HSC 2013 MX2 Marathon (archive) (6 Viewers)

study1234 · Jul 21, 2013

Re: HSC 2013 4U Marathon

Suppose that three points on the parabola y=x^2 have the property that their normal lines intersect at a common point. Show that the sum of their x-coordinates is 0.

Sy123 · Jul 22, 2013

Re: HSC 2013 4U Marathon

study1234 said:
Suppose that three points on the parabola y=x^2 have the property that their normal lines intersect at a common point. Show that the sum of their x-coordinates is 0.

Let the points be A(a,a^2), B(b,b^2), C(c,c^2)
Find through differentiating that nromal to curve at A is:

x+2ay=2a^3+a

Intersection of normals at A and B are

(-2ab(a+b), a^2+ab+b^2+1/2)

Intersections of normals at A and C are

(-2ac(a+c), a^2+ac+c^2+1/2)

Equating x-y coordinates we get that a+b+c=0, hence proof is complete.

============================

$$Prove that for some integer$ \ n$

$e^{\binom{n}{2}} > n!$

RealiseNothing · Jul 22, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$$Prove that for some integer$ \ n$

$e^{\binom{n}{2}} > n!$

$\frac{n-1}{2} \geq ln(n)$

$\frac{n(n-1)}{2} \geq ln(n^n)$

$e^{\frac{n(n-1)}{2}} \geq n^n$

$e^{\binom{n}{2}} \geq n^n > n!$

Sy123 · Jul 22, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
$\frac{n-1}{2} \geq ln(n)$

$\frac{n(n-1)}{2} \geq ln(n^n)$

$e^{\frac{n(n-1)}{2}} \geq n^n$

$e^{\binom{n}{2}} \geq n^n > n!$

The first inequality isn't always true for all positive integers n.

RealiseNothing · Jul 22, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
The first inequality isn't always true for all positive integers n.

It works for integers $\geq 4$

The cases of 1, 2, 3 can just be manually tested I guess. At the time I originally had $\frac{n(n+1)}{2}$ which works for all integers, until I realised it's n-1 not n+1, and thought it still worked without actually testing it lol.

nightweaver066 · Jul 22, 2013

Re: HSC 2013 4U Marathon

$\binom{n}{2} = \frac{n!}{(n-2)!2!}=\frac{n}{2}(n-1) = \frac{n}{2}(1 + n) - n = 1 + 2 + ... + (n - 1) > ln(1) + ln(2) + ... + ln(n-1) = ln((n-1)! \; (x > lnx)$

$\therefore e^{\binom{n}{2}} > (n-1)!$

so close.

Trebla · Jul 23, 2013

Re: HSC 2013 4U Marathon

Hint: first prove that for x > 1

x - 1 > ln x

then go from there

RealiseNothing · Jul 23, 2013

Re: HSC 2013 4U Marathon

Trebla said:
Hint: first prove that for x > 1

x - 1 > ln x

then go from there

When $x=1$ LHS=RHS

differentiating both sides gives LHS'=1 and RHS'=1/x so for x>1 LHS>RHS

then apply what nightweaver did, and the solution is done

Trebla · Jul 23, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
When $x=1$ LHS=RHS

differentiating both sides gives LHS'=1 and RHS'=1/x so for x>1 LHS>RHS

then apply what nightweaver did, and the solution is done

lol you can't differentiate both sides of an inequality

RealiseNothing · Jul 23, 2013

Re: HSC 2013 4U Marathon

Trebla said:
lol you can't differentiate both sides of an inequality

no not like that

like, if you differentiate x-1 you get 1, and if you differentiate ln(x) you get 1/x, so x-1 increases faster than ln(x) for x>1, but at x=1 they are equal, thus for x>1, x-1>ln(x)

Trebla · Jul 23, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
no not like that

like, if you differentiate x-1 you get 1, and if you differentiate ln(x) you get 1/x, so x-1 increases faster than ln(x) for x>1, but at x=1 they are equal, thus for x>1, x-1>ln(x)

Oh right. I would normally just show that the maximum value of y = ln x - x is -1.

Sy123 · Jul 23, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
It works for integers $\geq 4$

The cases of 1, 2, 3 can just be manually tested I guess. At the time I originally had $\frac{n(n+1)}{2}$ which works for all integers, until I realised it's n-1 not n+1, and thought it still worked without actually testing it lol.

ah ok, well done.

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$$The sequence$ \ \{f_k \} \ \ $is defined by$ \ \ f_k=f_{k-1} + f_{k-2} \ \ k \geq 2$

$f_0 =0 \ \ f_1 = 1$

$Assuming that the ratio$ \ \frac{f_{k+1}}{f_k} \ \ $approaches a finite limit as$ \ k \ $increases without bound$

$$Prove that$ \ \ \lim_{n \to \infty} \frac{f_{n+1}}{f_n} = \frac{1+\sqrt{5}}{2}$

RealiseNothing · Jul 23, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
ah ok, well done.

=====

$$The sequence$ \ \{f_k \} \ \ $is defined by$ \ \ f_k=f_{k-1} + f_{k-2} \ \ k \geq 2$

$f_0 =0 \ \ f_1 = 1$

$Assuming that the ratio$ \ \frac{f_{k+1}}{f_k} \ \ $approaches a finite limit as$ \ k \ $increases without bound$

$$Prove that$ \ \ \lim_{n \to \infty} \frac{f_{n+1}}{f_n} = \frac{1+\sqrt{5}}{2}$

Let $\frac{f_{n+1}}{f_n} = X$

Now using $f_k=f_{k-1} + f_{k-2}$

$\frac{f_{n+1}}{f_n} = 1 + \frac{f_{n-1}}{f_n}$

Therefore:

$X = 1 + \frac{1}{X}$

$X^2-X-1=0$

$X=\frac{1+\sqrt{5}}{2}$

RealiseNothing · Jul 23, 2013

Re: HSC 2013 4U Marathon

Sy123 · Jul 23, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:

$s=l\theta$

$v=\frac{ds}{dt} = l \frac{d\theta}{dt}$

$\frac{d \left(\frac{1}{2}v^2)}{d\theta} \cdot \frac{d\theta}{dv} = \frac{d\left(\frac{1}{2}v^2 \right )}{dv} = v$

$\therefore l \frac{d \theta}{dt} = \frac{d}{d\theta} \left( \frac{1}{2}v^2 \right) \frac{d \theta}{dv}$

$\frac{dv}{dt} = \frac{d^2s}{dt^2} = \frac{1}{l} \frac{d}{d\theta} \left(\frac{1}{2}v^2 \right )$

b) Resolve all forces with respect to the perpendicular axes of the Tension and tangential acceleration vectors, we find that

$ma = -mg \sin \theta$

Where a is the acceleration, the decomposed gravity vector opposes the tangential acceleration

$a = -g\sin \theta$

But from part (a), $a= \frac{d^2s}{dt^2} = \frac{1}{l}\frac{d}{d\theta} \left(\frac{1}{2}v^2 \right ) = -g\sin \theta$

===========================

$The recurrence sequence$ \{u_n \} \ \ $is defned by the recurrence relation$

$u_n = \sqrt{3 u_{n-1}}, \ \ u_1 = 1$

$$Prove that$ \ u_n < 3$

$$Prove that$ \ \ u_{n+1} > u_{n}$

Sy123 · Jul 23, 2013

Re: HSC 2013 4U Marathon

$$A right circular hollow cone has surface area$ \ \ A \ \ $and volume$ \ V$

$The surface area of the cone remains constant while the radius and height vary$

$Show that the maximum volume is given by$

$V_{max} = \sqrt{\frac{A^3}{72\pi}}$

study1234 · Jul 24, 2013

Re: HSC 2013 4U Marathon

Consider the ellipse x^2/100 + y^2/25 = 1, and let P be the point (6,4) on the ellipse. If the normal at P meets the major axis at G, and OH is the perpendicular drawn from the origin to the tangent at P, show that PG x OH = 25.

Sy123 · Jul 24, 2013

Re: HSC 2013 4U Marathon

$Define the following recurrence$

$H_1= 1 \ \ H_n = H_{n-1} + \frac{1}{n}$

$i) Prove that$ \ \ H_n \rightarrow \infty \ \ $as$ \ n \ $increases without bound$

$$ii) Prove through inequalities that the limit$ \ \ \gamma = \lim_{n\to \infty} \left(H_n - \ln n \right ) \ \ $exists$

RealiseNothing · Jul 24, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$Define the following recurrence$

$H_1= 1 \ \ H_n = H_{n-1} + \frac{1}{n}$

$i) Prove that$ \ \ H_n \rightarrow \infty \ \ $as$ \ n \ $increases without bound$

$$ii) Prove through inequalities that the limit$ \ \ \gamma = \lim_{n\to \infty} \left(H_n - \ln n \right ) \ \ $exists$

Very similar to last year's BOS trial question 15 by carrot.

Sy123 · Jul 24, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
Very similar to last year's BOS trial question 15 by carrot.

Well it is given that E_n approaches a finite limit, and

H_n - ln(n) = E_n + 1/n, so the result is practically given, for my one it is not given.

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