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Re: HSC 2016 4U Marathon Yeah I probably wouldn't do it in the above way for that Q in a HSC exam (I was just showing it as an interesting and more general result). And yeah, you can (and are probably expected to) do it via something like the binomial theorem (or you could try trinomial theorem).

Re: MATH1231/1241/1251 SOS Thread $\noindent Note $\mathbf{x}\equiv \begin{pmatrix}x_1 \\ x_2 \\ x_3\end{pmatrix}\in \mathbb{R}^3$ is in $S$ iff $A\mathbf{x} = \mathbf{0}_{\mathbb{R}^{2}}$, where $A$ is the matrix $\begin{pmatrix}3 & -4 & 0 \\ 2 & 0 & 5 \end{pmatrix}$ and...

Re: MATH1231/1241/1251 SOS Thread Use double angle formula for sinh: sinh(2a) = 2sinh(a)cosh(a). Note cosh(arcsinh(u)) = sqrt(1 + sinh^2 (arcsinh(u)) = sqrt(1+u^2).

Note that A = 25000 – B, where A is the r.v. describing the value from investment A, and similarly for B. Therefore, by properties of variance (namely that an additive constant doesn't change the variance, and neither does the presence of a multiplicative factor of -1), we have Var(A) =...

Re: MATH1231/1241/1251 SOS Thread x = 2*sinh(u). Therefore, sinh(u) = x/2. Then take the inverse sinh of both sides. (It's basically the same procedure as if it had been x = 2*tan(u) or something instead – just need to utilise the inverse function.)

Re: MATH1231/1241/1251 SOS Thread Rewrite your answer in terms of x. You got an answer of u, and x = 2*sinh(u), so get u in terms of x from this.

Re: HSC 2016 4U Marathon $\noindent We are being asked to show the expansion is a \textsl{palindromic polynomial}. In fact, the product of any two palindromic polynomials is a palindromic polynomial, and so it follows by induction that $\left(P(x)\right)^n$ is a palindromic polynomial for any...

Re: MATH1231/1241/1251 SOS Thread I think he was just posting those to show the context.

Re: MATH1231/1241/1251 SOS Thread What was the context in which the Q. was asked? And did it say specifically to solve a characteristic equation etc., or did it simply say to solve the ODE? (So basically what did the Q. actually say?)

Re: MATH1231/1241/1251 SOS Thread I think he meant something like this? $\noindent Given: $y' + 2y = 10e^{2x}$.$ $\noindent Differentiate: $y'' + 2y' = 20e^{2x}$.$ $\noindent Subtract twice the original ODE from the second: $y'' + 2y' -2y' -4y = 0 \Rightarrow y'' = 4y$. So $y= c_1...

Re: HSC 2016 4U Marathon Is the exponent meant to be n? You wrote 2.

Re: MATH1231/1241/1251 SOS Thread You can solve these like second-order ODE's with constant coefficients. Any order ODE with constant coefficients can be solved via characteristic equation etc. $\noindent For any linear ODE, the general solution is a particular solution plus the general...

Q has half the GPE of P, not twice. This is because the GPE function is inversely proportional to the distance r (since U = - GmM/r). Q doesn't have negative half the GPE of P, just half. Despite having half the GPE of P, Q has a greater GPE than P, because GPE's here are negative (in other...

Q is higher up, so has a greater GPE than P. So the answer should be (A).

Yeah you can. If you draw a sketch, you should be able to identify the relevant rectangle and relevant "red area" to subtract from the rectangle area. The "red area" will be able to be computed similarly to above (integrating a function of y between y limits).

Not really, no. When we lift it up, we do work against gravity in order to move it up. Remembering that work done = change in energy, this work we do to it to move it up is transferred to the object as (gravitational) potential energy. It is clear that the higher we move it up, the more work...

$\noindent For the red area, it's like a normal integral except the curve we're finding the area `under' is of the form $x = f(y)$ (so the roles of $x$ and $y$ are reversed compared to when you find the area under a typical curve, like bounded by the $x$-axis). (Here, $f(y)=x = \sin y$, since $y...

Pretty much. Note also that intuitively if we lift an object higher and higher, we've done more work against gravity to get it up there, so more (gravitational) potential energy is being stored in it.

$\noindent GPE is given by $U = -\frac{GMm}{r}$, where $r$ is the distance of the object of mass $m$ from the planet of mass $M$'s centre of mass ($G$ being the universal gravitational constant). This is an increasing function of $r$ (for positive $r$ of course) so when an object is higher...

Yeah I did that method if you see my working. In your diagram, the red area needs to be subtracted from the overall rectangle's area in order to give us the black area (which is what we're after). The rectangle has area pi/2 and the red area is 1, so the answer is pi/2 – 1.