HSC 2015 MX2 Marathon (archive) (1 Viewer)

Sy123 · Apr 24, 2015

Re: HSC 2015 4U Marathon

A prime number is a positive integer greater than 1 that is divisible by only 1 and itself.

i) Show that (n! + 1) is not divisible by any number from 2 to n inclusive

ii) Hence show that the number of primes is infinite

Sy123 · Apr 25, 2015

Re: HSC 2015 4U Marathon

$\\ $Let the recurrence relation for the sequence$ \ x_n \ $be$ \ x_n = 3x_{n-1} - 2x_{n-2} , \ (n \geq 2) \ $with$ \ x_0 = 0 \ $and$ \ x_1 = 1 \\ \\ $Find a recurrence relation for the sequence$ \ y_n = x_n^2 \ $involving only linear forms of$ \ y_n, y_{n-1}, $etc.$$

Drsoccerball · Apr 25, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Let the recurrence relation for the sequence$ \ x_n \ $be$ \ x_n = 3x_{n-1} - 2x_{n-2} , \ (n \geq 2) \ $with$ \ x_0 = 0 \ $and$ \ x_1 = 1 \\ \\ $Find a recurrence relation for the sequence$ \ y_n = x_n^2$

What exacly is this asking

do we just square the original equation to have it in terms of y_n ?

Sy123 · Apr 25, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
What exacly is this asking do we just square the original equation to have it in terms of y_n ?

yes, but the equation must be only in terms of y_n, and the terms must be linear (so no sqrt(y_n), y_n^2 etc.)

DannyRRR · Apr 25, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
yes, but the equation must be only in terms of y_n, and the terms must be linear (so no sqrt(y_n), y_n^2 etc.)

(2n-1)^2

Drsoccerball · Apr 25, 2015

Re: HSC 2015 4U Marathon

$area bounded by the graph $ y=x^2(4-x)^2 $ and the x axis is rotated about x=0$

$using slices find the volume$

swagswagyoloswag · Apr 25, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
$area bounded by the graph $ y=x^2(4-x)^2 $ and the x axis is rotated about x=0$

$using slices find the volume$

$Note that f(x) = x^{2}(4-x)^{2} $ $ is symmetrical around x = 2$

$Drawing a horizontal slice, we have endpoints P(x_1,y) and Q(x_1,y)$

$\frac{x_1 + x_2}{2} = 2$

$x_1 + x_2 = 4$

$\delta V = \pi ((x_2)^{2} - (x_1)^{2}) \delta y$

$\delta V = \pi ((x_2 + x_1)(x_2 - x_1)) \delta y$

$\delta V = \pi (4)(x_2 - (4-x_2)) \delta y$

$\delta V = \pi (8)(x_2 - 2) \delta y$

$\delta V = \pi (8)(\sqrt{4-\sqrt{y}} \delta y$

$V = 8\pi \int_{0}^{16} \sqrt{4-\sqrt{y}} dx$

Drsoccerball · Apr 25, 2015

Re: HSC 2015 4U Marathon

swagswagyoloswag said:
$Note that f(x) = x^{2}(4-x)^{2} $ $ is symmetrical around x = 2$

$Drawing a horizontal slice, we have endpoints P(x_1,y) and Q(x_1,y)$

$\frac{x_1 + x_2}{2} = 2$

$x_1 + x_2 = 4$

$\delta V = \pi ((x_2)^{2} - (x_1)^{2}) \delta y$

$\delta V = \pi ((x_2 + x_1)(x_2 - x_1)) \delta y$

$\delta V = \pi (4)(x_2 - (4-x_2)) \delta y$

$\delta V = \pi (8)(x_2 - 2) \delta y$

$\delta V = \pi (8)(\sqrt{4-\sqrt{y}} \delta y$

$V = 8\pi \int_{0}^{16} \sqrt{4-\sqrt{y}} dx$

What if its not symmetrical?

Ekman · Apr 25, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
$area bounded by the graph $ y=x^2(4-x)^2 $ and the x axis is rotated about x=0$

$using slices find the volume$

ALT METHOD:

$y=x^2(4-x)^2 \Rightarrow \pm\sqrt{y}=4x-x^2 $ Note that we take the positive square root as its always positive between 0 to 4 (Which is the region we are rotating around) $$

$Let $x_{2} $be the outer radius and $x_{1} $be the smaller radius$

$\therefore x_{1,2}= 2\pm\sqrt{4-\sqrt{y}}$

$V=\pi\int_{0}^{16} (x_{2})^2 - (x_{1})^2 dy = 8\pi\int_{0}^{16} \sqrt{4-\sqrt{y}} dy = \frac{2048\pi}{15} $ (Via $u^2=4-\sqrt{y})$

Im sure there is a quicker way of integrating it besides that substitution, but this was the first thing that came to my mind.

swagswagyoloswag · Apr 25, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
What if its not symmetrical?

f(4-x) = f(x)
I'm not sure what another method would be if it wasn't symmetrical besides finding the roots. But if you find the roots they would be conjugates of each other (real coefficients, surd roots come in conjugates) which indicate a symmetrical property. I would use the dreaded shell method otherwise

Drsoccerball · Apr 25, 2015

Re: HSC 2015 4U Marathon

swagswagyoloswag said:
f(4-x) = f(x)
I'm not sure what another method would be if it wasn't symmetrical besides finding the roots. But if you find the roots they would be conjugates of each other (real coefficients, surd roots come in conjugates) which indicate a symmetrical property. I would use the dreaded shell method otherwise

But the question forces slices and there been questions like that you'd just have to do the long way

Sy123 · Apr 25, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
A prime number is a positive integer greater than 1 that is divisible by only 1 and itself.

i) Show that (n! + 1) is not divisible by any number from 2 to n inclusive

ii) Hence show that the number of primes is infinite

$\\ $i) Assume that$ \ n! + 1 = kP \ $where$ \ k \ $is some integer between$ \ n \ $and$ \ 2 \ $inclusive$ \\ \\ $So,$ \ 1 = kP - n! = k(P-M) \ $where$ \ M \ $is an integer because$ \ n! \ $contains$ \ k \ $as a factor$ \\ \\ $However this is a clear contradiction because it implies$ \ |k| = |P-M| = 1 \ $which contradicts the original assumption that$ \ 2 \leq k \leq n$

$\\ $ii) Assume that there are only a finite number of primes, and so there must be a prime that is the largest prime, call this largest prime$ \ p_M \\ \\ $Consider the number$ \ p_M! + 1 $, this number is either prime or not prime, if it is prime, then this contradicts the assumption that$ \ p_M \ $is the largest prime, therefore meaning there must be an infinite number of primes (1)$$

$\\ $However if$ \ p_M! + 1 \ $is not prime, then it must be divisible by a prime, however it cannot be divisible by any number between$ \ 2 \ $and$ \ p_M \ $ (from i), and hence it must be divisible by some prime$ \ q \ $where$ \ p_M < q < p_M! + 1 \\ \\ $But then again, this contradicts the original assumption that there is a largest prime$ \ p_M \ $and hence there must be an infinite number of primes (2)$ \\ \\ $Therefore, since$ \ p_M! + 1 \ $is either prime or not prime, and if it is prime, it implies an infinite number of primes (1), and if it is not prime, it still implies an infinite number of primes (2), then therefore there must be an infinite number of primes.$ \ \ \square$

Sy123 · Apr 25, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Let the recurrence relation for the sequence$ \ x_n \ $be$ \ x_n = 3x_{n-1} - 2x_{n-2} , \ (n \geq 2) \ $with$ \ x_0 = 0 \ $and$ \ x_1 = 1 \\ \\ $Find a recurrence relation for the sequence$ \ y_n = x_n^2 \ $involving only linear forms of$ \ y_n, y_{n-1}, $etc.$$

$\\ x_n^2 = 9x_{n-1}^2 + 4x_{n-2}^2 - 12x_{n-1}x_{n-2} \ (*)$

$\\ x_{n-1} = 3x_{n-2} - 2x_{n-3} \Rightarrow 2x_{n-3} = 3x_{n-2} - x_{n-1} \\ \Rightarrow 4x_{n-3}^2 = 9x_{n-2}^2 - 6x_{n-2}x_{n-1} + x_{n-1}^2 \\ \Rightarrow 6x_{n-1}x_{n-2} = x_{n-1}^2 + 9x_{n-2}^2 - 4x_{n-3}^2 \ (**)$

$2\times (**) \to (*)$

$\\ \Rightarrow \ x_n^2 = 7x_{n-1}^2 -14x_{n-2}^2 + 8x_{n-3}^2$

$\\ \Rightarrow y_n = 7y_{n-1} - 14y_{n-2} + 8y_{n-3}$

Sy123 · Apr 25, 2015

Re: HSC 2015 4U Marathon

$\\ $Consider the graphs$ \ y = x^n \ $and$ \ y = x \ $where$ \ n \ $is a positive integer greater than 1$ \\ \\ $i) Consider the point$ \ (t,t^n) \ $which lies on$ \ y = x^n \ $and$ \ 0 \leq t \leq 1 $, find a function of$ \ t \ $that represents the distance from this point to the line$ \ y = x \\ \\ $ii) Hence find the volume of the solid generated (in terms of$ \ n $), when the area between$ \ y = x^n \ $and$ \ y= x \ $is rotated about the line$ \ y = x$

Drsoccerball · Apr 26, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Consider the graphs$ \ y = x^n \ $and$ \ y = x \ $where$ \ n \ $is a positive integer greater than 1$ \\ \\ $i) Consider the point$ \ (t,t^n) \ $which lies on$ \ y = x^n \ $and$ \ 0 \leq t \leq 1 $, find a function of$ \ t \ $that represents the distance from this point to the line$ \ y = x \\ \\ $ii) Hence find the volume of the solid generated (in terms of$ \ n $), when the area between$ \ y = x^n \ $and$ \ y= x \ $is rotated about the line$ \ y = x$

My teacher didn't solve questions like this because it wasn't in the syllabus

Sy123 · Apr 26, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
My teacher didn't solve questions like this because it wasn't in the syllabus

The way I gave the question made it HSC worthy, the first part guides you through, just use the ordinary logic you know from Volumes.

InteGrand · Apr 26, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Consider the graphs$ \ y = x^n \ $and$ \ y = x \ $where$ \ n \ $is a positive integer greater than 1$ \\ \\ $i) Consider the point$ \ (t,t^n) \ $which lies on$ \ y = x^n \ $and$ \ 0 \leq t \leq 1 $, find a function of$ \ t \ $that represents the distance from this point to the line$ \ y = x \\ \\ $ii) Hence find the volume of the solid generated (in terms of$ \ n $), when the area between$ \ y = x^n \ $and$ \ y= x \ $is rotated about the line$ \ y = x$

i) Using the perpendicular distance to line formula, the required distance r of (t, tⁿ) to the line $y=x \Leftrightarrow x -y = 0$ is:

$r(t)=\frac{|t-t^n+0|}{\sqrt{1^2 + (-1)^2}}= \frac{t-t^n}{\sqrt{2}}$ (we can drop the absolute value signs as t ≥ tⁿ).

Note this is the radius of a typical disc rotated about the axis of rotation (the line y = x).

ii) For a given t, denote by (a, a) the closest point to (t, tⁿ) on the line y = x. Denote by $\ell \equiv \ell(a(t))$ the distance along the line $y=x$ we have travelled from the origin to get to the point $(a,a)$ . Note that $\ell \in [0,\sqrt{2}]$ and $\ell = \sqrt{2}a$ (by Pythagoras' Theorem).

We find a as a function of t.

If D denotes the distance from $(a,a)$ to $(t,t^n)$ , then $D^2 = (a-t)^2 + (a-t^n)^2$ .

To find the a that will minimise D, we need to minimise D², so we need the derivative of the RHS above with respect to a to be 0:

$2(a-t)+2(a-t^n)=0 \Rightarrow a(t) = \frac{t+t^n}{2}$ .

So $\ell = \sqrt{2} a = \sqrt{2}\times\frac{t+t^n}{2}=\frac{t+t^n}{\sqrt{2}}$ .

Consider a typical disc at a point $\ell \in [0,\sqrt{2}]$ , it has radius r, thickness $\mathrm{d}\ell$ . So its volume is given by $\mathrm{d}V=\pi r^2 \text{ d}\ell$ .

Total volume of the solid is:

$V=\int _{\ell = 0} ^{\ell = \sqrt{2}} \pi r^2 \text{ d}\ell$ .

Substitute $\ell = \frac{t+t^n}{\sqrt{2}}$ (this is one-to-one for $t\in [0,1]$ , so it is safe to use), $\mathrm{d}\ell = \frac{1+nt^{n-1}}{\sqrt{2}}\text{ d}t$ , $\ell = 0$ when $t=0$ , $\ell = \sqrt{2}$ when $t=1$ , and $r(t)=\frac{t-t^n}{\sqrt{2}}\Rightarrow r^2 (t)=\frac{(t-t^n)^2}{2}$ .

Then $V = \int _{t=0} ^{t=1} \pi \cdot \frac{(t-t^n)^2}{2} \cdot \frac{1+nt^{n-1}}{\sqrt{2}} \text{ d}t$ .

This is now easy (albeit somewhat tedious) to evaluate.

Drsoccerball · Apr 26, 2015

Re: HSC 2015 4U Marathon

$$ Find the coordinates of the foci for $ xy =8$

braintic · Apr 26, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
$$ Find the coordinates of the foci for $ xy =8$

This is just bookwork - memorising a result.

InteGrand · Apr 26, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
$$ Find the coordinates of the foci for $ xy =8$

We do it for the general case $xy=c^2$ , $c>0$ .

This is a rectangular hyperbola, so eccentricity is $e=\sqrt{2}$ .

If you consider a left-right opening hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ , 'a' represents the distance of the point where the hyperbola cuts its horizontal axis to the origin.

So in the case of $xy=c^2$ , the a we would use in "focal length = ae" is the distance of the point where the hyperbola cuts the line y = x to the origin. The point in question is $(c,c)$ , so the distance in question is $a=\sqrt{c^2 + c^2}=\sqrt{2}c$ .

Then our focal length d is given by $d=ae=\sqrt{2}c\times \sqrt{2} = 2c$ .

As the focus lies on the line y = x, we need the coordinates (f, f) of the point on this line which is d units from the origin. This is given by: $f^2 + f^2 = d^2 \Rightarrow 2f^2 = 2^2 c^2 \Rightarrow f^2 =2c^2\Rightarrow f = \sqrt{2}c$ .

So the focus is at $(\sqrt{2}c,\sqrt{2}c)$ . In your question, we had $c=\sqrt{8}=2\sqrt{2}$ . So the focus would be at $(4,4)$ .

(Obviously, there's also a focus at $(-\sqrt{2}c,-\sqrt{2}c)$ , or (-4,-4) in your example.)

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