Question

The area enclosed by the curve $[x+3y]=[x-2]$ where $xϵ[3,4)$ is (where $[.]$ denotes greatest integer function.)

• A. $\frac{2}{3}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. $1$

Answer 1

The correct option is B $\frac{1}{3}$

$\text{Given that:}$

$[x+3y]=[x-2]\text{where}x\in [3,4)$

$x\in [3,4)$

$\therefore x-2\in [1,2)$

$\therefore (x-2)=1$

So,

$[x+3y]=1$

that is for,

$1\le x+3y<2\&3\le x<4$

$\frac{1-x}{3}\le y<\frac{2-x}{3}\&3⩽x<4$

$\therefore \frac{1-x}{3}\le y$ & $y<\frac{2-x}{3}$ & $3⩽x<4$

$\therefore x+3y⩾1\&x+3y<2$

$\&3⩽x<4$

The plot is as follows:

$\therefore \text{Area}=\text{Area of}ABCD\text{- Area of}△ABP$

$\text{- Area of}△QCD$

= $1\left(\frac{2}{3}\right)-\frac{1}{2}\left(\frac{1}{3}\right)-\frac{1}{2}\left(\frac{1}{3}\right)$

$=\frac{2}{3}-\frac{1}{3}$

$\text{doired Area}=\frac{1}{3}\text{sq. units}$