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@Rathin

Will you be proud of me when I get that 20/80 in MATH1151??

$a.$ \ $To assess if a function is "increasing" or "decreasing" then we need to assess the derivative. f(x) = \frac{x^3 + 8x}{8} f'(x) = \frac{1}{8}(3x^2 + 8) $The derivative is a positive value for all real x in its domain. What does this mean?$ $b.$ \ f'(x) = \frac{1}{8}(3x^2 +...

Answer: (e^(x+1))/(x+1) + C (assuming e is a variable and not Euler's Number)

Re: Announcement from BOSTES - significant change to calculus courses learn it cuz

Re: UNSW chit chat thread 2016 As long as we answered it in the correct question order (as opposed to page number) it's correct right? lol

Most likely without, since they seem to distinguish "Entry Rank" from "ATAR"

tan(pi/2) is undefined since tanx = sinx/cosx and cosx = 0 when x = pi/2.

Actually laughed But seriously I am not a stalker

Here is a diagram. http://imgur.com/b7sVG7V Solution (highlight): Notice that the height 'd' of the building that we are required to find is just equal to 45 - x. Where x is the distance you get when you extend the top of the 2nd building to the horizontal line at the top. x can be found using...

I think you can justify it since the lower limit will be ln2 < lne = 1 And we know that the integral from ln2 to 1 of the original integrand has a finite value so we can just split it up to the integral from ln2 to 1 of f(x) + integral from 1 to infty of f(x) and apply the p-test.

\cot (x+y) = \frac{\cot x \cot y - 1}{\cot x + \cot y} $i.e., prove $ \tan (x+y) = \frac{\cot x + \cot y}{\cot x \cot y - 1} \ ($Reciprocating both sides$) $LHS$ = \tan (x+y) = \frac{\tan x + \tan y}{1 - \tan x \tan y} \ ($By the tan expansion$) = \frac{\cot x + \cot y}{\cot x \cot y - 1}...

Re: MX2 2016 Integration Marathon \int \frac{\sqrt{x}}{\sqrt{a^3 - x^3 }} dx $where a is a constant$

i am innocent

What how did I just randomly get mentioned

hello friend

plz what'd i do

My Uni friend from HK told me that HK doesn't really have these "A-level" exams anymore, and he told me that the difficulty has been toned down (in comparison to this paper to when he did them in 2014/15).

The ball comes to a rest when the velocity equals zero. Letting x = 0 means you're finding the time where the ball started.

Can also be done by defining f(t) = (x-1)ln(1-x) for t E (0,1)