Daily Graph 21-03-09

from the given equation

x²+y²-2<0

x²+y²<2

y<√2-x² or -√2-x²

from this graph is below

after this xy-1<0

y<1/x

for this the graph is

so joint graph is given

here the graph of y=1/x and y<√2-x2 or -√2-x2 touch each other at two points

nishant sir is the graph correct

CASE I when xy-1 > 0

then x² + y² - 2 <0
take a complex no. z = x+iy.. then lzl²-2<0
==> lzl<√2

now. xy-1>0
let z=rcosθ + risinθ
==> x=rcosθ and y=rsinθ
==> r².sin2θ > 2
==> sin2θ_max = 1 ==> r²_min>2
==> lzl_min>√2

So. no solution if xy-1>0

CASE II when xy-1<0

then w²+y²-2>0
==> lzl>√2 ==> lzl_min = √2+

now.. xy-1<0
==> r²sin2θ < 2
when sin2θ = 1 lzl<√2 so no solution..
when sin2θ < 0 then all values of lzl has solution

as sin2θ value decreases from 1 to 0 lzl increases from √2 to ∞

So. the solution is:
when sin2θ<0 then lzl>√2
and when sin2θ>0 then lzl value decreases from ∞ to √2+ as 2θ increases from [0,π/2) and reverse order when 2θ increases from (π/2, π]

case1

x²+y²-2>0

and xy-1<0

simultaneously

so

outside the circle and inside the hyperbola

case2

x²+y²-2<0

xy-1>0

so inside the circle and outside the hyperbola

which isnt possible so case 1 suffices