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HSC 2013 MX2 Marathon (archive) (1 Viewer)

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Sy123

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Re: HSC 2013 4U Marathon

This is a little easier:




 

seanieg89

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Re: HSC 2013 4U Marathon

Find the error (if any) in the following working:

 
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seanieg89

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Re: HSC 2013 4U Marathon

True or false: a function that is differentiable over the interval [a,b] is integrable over the interval [a,b].
 

seanieg89

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Re: HSC 2013 4U Marathon

Prove that odd functions have even derivatives and vice versa.

Is it possible for a function that is NOT odd to have an even derivative?
 

GoldyOrNugget

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Re: HSC 2013 4U Marathon

If a function is odd, then f(x)=-f(-x). Differentiating both sides, we get f'(x) = f'(-x) therefore the derivative is even.

If a function is even, then f(x)=f(-x). Differentiating both sides, we get f'(x) = -f'(-x) therefore the derivative is odd.

Yes. f(x)=k, k!=0 is one example. f(x)=x+1 is another.
 

seanieg89

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Re: HSC 2013 4U Marathon

If a function is odd, then f(x)=-f(-x). Differentiating both sides, we get f'(x) = f'(-x) therefore the derivative is even.

If a function is even, then f(x)=f(-x). Differentiating both sides, we get f'(x) = -f'(-x) therefore the derivative is odd.

Yes. f(x)=k, k!=0 is one example. f(x)=x+1 is another.
Good :).
 

Sy123

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Re: HSC 2013 4U Marathon

If a function is odd, then f(x)=-f(-x). Differentiating both sides, we get f'(x) = f'(-x) therefore the derivative is even.

If a function is even, then f(x)=f(-x). Differentiating both sides, we get f'(x) = -f'(-x) therefore the derivative is odd.

Yes. f(x)=k, k!=0 is one example. f(x)=x+1 is another.
Indeed any odd function plus a (non zero) constant is an example of a non-odd function with an even derivative.
 
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seanieg89

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Re: HSC 2013 4U Marathon






















EDIT: Done. Is this correct?
Essentially. The main point is that in the changes of variables theorem you need a 1-1 correspondence between your old variable and your new variable...a point often glossed over by teachers.
 

Sy123

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Re: HSC 2013 4U Marathon

Essentially. The main point is that in the changes of variables theorem you need a 1-1 correspondence between your old variable and your new variable...a point often glossed over by teachers.
Can you explain this what you mean by 1-1 correspondence?
I haven't heard of this rule with substitution.
 

seanieg89

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Re: HSC 2013 4U Marathon

A bijection (wikipedia the term).

The problem here is that as x ranges from -1 to 1, u hits every value in (0,1] TWICE instead of once. You resolved this by splitting the integral into two intervals on which u doesn't have this behaviour.

(Also, naively looking at what u is at the endpoints of the integration in x and making the substitution, using these u-values as the new endpoints is flawed for this reason. The mapping from x->u takes the interval [-1,1] to the interval [0,1], not to the single number 1.)
 

Sy123

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Re: HSC 2013 4U Marathon

A bijection (wikipedia the term).

The problem here is that as x ranges from -1 to 1, u hits every value in (0,1] TWICE instead of once. You resolved this by splitting the integral into two intervals on which u doesn't have this behaviour.

(Also, naively looking at what u is at the endpoints of the integration in x and making the substitution, using these u-values as the new endpoints is flawed for this reason. The mapping from x->u takes the interval [-1,1] to the interval [0,1], not to the single number 1.)
Ok yep I get it.
 

OMGITzJustin

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Re: HSC 2013 4U Marathon

We wish to construct three-letter words from 26 letter of the english alphabet. How many possibilities are there if:
The word must come before EGG in alphabetical order?
 

GoldyOrNugget

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Re: HSC 2013 4U Marathon

Consider 3-letter string starting with A, B, C, D. There are 4*26^2=2704. There are 26 for each EA, EB, EC, ED, EF, so 2704+26*5=2834. Then there's EGA,EGB,...,EGG, 7 of these. So 2841 in total.

(edit: assuming no counting mistake)
 
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Sy123

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Re: HSC 2013 4U Marathon

We wish to construct three-letter words from 26 letter of the english alphabet. How many possibilities are there if:
The word must come before EGG in alphabetical order?
The amount of words starting with A, B, C, D is:

The amount of words starting with E, middle of A, B, C, D, E or F is

The amount of words starting with E, middle of G before EGG is 6.

So final answer:



======









 

Sy123

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Re: HSC 2013 4U Marathon

Consider 3-letter string starting with A, B, C, D. There are 4*26^2=2704. There are 26 for each EA, EB, EC, ED, EF, so 2704+26*5=2834. Then there's EGA,EGB,...,EGG, 7 of these. So 2841 in total.

(edit: assuming no counting mistake)
You are missing EE :p
 

braintic

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Re: HSC 2013 4U Marathon

Since sin^2(0) = cos^2 (90), sin^2 (3) = cos^2 (87), sin^2 (6) = cos^2 (84), sin^2 (93) = cos^2 (177), ...
Then the sum of sin^2 is equal to the sum of cos^2.

But the sum of the (sum of sin^2 and cos^2) is the sum of (1) which is 60.
So the answer is 30.
 
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