HSC 2013 MX2 Marathon (archive) (6 Viewers)

asianese · Apr 29, 2013

Re: HSC 2013 4U Marathon

AHAHHAA L0L

;explode;

Sy123 · Apr 29, 2013

Re: HSC 2013 4U Marathon

$Evaluate$

$I=\int_0^{\frac{\pi}{2}} \frac{\sin^2 x}{1+\sin(2x)} \ dx$

HeroicPandas · Apr 29, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$Find$

$\int_{-9001}^{9001} \sin(x^3) e^{-\pi^{77} x^{14}} \tan^{-1}(\ln(x^2)) (x^{191919923134}-1) \ dx$

That is an odd question..... wth..

Sy123 said:
$Evaluate$

$I=\int_0^{\frac{\pi}{2}} \frac{\sin^2 x}{1+\sin(2x)} \ dx$

sorry i made a mistake...

continuing from where i was...

$I = \frac{1}{2} \int_{0}^{\pi/2} \frac{dx}{(sinx + cosx)^2} = \frac{1}{2} \int_{0}^{\pi/2} \frac{dx}{(\sqrt{2}sin(x + \pi/4))^2} \\ = \frac{1}{4} \int_{0}^{\pi/2}\frac{dx}{sin^2 (x+\pi/4)} \\ = \frac{1}{4} \int_{0}^{\pi/2} cosec^2 (x+\pi/4)dx \\ =\frac{1}{4}\left [ -cot(x+ \pi/4) \right ]_0 ^{\pi/2} \\ = -\frac{1}{4}\left [ cot(\frac{3\pi}{4}) - cot(\frac{\pi}{4}) \right ] \\ = -\frac{1}{4} \left ( -1 - 1 \right ) \\ = \frac{1}{2}$

Sy123 · Apr 29, 2013

Re: HSC 2013 4U Marathon

HeroicPandas said:
That is an odd question..... wth..

use the property that $\int_{0}^{a}f(x)dx = \int_{0}^{a}f(a-x)dx$

$I = \int \frac{sin^2 x dx}{(sinx + cosx)^2} \\ $Use the property \\ \therefore I= $\int \frac{cos^2 x dx}{(cosx + sinx)^2} \\ \\ 2I = \int \frac{dx}{(sinx + cosx)^2} \\ I = \frac{1}{2} \int \frac{dx}{1 + 2sinxcosx} \\ \therefore I = \frac{1}{2} . \frac{sinx}{cosx + sinx} + C ($by using t = tan(x/2), but too lazy to post it up...)$

YEp something like that, nice work.

Carrotsticks · Apr 29, 2013

Re: HSC 2013 4U Marathon

HeroicPandas said:
That is an odd question..... wth..

i c wut u did thar

Drongoski · Apr 30, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
YEp something like that, nice work.

I tried out the t-method as suggested but did not end up with a nice expression. Maybe I made a mistake along the way. But we can integrate without the t-method this way:

$\int \frac {1}{1+2sinxcosx} dx = \int \frac {sec^2 x}{sec^2 x (1+2sinxcosx)} dx = \int \frac {dtanx}{sec^2 x + 2tan x} \\ \\ = \int \frac {dtan x}{tan^2 x + 2tan x + 1} = \int \frac {1}{(1+tanx)^2}d(1+tanx) = - \frac {1}{1+tanx} + C$

$= - \frac {cos x}{sin x + cos x} + C$ if you prefer.

seanieg89 · Apr 30, 2013

Re: HSC 2013 4U Marathon

Let a < b be two consecutive roots of a polynomial P(x), (so P is nonzero between a and b). Suppose these roots are each of multiplicity 1, so that:

P(x)=(x-a)(x-b)g(x)

for some polynomial g(x) with g(a),g(b) != 0 by the factor theorem.

1. Prove that g(a) and g(b) have the same sign.

2. Prove there is a real number c between a and b with f'(c)=0.

3. Prove that (2) is true even if we drop the assumption that the roots a and b are of multiplicity 1.

4. Draw a picture and give a geometric description of this result.

(This is a special case of the Mean Value theorem, one of the most important theorems in differential calculus. The general proof of the MVT is beyond the scope of MX2, but we know enough about polynomials to be able to prove this special case.)

Sy123 · Apr 30, 2013

Re: HSC 2013 4U Marathon

$i) Prove that$

$\lim_{z \to 0} \int_z^{\pi/2}\ln(\sin x) \ dx = \lim_{z \to 0} \frac{1}{2}\int_z^{\pi/2} \ln(\sin(2x)) \ dx - \frac{\pi}{4} \ln 2$

$ii) Explain why$

$\lim_{z \to 0} \int_z^{\pi/2} \ln(\sin(2x)) \ dx = \frac{1}{2} \lim_{z \to 0} \int_z^{\pi} \ln(\sin x) \ dx$

$iii) Hence evaluate$

$I = \lim_{z \to 0} \int_z^{\pi/2} \ln(\sin x) \ dx$

Sy123 · Apr 30, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
Let a < b be two consecutive roots of a polynomial P(x), (so P is nonzero between a and b). Suppose these roots are each of multiplicity 1, so that:

P(x)=(x-a)(x-b)g(x)

for some polynomial g(x) with g(a),g(b) != 0 by the factor theorem.

1. Prove that g(a) and g(b) have the same sign.

2. Prove there is a real number c between a and b with f'(c)=0.

3. Prove that (2) is true even if we drop the assumption that the roots a and b are of multiplicity 1.

4. Draw a picture and give a geometric description of this result.

(This is a special case of the Mean Value theorem, one of the most important theorems in differential calculus. The general proof of the MVT is beyond the scope of MX2, but we know enough about polynomials to be able to prove this special case.)

1.

$P'(x) = (2x-(a+b))g(x) + g'(x) (x-a)(x-b)$

If a and b are consecutive roots, then it is easy to see that P'(a) and P'(b) are opposite in sign by drawing a picture. They can't be zero since they have roots of multiplicity of 1.

$P'(a) = (a-b) g(a)$

$P'(b) = (b-a) g(b)$

Since a < b, let us denote a sign (positive or negative) as (1) and the other as (2)

So IF P'(a) has sign (1), and since P'(a) = (a-b) g(a), since a < b then g(a) has sign (2)
Since P'(b) is opposite in sign, it will have sign (2)

IF P'(b) has sign (2), since b-a > 0, then g(b) has sign (2)

Therefore g(a) and g(b) have the same sign.

2.

(I assume you mean P'(c) = 0 )

Since P'(a) and P'(b) have OPPOSITE gradient, therefore there must exist a point (since P is a polynomial and is therefore continuous) that P'(c) is zero.

3.

IF the multiplicity of a and/or b are greater than 1, then the P'(a) and/or P'(b) are zero. But that doesn't matter since for some value, P'(a+) (some value to the right of a), will either be positive or negative, therefore the polynomial is either increasing or decreasing.
No matter what the orientation, the curve must curve back to zero as b is a root as well. Since it must curve back there must exist some point where the curve has zero gradient. (easier explained by a picture).

4.

*coming soon*

======

seanieg89 · Apr 30, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
1.

$P'(x) = (2x-(a+b))g(x) + g'(x) (x-a)(x-b)$

If a and b are consecutive roots, then it is easy to see that P'(a) and P'(b) are opposite in sign by drawing a picture. They can't be zero since they have roots of multiplicity of 1.

$P'(a) = (a-b) g(a)$

$P'(b) = (b-a) g(b)$

Since a < b, let us denote a sign (positive or negative) as (1) and the other as (2)

So IF P'(a) has sign (1), and since P'(a) = (a-b) g(a), since a < b then g(a) has sign (2)
Since P'(b) is opposite in sign, it will have sign (2)

IF P'(b) has sign (2), since b-a > 0, then g(b) has sign (2)

Therefore g(a) and g(b) have the same sign.

2.

(I assume you mean P'(c) = 0 )

Since P'(a) and P'(b) have OPPOSITE gradient, therefore there must exist a point (since P is a polynomial and is therefore continuous) that P'(c) is zero.

3.

IF the multiplicity of a and/or b are greater than 1, then the P'(a) and/or P'(b) are zero. But that doesn't matter since for some value, P'(a+) (some value to the right of a), will either be positive or negative, therefore the polynomial is either increasing or decreasing.
No matter what the orientation, the curve must curve back to zero as b is a root as well. Since it must curve back there must exist some point where the curve has zero gradient. (easier explained by a picture).

4.

*coming soon*

======

Good, although your answer to 3. is more of a plausibility argument than a proof. The point is to do it in a similar way to to 2, using that a continuous function can only change sign on an interval if it possesses a root on this interval.

HeroicPandas · Apr 30, 2013

Re: HSC 2013 4U Marathon

Drongoski said:
I tried out the t-method as suggested but did not end up with a nice expression. Maybe I made a mistake along the way. But we can integrate without the t-method this way:

$\int \frac {1}{1+2sinxcosx} dx = \int \frac {sec^2 x}{sex^2 x (1+2sinxcosx)} dx \\ \\ = \int \frac {1}{(1+tanx)^2}d(1+tanx) = - \frac {1}{1+tanx} + C$

$= - \frac {cos x}{sin x + cos x} + C$ if you prefer.

wow! very nice! thank u for sharing

study1234 · Apr 30, 2013

Re: HSC 2013 4U Marathon

Sy123 · May 1, 2013

Re: HSC 2013 4U Marathon

$The polynomial$ \ \ P(x) \ \ $has integer co-efficients$

$p+\sqrt{q} \ \ $is a root of the polynomial, where$ \ \ p, q\ \ $are rational$$

$Prove that$ \ \ p-\sqrt{q} \ \ $must also be a root of the polynomial$

seanieg89 · May 1, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$The polynomial$ \ \ P(x) \ \ $has integer co-efficients$

$p+\sqrt{q} \ \ $is a root of the polynomial, where$ \ \ p, q\ \ $are rational$$

$Prove that$ \ \ p-\sqrt{q} \ \ $must also be a root of the polynomial$

(You must also assume that q is not the square of a rational number.)

seanieg89 · May 1, 2013

Re: HSC 2013 4U Marathon

Am interested in seeing peoples solutions actually. This is a v.good question because it is not very technically demanding at all, but if you don't properly understand the relevant part of the MX2 course you probably won't have any chance of solving it. ie it slay rote learners.

Sy123 · May 1, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
(You must also assume that q is not the square of a rational number.)

Ah yeah oops

Sy123 · May 1, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$The polynomial$ \ \ P(x) \ \ $has integer co-efficients$

$p+\sqrt{q} \ \ $is a root of the polynomial, where$ \ \ p, q\ \ $are rational and q is not the square of a rational$$

$Prove that$ \ \ p-\sqrt{q} \ \ $must also be a root of the polynomial$

Fixed.

RealiseNothing · May 1, 2013

Re: HSC 2013 4U Marathon

You'd also have to note the polynomial is not monic. Though I guess this is a consequence of $p+\sqrt{q}$ being a root.

seanieg89 · May 1, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
You'd also have to note the polynomial is not monic. Though I guess this is a consequence of $p+\sqrt{q}$ being a root.

Huh? The result is still true for monic polynomials...and having a root of the form p+q^{1/2} is perfectly possible for monic polynomials.

RealiseNothing · May 1, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
Huh? The result is still true for monic polynomials...and having a root of the form p+q^{1/2} is perfectly possible for monic polynomials.

Nevermind I derped and thought the root was rational after reading "p and q are rational" without considering the square root.

HSC 2013 MX2 Marathon (archive) (6 Viewers)

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