HSC 2013 MX2 Marathon (archive) (8 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.

=====

$$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.

HSC 2013 MX2 Marathon (archive) (8 Viewers)

Member

This too shall pass

what is that?It is Cowpea

This too shall pass

what is that?It is Cowpea

Well-Known Member

Administrator

what is that?It is Cowpea

Administrator

what is that?It is Cowpea

Administrator

This too shall pass

what is that?It is Cowpea

what is that?It is Cowpea

This too shall pass

This too shall pass

Member

This too shall pass

what is that?It is Cowpea

This too shall pass

Users Who Are Viewing This Thread (Users: 0, Guests: 8)