HSC 2015 MX1 Marathon (archive) (1 Viewer)

SpiralFlex · Apr 14, 2015

Re: HSC 2015 3U Marathon

InteGrand said:
$Can be many things probably.$

$One way is to make the denominator $D_n=10+20(n-1)=20n-10$ for $n=1,2,3,\cdots$, and to fit the numerators to a quadratic $N_n=an^2 + bn +c$.$

For n =3, you get 50.

Note: This can be done using knowledge of series/sequences in the 3U syllabus. No real fancy manipulation.

EDIT: Spoilers hardly work on the forum for some reason.

InteGrand · Apr 14, 2015

Re: HSC 2015 3U Marathon

SpiralFlex said:
For n =3, you get 50.

Note: This can be done using knowledge of series/sequences in the 3U syllabus. No real fancy manipulation.

EDIT: Spoilers hardly work on the forum for some reason.

Oops, misread the last denominator.

We can still fit them all to something of the form $an^2 + bn +c$ .

SpiralFlex · Apr 14, 2015

Re: HSC 2015 3U Marathon

InteGrand said:
Oops, misread the last denominator.

We can still fit them all to something of the form $an^2 + bn +c$ .

Yes the term on the top is indeed quadratic, I want a concrete values of a,b,c ^^

FrankXie · Apr 15, 2015

Re: HSC 2015 3U Marathon

only listing three terms, a pattern may not be clearly defined. like the denominators: 10, 30, 70, ... the next one could be 70+60=130, could also be 70+80=150, could even be something else

Carrotsticks · Apr 15, 2015

Re: HSC 2015 3U Marathon

Well, listing any finite number of terms may not clearly define a pattern.

SpiralFlex · Apr 15, 2015

Re: HSC 2015 3U Marathon

FrankXie said:
only listing three terms, a pattern may not be clearly defined. like the denominators: 10, 30, 70, ... the next one could be 70+60=130, could also be 70+80=150, could even be something else

Fair argument I'll give three more terms

Updated

Fizzy_Cyst · Apr 15, 2015

Re: HSC 2015 3U Marathon

Next value (I think) given below and hint for denominator and numerator

Next value will be 154/1270 -- hint for denominator 2^n -- hint for numerator is coefficient of n^2 is 7

FrankXie · Apr 15, 2015

Re: HSC 2015 3U Marathon

SpiralFlex said:
Fair argument I'll give three more terms

Updated

well, as Carrotsticks said, any finite number of terms may not uniquely determine a pattern.
but the formula $T_n=\frac{7n^2-5n}{20(2^n-1)}$ definitely gives one of the possible patterns

SpiralFlex · Apr 15, 2015

Re: HSC 2015 3U Marathon

FrankXie said:
well, as Carrotsticks said, any finite number of terms may not uniquely determine a pattern.
but the formula $T_n=\frac{7n^2-5n}{20(2^n-1)}$ definitely gives one of the possible patterns

Yes that's the form I was looking for

SpiralFlex · Apr 15, 2015

Re: HSC 2015 3U Marathon

Fizzy_Cyst said:
Next value (I think) given below and hint for denominator and numerator

Next value will be 154/1270 -- hint for denominator 2^n -- hint for numerator is coefficient of n^2 is 7

Yes

leehuan · Apr 16, 2015

Re: HSC 2015 3U Marathon

$A\quad particle\quad is\quad moving\quad in\quad a\quad straight\quad line\quad with\quad displacement\quad x\quad metres\quad at\quad time\quad t\quad seconds.\\ The\quad acceleration\quad of\quad the\quad particle\quad is\quad given:\\ \ddot { x } =\frac { x }{ 2 } ({ x }^{ 2 }-1)$

$\\ Initially,\quad the\quad particle\quad is\quad at\quad the\quad origin\quad moving\quad at\quad v=\frac { 1 }{ 2 } m{ s }^{ -1 }$

$Find\quad an\quad expression\quad for\quad the\quad displacement\quad of\quad the\quad particle\quad at\quad time\quad t,\quad given\\ \frac { 2 }{ 1-{ x }^{ 2 } } =\frac { 1 }{ 1+x } +\frac { 1 }{ 1-x }$

Drsoccerball · Apr 16, 2015

Re: HSC 2015 3U Marathon

leehuan said:
$A\quad particle\quad is\quad moving\quad in\quad a\quad straight\quad line\quad with\quad displacement\quad x\quad metres\quad at\quad time\quad t\quad seconds.\\ The\quad acceleration\quad of\quad the\quad particle\quad is\quad given:\\ \ddot { x } =\frac { x }{ 2 } ({ x }^{ 2 }-1)$

$\\ Initially,\quad the\quad particle\quad is\quad at\quad the\quad origin\quad moving\quad at\quad v=\frac { 1 }{ 2 } m{ s }^{ -1 }$

$Find\quad an\quad expression\quad for\quad the\quad displacement\quad of\quad the\quad particle\quad at\quad time\quad t,\quad given\\ \frac { 2 }{ 1-{ x }^{ 2 } } =\frac { 1 }{ 1+x } +\frac { 1 }{ 1-x }$

This is 2unit

braintic · Apr 16, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
This is 2unit

No it's not.

Crisium · Apr 16, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
This is 2unit

2 Unit Topic

3 Unit Difficulty

photastic · Apr 16, 2015

Re: HSC 2015 3U Marathon

Crisium said:
2 Unit Topic

3 Unit Difficulty

Its a 3 unit topic, 2 unit only know motion with time.

Crisium · Apr 16, 2015

Re: HSC 2015 3U Marathon

photastic said:
Its a 3 unit topic, 2 unit only know motion with time.

I haven't done the topic yet huehuehue, I was just assuming that was the case

Is it hard though?

photastic · Apr 16, 2015

Re: HSC 2015 3U Marathon

Crisium said:
I haven't done the topic yet huehuehue, I was just assuming that was the case

Is it hard though?

Same level as 2u motion.

Chris_S · Apr 22, 2015

Re: HSC 2015 3U Marathon

By using the substitution u²=x+2, find ∫(x-2)/√(x+2) dx

leehuan · Apr 22, 2015

Re: HSC 2015 3U Marathon

${ u }^{ 2 }=x+2\quad \Rightarrow \quad 2udu=dx\\ \int { \frac { x-2 }{ \sqrt { x+2 } } dx } \\ =\int { \frac { { u }^{ 2 }-4 }{ u } 2udu } \\ =2\int { \left( { u }^{ 2 }-4 \right) du } \\ =2\left( \frac { { u }^{ 3 } }{ 3 } -4u \right) +C\\ =\frac { 2\sqrt { { \left( x+2 \right) }^{ 3 } } }{ 3 } -8\sqrt { x+2 } +C$

New question, since it appears not many students have done 3u appns of calc.
$Find\quad an\quad inverse\quad function\quad for\quad f\left( x \right) ={ e }^{ x+\frac { 1 }{ x } }\quad for\quad x\ge 1$

InteGrand · Apr 22, 2015

Re: HSC 2015 3U Marathon

leehuan said:
${ u }^{ 2 }=x+2\quad \Rightarrow \quad 2udu=dx\\ \int { \frac { x-2 }{ \sqrt { x+2 } } dx } \\ =\int { \frac { { u }^{ 2 }-4 }{ u } 2udu } \\ =2\int { \left( { u }^{ 2 }-4 \right) du } \\ =2\left( \frac { { u }^{ 3 } }{ 3 } -4u \right) +C\\ =\frac { 2\sqrt { { \left( x+2 \right) }^{ 3 } } }{ 3 } -8\sqrt { x+2 } +C$

New question, since it appears not many students have done 3u appns of calc.
$Find\quad an\quad inverse\quad function\quad for\quad f\left( x \right) ={ e }^{ x+\frac { 1 }{ x } }\quad for\quad x>0$

$The function $f$ is not one-to-one on the interval $(0,\infty)$, so an inverse does not exist for this domain, although we can restrict the domain to make an inverse exist.$

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