HSC 2015 MX1 Marathon (archive) (3 Viewers)

laters · Jun 11, 2015

Re: HSC 2015 3U Marathon

InteGrand said:
Maybe I've misinterpreted the Q or screwed up some working out, but I'm not sure why the answer is $\binom{n-1}{k-1}\binom{10}{k}$ .

Say we had k = 2 (so two different numbers from 0-9 used), and n = 3 (three-digit password).

Let's just say for now the two distinct digits used were 1 and 2. Since n = 3, these are the possibilities: 112, 121, 211, 221, 212, 122. That's 6 possibilities.

But we could do the same for all possible $\binom{10}{2}$ choices of two distinct digits. So it seems like the total possible ways for n = 3, k = 2 is: $6\binom{10}{2}$ . But this doesn't agree with the answer, since $6\neq \binom{3-1}{2-1}=2$ .

Edit: My bad, Q said combinations, not permutations

Does this mean the question was only asking for the number of combinations but not the number of passwords that could be tried?

InteGrand · Jun 11, 2015

Re: HSC 2015 3U Marathon

laters said:
Does this mean the question was only asking for the number of combinations but not the number of passwords that could be tried?

Yes, so basically the number of possible COMBINATIONS of k distinct digits from 0-9, given that a total of n digits are used.

Sy123 · Jun 18, 2015

Re: HSC 2015 3U Marathon

$\\ $i) Expand using the Binomial Theorem$ \ \left(1 + \frac{1}{k+1} \right)^{k+1} \ $for positive integer$ \ k \ $and prove that$ \ \left(1 + \frac{1}{k+1} \right)^{k+1} > 2 \\ \\ $ii) Hence or otherwise, prove using mathematical induction$ \ \left(\frac{n+1}{2} \right)^n \geq n! \ $for integers$ \ n \geq 1$

leehuan · Jun 19, 2015

Re: HSC 2015 3U Marathon

Dodgy (i)
$\\ { \left( 1+\frac { 1 }{ k+1 } \right) }^{ k+1 }\\ =\left( \begin{matrix} k+1 \\ 0 \end{matrix} \right) +\left( \begin{matrix} k+1 \\ 1 \end{matrix} \right) \frac { 1 }{ k+1 } +\left( \begin{matrix} k+1 \\ 2 \end{matrix} \right) { \left( \frac { 1 }{ k+1 } \right) }^{ 2 }+\dots +\left( \begin{matrix} k+1 \\ k+1 \end{matrix} \right) { \left( \frac { 1 }{ k+1 } \right) }^{ k+1 }\\ We\quad note\quad that\quad \frac { d }{ dk } { \left( 1+\frac { 1 }{ k+1 } \right) }^{ k+1 }=\frac { 1 }{ k+1 } { \left( 1+\frac { 1 }{ k+1 } \right) }^{ k }\left[ \left( k+2 \right) \log { \left( \frac { 1 }{ k+1 } +1 \right) } -1 \right] \\ (Method:\quad Convert\quad into\quad { e }^{ \log { f(x) } }\quad and\quad use\quad product\quad \& \quad chain\quad rule)$

$\\ When\quad x>0,\quad we\quad note\quad that\quad all\quad terms\quad are\quad positive\quad or\quad equal\quad to\quad 0\\ (Equality\quad holding\quad when\quad x\rightarrow \infty )\\ Hence\quad \frac { d }{ dk } { \left( 1+\frac { 1 }{ k+1 } \right) }^{ k+1 }>0\quad for\quad k>0\quad so\quad { \left( 1+\frac { 1 }{ k+1 } \right) }^{ k+1 }\quad increases\quad for\quad k>0\\ Considering\quad the\quad lowest\quad possible\quad value\quad for\quad k,\quad k=1:\\ { \left( 1+\frac { 1 }{ k+1 } \right) }^{ k+1 }={ \left( 1+\frac { 1 }{ 1+1 } \right) }^{ 1+1 }=\frac { 9 }{ 4 } >2\\ Hence,\quad it\quad follows\quad that\quad { \left( 1+\frac { 1 }{ k+1 } \right) }^{ k+1 }>2\quad for\quad all\quad k=2\quad (value\quad will\quad continue\quad to\quad increase)$

Sy123 · Jun 19, 2015

Re: HSC 2015 3U Marathon

leehuan said:
Dodgy (i)
$\\ { \left( 1+\frac { 1 }{ k+1 } \right) }^{ k+1 }\\ =\left( \begin{matrix} k+1 \\ 0 \end{matrix} \right) +\left( \begin{matrix} k+1 \\ 1 \end{matrix} \right) \frac { 1 }{ k+1 } +\left( \begin{matrix} k+1 \\ 2 \end{matrix} \right) { \left( \frac { 1 }{ k+1 } \right) }^{ 2 }+\dots +\left( \begin{matrix} k+1 \\ k+1 \end{matrix} \right) { \left( \frac { 1 }{ k+1 } \right) }^{ k+1 }\\ We\quad note\quad that\quad \frac { d }{ dk } { \left( 1+\frac { 1 }{ k+1 } \right) }^{ k+1 }=\frac { 1 }{ k+1 } { \left( 1+\frac { 1 }{ k+1 } \right) }^{ k }\left[ \left( k+2 \right) \log { \left( \frac { 1 }{ k+1 } +1 \right) } -1 \right] \\ (Method:\quad Convert\quad into\quad { e }^{ \log { f(x) } }\quad and\quad use\quad product\quad \& \quad chain\quad rule)$

$\\ When\quad x>0,\quad we\quad note\quad that\quad all\quad terms\quad are\quad positive\quad or\quad equal\quad to\quad 0\\ (Equality\quad holding\quad when\quad x\rightarrow \infty )\\ Hence\quad \frac { d }{ dk } { \left( 1+\frac { 1 }{ k+1 } \right) }^{ k+1 }>0\quad for\quad k>0\quad so\quad { \left( 1+\frac { 1 }{ k+1 } \right) }^{ k+1 }\quad increases\quad for\quad k>0\\ Considering\quad the\quad lowest\quad possible\quad value\quad for\quad k,\quad k=1:\\ { \left( 1+\frac { 1 }{ k+1 } \right) }^{ k+1 }={ \left( 1+\frac { 1 }{ 1+1 } \right) }^{ 1+1 }=\frac { 9 }{ 4 } >2\\ Hence,\quad it\quad follows\quad that\quad { \left( 1+\frac { 1 }{ k+1 } \right) }^{ k+1 }>2\quad for\quad all\quad k=2\quad (value\quad will\quad continue\quad to\quad increase)$

That works (if you did your differentiation correctly), but I would first consider the function f(x) = (1 + 1/(x+1))^{x+1} where x is among the reals, and differentiate that since you can't differentiate over the integers!

The method I was looking for is:

$\left(1 + \frac{1}{k+1} \right)^{k+1} = \binom{k+1}{0} + \binom{k+1}{1} \frac{1}{k+1} + \dots + \binom{k+1}{k+1} \frac{1}{(k+1)^{k+1}}$

$\\ \binom{k+1}{0} = 1 \\ \\ \binom{k+1}{1}\frac{1}{k+1} = (k+1) \cdot \frac{1}{k+1} = 1$

$\\ \therefore \left(1 + \frac{1}{k+1} \right)^{k+1} = 1 + 1 + C = 2 + C \ $(for some$ \ C \ $greater than zero (it is the remaining terms of the series))$$

$\\ \therefore \left(1 + \frac{1}{k+1} \right)^{k+1}= 2 + C > 2$

leehuan · Jun 19, 2015

Re: HSC 2015 3U Marathon

Oh right! I had considered the (k+1 0) component to equal 1 but I overlooked (k+1 1) = (k+1) which cancels out with the 1/(k+1). I was thinking that differentiating was a flaw considering that Z is not continuous.

leehuan · Jun 19, 2015

Re: HSC 2015 3U Marathon

$(ii)\quad When\quad n=1\\ LHS={ \left( \frac { 1+1 }{ 2 } \right) }^{ 1 }\\ =1\\ =1!\\ \ge RHS\\ Hence\quad the\quad statement\quad is\quad true\quad for\quad n=1\\ Assuming\quad true\quad for\quad n=k\quad gives\\ { \left( \frac { k+1 }{ 2 } \right) }^{ k }\ge k!\quad \quad [k\ge 1]\\ \frac { { \left( k+1 \right) }^{ k } }{ { 2 }^{ k } } \ge k!\quad$

$\\ When\quad n=k+1\\ RTP:\quad { \left( \frac { k+2 }{ 2 } \right) }^{ k+1 }\ge \left( k+1 \right) !\\ From\quad part\quad (i)\\ { \left( \frac { k+2 }{ k+1 } \right) }^{ k+1 }>2\\ \therefore { \left( k+2 \right) }^{ k+1 }>2{ \left( k+1 \right) }^{ k+1 }\\ From\quad assumption\\ { \left( k+1 \right) }^{ k }\ge { 2 }^{ k }k!\\ \therefore 2{ \left( k+1 \right) }^{ k+1 }\ge { 2 }^{ k+1 }{ \left( k+1 \right) }!\\ \therefore { \left( k+2 \right) }^{ k+1 }\ge { 2 }^{ k+1 }{ \left( k+1 \right) }!$

$\therefore { \left( \frac { k+2 }{ 2 } \right) }^{ k+1 }\ge \left( k+1 \right) !\\ \therefore The\quad statement\quad is\quad true\quad for\quad n=k+1\quad if\quad it's\quad true\quad for\quad n=k\\ \therefore Since\quad the\quad statement\quad is\quad true\quad for\quad n=1,\quad it\quad is\quad true\quad for\quad n=2,3,\dots \\ i.e.\quad by\quad the\quad principle\quad of\quad mathematical\quad induction\quad it\quad is\quad true\quad for\quad all\quad integers\quad n\ge 1$

psyc1011 · Jun 19, 2015

Re: HSC 2015 3U Marathon

dunjaaa said:
Probability question View attachment 32069

What is the most likely number of defective shirts PER BOX?

Sy123 · Jun 19, 2015

Re: HSC 2015 3U Marathon

davidgoes4wce · Jun 21, 2015

Re: HSC 2015 3U Marathon

(2-3n)^9 Find the coefficient of the 3rd term in the expansion.

My working out:

9C2 x 3^7 x (2n)^2 = 314,298 n^2

could someone confirm with me?

leehuan · Jun 21, 2015

Re: HSC 2015 3U Marathon

$General\quad term\quad of\quad { \left( 2-3n \right) }^{ 9 }\quad is\quad given\\ { T }_{ k+1 }=\left( \begin{matrix} 9 \\ k \end{matrix} \right) { 2 }^{ 9-k }{ \left( -3n \right) }^{ k }\\ So\quad for\quad the\quad third\quad term\quad we\quad have\\ { T }_{ 3 }=\left( \begin{matrix} 9 \\ 2 \end{matrix} \right) { 2 }^{ 7 }{ \left( -3n \right) }^{ 2 }$

leehuan · Jun 21, 2015

Re: HSC 2015 3U Marathon

If nobody does this projectile motion question, I'll go through it one step at a time...

(i)
$\ddot { x } =0\\ \dot { x } =V\cos { \theta } \\ x=Vt\cos { \theta } \\ \ddot { y } =-g\\ \dot { y } =V\sin { \theta } -gt\\ y=Vt\sin { \theta } -\frac { g{ t }^{ 2 } }{ 2 }$

Remember to actually show x' = C and y = -1/2 gt^2 + Vtsin(theta) + C etc in the exam

leehuan · Jun 22, 2015

Re: HSC 2015 3U Marathon

(ii)
$\\ t=\frac { x }{ V\cos { \theta } } \\ Sub\quad into\quad y=f(t)\quad to\quad get\\ y=x\tan { \theta } -\frac { g{ x }^{ 2 } }{ 2{ V }^{ 2 } } \sec ^{ 2 }{ \theta } \\ As\quad the\quad particle\quad passes\quad \left( d\cos { \alpha } ,d\sin { \alpha } \right) \\ d\sin { \alpha } =d\cos { \alpha } \tan { \theta } -\frac { g{ d }^{ 2 }\cos ^{ 2 }{ \alpha } }{ 2{ V }^{ 2 } } \sec ^{ 2 }{ \theta }$

$\\ \frac { g{ d }\cos ^{ 2 }{ \alpha } }{ 2{ V }^{ 2 } } \sec ^{ 2 }{ \theta } =\cos { \alpha } \tan { \theta } -\sin { \alpha } \\ d=\frac { 2{ V }^{ 2 }\cos ^{ 2 }{ \theta } }{ g\cos ^{ 2 }{ \alpha } } \left( \cos { \alpha } \tan { \theta } -\sin { \alpha } \right) \\ d=\frac { 2{ V }^{ 2 } }{ g\cos { \alpha } } \left( \sin { \theta } \cos { \theta } -\cos ^{ 2 }{ \theta } \tan { \alpha } \right) \\$

leehuan · Jun 24, 2015

Re: HSC 2015 3U Marathon

(iii)
$\\ d=\frac { 2{ V }^{ 2 } }{ g\cos { \alpha } } \left( \frac { 1 }{ 2 } \sin { 2\theta } -\tan { \alpha } \left( \frac { 1 }{ 2 } +\frac { 1 }{ 2 } \cos { 2\theta } \right) \right) \\ \frac { dd }{ d\theta } =\frac { 2{ V }^{ 2 } }{ g\cos { \alpha } } \left( \cos { 2\theta } +\sin { 2\theta } \tan { \alpha } \right) \\ $To maximise $ d $, let $ \frac { dd }{ d\theta } =0\\ \frac { 2{ V }^{ 2 } }{ g\cos { \alpha } } \left( \cos { 2\theta } +\sin { 2\theta } \tan { \alpha } \right) =0$

$\\ \cos { 2\theta } \cos { \alpha } +\sin { 2\theta } \sin { \alpha } =0\\ \cos { \left( 2\theta -\alpha \right) } =0\\ 2\theta -\alpha =\frac { \pi }{ 2 } \\ \theta =\frac { \pi }{ 4 } +\frac { \alpha }{ 2 } \\ \therefore $ when $\theta =\frac { \pi }{ 4 } +\frac { \alpha }{ 2 } $ , $d$ is maximised.$

I couldn't be bothered doing a second derivative test..

leehuan · Jun 25, 2015

Re: HSC 2015 3U Marathon

$\\ $We have $ \\ { d }_{ max }=\frac { 2{ V }^{ 2 } }{ g\cos { \alpha } } \left( \frac { 1 }{ 2 } \sin { 2\theta } -\tan { \alpha } \left( \frac { 1 }{ 2 } +\frac { 1 }{ 2 } \cos { 2\theta } \right) \right) \\ =\frac { { V }^{ 2 } }{ g\cos { \alpha } } \left( \sin { \left( \frac { \pi }{ 2 } +\alpha \right) } -\tan { \alpha } \left( 1+\cos { \left( \frac { \pi }{ 2 } +\alpha \right) } \right) \right) \\ =\frac { { V }^{ 2 } }{ g\cos { \alpha } } \left( \cos { \alpha } -\tan { \alpha } \left( 1-\sin { \alpha } \right) \right)$

$\\ =\frac { { V }^{ 2 } }{ g\cos { \alpha } } \left( \frac { \cos ^{ 2 }{ \alpha } -\sin { \alpha } +\sin ^{ 2 }{ \alpha } }{ \cos { \alpha } } \right) \\ =\frac { { V }^{ 2 } }{ g } \left( \frac { 1-\sin { \alpha } }{ \cos ^{ 2 }{ \alpha } } \right) \\ =\frac { { V }^{ 2 } }{ g } \left( \frac { 1-\sin { \alpha } }{ 1-\sin ^{ 2 }{ \alpha } } \right) \\ =\frac { { V }^{ 2 } }{ g\left( 1+\sin { \alpha } \right) }$

$\\ $Q.E.D.$$

davidgoes4wce · Jun 26, 2015

Re: HSC 2015 3U Marathon

This has been doing my head in for an hour . Find the greatest coefficient of (4x-3y)^9

I want to do it the Tk+1/Tk method

leehuan · Jun 26, 2015

Re: HSC 2015 3U Marathon

(iv)
$\\ In\quad the\quad expansion\quad of\quad { \left( 4x-3y \right) }^{ 9 }\\ { T }_{ k+1 }=\left( \begin{matrix} 9 \\ k \end{matrix} \right) { \left( 4x \right) }^{ 9-k }{ \left( -3y \right) }^{ k }\\ { T }_{ k }=\left( \begin{matrix} 9 \\ k-1 \end{matrix} \right) { \left( 4x \right) }^{ 10-k }{ \left( -3y \right) }^{ k-1 }$

$\\ Comparing\quad magnitude\quad (absolute\quad value)\quad of\quad coefficients:\\ \frac { { T }_{ k+1 } }{ { T }_{ k } } =\frac { \left( \begin{matrix} 9 \\ k \end{matrix} \right) { 4 }^{ 9-k }{ 3 }^{ k } }{ \left( \begin{matrix} 9 \\ k-1 \end{matrix} \right) { 4 }^{ 10-k }{ 3 }^{ k-1 } } \\ =\frac { \frac { 9! }{ k!\left( 9-k \right) ! } { 4 }^{ 9-k }{ 3 }^{ k-1 } }{ \frac { 9! }{ \left( k-1 \right) !\left( 10-k! \right) } { 4 }^{ 9-k }{ 3 }^{ k-1 } } \times \frac { 3 }{ 4 } \\ =\frac { \frac { 1 }{ k\left( k-1 \right) !\left( 9-k \right) ! } }{ \frac { 1 }{ \left( k-1 \right) !\left( 10-k \right) \left( 9-k \right) ! } } \times \frac { 3 }{ 4 } \\ =\frac { 3\left( 10-k \right) }{ 4k } \\ =\frac { 30-3k }{ 4k }$

$\\ For\quad the\quad coefficient\quad of\quad { T }_{ k+1 }>{ T }_{ k }\\ i.e.\quad the\quad coefficient\quad of\quad \frac { { T }_{ k+1 } }{ { T }_{ k } } >1\\ i.e.\quad \frac { 30-3k }{ 4k } >1\\ 30-3k>4k\quad (since\quad k>0\quad here)\\ 7k<30\\ k<4\frac { 2 }{ 7 } \\ Hence,\quad when\quad k=1,2,3,4,\quad { T }_{ k+1 }>{ T }_{ k }\\ and\quad when\quad k=5,6,7,8,9,\quad { T }_{ k+1 }<{ T }_{ k }\\ So\quad the\quad greatest\quad coefficient\quad occurs\quad when\quad k=4,\quad i.e.\quad { T }_{ 5 }\\ So\quad the\quad greatest\quad coefficient\quad is\quad \left( \begin{matrix} 9 \\ 4 \end{matrix} \right) { 4 }^{ 9-4 }{ \left( -3 \right) }^{ 4 }=10450944$

davidgoes4wce · Jun 26, 2015

Re: HSC 2015 3U Marathon

many thanks for the answer lee huan.

One question I see in your solution, you write the solution (9-k!), should it be (9-k)! ?

davidgoes4wce · Jun 26, 2015

Re: HSC 2015 3U Marathon

I think the thing that confused me initially with this question (4x-3y)^9, was I was treating the (-3y) as the variable substituted in the equation, when I should have used the 'magnitude' absolute value.

Thanks for clearing this up leehuan.

davidgoes4wce · Jun 27, 2015

Re: HSC 2015 3U Marathon

Find the coefficient of x^2 in (x+(1/x))^3 (x+1)^5

i get an answer of 55 but the solution says 92.0

HSC 2015 MX1 Marathon (archive) (3 Viewers)

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