HSC 2015 MX1 Marathon (archive) (1 Viewer)

photastic · Oct 4, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
$$1. $\int \frac{1+\cos{x}}{1-\cos{x}} \, \mathrm{d}x$
$$2. $\int \frac{\cos{\frac{1}{x}}}{x^2} \, \mathrm{d}x$
$$3. Evaluate $ \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \frac{1}{3^{m+n}}$
$$4. Find $\int \frac{1+2\sqrt{x}}{2\sqrt{x}(x+\sqrt{x})} \, \mathrm{d}x$ using the substitution $u=x+\sqrt{x}$
$$5. $\int \frac{1+\cot{x}}{1-\cot{x}} \, \mathrm{d}x$
$$6. $\int^\frac{\pi}{2}_0 \sqrt{1-\sin{x}} \, \mathrm{d}x$

Take 2 and 3 back to the 2U marathon

rand_althor · Oct 4, 2015

Re: HSC 2015 3U Marathon

photastic said:
Take 2 and 3 back to the 2U marathon

Oh are they 2U? I'm just taking questions from old 3U marathons so idk.

Paradoxica · Oct 4, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
$$1. $\int \frac{1+\cos{x}}{1-\cos{x}} \, \mathrm{d}x$
$$2. $\int \frac{\cos{\frac{1}{x}}}{x^2} \, \mathrm{d}x$
$$3. Evaluate $ \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \frac{1}{3^{m+n}}$
$$4. Find $\int \frac{1+2\sqrt{x}}{2\sqrt{x}(x+\sqrt{x})} \, \mathrm{d}x$ using the substitution $u=x+\sqrt{x}$
$$5. $\int \frac{1+\cot{x}}{1-\cot{x}} \, \mathrm{d}x$
$$6. $\int^\frac{\pi}{2}_0 \sqrt{1-\sin{x}} \, \mathrm{d}x$

$$3. Evaluate $ \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \frac{1}{3^{m+n}}$

$$Expand out one of the summands. This results in an infinite number of infinite summands that start from one integer greater than the previous summand, beginning from zero. Each of these is a geometric series with common ratio $ \frac{1}{3}. $ The one starting from zero sums to $\frac{3}{2},$ and each one after that is identical to $\frac{1}{3} $ the previous. Collecting the entire series together and factoring out $\frac{3}{2}$ from each of the infinite terms results in another summand identical to the one starting from zero. This evaluates to $ \left( \frac{3}{2}\right )^2 = \frac{9}{4}$

photastic · Oct 4, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
Oh are they 2U? I'm just taking questions from old 3U marathons so idk.

Doable with 2U but perhaps we can classify them as harder 2U

kawaiipotato · Oct 4, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
Some more:
$$1. $\sum_{k=2}^n \frac{1}{\log_k u}$
$$2. Evaluate $\int_{\frac{1}{2}}^{2} \frac{\ln x}{1+x^2} \, \mathrm{d}x $ using the substitution $u=\frac{1}{x}$
$$3. $\lim_{c\to\infty}\int^c_{-c}\frac{\mathrm{d}x}{e^x+e^{-x}}$

2. $x = \frac{1}{u}$
$dx = \frac{-1}{u^2 } du$

$\therefore $ I $ = \int_{\frac{1}{2}}^{2} \frac{ln(\frac{1}{u})}{u^2 (1+ \frac{1}{u^2 }) } du$

$= \int_{\frac{1}{2}}^{2} \frac{ln(\frac{1}{u})}{u^2 + 1} du$

$= \int_{\frac{1}{2}}^{2} \frac{ln(\frac{1}{x})}{x^2 + 1} dx$

$\therefore $ I $ + $ I $ = 2I = 0$

$I = 0$

3. $= \int_{- \infty}^{\infty} \frac{e^x }{e^{2x} + 1}$

$= \tan ^{-1}(e^x) |_{- \infty}^{\infty}$

$= \pi$

.
.
.

1. $= \frac{1}{log_{2}u} + \frac{1}{log_{3}u} + ... + \frac{1}{log_{n}u}$

$= \frac{\ln 2}{\ln u} + \frac{\ln 3}{\ln u} + ... + \frac{\ln n}{\ln u}$

$= \frac{\ln (1*2*3*4*...*n)}{\ln u}$

$= \frac{\ln (n!)}{\ln u}$

$= log_{u}(n!)$

Paradoxica · Oct 4, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
Some more:
$$1. $\sum_{k=2}^n \frac{1}{\log_k u}$
$$2. Evaluate $\int_{\frac{1}{2}}^{2} \frac{\ln x}{1+x^2} \, \mathrm{d}x $ using the substitution $u=\frac{1}{x}$
$$3. $\lim_{c\to\infty}\int^c_{-c}\frac{\mathrm{d}x}{e^x+e^{-x}}$

$$1. $\sum_{k=2}^n \frac{1}{\log_k u} $By the log laws, the base can be moved up to the top. This series is divergent, and has no finite sum for any valid, finite value of u.$

Edit: Whoops, mistook n as infinity.

rand_althor · Oct 4, 2015

Re: HSC 2015 3U Marathon

kawaiipotato said:
3. $= \int_{- \infty}^{\infty} \frac{e^x }{e^{2x} + 1}$

$= \tan ^{-1}(e^x + 1) |_{- \infty}^{\infty}$

$= \pi$

You've made a mistake with this one. The others are all good.

rand_althor · Oct 4, 2015

Re: HSC 2015 3U Marathon

Here's a question Carrotsticks posted before:

Paradoxica · Oct 5, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
Here's a question Carrotsticks posted before:

Are you sure that negative in front of the "2" should be there? I'm not getting a negative in the exponent for the exponential, abd that means I won't have a limiting x value for large values of t.

davidgoes4wce · Oct 5, 2015

Re: HSC 2015 3U Marathon

I was a bit confused about the steps involved in Kinney-Lewis going from the bracket term to the last line.

InteGrand · Oct 5, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce said:
I was a bit confused about the steps involved in Kinney-Lewis going from the bracket term to the last line.

$We have $f(K)=(K -a)(K-b) \geq 0$, where in this case $a = -\sqrt{2}$, and $b = 1+\sqrt{2} > a$. In general, if we plot the parabola $f(K) = (K-a)(K-b)$ where $a<b$, we see the parabola hits the horizontal axis at $a$ and $b$ ($a$ is left of $b$ as $a<b$ and the parabola is upwards facing) and from the graph, clearly $f(K) \geq 0$ (graph is on or above the horizontal axis) if and only if $K \leq a $ or $K \geq b$. This is precisely the solution given for our particular $a$ and $b$ in that question.$

rand_althor · Oct 5, 2015

Re: HSC 2015 3U Marathon

kawaiipotato said:
3. $= \int_{- \infty}^{\infty} \frac{e^x }{e^{2x} + 1}$

$= \tan ^{-1}(e^x) |_{- \infty}^{\infty}$

$= \pi$

Correct answer is $\frac{\pi}{2}$ . I think you're mistaking $\tan ^{-1}(e^{-\infty})$ for $\tan ^{-1}(-e^{\infty})$ .

InteGrand · Oct 5, 2015

Re: HSC 2015 3U Marathon

Avoid writing something like e^∞ in the exam by the way.

rand_althor · Oct 5, 2015

Re: HSC 2015 3U Marathon

$$1. Let $f(x)$ be a function such that $f(xy) = f(x)+f(y)$. Prove by induction that $f(a^n) = n \times f(a)$ for all $n\geq1$.$
$$2. (a) Consider the circle $x^2+(y-k)^2=1$. For what value(s) of $k$ is the curve $y=x$ a tangent to the circle? \\ \indent (b) A ball of radius $1$ is dropped inside the curve $y=|x|$. Find the area bounded by the ball and the curve.$

rand_althor · Oct 5, 2015

Re: HSC 2015 3U Marathon

InteGrand said:
Avoid writing something like e^∞ in the exam by the way.

Would you write it as a limit?

InteGrand · Oct 5, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
$$1. Let $f(x,y$) be a function such that $f(xy) = f(x)+f(y)$. Prove by induction that $f(a^n) = n \times f(a)$ for all $n\geq1$.$
$$2. (a) Consider the circle $x^2+(y-k)^2=1$. For what value(s) of $k$ is the curve $y=x$ a tangent to the circle? \\ \indent (b) A ball of radius $1$ is dropped inside the curve $y=|x|$. Find the area bounded by the ball and the curve.$

$The first Q. shouldn't say $f(x,y)$ at the start, as that means that it is a function of two variables, when it is not.$

rand_althor · Oct 5, 2015

Re: HSC 2015 3U Marathon

InteGrand said:
$The first Q. shouldn't say $f(x,y)$ at the start, as that means that it is a function of two variables, when it is not.$

Haha copied the question. Wasn't sure about the notation but no one pointed it out at the time. Would it just be f(x)?

InteGrand · Oct 5, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
Would you write it as a limit?

$Yes, e.g. $\lim _{x\to \infty}\left( \tan^{-1} \left(\mathrm{e}^x \right)\right)$.$

Paradoxica · Oct 5, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
$$1. Let $f(x)$ be a function such that $f(xy) = f(x)+f(y)$. Prove by induction that $f(a^n) = n \times f(a)$ for all $n\geq1$.$
$$2. (a) Consider the circle $x^2+(y-k)^2=1$. For what value(s) of $k$ is the curve $y=x$ a tangent to the circle? \\ \indent (b) A ball of radius $1$ is dropped inside the curve $y=|x|$. Find the area bounded by the ball and the curve.$

$Please clarify 2.(b), as it does not make sense. A 3-D object and a 1-D object cannot have an area of intersection.$

InteGrand · Oct 5, 2015

Re: HSC 2015 3U Marathon

Paradoxica said:
$Please clarify 2.(b), as it does not make sense. A 3-D object and a 1-D object cannot have an area of intersection.$

I think it's a circle of radius 1. (Incidentally, a mathematical "ball" need not be 3-D: https://en.wikipedia.org/wiki/Ball_(mathematics))

HSC 2015 MX1 Marathon (archive) (1 Viewer)

Well-Known Member

Active Member

-insert title here-

Well-Known Member

Well-Known Member

-insert title here-

Active Member

Active Member

-insert title here-

Well-Known Member

Well-Known Member

Active Member

Well-Known Member

Active Member

Active Member

Well-Known Member

Active Member

Well-Known Member

-insert title here-

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)