HSC 2015 Maths Marathon (archive) (1 Viewer)

kawaiipotato · Oct 18, 2015

Re: HSC 2015 2U Marathon

If v*a > 0 particle is increasing speed.
If v*a < 0 particle is decreasing in speed.
Makes intuitive sense because if v <0 it wants to travel left. If a < 0 then it wants to increase velocity in the NEGATIVE direction hence allowing it to become faster travelling left. So if v <0 and a>0 then it wants to increase velocity in the positive direction, hence making v approach zero and then positive (which is why it slows down when going left. Eventually making it stop and go right)
v*a = -2*3=-6 < 0

Mr_Kap · Oct 18, 2015

Re: HSC 2015 2U Marathon

kawaiipotato said:
If v*a > 0 particle is increasing speed.
If v*a < 0 particle is decreasing in speed.
Makes intuitive sense because if v <0 it wants to travel left. If a < 0 then it wants to increase velocity in the NEGATIVE direction hence allowing it to become faster travelling left. So if v <0 and a>0 then it wants to increase velocity in the positive direction, hence making v approach zero and then positive (which is why it slows down)
v*a = -2*3=-6 < 0

Oh shit. I COMPLETELY FORGOT about that and it was so obvious. Luckily i did this paper today.

Thanks so much!

qwert73 · Oct 18, 2015

Re: HSC 2015 2U Marathon

The population W of Williamtown is increasing exponentially according to the equation W =W0e^0.02 while the population H of Hectorville is decreasing exponentially according to the equation H = H0e^-0.01
If the current populations of Williamtown and Hectorville are 8000 and 12000 respectively, how long, to the nearest year, will it be before their population are the same?

rand_althor · Oct 18, 2015

Re: HSC 2015 2U Marathon

qwert73 said:
The population W of Williamtown is increasing exponentially according to the equation W =W0e^0.02 while the population H of Hectorville is decreasing exponentially according to the equation H = H0e^-0.01
If the current populations of Williamtown and Hectorville are 8000 and 12000 respectively, how long, to the nearest year, will it be before their population are the same?

$\begin{align*}W&=W_0e^{0.02t} \\H&=H_0e^{-0.01t} \\\textup{When }t&=0 \to W=8000,~H=12000 \\8000&=W_0e^0 \to W_0=8000 \\12000&=H_0e^0 \to H_0=12000 \\\therefore W&=8000e^{0.02t},~H=12000e^{-0.01t} \\\textup{Populations are equal when }W&=H \\8000e^{0.02t}&=12000e^{-0.01t} \\e^{0.03t}&=\frac{3}{2} \\0.03t&=\ln{\frac{3}{2}} \\t&=\frac{\ln\frac{3}{2}}{0.03} \\&=13.52 \textup{ years} \\&\approx 14\textup{ years}\end{align*}$

photastic · Oct 18, 2015

Re: HSC 2015 2U Marathon

Geppetto’s wish to build Pinocchio requires 5 years of regular deposits at the beginning of every month. He decides to deposit $100 on the first month with the Pleasure Island Bank, who pays interest of 6% p.a. compounded monthly. However, Geppetto’s desire to build Pinocchio is unexpectedly hard such that at the beginning of every month Geppetto makes a deposit that is 5% less than the previous deposit. At the same time, Pleasure Island Bank decides to tax 1% of Geppetto’s balance of his account at the end of month in order for them to transform more bad boys into donkeys.

Find the balance of the account at the end of the 5th year, correct to the nearest cent.

rand_althor · Oct 19, 2015

Re: HSC 2015 2U Marathon

photastic said:
Geppetto’s wish to build Pinocchio requires 5 years of regular deposits at the beginning of every month. He decides to deposit $100 on the first month with the Pleasure Island Bank, who pays interest of 6% p.a. compounded monthly. However, Geppetto’s desire to build Pinocchio is unexpectedly hard such that at the beginning of every month Geppetto makes a deposit that is 5% less than the previous deposit. At the same time, Pleasure Island Bank decides to tax 1% of Geppetto’s balance of his account at the end of month in order for them to transform more bad boys into donkeys.

Find the balance of the account at the end of the 5th year, correct to the nearest cent.

Is the answer $1531.63?

rand_althor · Oct 19, 2015

Re: HSC 2015 2U Marathon

$The region bounded by the parabola $y=x^2$ and the circle $x^2+y^2=12$ is shown. The parabola and circle intersect at P and Q.$

$$(i)~~~P and Q have the same y-coordinate. Find its value.$ \\\indent $(ii)~~Find the exact value of the solid generated when the shaded region is rotated about the y-axis.$

davidgoes4wce · Oct 19, 2015

Re: HSC 2015 2U Marathon

rand_althor said:
$The region bounded by the parabola $y=x^2$ and the circle $x^2+y^2=12$ is shown. The parabola and circle intersect at P and Q.$

$$(i)~~~P and Q have the same y-coordinate. Find its value.$ \\\indent $(ii)~~Find the exact value of the solid generated when the shaded region is rotated about the y-axis.$

someone could double confirm with me

Mr_Kap · Oct 19, 2015

Re: HSC 2015 2U Marathon

seems correct...thats what i would have done

rand_althor · Oct 19, 2015

Re: HSC 2015 2U Marathon

davidgoes4wce said:
someone could double confirm with me

Yep that's the correct answer.

rand_althor · Oct 19, 2015

Re: HSC 2015 2U Marathon

$$Consider the function $y=\frac{x^2}{1+x^2}.$ \\(i) ~Show that $f''(x)=\frac{2(1-3x^2)}{(1+x^2)^3}$. \\(ii) For what values of $x$ is the function concave up?$

Crisium · Oct 19, 2015

Re: HSC 2015 2U Marathon

Mr_Kap said:
seems correct...thats what i would have done

rand_althor said:
Yep that's the correct answer.

Yeah it's correct

Just a heads up though, markers (apparently) prefer if you draw an arrow to the x^2 = 4 saying "Reject/Disregard Since x has no real solutions (discriminant < 0)"

photastic · Oct 19, 2015

Re: HSC 2015 2U Marathon

rand_althor said:
Is the answer $1531.63?

Bingo

davidgoes4wce · Oct 19, 2015

Re: HSC 2015 2U Marathon

rand_althor said:
$$Consider the function $y=\frac{x^2}{1+x^2}.$ \\(i) ~Show that $f''(x)=\frac{2(1-3x^2)}{(1+x^2)^3}$. \\(ii) For what values of $x$ is the function concave up?$

again someone could confirm that would be great.

rand_althor · Oct 19, 2015

Re: HSC 2015 2U Marathon

davidgoes4wce said:
again someone could confirm that would be great.

That is correct.

$$(1) Show that if $y=\frac{e^x+e^{-x}}{2}$, then $y''=\sqrt{1+(y')^2}$. \\\indent (2) For what values of $k$ is the quadratic $kx^2-2x\sqrt{6}+k+1$ positive definite?$

Mr_Kap · Oct 20, 2015

Re: HSC 2015 2U Marathon

My friend just sent me this question and for some reason i can't do it, and its really bothering me.

davidgoes4wce · Oct 20, 2015

Re: HSC 2015 2U Marathon

rand_althor said:
That is correct.

$$(1) Show that if $y=\frac{e^x+e^{-x}}{2}$, then $y''=\sqrt{1+(y')^2}$. \\\indent (2) For what values of $k$ is the quadratic $kx^2-2x\sqrt{6}+k+1$ positive definite?$

Part (1) to that question

davidgoes4wce · Oct 20, 2015

Re: HSC 2015 2U Marathon

rand_althor said:
That is correct.

$$(1) Show that if $y=\frac{e^x+e^{-x}}{2}$, then $y''=\sqrt{1+(y')^2}$. \\\indent (2) For what values of $k$ is the quadratic $kx^2-2x\sqrt{6}+k+1$ positive definite?$

Not sure what you mean for Q2 for 'positive definite'. Im assuming 2 roots?

Mr_Kap · Oct 20, 2015

Re: HSC 2015 2U Marathon

davidgoes4wce said:
Not sure what you mean for Q2 for 'positive definite'. Im assuming 2 roots?

positive definite means it is a positive parabola with no root.

EDIT: no REAL roots

rand_althor · Oct 20, 2015

Re: HSC 2015 2U Marathon

Mr_Kap said:
My friend just sent me this question and for some reason i can't do it, and its really bothering me.

$$Since the point $P(x_1,y_1)$ lies on $y=\sqrt{x-4}$, it can be expressed as $P(x_1,\sqrt{x_1-4})$. Using $m=\frac{y_2-y_1}{x_2-x_1}$, the gradient of $OP$ is $m=\frac{\sqrt{x_1-4}}{x_1}$. Since $OP$ is the tangent to $y=\sqrt{x-4}$ at $x=x_1$, you can differentiate the curve to find the gradient of $OP$ at $x=x_1$. $y=\sqrt{x-4}\to y'=\frac{1}{2\sqrt{x-4}}=\frac{1}{2\sqrt{x_1-4}}$. These two expressions for the gradient of $OP$ are equivalent, so solving $\frac{\sqrt{x_1-4}}{x_1}=\frac{1}{2\sqrt{x_1-4}}$, you should get $x_1=8$.$

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