Question Number 85540 by M±th+et£s last updated on 22/Mar/20
![find the range y=((x+[x])/(1−[x]+x))](https://www.tinkutara.com/question/Q85540.png)
$${find}\:{the}\:{range} \\ $$$${y}=\frac{{x}+\left[{x}\right]}{\mathrm{1}−\left[{x}\right]+{x}} \\ $$
Answered by mr W last updated on 23/Mar/20
![for x≥0: let x=n+d with 0≤d<1 y=((x+[x])/(1−[x]+x))=((2n+d)/(1+d))=1+((2n−1)/(1+d)) for n=0: y=1−(1/(1+d))≥1−(1/1)=0 y=1−(1/(1+d))<1−(1/2)=(1/2) ⇒0≤y<(1/2) for n≥1: y=1+((2n−1)/(1+d))≤1+((2n−1)/1)=2n y=1+((2n−1)/(1+d))>1+((2n−1)/(1+1))=n+(1/2) ⇒n+(1/2)<y≤2n ⇒1.5<y≤2 ⇒2.5<y≤4 ⇒3.5<y≤6 ⇒4.5<y≤8 ... ⇒y∈[0, (1/2)) ∨ (1.5, 2] ∨ (2.5, +∞) ...(i) for x<0: let x=−n−d with 0≤d<1 y=((x+[x])/(1−[x]+x))=1−((2n+1)/(1−d)) y=1−((2n+1)/(1−d))≤1−((2n+1)/1)=−2n y=1−((2n+1)/(1−d))>1−((2n+1)/0)=−∞ ⇒−∞<y≤2n ⇒y∈(−∞,0] ...(ii) from (i) and (ii) we get range y∈(−∞, (1/2)) ∨ ((3/2), 2] ∨ ((5/2), +∞)](https://www.tinkutara.com/question/Q85650.png)
$${for}\:{x}\geqslant\mathrm{0}: \\ $$$${let}\:{x}={n}+{d}\:{with}\:\mathrm{0}\leqslant{d}<\mathrm{1} \\ $$$${y}=\frac{{x}+\left[{x}\right]}{\mathrm{1}−\left[{x}\right]+{x}}=\frac{\mathrm{2}{n}+{d}}{\mathrm{1}+{d}}=\mathrm{1}+\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{1}+{d}} \\ $$$${for}\:{n}=\mathrm{0}: \\ $$$${y}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{d}}\geqslant\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}}=\mathrm{0} \\ $$$${y}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{d}}<\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{0}\leqslant{y}<\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${for}\:{n}\geqslant\mathrm{1}: \\ $$$${y}=\mathrm{1}+\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{1}+{d}}\leqslant\mathrm{1}+\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{1}}=\mathrm{2}{n} \\ $$$${y}=\mathrm{1}+\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{1}+{d}}>\mathrm{1}+\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{1}+\mathrm{1}}={n}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{n}+\frac{\mathrm{1}}{\mathrm{2}}<{y}\leqslant\mathrm{2}{n} \\ $$$$\Rightarrow\mathrm{1}.\mathrm{5}<{y}\leqslant\mathrm{2} \\ $$$$\Rightarrow\mathrm{2}.\mathrm{5}<{y}\leqslant\mathrm{4} \\ $$$$\Rightarrow\mathrm{3}.\mathrm{5}<{y}\leqslant\mathrm{6} \\ $$$$\Rightarrow\mathrm{4}.\mathrm{5}<{y}\leqslant\mathrm{8} \\ $$$$… \\ $$$$\Rightarrow{y}\in\left[\mathrm{0},\:\frac{\mathrm{1}}{\mathrm{2}}\right)\:\vee\:\left(\mathrm{1}.\mathrm{5},\:\mathrm{2}\right]\:\vee\:\left(\mathrm{2}.\mathrm{5},\:+\infty\right)\:\:\:…\left({i}\right) \\ $$$${for}\:{x}<\mathrm{0}: \\ $$$${let}\:{x}=−{n}−{d}\:{with}\:\mathrm{0}\leqslant{d}<\mathrm{1} \\ $$$${y}=\frac{{x}+\left[{x}\right]}{\mathrm{1}−\left[{x}\right]+{x}}=\mathrm{1}−\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{1}−{d}} \\ $$$${y}=\mathrm{1}−\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{1}−{d}}\leqslant\mathrm{1}−\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{1}}=−\mathrm{2}{n} \\ $$$${y}=\mathrm{1}−\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{1}−{d}}>\mathrm{1}−\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{0}}=−\infty \\ $$$$\Rightarrow−\infty<{y}\leqslant\mathrm{2}{n} \\ $$$$\Rightarrow{y}\in\left(−\infty,\mathrm{0}\right]\:\:\:…\left({ii}\right) \\ $$$$ \\ $$$${from}\:\left({i}\right)\:{and}\:\left({ii}\right)\:{we}\:{get}\:{range} \\ $$$${y}\in\left(−\infty,\:\frac{\mathrm{1}}{\mathrm{2}}\right)\:\vee\:\left(\frac{\mathrm{3}}{\mathrm{2}},\:\mathrm{2}\right]\:\vee\:\left(\frac{\mathrm{5}}{\mathrm{2}},\:+\infty\right) \\ $$
Commented by M±th+et£s last updated on 23/Mar/20
$${thank}\:{you}\:{so}\:{much}\:{sir} \\ $$