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#2 Re: Ganesh's Puzzles » Brain Teasers » 2010-05-18 06:29:35

BO

I am also getting 2.
sorry about the bad latexing.

#4 Re: Help Me ! » Unbounded function » 2010-05-17 06:22:00

BO
JaneFairfax wrote:

I suppose we could do that if we stated the problem a bit differently:



I have succeeded in proving from “first principles” that

for such a function, which may, or may not, be an intermediate result that is required.

By the way, last night, while thinking in my sleep (yes, thinking in my sleep tongue) about uniform continuity of real-valued functions, I remembered this thread, and then I conjectured if a function whose first and second derivatives are positive cannot be uniformly continuous. I might come up with a proof soon. dizzy

is non-negative for all values of x. Now if
is positive for all values of x, then
is a monotone increasing function, which in turn implies that
is unbounded above. so there must be some
for which

#5 Re: Help Me ! » Unbounded function » 2010-05-17 06:03:14

BO

Oh, extremely sorry, I didn't notice that you were not talking about the topic problem.

#7 Re: Ganesh's Puzzles » Level of Difficulty - III Questions. » 2010-05-16 19:26:06

BO

1.from sine rule we get sinA+sinC=2sinB, or cos([A-C]/2)=2cos(B/2)
tan(A/2)+tan(c/2)=cos(B/2)/[cos (A/2) cos (C/2)]=2cos(B/2)/[sin(B/2)+2cos(B/2)]=2/[1+cot (B/2)]

2.the radius of the circle=a*tan30=a/3
Then the diameter of the square=2a/3
area=(2a/3 sin45)^2=a^2/9

#8 Re: Help Me ! » Unbounded function » 2010-05-16 18:16:02

BO

f'(x)-ε<[f(x+h)-f(x)]/h<f'(x)+ε
f(x+h)>f(x)+f'(x)-ε
Since f'(x) is always +ve, f is a monotone increasing function.
Similarly f"(x) is a monotone increasing function.
Now, it is easy to prove that f(x+c)-f(x)=c f'(x+t), where 0<t<c, for any c and x.
We have f'(a)> (f"(a)-ε)*k+f'(a-k)=m for a point a in R.
Then, when x>a, f(x+nc)>ncm+f(x), which is unbounded above. Hence proved.

Jane Fairfax, consider the function f(x)=x^3 which is differentiable at every point of R. f'(x) is positive and differentiable at every point and so is f"(x), but f'(x) does not tend to 0 as x tends to infinity.

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