Question

Find the range of $\ln [{\left\{x\right\}}^2+3\left\{x\right\}+2]$ where $\left\{x\right\}$ is a fractional function.

Answer 1

As $\ln x$ is strictly increasing function it is sufficient to find range of $\ln [{\left\{x\right\}}^2+3\left\{x\right\}+2]$

$\ln [{\left\{x\right\}}^2+3\left\{x\right\}+2]={(\left\{x\right\}+\frac{3}{2})}^2+2-\frac{9}{4}$

$={(\left\{x\right\}+\frac{3}{2})}^2+\frac{8-9}{4}$

$={(\left\{x\right\}+\frac{3}{2})}^2-\frac{1}{4}$

Since $0\le \left\{x\right\}<1$

$\frac{3}{2}\le (\left\{x\right\}+\frac{3}{2})<\frac{5}{2}$

$\frac{9}{4}\le {(\left\{x\right\}+\frac{3}{2})}^2<\frac{25}{4}$

$\Rightarrow \frac{9}{4}-\frac{1}{4}\le {(\left\{x\right\}+\frac{3}{2})}^2-\frac{1}{4}<\frac{25}{4}-\frac{1}{4}$

$\Rightarrow \frac{9-1}{4}\le {(\left\{x\right\}+\frac{3}{2})}^2-\frac{1}{4}<\frac{25-1}{4}$

$\Rightarrow \frac{1}{4}\le {(\left\{x\right\}+\frac{3}{2})}^2-\frac{1}{4}<\frac{24}{4}$

$\Rightarrow 2\le {(\left\{x\right\}+\frac{3}{2})}^2-\frac{1}{4}<6$

$\Rightarrow \ln 2\le \ln ({\left\{x\right\}}^2+3\left\{x\right\}+2)<\ln 6$

Hence Range$\in [\ln 2,\ln 6)$