Let the points be A(a,a^2), B(b,b^2), C(c,c^2)Suppose that three points on the parabola y=x^2 have the property that their normal lines intersect at a common point. Show that the sum of their x-coordinates is 0.
The first inequality isn't always true for all positive integers n.
It works for integersThe first inequality isn't always true for all positive integers n.
When LHS=RHSHint: first prove that for x > 1
x - 1 > ln x
then go from there
lol you can't differentiate both sides of an inequalityWhen LHS=RHS
differentiating both sides gives LHS'=1 and RHS'=1/x so for x>1 LHS>RHS
then apply what nightweaver did, and the solution is done
no not like thatlol you can't differentiate both sides of an inequality
Oh right. I would normally just show that the maximum value of y = ln x - x is -1.no not like that
like, if you differentiate x-1 you get 1, and if you differentiate ln(x) you get 1/x, so x-1 increases faster than ln(x) for x>1, but at x=1 they are equal, thus for x>1, x-1>ln(x)
ah ok, well done.It works for integers
The cases of 1, 2, 3 can just be manually tested I guess. At the time I originally had which works for all integers, until I realised it's n-1 not n+1, and thought it still worked without actually testing it lol.
Letah ok, well done.
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Very similar to last year's BOS trial question 15 by carrot.
Well it is given that E_n approaches a finite limit, andVery similar to last year's BOS trial question 15 by carrot.