HSC 2014 MX2 Marathon ADVANCED (archive) (2 Viewers)

dunjaaa · May 4, 2014

Re: HSC 2014 4U Marathon - Advanced Level

oh dear lord, cant imagine what the other questions are gonna be like

mreditor16 · May 4, 2014

Re: HSC 2014 4U Marathon - Advanced Level

hi guys, a quick integration question - soz don't know how to use latex

integral of root ( x / 1-x ) dx

from 0 to 0.5

Kurosaki · May 4, 2014

Re: HSC 2014 4U Marathon - Advanced Level

mreditor16 said:
hi guys, a quick integration question - soz don't know how to use latex

integral of root ( x / 1-x ) dx

from 0 to 0.5

$\int^{0.5}_0\sqrt{\frac{x}{1-x}}dx=\int^{0.5}_0\frac{x}{\sqrt{x-x^2}}dx$
Is that enough of a pointer to get you started?

Edit: Consider the derivative of $\sqrt{x-x^2}$ . It's $\frac{1-2x}{2\sqrt{x-x^2}}=\frac{\frac{1}{2}-x}{\sqrt{x-x^2}}$
So then, $\int^{0.5}_{0}\frac{x}{\sqrt{x-x^2}}dx=-\int^{0.5}_0\frac{0.5-x-0.5}{\sqrt{x-x^2}}dx=-\int^{0.5}_0\frac{0.5-x}{\sqrt{x-x^2}}dx + \int^{0.5}_0\frac{0.5}{\sqrt{x-x^2}}dx$

mreditor16 · May 4, 2014

Re: HSC 2014 4U Marathon - Advanced Level

nup still don't see it - help me out?

mreditor16 · May 4, 2014

Re: HSC 2014 4U Marathon - Advanced Level

dw i got it anyway - i needed to substitute x = sin^2 theta. thanks anyways!

dunjaaa · May 4, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Screen shot 2014-05-04 at 6.09.19 PM.png

without substitution

seanieg89 · May 4, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Induction is enough Sy, I posted a proof of this in the jensen's article I wrote a little while ago. Will post the proof again here when I get home...it is quite short.

Sent from my GT-I9000 using Tapatalk 2

Davo_01 · May 4, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
Yea I quickly wrote the question up, I thought with part(ii) it would be straightforward induction, but I don't think it is.

Here is one (I did this one)

$\\ $Let$ \ a,b,c \ $be non-negative reals such that$ \ a^2+b^2+c^2 = 8 \\ \\ $Prove that$ \ (4-a)(4-b)(4-c) + abc \geq 16$

I tried it with induction, heres what i did:
From part i, we know it is true for n=2.

Let us assume true for n=k
$\sum_{j=1}^k \omega_j f(x_j) \geq f(\sum_{j=1}^k \omega_j x_j)$

Prove true for n=k+1
$RHS= f(w_{k+1} x_{k+1} +\frac{1-w_{k+1}}{1-w_{k+1}}\sum_{j=1}^k w_j x_j)$

$\leq w_{k+1} f(x_{k+1})+(1-w_{k+1})f(\dfrac{\sum_{j=1}^k w_j x_j}{1-w_{k+1}})$

$=w_{k+1}f( x_{k+1})+(1-w_{k+1})f(\dfrac{\sum_{j=1}^k w_j x_j}{\sum_{j=1}^k w_j} )$

Letting $\omega_j=\dfrac{w_j}{\sum_{j=1}^k w_j}$ , and hence using assumption...

$\leq w_{k+1} f(x_{k+1})+(1-w_{k+1})(\dfrac{\sum_{j=1}^k \omega_j f(x_j)}{ \sum_{j=1}^k w_j})$

$=\sum_{j=1}^{k+1} w_j f(x_j) =LHS$

Hence proven by Induction

seanieg89 · May 4, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Davo_01 said:
I tried it with induction, heres what i did:
From part i, we know it is true for n=2.

Let us assume true for n=k
$\sum_{j=1}^k \omega_j f(x_j) \geq f(\sum_{j=1}^k \omega_j x_j)$

Prove true for n=k+1
$RHS= f(w_{k+1} x_{k+1} +\frac{1-w_{k+1}}{1-w_{k+1}}\sum_{j=1}^k w_j x_j)$

$\leq w_{k+1} f(x_{k+1})+(1-w_{k+1})f(\dfrac{\sum_{j=1}^k w_j x_j}{1-w_{k+1}})$

$=w_{k+1}f( x_{k+1})+(1-w_{k+1})f(\dfrac{\sum_{j=1}^k w_j x_j}{\sum_{j=1}^k w_j} )$

Letting $\omega_j=\dfrac{w_j}{\sum_{j=1}^k w_j}$ , and hence using assumption...

$\leq w_{k+1} f(x_{k+1})+(1-w_{k+1})(\dfrac{\sum_{j=1}^k \omega_j f(x_j)}{ \sum_{j=1}^k w_j})$

$=\sum_{j=1}^{k+1} w_j f(x_j) =LHS$

Hence proven by Induction

Yep, this is pretty much correct. The only thing you have to be wary of is a potential division by zero when dividing by sum of the first k w's. But this is not a big deal.

The article I mentioned earlier is attached anyway.

View attachment jensen.pdf

one error correction: replace the words "by convexity" with "since f'' >= 0" in the proof of theorem 1.3.

Sy123 · May 7, 2014

Re: HSC 2014 4U Marathon - Advanced Level

This is a good question

$\\ $Let$ \ f(x) \ $be a function on$ \ a \leq x \leq b \ $and differentiable on$ \ a < x < b, \ $and also $ \ f' \ $is non-constant$ \\ \\ $Prove that there exists some$ \ a \leq c \leq b \ $such that$ \\ \\ |f'(c)| > \left|\frac{f(b) - f(a)}{b-a} \right|$

RealiseNothing · May 8, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Trebla said:
Sy123, I just wanna say SCREW YOU because this question (worded slightly differently) was a potential consideration for this year's BoS trials

Let me guess, 11a)???

TL1998 · May 8, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Don't know how to use Latex so....

Given that f'(1) = 2, f(1) = 1, evaluate the following the limits

A) Limit x->1 ( (f(x)-x^2*f(1)) / sin(x-1) )

B) Lim x->0 ( (f(2-cosx)-f(1)) / x^2)

Sy123 · May 8, 2014

Re: HSC 2014 4U Marathon - Advanced Level

TL1998 said:
Don't know how to use Latex so....

Given that f'(1) = 2, f(1) = 1, evaluate the following the limits

A) Limit x->1 ( (f(x)-x^2*f(1)) / sin(x-1) )

B) Lim x->0 ( (f(2-cosx)-f(1)) / x^2)

$\\ $A)$ \ \lim_{x \to 1} \frac{f(x) - x^2 f(1)}{\sin (x-1)} \\ h=x-1 \\ \Rightarrow \ \lim_{h \to 0} \frac{f(1+h) - (1+h)^2}{\sin h} = \lim_{h \to 0} \frac{\frac{f(1+h) - 1}{h} - h +2}{\frac{\sin h}{h}} \\ \\ \frac{\lim_{h \to 0} \frac{f(1+h) - f(1)}{h} - h + 2}{\lim_{h \to 0} \frac{\sin h}{h}} = \frac{f'(1) + 2}{1} = \fbox{4}$

$\\ $B)$ \ \lim_{x \to 0} \frac{f(2- \cos x) - f(1)}{x^2} \\ h = 1-\cos (x) \\ \lim_{ h\to 0} \frac{f(1+h) - f(1)}{(\cos^{-1}(1-h))^2} = f'(1) \underbrace{\lim_{ h \to 0} \frac{h}{(\cos^{-1} (1-h))^2}}_{L} = 2 \cdot \frac{1}{2} = \fbox{1} \\ \\ \cos z = 1-h \\ \Rightarrow \ L = \lim_{z \to 0} \frac{1-\cos z}{z^2} = \lim_{z \to 0} \left(\frac{sin z}{z} \right)^2 \lim_{z \to 0} \frac{1}{1+\cos z} = 1^2 \cdot \frac{1}{1+1} = \frac{1}{2}$

Sy123 · May 8, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$\\ $The numbers$ \ -1 \leq x_1, x_2, \dots, x_{n-1}, x_n \leq 1 \ \ $satisfy$ \ \ \sum_{i=1}^n x_i^3 = 0 \\ \\ $Prove that$ \ \ \sum_{i=1}^n x_i \leq \frac{n}{3}$

TL1998 · May 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

given that a>0, and b>0,
prove that Integral(a->0) ln(x+1)dx + Integral(b->0) (e^x-1)dx is greater or equal to ab

Sy123 · May 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
This is a good question

$\\ $Let$ \ f(x) \ $be a function on$ \ a \leq x \leq b \ $and differentiable on$ \ a < x < b, \ $and also $ \ f' \ $is non-constant$ \\ \\ $Prove that there exists some$ \ a \leq c \leq b \ $such that$ \\ \\ |f'(c)| > \left|\frac{f(b) - f(a)}{b-a} \right|$

$\\ $Since$ \ f' \ $is non-constant, there must be some real$ \ c \in (a,b) \ $so that$ \ f'(c) \geq f'(x) \ $or$ \ f'(c) \leq f'(x) \ $for$ \ x \in (a,b) \\ \\ $By comparing areas$ \\ \Rightarrow \ \left|\int_a^b f'(c) \ dx\right| \geq \left| \int_a^b f'(x) \ dx \right| \\ \\ \Rightarrow \ f'(c) |b-a| \geq |f(b) - f(a)| \\ \\ \Rightarrow \ f'(c) \geq \left| \frac{f(b) - f(a)}{b-a} \right|$

Sy123 said:
$\\ $The numbers$ \ -1 \leq x_1, x_2, \dots, x_{n-1}, x_n \leq 1 \ \ $satisfy$ \ \ \sum_{i=1}^n x_i^3 = 0 \\ \\ $Prove that$ \ \ \sum_{i=1}^n x_i \leq \frac{n}{3}$

$\\ $Let$ \ y_i = 1 + x_i \ \Rightarrow \ 0 \leq y_{i} \leq 2 \\ \\ \sum_{i=1}^n (y_i-1)^3 = 0 \\ \\ \Rightarrow \ (*) \sum_{i=1}^n (y_i^3 - 3y_i^2 + 3y_i) = n \\ \\ $Prove$ \ \sum y_i \leq \frac{4n}{3} \\ \\ \frac{4}{3} \times (*) \Rightarrow \frac{4}{3} \sum_{i=1}^n (y_i^3 - 3y_i^2 + 3y_i) = \frac{4}{3}n \\ \\ $Show that$ \ f(x) = \frac{4}{3} (x^3 - 3x^2 + 3x) -x\geq 0 \ $for$ \ 0 \leq x \leq 2 \\ \\ f'(x) = 4x^2 - 8x + 3 > 0 \ $as$ \ \triangle = 16 > 0 \\ \\ \therefore f \ $is monotonic inc$ \\ \\ f(x) \geq f(0) \\ f(x) \geq 0 \\ \therefore \ \frac{4}{3}(x^3 - 3x^2 + 3x) \geq x \\ $Substitute$ \ x = y_i \ $sum both sides from 1 to n$ \\ \\ \Rightarrow \ \sum y_i \leq \frac{4}{3}n \\ \\ \Rightarrow \ \sum (y_i-1) \leq \frac{n}{3} \\ \\ \Rightarrow \ \sum x_i \leq \frac{n}{3}$

Sy123 · May 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$\\ $Find the minimum value of the function$ \ f(x) = \int_0^{\frac{\pi}{4}} |\tan t - x| \ dt$

TL1998 · May 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ $A)$ \ \lim_{x \to 1} \frac{f(x) - x^2 f(1)}{\sin (x-1)} \\ h=x-1 \\ \Rightarrow \ \lim_{h \to 0} \frac{f(1+h) - (1+h)^2}{\sin h} = \lim_{h \to 0} \frac{\frac{f(1+h) - 1}{h} - h +2}{\frac{\sin h}{h}} \\ \\ \frac{\lim_{h \to 0} \frac{f(1+h) - f(1)}{h} - h + 2}{\lim_{h \to 0} \frac{\sin h}{h}} = \frac{f'(1) + 2}{1} = \fbox{4}$

$\\ $B)$ \ \lim_{x \to 0} \frac{f(2- \cos x) - f(1)}{x^2} \\ h = 1-\cos (x) \\ \lim_{ h\to 0} \frac{f(1+h) - f(1)}{(\cos^{-1}(1-h))^2} = f'(1) \underbrace{\lim_{ h \to 0} \frac{h}{(\cos^{-1} (1-h))^2}}_{L} = 2 \cdot \frac{1}{2} = \fbox{1} \\ \\ \cos z = 1-h \\ \Rightarrow \ L = \lim_{z \to 0} \frac{1-\cos z}{z^2} = \lim_{z \to 0} \left(\frac{sin z}{z} \right)^2 \lim_{z \to 0} \frac{1}{1+\cos z} = 1^2 \cdot \frac{1}{1+1} = \frac{1}{2}$

BTw sy you made a silly in your response to my previous question, part A.

you wrote -h+2 but it should be -h-2. But it doesn't really matter

seanieg89 · May 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ $Since$ \ f' \ $is non-constant, there must be some real$ \ c \in (a,b) \ $so that$ \ f'(c) \geq f'(x) \ $or$ \ f'(c) \leq f'(x) \ $for$ \ x \in (a,b) \\ \\ $By comparing areas$ \\ \Rightarrow \ \left|\int_a^b f'(c) \ dx\right| \geq \left| \int_a^b f'(x) \ dx \right| \\ \\ \Rightarrow \ f'(c) |b-a| \geq |f(b) - f(a)| \\ \\ \Rightarrow \ f'(c) \geq \left| \frac{f(b) - f(a)}{b-a} \right|$

I don't really buy this solution...

Is the first sentence saying that f' must attain a maximum or minimum on the interval [a,b]? Why is this true if we don't know whether or not f' is continuous?

The work that follows also seems to depend on signs a bit, though depending on what exactly you mean by your first line this might be okay.

Sy123 · May 12, 2014

Re: HSC 2014 4U Marathon - Advanced Level

An attempt at a more formal solution:

$\\ f \ $is differentiable on$ \ (a,b) \ $using this fact, prove that$ \ f' \ $is continuous on$ \ (a,b) \\ \\ $For$ \ f' \ $to be continuous on$ \ (a,b) \ $then$ \\ \\ \lim_{x \to c} f'(x) = f'( c) \ $for all$ \ c \in (a,b) \\ \\ \lim_{x \to c} \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \ $the inner limit exists as$ \ f \ $is differentiable on$ \ (a,b) \\ \\ h = x-c \\ \\ \Rightarrow \ \lim_{h \to 0} \frac{f(h + (h+c)) - f(h+c)}{h} = \lim_{h \to 0} f'(h+c) = f'(c)$

$\therefore \ \lim_{x \to c} f'(x) = f'(c) \ $for all$ \ c \in (a,b) \\ \\ $Due to the Extreme Value Theorem, if a maximum value of the function$ \ f' \ $on$ \ (a,b) \ $exists, then it exists inside the interval$ \ (a,b) \ $i.e. there is some$ \ f'(c) \geq f'(x) \ $true for some$ \ c \in (a,b) \\ \\ $Implying$ \ \int_a^b f'(c) \ dx \geq \int_a^b f'(x) \ dx \\ \\ $However, if a maximum does not exist in the interval$ \ (a,b) \ $for$ \ f' \ $then either$ \\ \\ L_b = \lim_{x \to b} f'(x) \geq f'(x) \ $or$ \ L_a = \lim_{x \to a} f'(x) \geq f'(x) \\ \\ \Rightarrow \ \int_a^bL_b \ dx > \int_a^bf'(x) \ dx \ $since$ \ f' \ $is non-constant$ \\ \\ \therefore \ L_b (b-a) > f(b) - f(a) \$

$$Since the inequality is strict, by bringing$ \ x \ $arbitrarily close to$ \ b \ $we can still choose some$ \ c \ $so that the strict inequality still holds, meaning further that$ \ f'(c) (b-a) > f(b) - f(a)$

HSC 2014 MX2 Marathon ADVANCED (archive) (2 Viewers)

Active Member

Well-Known Member

True Fail Kid

Well-Known Member

Well-Known Member

Active Member

Well-Known Member

Active Member

Well-Known Member

This too shall pass

what is that?It is Cowpea

Member

This too shall pass

This too shall pass

Member

This too shall pass

This too shall pass

Member

Well-Known Member

This too shall pass

Users Who Are Viewing This Thread (Users: 0, Guests: 2)