62[pi/4,pi]
Thing geomtrically!
and then go from -ve side.. from - pi to 0
did you figure it out?
62
Lokesh Verma
·2009-04-04 05:23:49
let the roots be a<=b<=c
so we want a+b>c for the two smallest roots...
and that each root is greater than zero...
abhee i cant figure out something immediately...
62
Lokesh Verma
·2009-04-04 05:24:33
asish.. we want everything outside 0 to pi/4 because we dont want a root!
33Ok Cipher..
here is it something algebraic.
Question is equivalent to:
Find α so that no solution exist to
y=x-1 ...(corresponding to arg(z-1)=450) ...(y>0)* ...(so x>1)* because arg(z-1)=450 not ±450
and y=xtanα+tanα ....(corresponding to arg(z+1)=α) .(y>0)* (so x>-1)*
so they can only intersect for x>1 and y>0
*( arg(z-1)=450 doesnot include 1,0 nor arg(z+1) include -1,0)
x=(1+tanα)/(1-tanα) ...(notice here we loose the case tanα=1 we will take care of it separately..) also
y=2tanα/(1-tanα)
Now it is clear that for they to intersect they have to do so for x≥1 and y≥0
x<1
(1+tanα)/(1-tanα)<1
tan>1 or tanα<0
so α belongs to (-π,0)U(π/4,π)
but we see for tanα=1 and tanα=0 it also satisfies (see *)
so α belongs to (-π,0]U[π/4,π] option (a)
33
Abhishek Priyam
·2009-04-04 11:21:09
but it is better to do it graphically
red one is fixed for arg(z-1)=450
blue line is varied for different α line I being parallel to red line.
you can easily see that any * marked line can satisfy given condition.
hence the answer.
11good work abhi
but 1st becomes quite simple if u put Z=2+i
jab paper mein dimaag na chale
to best idea[50]
1got it, thanks everyone!!
