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#### dcc

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##### SUPER-FUN-HAPPY-MATHS-TIME
« on: April 11, 2009, 11:50:19 am »
+2
Maths:

1.)  Find $\dfrac{\sin(5x)}{\sin(x)}$ in terms of $\cos$.

Solution 1 - Over9000
Solution 2 - dcc

2.) Show that $\arctan \left(2\right) = \arctan \left(\frac{3}{4}\right) + \arctan \left(\frac{1}{2}\right)$.

Solution 1 - TrueTears
Solution 2 - Neobeo
Solution 3 - dcc

3.) Show $\int_0^\pi e^{-2t} \sin^3(t) \ dt = \frac{48}{65} \left( \dfrac{e^{2\pi} + 1}{e^{2\pi} - 1} \right) \int_0^{\pi} e^{-2t} \sin^2 (t) \ dt$

4.) Show that $\dfrac{\sin(ax)}{\sin(x)} = \dfrac{\sin\left(x\left(a + 1\right)\right) + \sin\left(x\left(a - 1\right)\right)}{\sin(2x)}$ (or perhaps the even more general result for $\dfrac{\sin(ax)}{\sin(bx)}$).

Solution 1 - /0

5.) For a real number $a$, evaluate $\int_0^a |x(x - a^2)|\ dx$.

Solution 1 - coblin

6.) Find the minimum area of the part bounded by the parabola $y=a^3x^2-a^4x\ (a>0)$ and the line $y = x$.
(Source: 1963 Tokyo Metropolitan University entrance exam)

Solution 1 - Neobeo
Solution 2 - TrueTears

7.) Evaluate $\int_0^1 \frac{1}{1+e^x}\ dx$
(Source: 2008 Miyazaki University entrance exam/Agriculture)

Solution 1 - kamil9876

8.) Evaluate $\,\int_0^1\;\frac{x-1}{\ln\,x}\;dx\,$
(Source: Wikipedia)

Solution 1 - TrueTears
Solution 2 - dcc

9.) If you break a stick into 3 pieces what is the probability that the 3 pieces can form a triangle?
(Source: Neobeo)

Solution 1 - /0
Solution 2 - kamil9876

10.) Show (without calculus) that the minimum value of $f(x) = 4x^2 + \dfrac{64}{x} + 17\: (x > 0)$ is $65$.

Solution 1 - Damo17

11.) Find the maximum value of $a(x) = 3f(x) + 4g(x) + 10h(x)$ given that $f(x)^2 + g(x)^2 + h(x)^2 \leq 9$.

Solution 1 - humph

12.) Prove that $\log_{2} 5$ is irrational.
(Source: TrueTears)

Solution 1 - Over9000

13.) Show that $0.9\dot{9} = 1$.
(Source: Damo17)

Solution 1 - dcc
Solution 2 - Over9000
Solution 3 - golden

14.) Let N be the positive integer with 2008 decimal digits, all of them 1. That is, $N = 1111...1111$, with 2008 occurrences of the digit 1. Find the 1005th digit after the decimal point in the decimal expansion of $\sqrt{N}$.
(Source: /0 - Melbourne University/BHP Billiton Maths Competition 2008)

Solution 1 - Over9000

15.) Neobeo is walking around in Luna Park, and notices an alleyway called 'Infinite Ice Cream'.  Neobeo notes that the 'Infinite Ice-Cream' appears to possess an infinitely large number of people selling ice-cream.  Upon walking outside any particular shop, Neobeo feels a huge compulsion to purchase an ice-cream.  For every shop that Neobeo visits, he is $37\%$ less likely to purchase an ice-cream then at the previous shop.  After purchasing an ice-cream, Neobeo leaves Luna Park.  What is the probability of Neobeo purchasing an ice-cream at the second shop in 'Infinite Ice Cream'?
(Source: IRC & the recesses of my brain)

Solution 1 - golden

16.) Find $\lim_{r \to 0} \dfrac{e^{rx} - 1}{r}$.

Solution 1 - TrueTears
Solution 2 - kamil9876
Solution 3 - dcc

17.) Find $\lim_{x \to 0}\frac {e^{x^2} - e^{3x}}{\sin(\frac {x^2}{2}) - \sin x}$

18.) Find $\lim_{x \to \infty} \frac{1}{x} \left(\dfrac{a^x + a^2 + a}{a - 1}\right)^{\frac{1}{x}}$.
Do not use L'Hospital's rule to evaluate this, because that is boring.  Try and use limit laws and properties, rather then mindless derivatives.

MORE TO COME, I WILL EDIT THIS POST.
« Last Edit: December 20, 2010, 12:46:10 pm by dcc »

#### enwiabe

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##### Re: SUPER-FUN-HAPPY-MATHS-TIME
« Reply #1 on: April 11, 2009, 12:10:07 pm »
0
Wow... those look like some good questions. Happy hunting!

#### TrueTears

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##### Re: SUPER-FUN-HAPPY-MATHS-TIME
« Reply #2 on: April 11, 2009, 01:57:22 pm »
0
2) I didn't use the information z = 15+ 30i but here's my working:

Let $tan^{-1}(\frac{3}{4}) = a$ and $tan^{-1}(\frac{1}{2}) = b$

$tan(a) = \frac{3}{4}$ and $tan(b) = \frac{1}{2}$

$tan(a+b) = 2$

$tan(a+b) = \frac{tan(a) + tan(b)}{1-tan(a)tan(b)} = \frac{\frac{3}{4} + \frac{1}{2}}{1 - \frac{3}{4} \times \frac{1}{2}} = 2$
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#### Over9000

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##### Re: SUPER-FUN-HAPPY-MATHS-TIME
« Reply #3 on: April 11, 2009, 03:12:49 pm »
0
This way is probabaly way too long to be acceptable but anyway, for question 1

$\frac{\sin5x}{sinx} = \frac{\sin(x+4x)}{sinx}$
$\frac{\sin(x+4x)}{sinx} = \frac{\sinx \cos4x + \cosx sin4x}{sinx}$
$\frac{\sinx \cos4x + \cosx \sin4x}{sinx}$ = $\cos4x + \frac{\cosx \sin4x}{sinx}$
lets focus on $\cosx \sin4x$ for now
$sin4x= 2\sin2x \cos2x$
$2\sin2x \cos2x= 4\sinx \cosx \times (2\cos^{2}x)$
$4\sinx \cosx \times (2\cos^{2}x)= 8\sinx \cos^{3}x - 4\sinx \cosx$
so $\sin4x = 8\sinx \cos^{3}x - 4\sinx \cosx$
so $\cosx \sin4x = 8\sinx \cos^{4}x - 4\sinx \cos^{2}x$
that means $\frac{\cosx \sin4x}{\sinx}$ = $8\cos^{4}x - 4\cos^{2}x$
$8\cos^{4}x - 4\cos^{2}x = 4(2\cos^{4}x - \cos^{2}x)$
$4(2\cos^{4}x - \cos^{2}x) = 4(\cos^{2}x(2\cos^{2}x -1)$
$4(\cos^{2}x(2\cos^{2}x -1) = 4(\cos^{2}x (\cos2x))$
$4(\cos^{2}x (\cos2x)) = 4(\frac{1}{2}\cos2x + \frac{1}{2} \times (\cos2x))$
$4(\frac{1}{2}\cos2x + \frac{1}{2} \times (\cos2x))$ = $4(\frac{1}{2}cos^{2}2x + \frac{1}{2}\cos2x)$
$4(\frac{1}{2}\cos^{2}2x + \frac{1}{2}\cos2x) = 2\cos^{2}2x + 2\cos2x$
however $2\cos^{2}2x = \cos4x +1$
so $2\cos^{2}2x + 2\cos2x = \cos4x +2\cos2x +1$
now we add the \cos4x we got all the way back at the top and we get
$2\cos4x +2\cos2x +1$
$\cos2x = 2cos^{2}x - 1$and $\cos4x = 2\cos^{2}2x -1$
so $2\cos4x +2\cos2x +1 = 4\cos^{2}2x -2 + 4\cos^{2}x - 2 + 1$
$4\cos^{2}2x -2 + 4\cos^{2}x - 2 + 1 = 4(2\cos^{2}x -1)^{2} + 4\cos^{2}x -3$
$4(2cos^{2}x -1)^{2} + 4cos^{2}x -3 = 4(4cos^{4}x - 4cos^{2}x + 1) + 4cos^{2}x -3$
$4(\4cos^{4}x - 4\cos^{2}x + 1) + 4\cos^{2}x -3 = 16\cos^{4}x - 16\cos^{2}x + 4 + 4\cos^{2}x -3$
$16\cos^{4}x - 16\cos^{2}x + 4 + 4\cos^{2}x -3 = 16\cos^{4}x -12\cos^{2}x +1$
Therefore to sum up $\frac{\sin5x}{sinx} = 16\cos^{4}x -12\cos^{2}x +1$
« Last Edit: April 11, 2009, 03:30:28 pm by Over9000 »
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#### dcc

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##### Re: SUPER-FUN-HAPPY-MATHS-TIME
« Reply #4 on: April 11, 2009, 04:18:41 pm »
0
Alternate solution for Q1:

Let $\phi =\dfrac{\sin(5x)}{\sin(x)} = \dfrac{e^{5ix} - e^{-5ix}}{e^{ix} - e^{-ix}} = \dfrac{\left(e^{5ix} - e^{-5ix}\right)\left(e^{ix} + e^{-ix}\right)}{e^{2ix} - e^{-2ix}} = \dfrac{e^{6ix} - e^{-6ix} + e^{4ix} - e^{-4ix}}{e^{2ix} - e^{-2ix}} = \dfrac{\sin(6x) + \sin(4x)}{\sin(2x)}$

$\phi = \dfrac{\sin(4x)\cos(2x) + \cos(4x)\sin(2x) + 2\sin(2x)\cos(2x)}{\sin(2x)} = 2\cos^2(2x) + \cos(4x) + 2\cos(2x) = \boxed{2\cos(4x) + 2\cos(2x) + 1}$

Note: $\mbox{cis}(x) = e^{ix}$.
« Last Edit: April 11, 2009, 04:22:36 pm by dcc »

#### dcc

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##### Re: SUPER-FUN-HAPPY-MATHS-TIME
« Reply #5 on: April 11, 2009, 04:28:33 pm »
0
NEW QUESTION POSTED.

#### kamil9876

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##### Re: SUPER-FUN-HAPPY-MATHS-TIME
« Reply #6 on: April 11, 2009, 06:14:06 pm »
0
I've attatched the trick to number7, rest is trivial.
Voltaire: "There is an astonishing imagination even in the science of mathematics ... We repeat, there is far more imagination in the head of Archimedes than in that of Homer."

#### dcc

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##### Re: SUPER-FUN-HAPPY-MATHS-TIME
« Reply #7 on: April 11, 2009, 08:14:59 pm »
0
I've attatched the trick to number7, rest is trivial.

How hard would it be to just go the extra yard and finish it off?

#### /0

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##### Re: SUPER-FUN-HAPPY-MATHS-TIME
« Reply #8 on: April 11, 2009, 08:32:44 pm »
0
Question 4)

$\frac{\sin{(ax)}}{\sin{(x)}} = \frac{\sin{(x(a+1))}+\sin{(x(a-1))}}{\sin{(2x)}}, \ \ \sin{(x)}, \sin{(2x)} \neq 0$

$\frac{\sin{(2x)}\sin{(ax)}}{\sin{(x)}} = \sin{(x(a+1))}+\sin{(x(a-1))}$

$2\cos{(x)}\sin{(ax)}=\sin{(x(a+1))}+\sin{(x(a-1))}$

$\sin{(ax)}\cos{(x)}=\frac{\sin{(x(a+1))}+\sin{(x(a-1))}}{2}$

The rest follows from the identity

$\sin{\theta}\cos{\alpha}=\frac{\sin{(\theta+\alpha)}+\sin{(\theta-\alpha)}}{2}$

All steps are reversible if we assume $\sin{(x)}, \sin{(2x)} \neq 0$

For the more general case, the cheap way to solve is to simply say:

$\sin{(ax)} = \frac{\sin{(x)}}{\sin{(2x)}}(\sin{(x(a+1))}+\sin{(x(a-1))})$

$\sin{(bx)} = \frac{\sin{(x)}}{\sin{(2x)}}(\sin{(x(b+1))}+\sin{(x(b-1))})$

$\therefore \frac{\sin{(ax)}}{\sin{(bx)}}=\frac{\sin{(x(a+1))}+\sin{(x(a-1))}}{\sin{(x(b+1))}+\sin{(x(b-1))}}$
« Last Edit: April 11, 2009, 09:30:46 pm by /0 »

#### kamil9876

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##### Re: SUPER-FUN-HAPPY-MATHS-TIME
« Reply #9 on: April 11, 2009, 09:16:20 pm »
0
I've attatched the trick to number7, rest is trivial.

How hard would it be to just go the extra yard and finish it off?

It was all in the spirit of mathematics:

"A physicist and engineer and a mathematician were sleeping in a hotel room when a fire broke out in one corner of the room. Only the engineer woke up he saw the fire, grabbed a bucket of water and threw it on the fire and the fire went out, then he filled up the bucket again and threw that bucketfull on the ashes as a safety factor, and he went back to sleep. A little later, another fire broke out in a different corner of the room and only the physicist woke up. He went over measured the intensity of the fire, saw what material was burning and went over and carefully measured out exactly 2/3 of a bucket of water and poured it on, putting out the fire perfectly; the physicist went back to sleep. A little later another fire broke out in a different corner of the room. Only the mathematician woke up. He went over looked at the fire, he saw that there was a bucket and he noticed that it had no holes in it; he turned on the faucet and saw that there was water available. He, thus, concluded that there was a solution to the fire problem and he went back to sleep."

Voltaire: "There is an astonishing imagination even in the science of mathematics ... We repeat, there is far more imagination in the head of Archimedes than in that of Homer."

#### Collin Li

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##### Re: SUPER-FUN-HAPPY-MATHS-TIME
« Reply #10 on: April 11, 2009, 09:35:30 pm »
0
$\int x(x - a^2)\ dx = \frac{1}{3}x^3 - \frac{1}{2}x^2a^2$

When $a < 0$, $|x(x-a^2)| = x(x-a^2)\; \therefore \int_0^a |x(x - a^2)|\ dx = \frac{1}{3}a^3 - \frac{1}{2}a^4$

When $a > 1$, $|x(x-a^2)| = -x(x-a^2)\; \therefore \int_0^a |x(x - a^2)|\ dx = -\frac{1}{3}a^3 + \frac{1}{2}a^4$

When $0 \leq a \leq 1$, $|x(x-a^2)|$ is negative from $0 < x < a^2$ and positive from $a^2 < x < a$

$\therefore \int_0^a |x(x - a^2)|\ dx = -\int_0^{a^2} x(x - a^2)\ dx + \int_{a^2}^a x(x - a^2)\ dx$

$= \left(- \frac{1}{3}a^6 + \frac{1}{2}a^6\right) + \left(\frac{1}{3}a^3 - \frac{1}{2}a^4\right) + \left(-\frac{1}{3}a^6 + \frac{1}{2}a^6\right)$

$= \frac{1}{3}a^6 + \frac{1}{3}a^3 - \frac{1}{2}a^4$

#### Neobeo

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##### Re: SUPER-FUN-HAPPY-MATHS-TIME
« Reply #11 on: April 12, 2009, 12:34:16 pm »
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6.) Find the minimum area of the part bounded by the parabola $y=a^3x^2-a^4x\ (a>0)$ and the line $y = x$.
(Source: 1963 Tokyo Metropolitan University entrance exam)

It is clear that $y=x$ lies above $y=a^3x^2-a^4x$ in the region, so we define:
$f(a,x)=x-(a^3x^2-a^4x)=(a^4+1)x-a^3x^2$

Also define $b>0$ to be the upper terminal, such that $f(a,0)=f(a,b)=0$

Minimising area, we get
\begin{align*}0&=\frac{d}{da}\int_0^bf(a,x)dx\\&=\int_0^b\frac{\partial}{\partial a}f(a,x)dx\\&=\int_0^b4a^3x-3a^2x^2dx\\&=2a^3b^2-a^2b^3=a^2b^2(2a-b) \implies\mbox{ Minimum value when }b=2a\end{align*}

Backsolving, $0=f(a,2a)=2a^5+2a-4a^5 \implies a=1,b=2$

Then the integral for the area under the curve is simply:
$\int_0^2 2x-x^2dx=\frac{4}{3}$
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#### Neobeo

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##### Re: SUPER-FUN-HAPPY-MATHS-TIME
« Reply #12 on: April 12, 2009, 01:03:58 pm »
+1
2.) Show that $\arctan \left(2\right) = \arctan \left(\frac{3}{4}\right) + \arctan \left(\frac{1}{2}\right)$ given $z = 15 + 30i$.

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#### TrueTears

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##### Re: SUPER-FUN-HAPPY-MATHS-TIME
« Reply #13 on: April 12, 2009, 01:33:55 pm »
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6) let $y1(x) = a^3x^2 - a^4x$

and $y2(x) = x$

(Say $a = 1$)

after sketching the graphs of $y1(x) = x^2 - x$and $y2(x) = x$

The intersection points in terms of a are $a^3x^2 - a^4x = x \implies x = \frac{a^4+1}{a^3} , 0$

These are the integration limits so:

$A = \int_0^{\frac{a^4+1}{a^3}} [x-(a^3x^2-a^4x)]dx = \int_0^{\frac{a^4+1}{a^3}}[x(1+a^4)-a^3x^2]dx$

$= [\frac{x^2}{2}(1+a^4) - \frac{a^3}{3}x^3]_0^{\frac{a^4+1}{a^3}}$

$= \frac{(a^4+1)^3}{6a^6}$

To find the minimum area we require $\frac{dA}{da}= 0$

$\frac{dA}{da} = \frac{2(a^4+1)^2}{a^3} - \frac{(a^4+1)^3}{a^7} = 0$

solving for a yields $a = 1$

Therefore the intersection points become $x = 0 , 2$

$\boxed{\int_0^2 [x-(1x^2-1x)] dx = \frac{4}{3}}$

« Last Edit: April 12, 2009, 02:04:49 pm by TrueTears »
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#### dcc

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2.) Show that $\arctan \left(2\right) = \arctan \left(\frac{3}{4}\right) + \arctan \left(\frac{1}{2}\right)$
Note that $(4 + 3i)(2 + i) = 5 + 10i$$\arg(4 + 3i) = \arctan \left(\frac{3}{4}\right),\, \arg(2 + i) = \arctan \left(\frac{1}{2} \right),\, \arg(5 + 10i) = \arctan \left( 2 \right)$.
Therefore $\arg((4 + 3i)(2 + i)) = \arg(4 + 3i) + \arg(2 + i) = \arg(5 + 10i)$, which completes the proof.