HSC 2015 MX1 Marathon (archive) (4 Viewers)

Drsoccerball · Feb 23, 2015

Re: HSC 2015 3U Marathon

iStudent said:
L'hopital's?

That's kind of wrong, since your denominator "x" approaches 0 too.

Even though not in the syllabus:
$\frac{cosx -1}{x} $ differenciate top and bottom $$

$\frac{-sinx}{1} Lim_{x \to 0} =0 []$

FrankXie · Feb 23, 2015

Re: HSC 2015 3U Marathon

the question was asking you prove the derivative of sine is cosine, but you actually admit that the derivative of cosine is sine when you use l'hospital rule. this is obviously not allowed.
a method inside the scope of high school is: using double angle $1-\cos x=2\sin^2\frac{x}{2}, \lim\frac{1-\cos x}{x}=\lim\frac{\sin \frac{x}{2}}{\frac{x}{2}}\times\lim\sin\frac{x}{2}=1\times0=0$
meanwhile, from the beginning why not use difference to product $\sin(x+h)-\sin x=2\cos\frac{2x+h}{2}\sin\frac{h}{2}, \lim\frac{\sin(x+h)-\sin x}{h}=\lim\frac{\sin\frac{h}{2}}{\frac{h}{2}}\times\lim\cos(x+\frac{h}{2})=1\times\cos x=\cos x$

Trebla · Feb 23, 2015

HSC 2015 3U Marathon

Isn't the proof directly in the syllabus document?

braintic · Feb 24, 2015

Re: HSC 2015 3U Marathon

FrankXie said:
the question was asking you prove the derivative of sine is cosine, but you actually admit that the derivative of cosine is sine when you use l'hospital rule. this is obviously not allowed.
a method inside the scope of high school is: using double angle $1-\cos x=2\sin^2\frac{x}{2}, \lim\frac{1-\cos x}{x}=\lim\frac{\sin \frac{x}{2}}{\frac{x}{2}}\times\lim\sin\frac{x}{2}=1\times0=0$
meanwhile, from the beginning why not use difference to product $\sin(x+h)-\sin x=2\cos\frac{2x+h}{2}\sin\frac{h}{2}, \lim\frac{\sin(x+h)-\sin x}{h}=\lim\frac{\sin\frac{h}{2}}{\frac{h}{2}}\times\lim\cos(x+\frac{h}{2})=1\times\cos x=\cos x$

The proof comes out much more nicely when you use an alternative formula for the derivative:

View attachment Alt derivation of deriv of sinx.pdf

RealiseNothing · Feb 24, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
$$ as $ x \to 0 ,cos x \to 1 \therefore 1-1=0$

$\frac{0}{0}= $???$$

Drsoccerball · Feb 24, 2015

Re: HSC 2015 3U Marathon

RealiseNothing said:
$\frac{0}{0}= $???$$

Yeah its wrong but look at the other proof i used

SammyT123 · Feb 24, 2015

Re: HSC 2015 3U Marathon

How do I sketch y=logx/x

When I get to the y' step, I let

U = logx

Du=1/x dx

Y'=udu

This is somehow wrong :/

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FrankXie · Feb 24, 2015

Re: HSC 2015 3U Marathon

SammyT123 said:
How do I sketch y=logx/x
When I get to the y' step, I let
U = logx
Du=1/x dx
Y'=udu
This is somehow wrong :/

i don't know what were you doing, but did you try quotient rule because y is a quotient of two functions

SammyT123 · Feb 24, 2015

HSC 2015 3U Marathon

FrankXie said:
i don't know what were you doing, but did you try quotient rule because y is a quotient of two functions

I can do it with the quotient rule. But what is wrong when I substitute u=logx

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Drsoccerball · Feb 24, 2015

Re: HSC 2015 3U Marathon

SammyT123 said:
How do I sketch y=logx/x

When I get to the y' step, I let

U = logx

Du=1/x dx

Y'=udu

This is somehow wrong :/

Sent from my iPhone using Tapatalk

Like mentioned in the 4u marathon you have to sketch logx and x and divide the ordinates of logx by x. However if you're still keen to use calculus :
$\frac{logx=u}{x=v} u'=\frac{1}{x} v'=1$

$\therefore y'= \frac{1-logx}{x^2} $ Find stationary point, find asymptotes sub in values close to the asymptote to get general idea... $$

SammyT123 · Feb 24, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
Like mentioned in the 4u marathon you have to sketch logx and x and divide the ordinates of logx by x. However if you're still keen to use calculus :
$\frac{logx=u}{x=v} u'=\frac{1}{x} v'=1$

$\therefore y'= \frac{1-logx}{x^2} $ Find stationary point, find asymptotes sub in values close to the asymptote to get general idea... $$

Thanks so much for that

Just wondering , what exactly did i do wrong when I used "u substitution" . This qs was in a school exam and I didn't have the answers at the time, so I couldn't just change my method to quotient rule for no reason unless I knew this was wrong.

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Silly Sausage · Feb 24, 2015

Re: HSC 2015 3U Marathon

SammyT123 said:
Thanks so much for that

Just wondering , what exactly did i do wrong when I used "u substitution" . This qs was in a school exam and I didn't have the answers at the time, so I couldn't just change my method to quotient rule for no reason unless I knew this was wrong.

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Should be $\frac{du}{dx}=\frac{1}{x}$ not $du=\frac{1}{x}dx$ . Don't mistake this for u substitution which is for integration, the "u" used for the quotient rule is just a variable in the general form for the QR $\frac{u'v-v'u}{v^2}$ .

SammyT123 · Feb 24, 2015

Re: HSC 2015 3U Marathon

Silly Sausage said:
Should be $\frac{du}{dx}=\frac{1}{x}$ not $du=\frac{1}{x}dx$ . Don't mistake this for u substitution which is for integration, the "u" used for the quotient rule is just a variable in the general form for the QR $\frac{u'v-v'u}{v^2}$ .

That clears everything . You can't use u substitution for differentiation, only integration. Thanks

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Sy123 · Feb 26, 2015

Re: HSC 2015 3U Marathon

$\\ $Show that$ \ \binom{k}{k} + \binom{k+1}{k} + \dots + \binom{n-1}{k} = \binom{n}{k+1} \ $for$ \ 0 \leq k \leq (n-1) \ $for integers$ \ k,n$

authenticity · Feb 26, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
A group consists of 9 people. These 9 people are to be seated around a circular table. The group of 9 people consists of 4 students 3 teachers and 2 parents. How many possible arrangements are there if all the students sit together, all the teachers sit together and none of the students sit next to the teachers?

For this Q, what's the answer?

laters · Feb 26, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
A group consists of 9 people. These 9 people are to be seated around a circular table. The group of 9 people consists of 4 students 3 teachers and 2 parents. How many possible arrangements are there if all the students sit together, all the teachers sit together and none of the students sit next to the teachers?

authenticity said:
For this Q, what's the answer?

I got 4! x 3! x 2 = 288 arrangements?

Ekman · Feb 26, 2015

Re: HSC 2015 3U Marathon

laters said:
I got 4! x 3! x 2 = 288 arrangements?

Correct!

Drsoccerball · Feb 26, 2015

Re: HSC 2015 3U Marathon

Sy123 said:
$\\ $Show that$ \ \binom{k}{k} + \binom{k+1}{k} + \dots + \binom{n-1}{k} = \binom{n}{k+1} \ $for$ \ 0 \leq k \leq (n-1) \ $for integers$ \ k,n$

Simplify each expresion and use induction right?

Sy123 · Feb 26, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
Simplify each expresion and use induction right?

How would you simplify the LHS?

Drsoccerball · Feb 26, 2015

Re: HSC 2015 3U Marathon

Sy123 said:
How would you simplify the LHS?

Its becomes:
$1+ k+1 + \frac{(k+2)(k+1)}{2!} +... \frac{(k+n)(k+n-1)}{n!}$

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