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#2 Re: Jai Ganesh's Puzzles » Brain Teasers » 2010-05-18 06:29:35
I am also getting 2.
sorry about the bad latexing.
#4 Re: Help Me ! » Unbounded function » 2010-05-17 06:22:00
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
) 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.
#5 Re: Help Me ! » Unbounded function » 2010-05-17 06:03:14
Oh, extremely sorry, I didn't notice that you were not talking about the topic problem.
#6 Re: Jai Ganesh's Puzzles » Level of Difficulty - III Questions. » 2010-05-16 19:40:23
3. tan θ=1/2
#7 Re: Jai Ganesh's Puzzles » Level of Difficulty - III Questions. » 2010-05-16 19:26:06
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
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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