statement1: let f:[0,4]-> R be a cont fn and diff in (0,4) then there exist some 'a' and 'b' in (0,4) where f2(4)-f2(0)=8f(a)f'(b).
statement2: LMVT
62statement 1 is correct...
statement 2 is only partial explanation
This is a bit dicey.. bcos even when LMVT is a part of the explanation.. it is nto complete...
Having said this .. i would go with the fact that statement 2 is the correct reason of the assertion!
33yahi to option a and b i.e. correct and incorrect explanation me hi to pata nahi chalta hai...
so this half xplaination can be taken as xplanation??
waise i got this wrong in test as there was a slight change in question and took it this way
the question was:
=8f(a)f'(a). both were a and i didn,t took notice of it....so in this case st 1 is wrong naa?
62i have a gut feeling that it will be true still!!
But i have no reason right now ...
33that means for a same point LMVT is satisfied and same point has mean value of f(0) and f(4)???
62no i meant that the assertion will not be explained correctly though..
but as i said i have no proof right now.! for the first part.. it is only a feeling that is there bcos of continuity!!
33f(x)={2x+1, x ε [-1,0)
{2x , x = 0
{2x-1 ,x ε (0,1]
st 1: f is bounded but never reaches its max or min
st 2: f is discont at x=0
621 is true.. (this seems obvious to me)
2 is not the reason.. (Why i am saying this is bcos a function being discontinuous at a point is never the reason for the function not attaining its maxima or minima!!!
f(x)={2x-100, x ε [-1,0)
{2x , x = 0
{2x-1 ,x ε (0,100]
It is discontinuous,... but it reaches its max and minima!
33oh!!!! phase test ka kitna answer galat niklega.....
chalo thik hi hai mujhe isme +4 milega.
62yaar this question is again dicey... i am sure these ppl may come up with a justification that this is the corrrect soln and the 2nd statement is the correct reason... but i guess they should correct it!
62not sure yet... actually i forgot this question! i will see it again..
621 is true.. (this seems obvious to me)
2 is not the reason.. (Why i am saying this is bcos a function being discontinuous at a point is never the reason for the function not attaining its maxima or minima!!!
f(x)={2x-100, x ε [-1,0)
{2x , x = 0
{2x-1 ,x ε (0,100]
for this question ... my explanation to me seems very vrey justified...
but lets see teh final thing that fiitjee says..
may be they are actually making a mistake!