HSC 2015 Maths Marathon (archive) (4 Viewers)

VBN2470 · Sep 25, 2015

Re: HSC 2015 2U Marathon

NEW QUESTION:

Find the exact value of $m$ for which the graphs $y=mx$ and $y=\dfrac{e^x}{x}$ have a common tangent at their point of contact.

Drsoccerball · Sep 25, 2015

Re: HSC 2015 2U Marathon

VBN2470 said:
NEW QUESTION:

Find the exact value of $m$ for which the graphs $y=mx$ and $y=\dfrac{e^x}{x}$ have a common tangent at their point of contact.

Hint: Only way they can have a common tangent is if the graph y=mx is the tangent.

BlueGas · Sep 25, 2015

Re: HSC 2015 2U Marathon

Drsoccerball said:
$$ Solve for $ x : (x-2)^{2x^2-8} =1$

rand_althor said:
If 2f(x)+3f(-x)=15-4x, find f(x).

Can someone show the working out for these two questions?

Flop21 · Sep 25, 2015

Re: HSC 2015 2U Marathon

Drsoccerball said:
$$ Solve for $ x : (x-2)^{2x^2-8} =1$

Can someone do the working out, I'm getting a different answer (-2 and 2).

Drsoccerball · Sep 25, 2015

Re: HSC 2015 2U Marathon

Flop21 said:
Can someone do the working out, I'm getting a different answer (-2 and 2).

If you substitute your answers you get 0^0 = 1 ... which is not true. Ill post my solution now

Drsoccerball · Sep 25, 2015

Re: HSC 2015 2U Marathon

$(x-2)^{2x^2-8} =1$

$(2x^2-8)ln(x-2) = 0$

$x= 3, -2 $ Since $ x $ can't $ equal to 2$
Should've done this part from the beginning :
$Now notice that the RHS=1 and the power of the LHS is even therefore inside the brackets can either = 1,-1$

$\therefore 1 $ is also a solution $$

$1,3, -2$

VBN2470 · Sep 25, 2015

Re: HSC 2015 2U Marathon

Basic Idea is $a^{0}=1$ for $a \neq{0}$ , $1^{n}=1$ for all integers $n$ and $(-1)^{n}=1$ for all even integers $n$ .

Pro12 · Sep 25, 2015

Re: HSC 2015 2U Marathon

I need exam tips pls tell me I fucking need it Appreciate LOLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLL

Drsoccerball · Sep 25, 2015

Re: HSC 2015 2U Marathon

BlueGas said:
Can someone show the working out for these two questions?

Usually i would do this :

$f(x) =a_n x^n +a_{n-1}x^{n-1} +...+a_1 x +a_0$

$$ However since it is linear $ f(x) =a_1x +a_0$

$5a_0 -a_1 x = 15-4x$

$a_1= 4, a_0=3$

$\therefore f(x)=3+4x$

Drsoccerball · Sep 25, 2015

Re: HSC 2015 2U Marathon

Pro12 said:
I need exam tips pls tell me I fucking need it Appreciate LOLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLL

Take it slowly and be careful of silly mistakes by checking straight after you do it. Also don't just stare at the question do something to trigger things in your brain. If you end up giving up write everything you know about the question on the paper you're bound to get some marks.

Drsoccerball · Sep 25, 2015

Re: HSC 2015 2U Marathon

Task 1 last question from my school when i was in 2 unit (plz non 2unit people give others a chance) :

$$ Let $ f(x) $ be a function that has a minimum stationary point at $ x=a.$

$$ Prove that the curve $ y= \frac{1}{f(x)} $ has a maximum stationary point at $ x=a$

astroman · Sep 25, 2015

Re: HSC 2015 2U Marathon

Pro12 said:
I need exam tips pls tell me I fucking need it Appreciate LOLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLL

Start from the first question, don't be tempted to go straight to the big mark questions, the exam is set from easy to hard, starting from easy will get ur brain prepped for the rapage that awaits you.

Drsoccerball · Sep 25, 2015

Re: HSC 2015 2U Marathon

astroman said:
Start from the first question, don't be tempted to go straight to the big mark questions, the exam is set from easy to hard, starting from easy will get ur brain prepped for the rapage that awaits you.

Not true. Question 15 tends to be the hardest while question 16 is similar in difficulty to question 14.

astroman · Sep 25, 2015

Re: HSC 2015 2U Marathon

Drsoccerball said:
Not true. Question 15 tends to be the hardest while question 16 is similar in difficulty to question 14.

same shit, u get what i'm trying to say.

keepLooking · Sep 25, 2015

Re: HSC 2015 2U Marathon

Drsoccerball said:
Task 1 last question from my school when i was in 2 unit (plz non 2unit people give others a chance) :

$$ Let $ f(x) $ be a function that has a minimum stationary point at $ x=a.$

$$ Prove that the curve $ y= \frac{1}{f(x)} $ has a maximum stationary point at $ x=a$

I knew that was familiar, I found it in the BOS trials. The trial paper I thoroughly struggled in.

Drsoccerball · Sep 25, 2015

Re: HSC 2015 2U Marathon

keepLooking said:
I knew that was familiar, I found it in the BOS trials. The trial paper I thoroughly struggled in.

No wayyyyyyy my teacher took it from the BOS trials :L ?....
I thought he made it

Drsoccerball · Sep 25, 2015

Re: HSC 2015 2U Marathon

Drsoccerball said:
Task 1 last question from my school when i was in 2 unit (plz non 2unit people give others a chance) :

$$ Let $ f(x) $ be a function that has a minimum stationary point at $ x=a.$

$$ Prove that the curve $ y= \frac{1}{f(x)} $ has a maximum stationary point at $ x=a$

$f'(a) = 0, f''(a) > 0$

$g(x) =\frac{1}{f(x)}$

$g'(x) = \frac{-f'(x)}{(f(x))^2}$

$g'(a)= 0$

$\therefore $ S.P at $ x=a$

$g''(x)= \frac{-f''(x)f(x) +2f'(x)}{f(x)^3}$

$g''(a)=\frac{-f''(a)}{f(a)^2}$

$\therefore g''(a) <0$

$\therefore $ Max S.P at $ x=a$

[]

kawaiipotato · Sep 25, 2015

Re: HSC 2015 2U Marathon

Drsoccerball said:
Usually i would do this :

$f(x) =a_n x^n +a_{n-1}x^{n-1} +...+a_1 x +a_0$

$$ However since it is linear $ f(x) =a_1x +a_0$

$5a_0 -a_1 x = 15-4x$

$a_1= 4, a_0=3$

$\therefore f(x)=3+4x$

How did you know f(x) was linear? It could have been something like a cubic and the x^3 could just cancel each other out
edit: nevermind, the coefficients are different so there's no way x^3 could cancel out. Wondering if there's an alternative method to this question though. Tried doing x --> -x giving me f(x) + f(-x) = 6 but that's it

Drsoccerball · Sep 25, 2015

Re: HSC 2015 2U Marathon

kawaiipotato said:
How did you know f(x) was linear? It could have been something like a cubic and the x^3 could just cancel each other out
edit: nevermind, the coefficients are different so there's no way x^3 could cancel out. Wondering if there's an alternative method to this question though. Tried doing x --> -x giving me f(x) + f(-x) = 6 but that's it

If they had the same coefficients then as i said i'd do a series like i stated

rand_althor · Sep 25, 2015

Re: HSC 2015 2U Marathon

kawaiipotato said:
Wondering if there's an alternative method to this question though. Tried doing x --> -x giving me f(x) + f(-x) = 6 but that's it

$2f(x)+3f(-x)=15-4x$ (1).\\ Let $x=-x \to 2f(-x)+3f(x)=15+4x$ (2). $\\$Now solve these equations simultaneously$.$

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