HSC 2013 MX2 Marathon (archive) (2 Viewers)

asianese · May 20, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
$LHS=nx_i^2=RHS$

.

hahaha stop mocking his typo

Sy123 · May 20, 2013

Re: HSC 2013 4U Marathon

This is a little easier:

$Prove$

$\sum_{k=0}^{n} \binom{n}{k} \cos(k \theta) = 2^n \cos^n \left(\frac{\theta}{2} \right ) \cos \left(\frac{n\theta}{2} \right) \ \ \ \ \fbox{3}$

seanieg89 · May 21, 2013

Re: HSC 2013 4U Marathon

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

$I:=\int_{-1}^1 x^2e^{-x^2}\, dx \\ \\ $Let $u=x^2.\\ \\ \Rightarrow I=\int_1^1 \frac{xe^{-u}\, du}{2}= \int_1^1 \frac{e^{-u}\sqrt{u}}{2}\, du\\ \\ $as $u>0$ in our interval of integration. Hence $I$, having been rewritten as an integral over an interval of length $0$, must be equal to $0.$

seanieg89 · May 21, 2013

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 · May 21, 2013

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 · May 21, 2013

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 · May 21, 2013

Re: HSC 2013 4U Marathon

GoldyOrNugget said:
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 · May 21, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
Find the error (if any) in the following working:

$I:=\int_{-1}^1 x^2e^{-x^2}\, dx \\ \\ $Let $u=x^2.\\ \\ \Rightarrow I=\int_1^1 \frac{xe^{-u}\, du}{2}= \int_1^1 \frac{e^{-u}\sqrt{u}}{2}\, du\\ \\ $as $u>0$ in our interval of integration. Hence $I$, having been rewritten as an integral over an interval of length $0$, must be equal to $0.$

$Upon the substitution$

$u=x^2$

$The positive square root is taken of both sides (as evident in the working)$

$\sqrt{u} = |x|$

$\sqrt{u} = \begin{cases} x \ \ x\geq 0 \\ -x \ \ x<0 \end{cases}$

$By definition$

$Hence the proper manipulation would be$

$I = \int_{-1}^{1} x^2 e^{-x^2} \ dx = \int_{-1}^{0} x^2 e^{-x^2} \ dx + \int_0^1 x^2 e^{-x^2}$

$$Upon the substitution$ \ \ u=x^2$

$I = \int_1^0 \frac{-\sqrt{u} e^{-u} }{2} \ du + \int_0^1 \frac{\sqrt{u}e^{-u}}{2} \ du = 2\int_0^1 \frac{\sqrt{u}e^{-u}}{2} \ du$

EDIT: Done. Is this correct?

Sy123 · May 21, 2013

Re: HSC 2013 4U Marathon

GoldyOrNugget said:
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.

seanieg89 · May 21, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$Upon the substitution$

$u=x^2$

$The positive square root is taken of both sides (as evident in the working)$

$\sqrt{u} = |x|$

$\sqrt{u} = \begin{cases} x \ \ x\geq 0 \\ -x \ \ x<0 \end{cases}$

$By definition$

$Hence the proper manipulation would be$

$I = \int_{-1}^{1} x^2 e^{-x^2} \ dx = \int_{-1}^{0} x^2 e^{-x^2} \ dx + \int_0^1 x^2 e^{-x^2}$

$$Upon the substitution$ \ \ u=x^2$

$I = \int_1^0 \frac{-\sqrt{u} e^{-u} }{2} \ du + \int_0^1 \frac{\sqrt{u}e^{-u}}{2} \ du = 2\int_0^1 \frac{\sqrt{u}e^{-u}}{2} \ du$

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 · May 21, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
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 · May 21, 2013

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 · May 21, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
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 · May 22, 2013

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 · May 22, 2013

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)

Sy123 · May 23, 2013

Re: HSC 2013 4U Marathon

OMGITzJustin said:
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: $4 \cdot 26^2$

The amount of words starting with E, middle of A, B, C, D, E or F is $26 \cdot 6$

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

So final answer:

$\fbox{2866}$

======

$$What is the value of the sum$ \ \ \ \sum_{k=0}^{60} \sin^2 (3k)^{\circ}$

$$A) 31$

$$B) 30$

$$C) 32$

$D) An irrational value$

Sy123 · May 23, 2013

Re: HSC 2013 4U Marathon

GoldyOrNugget said:
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

hayabusaboston · May 23, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
The amount of words starting with A, B, C, D is: $4 \cdot 26^2$

The amount of words starting with E, middle of A, B, C, D, E or F is $26 \cdot 6$

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

So final answer:

$\fbox{2866}$

======

$$What is the value of the sum$ \ \ \ \sum_{k=0}^{60} \sin^2 (3k)^{\circ}$

$$A) 31$

$$B) 30$

$$C) 32$

$D) An irrational value$

Its the D

lolol

seanieg89 · May 23, 2013

Re: HSC 2013 4U Marathon

hayabusaboston said:
Its the D

lolol

no, its not.

braintic · May 23, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$$What is the value of the sum$ \ \ \ \sum_{k=0}^{60} \sin^2 (3k)^{\circ}$

$$A) 31$

$$B) 30$

$$C) 32$

$D) An irrational value$

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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