac{8\left\{x\right\}+12}{\left[x\right][x-2]}$

$\Rightarrow$ $(\left[x\right]-[x-2])(\left[x\right]+[x-2])=8\left\{x\right\}+12$

$\Rightarrow$ $(\left[x\right]-\left[x\right]+2)(\left[x\right]+\left[x\right]-2)=8\left\{x\right\}+12$ .......... $(\because [x+I]=\left[x\right]+I)$

$\Rightarrow 4\left(\right[x]-1)=8\left\{x\right\}+12$

$\Rightarrow \left[x\right]-4=2\left\{x\right\}$ ..... $\left(i\right)$

Now, as we know

$0\le \left\{x\right\}<1\Rightarrow 0\le 2\left\{x\right\}<2$

$\Rightarrow 0\le \left[x\right]-4<2\Rightarrow 4\le \left[x\right]<6$

$\Rightarrow \left[x\right]=4,5$
If $\left[x\right]=4\Rightarrow 2\left\{x\right\}=\left[x\right]-4$ $\Rightarrow \left\{x\right\}=0$ ........ $\left(ii\right)$
And if $\left[x\right]=5\Rightarrow 2\left\{x\right\}=\left[x\right]-4$ $\Rightarrow \left\{x\right\}=\frac{1}{2}$ ...... $\left(iii\right)$
From eqs. $\left(ii\right)$ and $\left(iii\right)$, we have

$x=\left[x\right]+\left\{x\right\}$
ie, $x=4+0=4$ ...... [Using Eq. $\left(ii\right)$]
and $x=5+\frac{1}{2}=\frac{11}{2}$ ........ [Using Eq. $\left(iii\right)$]

$\Rightarrow x\in \{4,\frac{11}{2}\}$

Hence, option A is correct.