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No. There exist functions that are differentiable somewhere but not continuously differentiable there. An example of such a function is found here: https://en.wikipedia.org/wiki/Differentiable_function#Differentiability_classes .

Re: HSC 2016 MX2 Combinatorics Marathon Ah I see. I (mis)interpreted it as asking for what middle length (out of the unit interval) should we ask for in order for that probability (Pr(|A – B| ≤ x) to be 1/3, and didn't read Paradoxica's solution too closely, and thought he was doing the same...

We used that assumption in the last equality. I.e. assuming that f'(a + ph) will tend to f'(a) as h -> 0. This essentially is using the assumption that f' is continuous at a (by definition of continuity). (In general it could happen that f'(a + t) has no limit even as t -> 0.)

Re: HSC 2016 MX2 Combinatorics Marathon I think braintic is asking for part (b) what length x (or L) of segment we should ask for in order for the probability that X ≤ L is 1/3, where X := |A – B|, where A and B are the r.v.'s corresponding to the positions of the two cuts. (In other words...

If we use L'Hospital's rule, we'll end up having to assume that f' is continuous at a after the differentiation step in order to take the limit as h -> 0 of the resulting expression. But we aren't given this fact, so we can't assume it, so can't really use L'Hospital like this.

So the ellipse condition is that the point P on the (upper-half of the) ellipse has y = \sqrt{1-\frac{x^2}{4}}. The squared distance as a function of x is then s(x) = (x-a)^2 + 1-\frac{x^2}{4}. This can be differentiated to help find the minimum: s^\prime (x) =...

That is indeed a valid method.

Re: First Year Uni Calculus Marathon Yeah it's basically squeeze law. We can use that to show that f'(0) = 0.

We can parameterise a point in the ellipse like x = t, y = sqrt(1 - (t^2/4)), for -2 <= t <= 2. (This only gets the top half of the ellipse but by symmetry, the distance will be same as considering the whole ellipse). We can then consider the squared distance of the point (a, 0) to a point on...

I used some stuff from your first two parts, like the fact that the second derivative is always positive.

Note that p4'(x) = p3(x) for all x. Also, p4"(x) = p2(x) for all x. We've shown that p2(x) is a positive definite quadratic, so that means the second derivative of p4 is always positive. This means the minimum of p4 will be a global min. Now, to find this min., set p4'(x) = p3(x) = 0. (There is...

Say the curve is r = a + sin(theta) = f(theta), say. Now, y = r.sin(theta) = f(theta).sin(theta) and similarly x = f(theta).cos(theta) Now, dy/d(theta) = f'(theta).sin(theta) + f(theta).cos(theta) = cos(theta).sin(theta) + (a + sin(theta)).cos(theta) = a.cos(theta) + sin(2*theta). Also...

Re: HSC Chemistry Marathon 2016 Had a closer look at the Q. The answer is simply 4*0.697*NA, where NA is Avogadro's number, which is on the HSC Chemistry formula sheet. This is because in 0.697 mol of silver sulfate, there are 4*0.697 mol of oxygen atoms (from the silver sulfate formula)...

Re: HSC Chemistry Marathon 2016 The formula for silver sulfate is Ag2SO4. Knowing this, this question becomes similar to the previous ones you posted here.

Re: First Year Uni Calculus Marathon Imagine Carrotsticks BOS trial last Q. of 4U paper 2016: 'Sketch the graphs of the Weierstrass and Dirichlet functions' :p.

That's a circle with diameter 6 on the y-axis. Centre is (0, 3), radius is 3. To derive this, you could replace sin(theta) with y/r, so r = a*(y/r) (in this case a = 6), so r^2 = ay, that is, x^2 + y^2 = ay. Now just moves things around and complete the square to get the circle's Cartesian...

Depends what you mean by 'better'.

Re: First Year Uni Calculus Marathon ^ I think you should try invoking the IVT to prove some things I think you're trying to prove.

Re: First Year Uni Calculus Marathon For one thing, what if f need not be differentiable?

Re: First Year Uni Calculus Marathon Is this thread's main intention as a Q&A, or as a challenge thread?