1chhoriye ! i think i will go and sleep :P :P :P
my mind is like the line of your fourth graph n wat ever you said is getting like the curve :'(
33 Answers
11hmmmmm... i think both the cases will have the case c=0 also.... which i took to be >0 [silly me [3]]
but bhaiya[any one]... pls rep......
62sky think in terms of graph...
then see the parabola..
the parabola goes upwards so there will be no root for large lamda
62For very different values of lamda... the graph will change like shown in the figure below (for case when a>0)
62sky you have urself said that for small lamda in first case
so that means that if lamda becomes large then it is not possible at all tha tthe graph cuts the x axis.. thus no real roots :)
1ax2+bx+c + λ =0
condition : two disticnt positive roots.
let they be α n β .
so, α+β>0 and αβ >0
=> b/a < 0 and c+λ / a >0
b +ve, a -ve => c+λ ...-ve => if λ>0, c<0 ; if λ<0, c can be <0 or>0. [so three cases]
b -ve, a +ve => c+λ ... +ve => if λ>0, c>0 or c<0 ; if λ<0, c>0 [again three cases here]
so is the ans six ??
62see it is said that for all lamda :)
but yeah dont stress your mind a lot :)
1sky you have urself said that for small lamda in first case
kaha boli bhaiya ?
62ye rone wala smiley was ur request :D
"Waisa smiley jab bhahut jor se rone ka man kar raha ho" :D
1yes akand was spot on i think
degree of divisor = 4
so degree of remainder = 2007 % 4 = 3
i dont know why noone paid attention to this
1for two distinct positive roots
first of all D > 0
then as sky said α+β > 0
and f(0) > 0
since not much is given about a b and c i dont get how to get the answer
13Can the first one be done with the help of Complex numbers???
1then acc to sky the constant term 1 would be cancelled and the reminder must come zero..................
I think so is it rite
1but in 2nd one,
it depends whether original has how many
roots ?
then only v can tell bout new eqn naaaaaaa ?
1(x2+1)(x2+x+1)
= (x2+1)(x3-1)/ (x-1)
x2007-1
= [(x3-1) + 1]669 - 1
=k (x3 -1)
try this way ....
ghum pachhe :P
1x2007 and maximum degree of divisor is x4....
so x4 can divide only till 2004...
so 2007=2004+3
so x3 remains....
soo degree is 3...
IS MY THINKIN crt?
11bhai write this in the form
[(x3)669-1]
[(x3)3 X 223-1]
u know that x3-1=(x-1)(x2+x+1)
[(x3)3 X 223-1](x-1)
-------------------
x2+1(x2+x+1)(x-1)
yaar last mein x ko kuch daal de
upper niche kaaatne ke baad
1oh
u mean
put any value of x
that remainder must be
then watever left ,say 1000 or 27 ? to be perfect cubes
wod b remainder?
is my thinkin crt ?