4 Unit Revising Marathon HSC '10 (3 Viewers)

hscishard · Apr 17, 2010

:O
What a way of saying that's wrong.

Lukybear · Apr 17, 2010

For sqrt(k+1) <= (8k + 5 )/6
Not really sure if its an applicable method:

Solving via quadratics:
$k = \frac{-44 \pm \sqrt (44^2 - 4 . (-11) . 64)}{128}\\ k = 0.1948 or -0.88$
Via graph and observation, k > 0.1948 for
hence sqrt(k+1) <= (8k + 5 )/6 for all n>=1

Using Induction
Assume True for n = k
$\sqrt 1 + \sqrt 2 + \sqrt 3 + ... + \sqrt k \leq \frac{(4k+1)}{6}. (k+1)\\$

Prove True for n= k+1
$\sqrt 1 + \sqrt 2 + \sqrt 3 + ... + \sqrt k + \sqrt (k+1) \leq \frac{4k^2+9k+5}{6}\\ = \frac{4k^2 + k}{6} + \frac{8k+5}{6}$

From assumption and proven above, i.e. via quadratics induction is true...

Flying_Mangina · Apr 17, 2010

intgrate x dx from limits of 1 to 2

adomad · Apr 17, 2010

$\int_{1}^{2}x.dx = \left [\frac{x^2}{2} \right ]^2 _1 = 1.5$

Question:
$$find the real numbers A and B so that$ \frac{A}{1+i}+\frac{B}{1+2i} =1$

omgmynameislong · Apr 17, 2010

a(1+2i)+b(1+i)=(1+i)(1+2i)

a+b+i(2a+b)=-1+3i

a+b=-1
2a+b=3

a=4
b=-5

richman92 · Apr 17, 2010

A and B complex or real ?

Lukybear · Apr 17, 2010

richman92 said:
A and B complex or real ?

thats interesting. Can they be complex?

nikkifc · Apr 17, 2010

nonsenseTM said:
why is that induction a 4u question?

Because it's not a standard induction question. It's from the 1990 4U HSC Paper. It was in Q8 I believe.

adomad · Apr 18, 2010

richman92 said:
A and B complex or real ?

real

adomad · Apr 18, 2010

omgmynameislong said:
a(1+2i)+b(1+i)=(1+i)(1+2i)

a+b+i(2a+b)=-1+3i

a+b=-1
2a+b=3

a=4
b=-5

ur supposed to post a new question now

omgmynameislong · Apr 18, 2010

integrate xlnx

cutemouse · Apr 18, 2010

omgmynameislong said:
integrate xlnx

lnx * (x^2)/2 - (x^2)/4

Find the general solutions of:

$cos4\theta+cos2\theta = \sqrt{2}cos^2\theta + \frac{1}{\sqrt{2}}sin2\theta$

Lukybear · Apr 19, 2010

any one?

shaon0 · Apr 19, 2010

Lukybear said:
any one?

Apparently not.

Trebla · Apr 19, 2010

Not completely confident but:

$\cos 4\theta + \cos 2\theta = \sqrt{2}\cos^2\theta +\dfrac{1}{\sqrt{2}}\sin 2\theta\\\\\cos (3\theta+\theta)+ \cos (3\theta-\theta) = \dfrac{2\cos^2\theta + \sin 2\theta}{\sqrt{2}}\\\\2\cos3\theta\cos\theta=\dfrac{2\cos^2\theta + 2\sin \theta\cos\theta}{\sqrt{2}}\\\\\cos\theta(\cos3\theta-\dfrac{1}{\sqrt{2}}\cos\theta-\dfrac{1}{\sqrt{2}}\sin\theta)=0\\\\\cos\theta(\cos3\theta-\cos\dfrac{\pi}{4}\cos\theta-\sin\dfrac{\pi}{4}\sin\theta)=0\\\\\cos\theta\left (\cos3\theta-\cos\left ( \theta-\frac{\pi}{4} \right ) \right )=0\\\\\cos\theta = 0 $ or $ \cos 3\theta=\cos \left( \theta-\frac{\pi}{4} \right )\\\\\therefore\theta = 2\pi n\pm\dfrac{\pi}{2} $ for integer $n$ when $\cos\theta = 0\\\\3\theta = 2k\pi\pm \left( \theta-\frac{\pi}{4} \right ) $ for integer $k$ when $\cos 3\theta=\cos \left( \theta-\frac{\pi}{4} \right )\\\\\Rightarrow 2\theta = 2k\pi- \dfrac{\pi}{4} $ or $4\theta = 2k\pi+ \dfrac{\pi}{4}\\\\$

$\therefore \theta = k\pi- \dfrac{\pi}{8} $ or $\theta = \dfrac{k\pi}{2}+ \dfrac{\pi}{16}$

cutemouse · Apr 20, 2010

Trebla said:
Not completely confident but:

$\cos 4\theta + \cos 2\theta = \sqrt{2}\cos^2\theta +\dfrac{1}{\sqrt{2}}\sin 2\theta\\\\\cos (3\theta+\theta)+ \cos (3\theta-\theta) = \dfrac{2\cos^2\theta + \sin 2\theta}{\sqrt{2}}\\\\2\cos3\theta\cos\theta=\dfrac{2\cos^2\theta + 2\sin \theta\cos\theta}{\sqrt{2}}\\\\\cos\theta(\cos3\theta-\dfrac{1}{\sqrt{2}}\cos\theta-\dfrac{1}{\sqrt{2}}\sin\theta)=0\\\\\cos\theta(\cos3\theta-\cos\dfrac{\pi}{4}\cos\theta-\sin\dfrac{\pi}{4}\sin\theta)=0\\\\\cos\theta\left (\cos3\theta-\cos\left ( \theta-\frac{\pi}{4} \right ) \right )=0\\\\\cos\theta = 0 $ or $ \cos 3\theta=\cos \left( \theta-\frac{\pi}{4} \right )\\\\\therefore\theta = 2\pi n\pm\dfrac{\pi}{2} $ for integer $n$ when $\cos\theta = 0\\\\3\theta = 2k\pi\pm \left( \theta-\frac{\pi}{4} \right ) $ for integer $k$ when $\cos 3\theta=\cos \left( \theta-\frac{\pi}{4} \right )\\\\\Rightarrow 2\theta = 2k\pi- \dfrac{\pi}{4} $ or $4\theta = 2k\pi+ \dfrac{\pi}{4}\\\\$

$\therefore \theta = k\pi- \dfrac{\pi}{8} $ or $\theta = \dfrac{k\pi}{2}+ \dfrac{\pi}{16}$

That's correct. Good job!

Next question:

$\text{If } h(n)=n^4+6n^2+9 \\ \text{(i) Show that } h(n+2)-h(n)=8(n+1)(n^2+2n+5) \\ \text{(ii) Hence prove by mathematical induction that } h(n) \text{ is divisible by 8 if } n \text{ is an odd positive integer}$

Rezen · May 12, 2010

$h(n+2)-h(n)=(n+2)^4+6(n+2)^2+9-n^4-6n^2-9\\=[(n+2)^2-n^2][(n+2)^2+n^2]+6[(n+2)-n][(n+2)+n] \\=(4n+4)(2n^2+4n+4)+12(2n+2)\\=8(n+1)[(n^2+2n+2)+3]\\=8(n+1)(n^2+2n+5)$

Now that took alot of time. Too lazy to do the induction part, but its fairly simple.

4 Unit Revising Marathon HSC '10 (3 Viewers)

hscishard

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adomad

HSC!!

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HSC!!

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