Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

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I am also getting 2.

sorry about the bad latexing.

JaneFairfax wrote:

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
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.

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

3. tan θ=1/2

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

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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