IIT JEE , AIEEE Forum - Mathematics , Integral Calculus - find the area bounded by the curve [mod x]+[mod y]=1

enthalpy

find the area bounded by the curve [mod x]+[mod y]=1

abhiforiit

ans ) is it [lXl] + [lYl] = 1 or just lXl + lYl = 1 here [ ] denotes the greatest integer function..

sahilxxx

if it is simply mod n not gint then ...

x+y=1 ... when x>0 n y>0 i.e. 1st quadrant

-x+y=1 ... when x<0 n y> 0 ... i.e. 2nd quadrant ...

-x-y=1 ... when x<0 n y<0 ... i.e. 3rd quadrant ....

x-y = 1 .... when x>0 n y<0 .... i.e. 4th quadrant ...

draw the graphs in different quads n u'll get the required graph ...

sahilxxx

okkk ... if it is gint then put gint to x n y and draw the graph ... if u know how to proceed with [y]=[x] types of functions ...

abhiforiit

ya dude ure right if there is no [ ] then the answer is 2 sq . units .since lXl + lYl = 1 forms a square of each side of length 2^1/2 units .......

venkat_tatolu

Answer is 8.
Sol) First lets take [x]+[y]=1.

so If [x] =0 then [y]=1 or If [x]=1 then [y]=

x belongs to [0,1) and y in [1,2) or x is in [1.2) and y in [0,1)

we get two squares of area each 1 unit in I quadrant.i.e the area of given figure in I quadrant is 2.

But there is modulus for x and y .So we can apply above argument in other 3 quadrants also.{since x,y may be -ve}

Total area is 4.2=8.

The area of [x]+[y]=n n is natural number is 4.(n+1) {short cut}

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