multiple choice: differentiation

2nd one is based on a property that that derivative of a function has the intermediate value propery (even though they may be discontinuous). Read up: http://en.wikipedia.org/wiki/Darboux%27s_theorem_(analysis)

\frac{f'(0) + f'(1)}{2} lies between f'(0) and f'(1)

So by the theorem there exists x_0 \in (0,1) such that f'(x_0) = \frac{f'(0) + f'(1)}{2}

So all options are correct.

oh i got my mistake

tnk u prophet sir :)

1
(d) f(x) = |x| us a counterexample
(a): is false: counterexample again is f(x) = |x| with k=1 (however it certainly implies continuity. Such functions are called Lipschitz continuous)

(c) By Sandwich Theorem we get that f'(x) = 0 for all x and so it is differentiable

(b) When |x-y|<1, we have |x-y|<|x-y|^{\frac{1}{2}}

So, f(x) = |x| again satisfies this condition when |x-y|<1(k=1 again) but is not differentiable at x=0. So its a counterexample

So c is the only correct option

|f(x)-f(y)|\leq k|x-y|

so \frac{|f(x)-f(y)|}{|x-y| }\leq k

now apply limit
so
|f^{!}(x)|\leq k

srry abt post #12 (forgot to edit it in a hurry....) :)

in the given question we cannot directly apply limit , so here we have to use sandwich theorem......

in 1st option there is no function to "bind it" from the left side so we cannot calculate left hand derivative.....but it won't be -k , earlier i made it bound with -ve side but that was wrong, so only problem with the 1st problem is that we don't have a function which can bind it from left, hence we cannot apply sandwich theorem, hence we cannot calculate its derivative.

in the given question we cannot directly apply limit , so here we have to use sandwich theorem......

in 1st option there is no function to "bind it" from the left side so we cannot calculate left hand derivative.....but it won't be zero , earlier i made it bound with -ve side but that was wrong, so only problem with the 1st problem is that we don't have a function which can bind it from left, hence we cannot apply sandwich theorem, hence we cannot calculate its derivative.

how -k?

Q1> a,c,d ?

i got that... in 1st option if u take the left hand derivative then... it comes out to be -k nd right hand derivative as k......so 1st option is wrong...

yup thats...what i am saying.....didn't saw that u hv already posted that...:)

in option a , after applying the limit we would get k , which is a finite +ve real no....so according to this logic option a should also be correct....nd in option c we would get 0 which is correct.......

|f(x)-f(y)|\leq k(x-y)^{m}

\frac{|f(x)-f(y)|}{|x-y|}\leq \frac{k(x-y)^{m}}{|x-y|}

apply limit x→ y on both sides

so f^{1}(x)\leq \lim_{x\rightarrow y}\frac{k(x-y)^{m}}{|x-y|}

now if limit on the Right side is finit and exists then f'(x) is diferentiable

ans for 1 is .c nd that of 2 is d