1y ans B???
bhaiya jus proved IVT part of Reason to be false....
Let f(x) be a continuous and differential fn.
(A) : There exists a 'c' in [2,6] such that (f(6))2 - (f(2))2 = 8f(c)f '(c)
(R) : By LMVT, f(6) - f(2) = 4f '(c) for some 'c' in [2,6] and
By IVT, f(6) + f(2) = 2f(c) for some 'c' in [2,6].
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24 Answers
11The answer is B (as mentioned by Celestine)
St. 1- True : take G(x)= F(x)2 and apply LMVT (as mentioned by Celestine)
St. 2-True : (a,b) is used.. but then if applicable in (a,b) it has to also be applicable in a greater set. (as mentioned by Nishant bhaiyya)
how do you justify that the value will hold for the same value of c!! (as mentioned by Nishant bhaiyya)
The 2 c's need not be same.
Hence answer is B.
11I am EXTREMELY Sorry.I made a typing error!![2][2][2]
It is By IVT f(6)+f(2)=2f(c) for some 'c' in [2,6]
I am very sorry!! I have made the change.
62@madeforiit.. I have already shown it is not true.
and celestine has given teh condition when it wud work !
1celestine y r u sayin this is true????
coz as bhaiya said
Take a very simple example when f(x)=1 is the function.. for all x.
f(6)=1 and f(2)=1
then 4. f(c)= 4 for all c
so 4f(c) ≠1+1 for any c in (2,6)
1can sum1 prove this : [provided it is rite]
f(6) + f(2) = 4f(c) for some 'c' in [2,6].
62but celestine.. how do you justify that the value will hold for the same value of c!!
there are 2 c's involved?
1@nish bhaiya.... k so if it cums as an assertn reason we must not b concerned abt the [].....
62in IVT, LMVT
(a,b) is used.. but then if applicable in (a,b) it has to also be applicable in a greater set.
9ans is B
if uve wrongly typed
By IVT, f(6) + f(2) = 4f(c) for some 'c' in [2,6].
in place of
By IVT, f(6) + f(2) = 2f(c) for some 'c' in [2,6].
1I hav one fundamental doubt......... In LMVT and IVT while defining range of c....
which of the fol is correct range.....
(a,b) OR [a,b] or is it that both are ok....
1yeah!!! cool....
Is this stat true : f(6) + f(2) = 2f(c) for c belonging (2,6) { IVT}
If yes then stat A can be easily proved
There exists a 'c' in [2,6] such that (f(6))2 - (f(2))2 = 8f(c)f '(c)
jus multiply the 2 statements.... :)
62Take a very simple example when f(x)=1 is the function.. for all x.
f(6)=1 and f(2)=1
then 4. f(c)= 4 for all c
so 4f(c) ≠1+1 for any c in (2,6)
!!!
1que1 y is this stat false : By IVT, f(6) + f(2) = 4f(c) for some 'c' in [2,6]. ????
Q2 wud the stat been rite had it been By IVT, f(6) + f(2) = 4f(c) for some 'c' in (2,6).
62what does your statement mean? sorry i was unable to comprehend it :(
1By IVT, f(6) + f(2) = 4f(c) for some 'c' in [2,6].
this stat is false ne.....
In this functn bcoz we dunno f(2) & f(6) here.....
Can ne be made using IVT???
OR IS JUST THE MISTAKE IN [2,6] which shud b (2,6)??
62I dont C why C should not be the answer..
R is false.. because the 2nd statement is untrue.
By IVT, f(6) + f(2) = 4f(c) for some 'c' in [2,6].
A is true! (I cant think of a proof immediately.. but am sure.. and I think this is a standard often asked question :)