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$\noindent The additive constants don't affect the derivative because of what I explained above (basically because derivative of $f(x)+C$ is just that of $f(x)$, for any differentiable function $f$, $C$ a constant). Then the derivative of $\sqrt{x-3}$ is just $\frac{1}{2\sqrt{x-3}}$, because...

$\noindent It'd be $y^\prime = \frac{1}{2\sqrt{x-3}}$ (following from the sum rule, derivative of a constant is 0, chain rule, and derivative of the square root function).$

$\noindent Note that $\frac{\mathrm{d}}{\mathrm{d}x}\left (x^{p}\right)= px^{p-1}$ for any real $p$. And surds can be put into that form. E.g. $ \frac{ \mathrm{d}}{ \mathrm{d}x} \left( \sqrt{x}\right) = \frac{\mathrm{d}}{\mathrm{d}x} \left( x^{\frac{1}{2}} \right) =...

It's because the discriminant is a perfect square. So if m is rational, the solutions will be rational as the discriminant is a perfect square and the coefficients are all rational, for m ≠ 5/3. (If m = 5/3, then the given equation is a linear one rather than quadratic, as the quadratic term...

Maybe leehuan meant when the m and n are particular numerical values rather than parameters? E.g. finding something like ∫sec3 x tan2 x dx.

Yeah, like relax a little?

Re: HSC 2016 4U Marathon Yes.

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent From part (a), we know that $\theta = \tan ^{-1}t + \frac{\pi}{4}$, for $t\geq 0$.$ $\noindent Then since the inverse tan function has range $0\leq \tan ^{-1} t< \frac{\pi}{2}$ for all $t\geq 0$, clearly...

Re: HSC 2016 4U Marathon By lateral force, Katebate most likely meant the friction force (which is a lateral force, i.e. up or down the bend).

In the absence of any further information, I'd guess it's just a typo or something.

The roots will be real if and only if the discriminant is non-negative (doesn't need to be strictly positive; when the discriminant is 0, the roots are real, but they're identical, i.e. double root).

Set the discriminant to 0 and solve for k.

We can also have k = -4.

Yes. You'll want (and need) the discriminant to be 0.

Re: HSC 2016 2U Marathon Well done!

Re: HSC 2016 2U Marathon NEW QUESTION $\noindent Show that the average of the numbers in a finite list of (at least two) numbers in arithmetic progression is equal to the average of the smallest and largest terms.$

The answer is just one-millionth of 2000 g, which is 2 mg.

Re: HSC 2016 3U Marathon $\noindent \textbf{ANOTHER QUESTION}$ $\noindent i) Write down the formula for $S_1 (n):=\sum _{k =1}^{n} k$, for $n\in \mathbb{Z}^{+}$.$ $\noindent ii) Let $p$ be a positive integer, and let $S_p (n) = \sum_{k = 1}^{n} k^{p}$, for $n\in \mathbb{Z}^{+}$. Show by...

Don't heaps of people memorise essays?

It's probably horrifyingly tedious most of the time. But in principle it'd work.