SimonM
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(Updated as far as #55) SimonM - 10.05.2009

STEP I:
1: Solution by DeanK22
2: Solution by sonofdot
3: Solution by SimonM
4: Solution by Unbounded
5: Solution by Unbounded
6: Solution by brianeverit
7: Solution by SimonM
8: Solution by Daniel Freedman
9: Solution by Unbounded
10: Solution by brianeverit
11: Solution by brianeverit
12: Solution by Unbounded
13: Solution by brianeverit
14: Solution by brianeverit


STEP II:
1: Solution by Unbounded
2: Solution by Daniel Freedman
3: Solution by Daniel Freedman
4: Solutiom by Aurel-Aqua
5: Solution by Daniel Freedman
6: Solution by SimonM
7: Solution by Daniel Freedman
8: Solution by Daniel Freedman
9: Solution by brianeverit
10: Solution by Daniel Freedman
11: Solution by Zuzuzu
12: Solution by Aurel-Aqua
13: Solution by brianeverit
14: Solution by brianeverit


STEP III:
1: Solution by sonofdot
2: Solution by Daniel Freedman
3: Solution by Daniel Freedman
4: Solution by SimonM
5: Solution by DeanK22
6: Solution by SimonM
7: Solution by DeanK22
8: Solution by DeanK22
9: Solution by brianeverit
10: Solution by Aurel-Aqua
11: Solution by SimonM
12: Solution by brianeverit
13: Solution by Aurel-Aqua
14: Solution by Aurel-Aqua


Solutions written by TSR members:
1987 - 1988 - 1989 - 1990 - 1991 - 1992 - 1993 - 1994 - 1995 - 1996 - 1997 - 1998 - 1999 - 2000 - 2001 - 2002 - 2003 - 2004 - 2005 - 2006 - 2007
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Glutamic Acid
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Sequel to "2001: A Space Odyssey"?

Sounds cool, but the solutions will be on Meikleriggs. No reason we can't have a separate set, though.
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Kyalimers
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(Original post by DeanK22)
... to do the papers - solutions only...

...
You should have a look at how the other threads were set out and try to achieve a similar style, coz at the moment, the thread title and the OP are ****.
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Oh I Really Don't Care
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(Original post by sohanshah)
OP is ****.
... a laughable comment.
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Kyalimers
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(Original post by DeanK22)
... a laughable comment.
I meant Original Post. But hey if you want it to mean Original Poster, feel free :yep: :tongue:
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Daniel Freedman
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STEP II, 2004, Question 2

Spoiler:
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No solutions to  \\ x^2 - \alpha |x| + 2 < 0 \implies b^2 - 4ac < 0

  \implies \alpha^2 - 8 < 0 \implies |\alpha| < 2\sqrt{2}


When  \alpha = 3, \ x^2 - 3|x| + 2 < 0 \implies (|x|-1)(|x|-2) < 0

 \implies 1 < x < 2 \mbox{ and } \therefore -2 < x < -1 due to symmetry.


 x^2 - \alpha |x| + 2 < 0 \implies (|x| - \frac{\alpha}{2})^2 - \frac{\alpha^2}{4} + 2 < 0

 \implies (|x| - \frac{\alpha}{2})^2 < t^2 where  t^2 = \frac{\alpha^2}{4} - 2

 \therefore -t < |x| - \frac{\alpha}{2} < t \implies -t + \frac{\alpha}{2} < |x| < t + \frac{\alpha}{2}

This interval has length 2t  \implies the sum of the lengths of the intervals is 4t due to symmetry as before.

 \therefore S = 4 \sqrt{\frac{\alpha^2}{4} - 2} = 2 \sqrt{\alpha^2 - 8} < 2\sqrt{\alpha^2} as  \alpha > 2\sqrt{2}

Graph attached (slightly dodgy)

EDIT: In the first part, you'd probably need to mention why the discriminant of this particular quadratic is still b^2 - 4ac (consider x < 0 and x > 0).
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Oh I Really Don't Care
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STEP I

Question 1:
Spoiler:
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 \displaystyle (3 + 2\sqrt{5})^3 = a + b\sqrt{5}



\displaystyle 3^3 + 3(3^2)(2\sqrt{5}) + 3(3)(2\sqrt{5})^2 + (2\sqrt{5})^3 = a + b\sqrt{5}

Equating surds and natural numbers; a = 207 and b = 94


ii)

Spoiler:
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 \displaystyle 99 - 70\sqrt{2} = c^3 - 3c^2(d\sqrt{2}) + 3c(d\sqrt{2})^2 - (d\sqrt{2}^3



\displaystyle \therefore c(c^2 + 6d^2) = 99 \; \; d(2d^2 + 3c^2) = 70

By condidering the prime factors; 70 = 7 x 2 x 5 and 99 = 3 x 3 x 11;

c = 3 and d = 2 (with numbers this small it is quite easy to see that these are the only solutions, however listing all combinations would be necccessary for larger).


iii)

Spoiler:
Show


 \displaystyle x^6 - 198x^3 + 1 = 0 \; let \; y \; = \; x^3



\displaystyle \therefore y^2 - 198y + 1 = 0 \Rightarrow (y-99)^2 = 99^2 - 1 = 9800



\displaystyle \therefore x^3 = 99 \pm 70\sqrt{2} \Rightarrow x = 3 \pm 2\sqrt{2}
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Daniel Freedman
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STEP II, 2004, Question 5

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 \displaystyle \int_0^{\pi} x\sin{x} \ \mathrm{d}x = -x\cos{x}\big|_0^{\pi} + \int_0^{\pi} \cos{x} \ \mathrm{d}x =  \pi + \sin{x}\big|_0^{\pi} = \pi

 \displaystyle \int_0^{\pi} x\cos{x} \ \mathrm{d}x = x\sin{x}\big|_0^{\pi} + \int_0^{\pi} \sin{x} \ \mathrm{d}x =  0 + -\cos{x}\big|_0^{\pi} = - 2

 \\ \displaystyle f(t) = t + \int_0^{\pi} f(x)\sin{(x+t)} \ \mathrm{d}x = \int_0^{\pi} f(x) \left( \sin{x}\cos{t} + \cos{x}\sin{t}\right) \ \mathrm{d}x

 \displaystyle \\ = t + \int_0^{\pi} f(x)\sin{x} \ \mathrm{d}x \cos{t} + \int_0^{\pi} f(x) \cos{x} \ \mathrm{d}x \sin{t} as required.

From expression (**),  f(t) = t + A\sin{t} + B\cos{t} and  f(x) = x + A\sin{x} + B\cos{x}

Substituting into (*) gives:

 \\ \displaystyle t + A\sin{t} + B\cos{t} = t + \int_0^{\pi} (x + A\sin{x} + B\cos{x})\sin{(x+t)} \ \mathrm{d}x  \\

\\ \implies A\sin{t} + B\cos{t} = \int_0^{\pi} (x + A\sin{x} + B\cos{x}) ( \sin{x}\cos{t} + \cos{x}\sin{t}) \ \mathrm{d} x \\

\\ \implies A\sin{t} + B\cos{t} = \int_0^{\pi} x \sin{x} \ \mathrm{d}x \cos{t} + \int_0^{\pi} x \cos{x} \ \mathrm{d}x \sin{t} \\

\\ + A \left( \int_0^{\pi} \sin^2{x} \ \mathrm{d}x \cos{t} + \int_0^{\pi} \sin{x}\cos{x} \ \mathrm{d}x \sin{t} \right) \\

\\ + B \left( \int_0^{\pi} \sin{x}\cos{x} \ \mathrm{d}x \cos{t} + \int_0^{\pi} \cos^2{x} \ \mathrm{d}x \sin{t} \right) \\

\\ = \pi \cos{t} - 2\sin{t} + A \left(\int_0^{\pi} \frac{1}{2} - \frac{1}{2} \cos{2x} \ \mathrm{d}x \cos{t} + \int_0^{\pi} \frac{1}{2}\sin{2x} \ \mathrm{d}x \sin{t} \right) \\

\\ + B \left( \int_0^{\pi} \frac{1}{2} \sin{2x} \ \mathrm{d} x \cos{t} + \int_0^{\pi} \frac{1}{2} + \frac{1}{2} \cos{2x} \ \mathrm{d}x \sin{t} \right)

 \\ \implies A\sin{t} + B\cos{t} = \pi \cos{t} - 2\sin{t} + A \left( [\frac{1}{2}x - \frac{1}{4}\sin{2x}]_0^{\pi} \cos{t} - \frac{1}{2}\cos{2x}\big|_0^{\pi} \sin{t} \right) \\

\\ + B \left( -\frac{1}{4}\cos{2x}\big|_0^{\pi} + [\frac{1}{2}x + \frac{1}{4}\sin{2x}]_0^{\pi} \sin{t} \right) \\

\\ \implies A\sin{t} + B\cos{t} = \sin{t}\left(\frac{B\pi}{2}-2\right) + \cos{t}\left(\frac{A\pi}{2}+\pi\  right) \\

\\ \implies A = \frac{B\pi}{2}-2, \ B = \frac{A\pi}{2}+\pi \\

\\ \implies B = 0, \ A = - 2
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Unbounded
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STEP I, 2004, Question 5

A:  u_a = 5(a-1) + 1
B:  u_b = 5(b-1) + 2
C:  u_c = 5(c-1) + 3
D:  u_d = 5(d-1) + 4
E:  u_e = 5e

where  a, b, c, d, e \in \mathbb{N}

Statement: The sum of any term in B with any term in C is a term in E
 u_b + u_c = 5(b-1) + 2 + 5(c-1) + 3 = 5(b-1+c-1) + 5
 = 5(b+c-1) = 5e as for all b, c there exists an e such that (b+c-1) = e
\therefore (u_b + u_c) \in E \ \ \square
Spoiler:
Show
With modular arithmetic, consider  u_b + u_c with modulus 5, we have  2 + 3 \equiv 0 \pmod{5} \implies u_b + u_c \equiv u_e


Statement: The square of any term in B is in D.
 u_b^2 = ((5(b-1)+2)^2 = 25(b-1)^2 + 20(b-1) + 4 = 5(5(b-1)^2+4(b-1)) + 4
 = 5(d-1)+4 as for all b, there exists a d such that  d-1 = 5(b-1)^2+4(b-1)
 \therefore u_b^2 \in D \ \ \square
Spoiler:
Show
With modular arithmetic, consider  u_b^2 with modulus 5, we have  (2)^2 \equiv 4 \pmod{5} \implies u_b^2 \equiv u_d


Statement: The square of any term in C is also in D.
 u_c^2 = ((5(c-1)+3)^2 = 25(c-1)^2 + 30(c-1) + 9
= 5(6(c-1)+1)+4 = 5(d-1)+4 as for all c, there exists a d such that  d-1 = 6(c-1)+1
 \therefore u_c^2 \in D \ \ \square
Spoiler:
Show
With modular arithmetic, consider  u_c^2 with modulus 5, we have  (-2)^2 \equiv 4 \pmod{5} \implies u_c^2 \equiv u_d


(i) Considering the equation  x^2 + 5y = 243723 with modulus 5:

 x^2 \equiv 3 \pmod{5}

Looking at quadratic residues with modulus 5:
 0^2 \equiv 0 \pmod{5}
1^2 \equiv 1 \pmod{5}
 2^2 \equiv 4 \pmod{5}
 3^2 \equiv 4 \pmod{5}
 4^2 \equiv 1 \pmod{5}

Therefore, there exists no integer such that it squares to have a residue of 3 (mod 5).  \square

(ii) Again looking at the equation with modulus 5:
 x^4 + 2y^4 \equiv 4 \pmod{5}

Similarly, looking at quartic residues with modulus 5:

 0^4 \equiv 0 \pmod{5}
1^4 \equiv 1 \pmod{5}
 2^4 \equiv 1 \pmod{5}
 3^4 \equiv 1 \pmod{5}
 4^4 \equiv 1 \pmod{5}

Considering positive integers (x^4,y^4), we have pairs with quartic residues (mod 5) of either (1,1), (0,0), (1,0), or (0,1), and so the residues modulo 5 of the pairs (x^4, 2y^4) are either (1,2), (0,0), (1,0) or (0,2). and summing the pairs in the brackets, none sum to 4 (mod 5), so there exist no pairs of integers x and y, such that  x^4 + 2y^4 = 26081974 \ \ \square
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Oh I Really Don't Care
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STEP III, Question 8;

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 \displaystyle \frac{dy}{dx} = f(x)y + \frac{g(x)}{y} \therefore y = u^2



\displaystyle \frac{du}{dx} = 2y\frac{dy}{dx}



\displaystyle 2y\frac{dy}{dx} = 2f(x)y^2 + 2g(x)



\displaystyle \frac{du}{dx} = 2f(x)u + 2g(x)

a linear differential equation in u

ii)

 \displaystyle \frac{dy}{dx} = f(x)y + \frac{g(x)}{y} = \frac{y}{x} - \frac{1}{y}



\displaystyle \Rightarrow f(x) = \frac{1}{x} , g(x) = -1



\displaystyle \frac{du}{dx} = \frac{2u}{x} - 2

 \displaystyle P(x) = \frac{-2}{x} \Rightarrow I(x) = e^{ \int P(x) dx } = x^{-2}

 \displaystyle x^{-2} u = \int -2x^{-2} dx = 2x^{-1} + C

 \frac{u}{x^2} = \frac{(2 + cx)}{x} \therefore y^2 = (2x + cx^2)
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sonofdot
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STEP I Question 2

i) Graph attached (I was lazy and didn't draw it myself...)

From graph:

\begin{array}{rl}

\displaystyle\int_0^a \sqrt{[x]} \, dx & = \displaystyle\int_0^1 \sqrt{0} \, dx + \int_1^2 \sqrt{1} \, dx + \cdots + \int_{a-1}^a \sqrt{a-1} \, dx \\ \br \\

& = \displaystyle\sum_{r=0}^{a-1} \sqrt{r} \end{array}

ii) Similarly,

\begin{array}{rl}

\displaystyle\int_0^a 2^{[x]} \, dx & = \displaystyle\int_0^1 2^0 \, dx + \int_1^2 2^1 \, dx + \cdots \int_{a-1}^a 2^{a-1} \, dx \\ \br \\

& = \displaystyle\sum_{r=0}^{a-1} 2^r = \frac{1 - 2^{a}}{1-2} = 2^a - 1 \end{array}
(sum of a GP)

iii) \begin{array}{rl}

\displaystyle\int_0^a 2^{[x]} \, dx & = \displaystyle\int_0^{[a]} 2^{[x]} \, dx + \int_{[a]}^{a-[a]} 2^{[x]} \, dx \\ \br \\

& \displaystyle = 2^{[a]} - 1 + (a-[a])2^{[a]} = \boxed{(a-[a]+1)2^{[a]} -1} \end{array}
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Daniel Freedman
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STEP II, 2004, Question 10

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i)

 \\ \bold{P:} \\

\\ \sum F = m a \implies -40\sin{30} - \frac{40\cos{30}}{5\sqrt{3}} = 4a \implies a = - 6

To find the time after which it stops,  v = u + at \implies 0 = 6-6t \implies t = 1

After it has stopped,  \sum F = ma \implies -40\sin{30} + \frac{40\cos{30}}{5\sqrt{3}} = 4a \implies a = -4

\\  \therefore \mbox{by } s = ut + \frac{1}{2}at^2, \\

\\ s_{p} = \left\{

\begin{array}{c l}

  6t - 3t^2 & 0 \leq t \leq 1 \\

  3 - 2(t-1)^2 & t &gt; 1

\end{array}

\right.

 \\ \implies s_{Q} = 6(t-T) - 3(t-T)^2 \mbox{ for } T \leq t \leq 1 + T

Note that it is necessary that  T &lt; 1 + \sqrt{\frac{3}{2}} , as at  T = 1 + \sqrt{\frac{3}{2}} , P has returned to its point of projection.

When they collide,  s_p = s_q

 \\ \implies 3 - 2(t-1)^2 = 6(t-T) - 3(t-T)^2 \\

\\ \implies t^2 + t(-6T-2) + 3T^2 + 6T + 1 = 0 \\

\\ \implies t = \frac{6T + 2 \pm \sqrt{(6T+2)^2 - 4(3T^2 + 6T + 1)}}{2} \\

\\ \implies t = \frac{6T + 2 \pm \sqrt{24T^2}}{2} = (3 \pm \sqrt{6})T + 1

We want smallest positive t, therefore  t = (3-\sqrt{6})T + 1 as required.

ii)

At  T = 1 + \sqrt{\frac{2}{3}}, \ t = 3\left(1+\sqrt{\frac{2}{3}}\righ  t) - \sqrt{6}\left(1+\sqrt{\frac{2}{3  }}\right) + 1 = 3 + \sqrt{6} - \sqrt{6} -2 + 1 = 2

After 2s, the total distance travelled by P is equal to  (3) + (3 - 2(1-1)^2) = 6 .

There is a constant resistive force of 4N, so the total work done is 24J.
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Oh I Really Don't Care
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... has anybidy got a latex issue??? I can't view it.
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Daniel Freedman
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STEP II, 2004, Question 3

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 \\ y = x(x+1)(x-2)^4 \\

\\ \implies \ln{y} = \ln{x} + \ln{(x+1)} + 4\ln{(x-2)} \\

\\ \implies \frac{1}{y}\frac{dy}{dx} = \frac{1}{x} + \frac{1}{x+1} + \frac{4}{x-2} \\

\\ \implies \frac{dy}{dx} = \frac{x(x+1)(x-2)^4}{x} + \frac{x(x+1)(x-2)^4}{x+1} + \frac{4x(x+1)(x-2)^4}{x-2} \\ 

\\ = (x+1)(x-2)^4 + x(x-2)^4 + 4x(x+1)(x-2)^3 \\

\\ = (x-2)^3((x+1)(x-2) + x(x-2) + 4x(x+1)) \\

\\ = (x-2)^3(x^2 - x - 2 + x^2 - 2x + 4x^2 + 4x) \\

\\ = (x-2)^3(6x^2 + x  - 2) \\

\\ = (x-2)^2(3x+2)(2x-1)

as required.

At a stationary point,  \frac{dy}{dx} = 0 \implies x = 2, \ \frac{1}{2}, \ -\frac{2}{3}

Therefore the coordinates of the stationary points are  (2,0), \ (\frac{1}{2}, \frac{243}{64}), \ (-\frac{2}{3}, - \frac{8192}{729})

 \frac{d^2 y}{dx^2} = 3(x-2)^2(6x^2+x-2) + (x-2)^3(12x+1)

At  x = \frac{1}{2}, \ \frac{d^2y}{dx^2} &lt; 0 \implies \mbox{maximum}

At  x = - \frac{2}{3}, \ \frac{d^2y}{dx^2} &gt; 0 \implies \mbox{minimum}

\implies \mbox{minimum at } (2,0)

(graph attached)

i)

 y^2 = x(x+1)(x-2)^4 \implies y = \pm \sqrt{x(x+1)(x-2)^4}

Therefore curve has same roots as C, is symmetrical about x-axis, and is not defined were -1 < x < 0

(graph attached)

ii)

 y = x^2(x^2+1)(x^2-2)^4

This is an even function, and so is symmetrical about the y axis. It has roots at  x=0, \ x = \pm \sqrt{2}

Let  u = x^2 \implies y = u(u+1)(u-2)^4 , which is curve C.

 \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} =  \frac{dy}{du} \times 2x = 0

Therefore it has stationary points at  x = 0, \ x^2 = \frac{1}{2}, \ x^2 = 2

(graph attached)
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Daniel Freedman
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Someone do II question 1. I can't be arsed typing it up (and it's a rubbish question).
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Unbounded
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STEP I, 2004, Question 9
the vertical velocity of the particle upwards is initially  u\sin \theta . noting that at the greatest height 'h', v = 0

 v^2 = u^2 +2as
 0 = u^2\sin^2 \theta -2gh \iff \boxed{h = \displaystyle\frac{u^2\sin^2 \theta}{2g}}

As the particle doesn't hit the top, \displaystyle\frac{u^2\sin^2 \theta}{2g} &lt; \frac{9}{10}d \iff u^2\sin^2 \theta &lt; \frac{9dg}{5} \ \ (\ast)

at point P, let the time be T:

Considering it horizontally:  d = u\cos \theta \cdot T \ \ (\ast \ast )

Considering it vertically:  \frac{d}{2} = u\sin \theta \cdot T - \frac{g}{2}T^2

 \iff u\cos \theta \cdot T = 2u\sin \theta \cdot T - gT^2

 \iff u\cos \theta = 2u\sin \theta - gT

 \iff T = \displaystyle\frac{u(2\sin \theta - \cos \theta)}{g}


Substitution back into  (\ast \ast ):

 \therefore d = \displaystyle\frac{u^2\sin 2\theta}{g} - \frac{u^2\cos^2 \theta}{g} \iff dg = u^2\sin 2\theta - u^2\cos^2 \theta


Substitution into  (\ast ) :

 u^2\sin^2 \theta &lt; \frac{9}{5} u^2(\sin 2\theta - cos^2\theta)

 \implies \frac{5}{9} \sin^2 \theta - 2\sin \theta \cos \theta + \cos^2 \theta &lt; 0

 \implies (\frac{5}{3} \sin \theta - \cos \theta)(\frac{1}{3} \sin \theta - \cos \theta) &lt; 0

considering roots of  (\frac{5}{3} \sin \theta - \cos \theta)(\frac{1}{3} \sin \theta - \cos \theta) = 0

 \frac{5}{3} \sin \theta - \cos \theta = 0 \implies \tan \theta = \frac{3}{5}

 \frac{1}{3} \sin \theta - \cos \theta = 0 \implies \tan \theta = 3

 \therefore (\frac{5}{3} \sin \theta - \cos \theta)(\frac{1}{3} \sin \theta - \cos \theta) &lt; 0 \implies \frac{3}{5} &lt; \tan \theta &lt; 3

 \therefore \boxed{\arctan \frac{3}{5} &lt; \theta &lt; \arctan 3} \ \ \square
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Oh I Really Don't Care
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(Original post by Erdős)
Or do we?
... going to do it again
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(Original post by Daniel Freedman)
Someone do II question 1. I can't be arsed typing it up (and it's a rubbish question).
STEP II, 2004, Question 1
(i)
 \sqrt{3x^2 + 1} + \sqrt{x} -2x -1 = 0

 \iff \sqrt{3x^2+1} = 2x + 1 - \sqrt{x}

 \implies 3x^2 +1 = 4x^2 - 4x\sqrt{x} + 5x - 2\sqrt{x} + 1

 \iff x^2 - 4x\sqrt{x} + 5x - 2\sqrt{x} = 0

Letting  x = y^2 :

\iff y^4 -4y^3 +5y^2 -2y =0

 \iff y(y-1)^2(y-2) = 0

 \iff y= 0, 1, 2

 \implies \boxed{x = 0, 1, 4} and checking the original equation, all values work.
(ii)
 \sqrt{3x^2+1} -2\sqrt{x} + x -1 = 0

 \implies 3x^2 + 1 = x^2 -4x\sqrt{x} + 2x + 4\sqrt{x} + 1

 \iff 2x^2 + 4x\sqrt{x} - 2x - 4\sqrt{x} = 0

Letting  x = y^2 and dividing by 2:

 \iff y^4 + 2y^3 -y^2-2y = 0

 \iff y(y-1)(y+1)(y+2) = 0

 \iff y = 0, 1, -1, -2

 \implies x = 0, 1, 4 and checking the original equation, all of these values work, except x = 4.  \therefore \boxed{x = 0, 1}
(iii)
 \sqrt{3x^2+1} -2\sqrt{x} -x +1 = 0

 \implies 3x^2 + 1 = x^2+4x\sqrt{x}+2x-4\sqrt{x}+1

 \iff 2x^2 -4x\sqrt{x} -2x +4\sqrt{x} = 0

Letting  x = y^2 and dividing by 2:

\iff y^4-2y^3-y^2+2y = 0

 \iff y(y-1)(y+1)(y-2) = 0

 \iff y = 0, 1, -1, 2

 \implies x = 0, 1, 4 and checking these in the original equation, x = 0, does not work.

 \therefore \boxed{x = 1, 4}
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III Question 1

Part One
\displaystyle\int_0^a \frac{\sinh x}{2 \cosh^2 x -1} \, dx

Let u=\cosh x \Rightarrow dx = \frac{du}{sinhx}

 \begin{array}{rl}

\therefore \displaystyle\int_0^a \frac{\sinh x}{2 \cosh^2 x -1} \, dx & = \displaystyle\int_1^{\cosh a} \frac{1}{2u^2 - 1} \, du \\

\br \\

& = \displaystyle\int_1^{\cosh a} \frac{1}{(\sqrt2 u +1)(\sqrt2 u -1)} \, du \\ \br \\

& = \displaystyle\frac{1}{2} \int_1^{\cosh a} \frac{1}{\sqrt2 u -1} - \frac{1}{\sqrt2 u+1} \, du \\ \br \\

& = \displaystyle\frac{1}{2\sqrt2} \left[ \ln (\sqrt2 u-1) - \ln (\sqrt2 u+1) \right]_1^{\cosh a} \\ \br \\

& = \displaystyle\frac{1}{2\sqrt2} \left[ \ln \left( \frac{\sqrt2 u-1}{\sqrt2 u+1} \right) \right]_1^{\cosh a} \\ \br \\

& = \displaystyle\boxed{\frac{1}{2 \sqrt2}\ln \left( \frac{\sqrt2 \cosh a -1}{\sqrt2 \cosh a +1} \right) + \frac{1}{2\sqrt2} \ln \left( \frac{\sqrt2 +1}{\sqrt2 -1} \right)}

\end{array}
Part Two
\displaystyle\int_0^a \frac{\cosh x}{1 + 2\sinh^2 x} \, dx

Let u = \sinh x \Rightarrow dx = \frac{du}{\cosh x}

 \begin{array}{rl}

\therefore\displaystyle\int_0^a \frac{\cosh x}{1 + 2\sinh^2 x} \, dx & = \displaystyle\int_0^{\sinh a} \frac{1}{1+2u^2} \, du \\ \br \\

& = \displaystyle\frac{1}{2} \int_0^{\sinh a} \frac{1}{\frac{1}{2} + u^2} \, du \\ \br \\

& = \displaystyle\frac{1}{\sqrt2} \left[ \arctan (\sqrt2 u) \right]_0^{\sinh a} \\ \br \\

& = \displaystyle \boxed{\frac{1}{\sqrt2} \arctan (\sqrt2 \sinh a)}\end{array}
Part Three
\displaystyle\int_0^{\infty} \frac{\cosh x - \sinh x}{1 + 2\sinh^2 x}  \, dx

\begin{array}{rl}

\cosh 2x & \equiv \cosh^2 x + \sinh^2 x \\

& \equiv 1 + 2\sinh^2 x \\

& \equiv 2\cosh^2 x - 1 \end{array}

 \begin{array}{rl}

\therefore \displaystyle\int_0^a \frac{\cosh x - \sinh x}{1 + 2\sinh^2 x}  \, dx

& = \displaystyle\int_0^a \frac{\cosh x}{1 + 2\sinh^2 x} \, dx

- \int_0^a \frac{\sinh x}{2\cosh^2 x - 1} \, dx \\  \br \\

& = \displaystyle\frac{1}{\sqrt2} \arctan (\sqrt2 \sinh a) - \frac{1}{2\sqrt2}\ln \left( \frac{\sqrt2 \cosh a -1}{\sqrt2 \cosh a +1} \right) \\ \br \\

& \, \, \, \, \displaystyle -  \frac{1}{2\sqrt2} \ln \left( \frac{\sqrt2 +1}{\sqrt2 -1} \right)}\end{array}

As \displaystyle a \rightarrow \infty , \, \sinh a \rightarrow \infty \Rightarrow \arctan (\sqrt2 \sinh a) \rightarrow \frac{\pi}{2}

As \displaystyle a \rightarrow \infty , \, \left( \frac{\sqrt2 \cosh a -1}{\sqrt2 \cosh a +1} \right) \rightarrow 1 \Rightarrow \ln \left( \frac{\sqrt2 \cosh a -1}{\sqrt2 \cosh a +1} \right) \rightarrow 0

\therefore \displaystyle\int_0^{\infty} \frac{\cosh x - \sinh x}{1 + 2\sinh^2 x}  \, dx = \boxed{\frac{\pi}{2\sqrt2} - \frac{1}{2\sqrt2} \ln \left( \frac{\sqrt2 +1}{\sqrt2 -1} \right)}
Part Four
\displaystyle\int_1^{\infty} \frac{1}{1+u^4} \, du

\cosh x - \sinh x \equiv e^{-x}

1 + 2 \sinh^2 x \equiv \cosh2x \equiv \frac{e^{2x} + e^{-2x}}{2}

\begin{array}{rl}

\therefore \displaystyle\int_0^a \frac{\cosh x - \sinh x}{1 + 2\sinh^2 x}  \, dx & = \displaystyle\int_0^a \frac{2e^{-x}}{e^{2x} + e^{-2x}} \, dx \\ \br \\

& = \displaystyle\int_0^a \frac{2e^x}{1 + e^{4x}} \, dx \end{array}

Let u=e^x \Rightarrow dx = \frac{du}{u}

\begin{array}{rl}

\therefore \displaystyle\int_0^a \frac{2e^x}{1 + e^{4x}} \, dx

& = \displaystyle\int_1^{e^a} \frac{2u}{1+u^4}\times \frac{1}{u} \, du \\ \br \\

& = 2 \displaystyle\int_1^{e^a} \frac{1}{1+u^4} \, du \end{array}

\begin{array}{rl}

\therefore \displaystyle\int_1^{\infty} \frac{1}{1+u^4} \, du

& = \frac{1}{2} \displaystyle\int_0^{\infty} \frac{\cosh x - \sinh x}{1 + 2\sinh^2 x}  \, dx \\

\br \\

& = \boxed{\frac{\pi}{4\sqrt2} - \frac{1}{4\sqrt2} \ln \left( \frac{\sqrt2 +1}{\sqrt2 -1} \right)}\end{array}
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rnd
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#20
I know it says answers only at the top of this thread but I'd like to ask a question. Sorry.

I just did 4i from paper 2 and out popped the required result but I didn't use/need the condition that \tan^3\alpha=\frac{a}{b}.

Any comments on why this condition is necessary?
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