1
madeforiit
·2009-02-15 23:24:19
y ans B???
bhaiya jus proved IVT part of Reason to be false....
9
Celestine preetham
·2009-02-16 08:14:25
i fore saw that in post no 12 ;)
11
rkrish
·2009-02-16 04:05:27
The 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.
11
rkrish
·2009-02-16 03:59:19
I 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
Lokesh Verma
·2009-02-15 23:37:28
@madeforiit.. I have already shown it is not true.
and celestine has given teh condition when it wud work !
1
madeforiit
·2009-02-15 23:37:12
celestine 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)
1
madeforiit
·2009-02-15 23:36:11
can sum1 prove this : [provided it is rite]
f(6) + f(2) = 4f(c) for some 'c' in [2,6].
62
Lokesh Verma
·2009-02-15 23:28:40
but celestine.. how do you justify that the value will hold for the same value of c!!
there are 2 c's involved?
1
madeforiit
·2009-02-15 23:25:34
@nish bhaiya.... k so if it cums as an assertn reason we must not b concerned abt the [].....
62
Lokesh Verma
·2009-02-15 23:20:36
in IVT, LMVT
(a,b) is used.. but then if applicable in (a,b) it has to also be applicable in a greater set.
9
Celestine preetham
·2009-02-15 23:20:29
ans 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].
1
madeforiit
·2009-02-15 22:49:58
I 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....
1
madeforiit
·2009-02-15 22:48:00
yeah!!! 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.... :)
62
Lokesh Verma
·2009-02-15 22:42:26
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)
!!!
1
madeforiit
·2009-02-15 22:37:25
que1 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).
62
Lokesh Verma
·2009-02-15 22:31:54
what does your statement mean? sorry i was unable to comprehend it :(
1
madeforiit
·2009-02-15 22:21:36
By 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)??
62
Lokesh Verma
·2009-02-15 21:42:31
I 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 :)
11
rkrish
·2009-02-15 08:17:57
No. Even I thought it to be (C)[2]