Question

The area of the region $\lfloor x\rfloor +\lfloor y\rfloor =1,-1\le x\le 1$ and $xy\le 1/2$

• A. rational
• B. $\frac{1}{2}(\frac{3}{2}+\ln 2)$
• C. $\frac{1}{2}(\frac{5}{2}+\ln 2)$
• D. Irrational

Answer 1

Answer: (D)

Irrational

NOTE: $xy\le 1/2$ is equation of region below the parabola $y\le 1/2x$ and above the co-ordinate axis (shown by blue)

$CASE-1:$ $x\in [-1,0)$

$\lfloor x\rfloor =-1$, so we want $\lfloor y\rfloor =2$

$\to$we want $y\in [2,3)$

$\to$This region is depited by full red square in image

$\to Are{a}_1=1\ast 1=1$

$CASE-2$ $x\in [0,1)$

$\lfloor x\rfloor =0$ , so we want $\lfloor y\rfloor =1$

$\to$we want $y\in [1,2)$

$\to$ so this case depicts a square (with lower-left vertex at $(0,1)$ and upper-right vertex at $(1,2)$ ) on co-ordinate axis, but note that $xy\le 1/2$ cuts of this square as shown in image

We can find the shaded red region by doing,

$Are{a}_2={\int }_1^2\frac{1}{2y}dy$ ,treating x as a function of y, ($x=1/2y$)

$\frac{1}{2}$ ${\int }_1^2\frac{1}{y}dy=\frac{1}{2}\left(ln\right(2)-ln(1\left)\right)=\frac{ln\left(2\right)}{2}$

Total area=$Are{a}_1+Are{a}_2=1+ln\left(2\right)/2=\frac{1}{2}(2+ln(2\left)\right)\to IRRATIONAL$

Note: ln(n) -> IRRATIONAL for all integers n$\ge$2