is it C?
OA?

YEs it is . Kindly explain

ok..very simple concept.
y(u-c)=0 & j(u-k)=0

it directly follows from the above equations that y=0 and j=0
out of all the possibilities,the only one correct it yj=0.only choice 4, y=0 is incomplete,only choice 5 , j=0 is also incomplete.both have to be zero.hence the answer c.choices 1,2 are outta question.

But u have neglected the fact that frm eq 1, u=c
& frm eq 2, u=k is another possibility.

From y(u-c)=0 , we can say that either y=0 or u=c.

event_horizon wrote:ok..very simple concept.
y(u-c)=0 & j(u-k)=0

it directly follows from the above equations that y=0 and j=0
out of all the possibilities,the only one correct it yj=0.only choice 4, y=0 is incomplete,only choice 5 , j=0 is also incomplete.both have to be zero.hence the answer c.choices 1,2 are outta question.

y(u-c) = 0 & j(u-k) =0
also,
c<k => c-k <0 .....this is the hint here.

but, the only link in 2 eqns is u, so we need to get a value for it.

therefore, in y(u-c)=0 ,assuming that y!= 0 => u =c

now => in j(u-k) = j(c-k)=0 => j =0 [my original answer ]

also, y can as well be 0 (contradicting our assumption above)
thus, y*j will definitely be 0. hence, the answer.