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Re: MX2 2015 Integration Marathon Nice solution. I think it would have saved some working though to just do the integrals as definite ones from the start and change the limits as you go for the substitutions.

Re: HSC 2015 3U Marathon $They merged the expressions in the brackets as they are raised to the same power, and then used difference of two squares.$ $$\left(1-x\right)^n \left(\frac{x+1}{x}\right)^n = \left( (1-x)\times \frac{x+1}{x}\right)^n$$ $$ = \left(\frac{1-x^2}{x}\right)^n$, as...

Re: HSC 2015 3U Marathon Yes, it is mathematically correct.

Re: HSC 2015 3U Marathon Yeah it's highly unlikely, especially in today's papers.

Re: HSC 2015 3U Marathon Couldn't the HSC technically ask it if they defined it? E.g. they could give a diagram and say what it is. Or they could just say "approximations are taken by using rectangles at the midpoints of sub-intervals" (with diagram etc.), in which case it's just using a...

Re: HSC 2015 3U Marathon Pretty much everything in maths is used a lot in physics.

I think you should know the radiance or intensity vs. wavelength graph (but don't need vs. frequency).

Re: HSC 2015 3U Marathon Q1) Yes – typo Q2) No.

Re: HSC 2015 3U Marathon Choose the coordinate axes so that the origin is at the vertex of the parabola, with the vertical axis being the axis of symmetry of the parabola (positive direction being downwards) and the horizontal axis being tangential to the parabola.

It's not, he just moved the minus sign to the numerator.

$No he didn't actually (assuming you're referring to integral95's answer). His denominator from the previous line becomes $-\sin x$ (another identity), so he just moved the minus sign of this to the numerator in his next line, which is valid to do (though I can understand why you could be...

$Actually we would need to divide that answer by $2! = 2$, since the two pairs are not ordered. I.e. $5\times 5 \times 4 \times 4$ considers these two matchups as \emph{different} matchups:$ $$\bullet$ (Alex, Alice) vs. (Ben, Cindy)$ $and$ $$\bullet$ (Ben, Cindy) vs. (Alex, Alice).$ $Clearly...

$It's not that, because this also counts cases where spouses are paired up with each other, which we said we cannot have.$

$If you want an algebraic proof, here is one:$ $$\sin \left(\frac{\pi}{2}+x \right) = \cos \left(\frac{\pi}{2}- \left(\frac{\pi}{2}+x \right) \right)$ (complementary angles trig. identity)$ $$=\cos \left(-x \right)$ $ $$=\cos x$ (cosine function is even).$ $(The identity is $\sin...

$This is a trigonometric identity that can be proved in many ways, e.g. the graph of $y=\sin \left(\frac{\pi}{2}+x \right)$ is that of $y=\sin x$ shifted left by $\frac{\pi}{2}$ units, which we observe is just the graph of $y=\cos x$, which means that $\cos x = \sin \left(\frac{\pi}{2}+x...

The question wording doesn't mean that a specific man or pair has to be in the game I think; I think it's just asking for the number of ways to choose people for a match if we need mixed doubles pairs and spouses can't be on the same team.

No, because to choose a pair, we don't choose 2 from 5. We choose 1 from 5 and one from another 5, which is more than doing 5C2.

Re: HSC 2015 3U Marathon Lol avoid the letter x for the difference between the roots (use d or something), since it's the variable in the equation too (it's not that big a deal but maybe some markers wouldn't like it).

$Yeah it's just luck that the guess of $4\times 4$ happens to be equal to the answer obtained by Cases 1 and 2 ($1\times 4 + 4\times 3$).$

$For the second question:$ $$g(x)+a-x=0$. Let $f(x) := g(x) + a-x$, so $f^\prime (x) = g^\prime (x) -1$ and the equation to solve is $f(x) = 0$.$ $The first guess is $x_0 = a$.$ $The next approximation is $x_1 = x_0 - \frac{f(x_0)}{f^\prime (x_0)}$$ $$\Rightarrow x_1 = a-...