HSC 2014 MX2 Marathon ADVANCED (archive) (1 Viewer)

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)$

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