HSC 2015 MX1 Marathon (archive) (1 Viewer)

rand_althor · Oct 6, 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.$

Question 1:
$\begin{align*}\textup{RTP }f(a^n)&=n\cdot f(a) \textup{ for }n\geq1\\ \textup{1. Show true for }n&=1 \\ \textup{LHS}&=f(a^1)=f(a) \\ \textup{RHS}&=1\times f(a)=f(a)=\textup{LHS} \\ \textup{Therefore, true for }n&=1. \end{align*}$

$\begin{align*}\textup{2. Assume true for }n&=k \\f(a^k)&=k\cdot f(a)\end{align*}$

$\begin{align*}\textup{3. Prove true for }n&=k+1 \\\textup{LHS}&=f(a^{k+1}) \\&=f(a\cdot a^k) \\&=f(a)+f(a^k) \textup{ since } f(xy)=f(x)+f(y) \\&=f(a)+k\cdot f(a) \textup{ (using the assumption)} \\&=(k+1)f(a) \\\textup{RHS}&=(k+1)f(a)=\textup{LHS} \\\textup{Therefore, true for }n&=k+1 \\\end{align*}$

I'll post a solution for question 2 tomorrow.

Drsoccerball · Oct 6, 2015

Re: HSC 2015 3U Marathon

is 2a answer k= $\pm \sqrt{2}$ ?

rand_althor · Oct 6, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
is 2a answer k= $\pm \sqrt{2}$ ?

Yep.

rand_althor · Oct 7, 2015

Re: HSC 2015 3U Marathon

If anyone is unsure what the scenario looks like, here you go.

Drsoccerball · Oct 7, 2015

Re: HSC 2015 3U Marathon

For the area question i got a very weird answer:

$\pi ( sin^{-1}( \frac{1}{\sqrt{2}} + \frac{k}{2}) +(\frac{1}{\sqrt{2}} + \frac{k}{2}) \sqrt{1-(\frac{1}{\sqrt{2}} + \frac{k}{2})^2} +2k(\frac{1}{\sqrt{2}} + \frac{k}{2}) -(\frac{1}{\sqrt{2}} + \frac{k}{2})^2)$

Drsoccerball · Oct 7, 2015

Re: HSC 2015 3U Marathon

I think I made a mistake somewhere in my working considering that substituting k = root 2 gives my a maths error. Whats the answer?

rand_althor · Oct 7, 2015

Re: HSC 2015 3U Marathon

The answer I have is $1-\frac{\pi}{4}$ . Note that in this scenario, the value of k is $\sqrt{2}$ . So in your area equation, the $\sin^{-1}{\left (\frac{1}{\sqrt{2}}+\frac{k}{2} \right )}$ is outside the domain of -1 to 1 for inverse sine functions. Can you explain what you did?

Drsoccerball · Oct 7, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
The answer I have is $1-\frac{\pi}{4}$ . Note that in this scenario, the value of k is $\sqrt{2}$ . So in your area equation, the $\sin^{-1}{\left (\frac{1}{\sqrt{2}}+\frac{k}{2} \right )}$ is outside the domain of -1 to 1 for inverse sine functions. Can you explain what you did?

Outer area minus inner area my integral was :
$\int_0^{\frac{1}{\sqrt{2}} + \frac{k}{2}}{(\sqrt{1-x^2}+k-x)}dx$

rand_althor · Oct 7, 2015

Re: HSC 2015 3U Marathon

.
Just so that we're clear, the area you're finding is enclosed in red.
I didn't integrate to get the answer, so I'm not sure about your solution. How did you integrate $\sqrt{1-x^2}$ ?

Drsoccerball · Oct 7, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
.
Just so that we're clear, the area you're finding is enclosed in red.
I didn't integrate to get the answer, so I'm not sure about your solution. How did you work integrate $\sqrt{1-x^2}$ ?

let u =sinx or cosx . Also I did the with integration it should've worked :/

Drsoccerball · Oct 7, 2015

Re: HSC 2015 3U Marathon

LOL did you just use triangle minus sector?

rand_althor · Oct 7, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
LOL did you just use triangle minus sector?

Yea pretty much. Initially tried what you did but then I saw the easier way.

InteGrand · Oct 7, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
Outer area minus inner area my integral was :
$\int_0^{\frac{1}{\sqrt{2}} + \frac{k}{2}}{(\sqrt{1-x^2}+k-x)}dx$

$That integral won't make sense if your upper limit is greater than 1 but you have a $\sqrt{1-x^2}$ in the integrand. Are you sure it's the correct integral?$

Paradoxica · Oct 7, 2015

Re: HSC 2015 3U Marathon

Pardon me, but aren't the limits of integration
$\pm \frac{1}{\sqrt2}$

davidgoes4wce · Oct 7, 2015

Re: HSC 2015 3U Marathon

A sample of three players is to be taken from a soccer team of eleven players. How many samples are possible?

A) 11^3/3^3
B) 11!/3!
C) (11x10x9)/(3x2x1)
D) 11x10x9

leehuan · Oct 7, 2015

Re: HSC 2015 3U Marathon

11C3 = 11!/(3!8!) = (11*10*9)/3!
as n!=n(n-1)!

rand_althor · Oct 7, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
$$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.$

2. (a)
$\begin{align*}y&=x\to y^2=x^2 \\x^2+(y-k)^2&=1 \\\textup{Solving simultaneously:}& \\y^2+(y-k^2)&=1 \\y^2+y^2-2ky+k^2-1&=0 \\2y^2-2ky+(k^2-1)&=0 \\\textup{Since the line }y&=x \textup{ is a tangent the discriminant is equal to zero} \\\Delta&=b^2-4ac \\(-2k)^2-4(2)(k^2-1)&=0 \\4k^2-8k^2+8&=0 \\4k^2&=8 \\k&=\pm \sqrt{2}\end{align*}$

2. (b)
$The ball will touch either branch of $y=|x|$ only once, i.e. it is a tangent to each branch. From part (a), we know that for this to happen $k=\pm \sqrt{2}$. Since the circle is above the x-axis, it's equation is then $x^2+(y-\sqrt{2})^2=1$. Simultaneously solving this equation and $y=x$, you find the point of contact between the circle and right branch of $y=|x|$ to be $A\left (\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2} \right )$. Similarly, the point of contact on the left branch of $y=|x|$ is $B\left (-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2} \right )$
$$The area between $y=|x|$ and the circle can be found by subtracting the area of segment BAD from triangle BAO. Now, $CA=CB=1$ as they are radii of the circle. $AB=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}=\sqrt{2}$. Using the cosine rule, $\angle{BCA}$ is $\frac{\pi}{2}$. Using $A=\frac{1}{2}r^2(\theta-\sin{\theta})$, the area of segment BAD is $A_1=\frac{\pi}{4}-\frac{1}{2}$ units$^2$. For triangle BAO, the base length is $\sqrt{2}$ and the height is $\frac{\sqrt{2}}{2}$ (ordinate of $A$). So using $A=\frac{1}{2}hb$, the area of triangle BAO is $A_2=\frac{1}{2}$ units$^2$.$
$$Thus, $A_{total}=A_2-A_1=\frac{1}{2}-\left (\frac{\pi}{4}-\frac{1}{2} \right )=\frac{1}{2}-\frac{\pi}{4}+\frac{1}{2}=1-\frac{\pi}{4}$ units$^2$.$

InteGrand · Oct 7, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
2. (a)
$\begin{align*}y&=x\to y^2=x^2 \\x^2+(y-k)^2&=1 \\\textup{Solving simultaneously:}& \\y^2+(y-k^2)&=1 \\y^2+y^2-2ky+k^2-1&=0 \\2y^2-2ky+(k^2-1)&=0 \\\textup{Since the line }y&=x \textup{ is a tangent the discriminant is equal to zero} \\\Delta&=b^2-4ac \\(-2k)^2-4(2)(k^2-1)&=0 \\4k^2-8k^2+8&=0 \\4k^2&=8 \\k&=\pm \sqrt{2}\end{align*}$

2. (b)
$The ball will touch either branch of $y=|x|$ only once, i.e. it is a tangent to each branch. From part (a), we know that for this to happen $k=\pm \sqrt{2}$. Since the circle is above the x-axis, it's equation is then $x^2+(y-\sqrt{2})^2=1$. Simultaneously solving this equation and $y=x$, you find the point of contact between the circle and right branch of $y=|x|$ to be $A\left (\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2} \right )$. Similarly, the point of contact on the left branch of $y=|x|$ is $B\left (-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2} \right )$
$$The area between $y=|x|$ and the circle can be found by subtracting the area of segment BAD from triangle BAO. Now, $CA=CB=1$ as they are radii of the circle. $AB=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}=\sqrt{2}$. Using the cosine rule, $\angle{BCA}$ is $\frac{\pi}{2}$. Using $A=\frac{1}{2}r^2(\theta-\sin{\theta})$, the area of segment BAD is $A_1=\frac{\pi}{4}-\frac{1}{2}$ units$^2$. For triangle BAO, the base length is $\sqrt{2}$ and the height is $\frac{\sqrt{2}}{2}$ (ordinate of $A$). So using $A=\frac{1}{2}hb$, the area of triangle BAO is $A_2=\frac{1}{2}$ units$^2$.$
$$Thus, $A_{total}=A_2-A_1=\frac{1}{2}-\left (\frac{\pi}{4}-\frac{1}{2} \right )=\frac{1}{2}-\frac{\pi}{4}+\frac{1}{2}=1-\frac{\pi}{4}$ units$^2$.$

Alternatively, the area is: area of square ABCD – area of quadrant = 1 – pi/4.

rand_althor · Oct 7, 2015

Re: HSC 2015 3U Marathon

rand_althor said:
$$Let $\alpha=\frac{m\pi}{n}$ where $m$ and $n$ are positive relatively prime integer numbers. Find $m+n$ if $0 \leq \alpha \leq \frac{\pi}{2}$ and $\sin{\alpha}=\frac{1-\sqrt{2}}{\sqrt{6-3\sqrt{2}}-\sqrt{2+\sqrt{2}}}$.$

$$Firstly, note that $\cos{x}=\sqrt{\frac{1+\cos{2x}}{2}}$ and $\sin{x}=\sqrt{\frac{1-\cos{2x}}{2}}$, using $\cos{2x}=1-2\sin^2{x}=2\cos^2{x}-1$.$

$\begin{align*}&\frac{1-\sqrt{2}}{\sqrt{6-3\sqrt{2}}-\sqrt{2+\sqrt{2}}}\\&=\frac{1-\sqrt{2}}{\sqrt{3}\sqrt{2-\sqrt{2}}-\sqrt{2+\sqrt{2}}} \\&=\frac{1-\sqrt{2}}{\sqrt{3}\sqrt{2-\sqrt{2}}-\sqrt{2+\sqrt{2}}} \times \frac{\sqrt{3}\sqrt{2-\sqrt{2}}+\sqrt{2+\sqrt{2}}}{\sqrt{3}\sqrt{2-\sqrt{2}}+\sqrt{2+\sqrt{2}}} \\&=\frac{\left (1-\sqrt{2} \right ) \left (\sqrt{3}\sqrt{2-\sqrt{2}}+\sqrt{2+\sqrt{2}} \right )}{\left (\sqrt{3}\sqrt{2-\sqrt{2}}\right )^2-\left (\sqrt{2+\sqrt{2}} \right )^2} \\&=\frac{\left (1-\sqrt{2} \right ) \left (\sqrt{3}\sqrt{2-\sqrt{2}}+\sqrt{2+\sqrt{2}} \right )}{3(2-\sqrt{2})-(2+\sqrt{2})} \\&=\frac{\left (1-\sqrt{2} \right ) \left (\sqrt{3}\sqrt{2-\sqrt{2}}+\sqrt{2+\sqrt{2}} \right )}{6-3\sqrt{2}-2-\sqrt{2}} \\\end{align*}$
$\begin{align*}&=\frac{\left (1-\sqrt{2} \right ) \left (\sqrt{3}\sqrt{2-\sqrt{2}}+\sqrt{2+\sqrt{2}} \right )}{4-4\sqrt{2}} \\&=\frac{1}{4}\left (\sqrt{3}\sqrt{2-\sqrt{2}}+\sqrt{2+\sqrt{2}} \right ) \\&=\frac{\sqrt{3}}{2}\cdot \sqrt{\frac{2-\sqrt{2}}{4}}+\frac{1}{2}\cdot \sqrt{\frac{2+\sqrt{2}}{4}} \\&=\frac{\sqrt{3}}{2} \cdot \sqrt{\frac{1-\frac{\sqrt{2}}{2}}{2}} + \frac{1}{2} \cdot \sqrt{\frac{1+\frac{\sqrt{2}}{2}}{2}} \\&=\cos{\frac{\pi}{6}}\cdot\sqrt{\frac{1-\cos{\frac{\pi}{4}}}{2}}+\sin{\frac{\pi}{6}}\cdot \sqrt{\frac{1+\cos{\frac{\pi}{4}}}{2}} \\&=\cos{\frac{\pi}{6}}\sin{\frac{\pi}{8}}+\sin{\frac{\pi}{6}}\cos{\frac{\pi}{8}} \\&=\sin{\frac{7\pi}{24}}\end{align*}$
$\sin{\alpha}=\sin{\frac{m\pi}{n}}=\sin{\frac{7\pi}{24}}$. $\therefore$ $m=7$, $n=24$. So $m+n=31$.$

Drsoccerball · Oct 7, 2015

Re: HSC 2015 3U Marathon

You are so bothered.

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