STEP 2003 Solutions Thread Watch

SimonM
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You can let someone else take control of the OP whenever.

STEP I:
1: Solution by Horizontal 8
2: Solution by Horizontal 8
3: Solution by Horizontal 8
4: Solution by Horizontal 8
5: Solution by Unbounded
6: Solution by Unbounded
7: Solution by Unbounded
8: Solution by Unbounded
9: Solution by Unbounded
10: Solution by Unbounded
11: Solution by Unbounded
12: Solution by Unbounded
13: Solution by AnonyMatt and Jkn
14: Solution by nota bene


STEP II:
1: Solution by tommm
2: Solution by tommm
3: Solution by Horizontal 8
4: Solution by tommm
5: Solution by Dadeyemi
6: Solution by Horizontal 8
7: Solution by SimonM
8: Solution by tommm
9: Solution by Elongar
10: Solution by TwoTwoOne
11: Solution by Rocious
12: Solution by SimonM
13: Solution by SimonM
14: Solution by TwoTwoOne


STEP III:
1: Solution by Adje
2: Solution by SimonM
3: Solution by Adje
4: Solution by tommm
5: Solution by SCE1912
6: Solution by SimonM
7: Solution by Daniel Freedman
8: Solution by Adje
9: Solution by brianeverit
10: Solution by tommm
11: Solution by brianeverit
12: Solution by brianeverit
13: Solution by toasted-lion
14: Solution by SimonM


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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Adjective
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III/1
Differentiating arcsin
 \displaystyle \frac{d}{dx} \arcsin \ \left(\frac{x+a}{x+b}\right)

 \displaystyle = \frac{(x+b) - (x+a)}{(x+b)^2\sqrt{1 - (\frac{x+a}{x+b})^2}}}

 \displaystyle = \frac{b-a}{(x+b)^2\sqrt{\frac{(x+a)^2 - (x + b)^2}{(x+b)^2}}}}

 \displaystyle = \frac{b-a}{(x+b)^2\frac{\sqrt{(x+a)^2 - (x + b)^2}}{x+b}}}}

 \displaystyle = \frac{b - a}{(x+a)\sqrt{[(x+b) + (x+a)][(x+b)-(x+a)]}}}

 = \displaystyle  \frac{b - a}{(x+b)\sqrt{b-a}\sqrt{a +b + 2x}}

 \displaystyle = \frac{\sqrt{b - a}}{(x+b)\sqrt{a+b+2x}}}}


Differentiating arcosh
 \displaystyle \frac{d}{dx} \text{arcosh} \ \left(\frac{x+b}{x+a}\right)

 \displaystyle = \frac{(x+a) - (x+b)}{(x+a)^2\sqrt{(\frac{x+b}{  x+a})^2 - 1}}}

 \displaystyle = \frac{a - b}{(x+a)\sqrt{[(x+b) + (x+a)][(x+b)-(x+a)]}}}

 = \displaystyle - \frac{b - a}{(x+a)\sqrt{b-a}\sqrt{a +b + 2x}}

 \displaystyle = - \frac{\sqrt{b - a}}{(x+a)\sqrt{a+b+2x}}}}


First integral
 \displaystyle \frac{d}{dx} \text{arcosh} \ \left(\frac{x+5}{x+1}\right)

 \displaystyle = - \frac{\sqrt{4}}{(x+1)\sqrt{6+2x}  }}}

 \displaystyle = - \frac{2}{(x+1)\sqrt{2(3+x)}}}}

 \displaystyle = - \frac{2}{\sqrt{2}}\frac{1}{{(x+1  )\sqrt{3+x}}}}}

 \displaystyle \Rightarrow \int \frac{1}{{(x+1)\sqrt{x+3}}}}} \ dx = - \frac{\sqrt{2}}{2}\text{arcosh} \ \left(\frac{x+5}{x+1}\right) + C


Second integral
 \displaystyle \frac{d}{dx} \arcsin \left(\frac{x-1}{x+3}\right)

 \displaystyle = \frac{\sqrt{4}}{(x+3)\sqrt{2+2x}  }}}

 \displaystyle = \frac{2}{\sqrt{2}}\frac{1}{{(x+3  )\sqrt{1+x}}}}}

 \displaystyle \Rightarrow \int \frac{1}{{(x+3)\sqrt{x+1}}}}} \ dx =  \frac{\sqrt{2}}{2}\arcsin \left(\frac{x-1}{x+3}\right) + C
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SimonM
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STEP III, Question 14

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The probability generating function for one standard die is \displaystyle p(x) = \frac{1}{6}x + \frac{1}{6}x^2 + \frac{1}{6}x^3 + \frac{1}{6}x^4 + \frac{1}{6}x^5 +\frac{1}{6}x^6 = \frac{1}{6}x(1+x+x^2+x^3+x^4+x^5  ) = \frac{1}{6}x(1-x)(1+x+x^2)(1-x+x^2)

The probability generating function for the sum of two will be the product of two (since they are independent) which gives us what we wanted

\displaystyle \frac{1}{6}(x+2x^2+2x^3+x^4) \frac{1}{6} (x+x^3+x^4+x^5+x^6+x^8) = \frac{1}{36}x^2(1+x)^2(1-x+x^2)^2(1+x+x^2)^2 as required

For our tetrahedral die, we have

\frac{1}{16}t^2(1+t)^2(1+t^2)^2

So we can have die \{(1,2,3,4),(1,2,3,4)\},\{(1,2,2  ,3),(1,3,3,5)\}
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SimonM
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Given that's what I've written down on my whiteboard, I'll assume that was a typo
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STEP I question 1

First part:
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Let f(n) = \displaystyle\sum_{r=-1}^n r^2 = pn^3+qn^2+rn+s

f(0)=s=1
f(-1)= -p+q-r+s =1 \implies -p+q-r= 0
f(1)=p+q+r+1 = 2 \implies p+q+r = 1
f(2)=8p+4q+2r+1=6 \implies 8p+4q+2r=5

Adding the second and third equations \implies \boxed{q=\frac{1}{2}}

Subbing back into other equations  \implies p+r=\frac{1}{2} and  8p+2r = 3

solving simultaneously  \implies \boxed{p=\frac{1}{3}} and \boxed{r=\frac{1}{6}}

so,  \displaystyle\sum_{r=-1}^n r^2 = \frac{n^3}{3}+\frac{n^2}{2}+\fra  c{n}{6}+1
 \implies \displaystyle\sum_{r=0}^n r^2 = \frac{n^3}{3}+\frac{n^2}{2}+\fra  c{n}{6} = \frac{n}{6}(n+1)(2n+1)


Second part:

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Let g(n) = \displaystyle\sum_{r=-2}^n r^3 = an^4+bn^3+cn^2+dn+e

 g(0) =e=-9
 g(-2) = 16a-8b+4c-2d-9=-8 \implies 16a-8b+4c-2d=1
 g(-1) = a-b+c-d-9=-9 \implies a-b+c-d=0
 g(1) = a+b+c+d-9=-8 \implies a+b+c+d=1
 g(2) = 16a+8b+4c+2d-9=0 \implies 16a+8b+4c+2d=9

 16a-8b+4c-2d=1
16a+8b+4c+2d=9

adding  \implies 32a+8c=10

 a-b+c-d=0
 a+b+c+d=1

adding  \implies a+c = \frac{1}{2}

Solving simultaneously gives  \boxed{a,c = \frac{1}{4}}

Substituting a and c into other equations gives:

 b+d = \frac{1}{2}
 8b+2d=4 \implies 4b+d=2

Subtracting  \implies 3b = 2-\frac{1}{2} = \frac{3}{2} \implies \boxed{b = \frac{1}{2}} \implies \boxed{d=0}

 \implies \displaystyle\sum_{r=-2}^n r^3 = \frac{n^4}{4}+\frac{n^3}{2}+\fra  c{n^2}{4}-9

 \implies \displaystyle\sum_{r=0}^n r^3 = \frac{n^4}{4}+\frac{n^3}{2}+\fra  c{n^2}{4} = \frac{n^2}{4}(n+1)^2

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Adjective
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III, 3
Denominator ≠ 0


Considering the expansion of  (1+x)^m + (1-x)^m , which is:

 \displaystyle 1 + mx + ^mC_2 x^2 + ^mC_3 x^3 + ^mC_4 x^4 + \cdots

 \displaystyle + 1 - mx + ^mC_2 x^2 - ^mC_3 x^3 + ^mC_4 x^4 - \cdots

All odd powers cancel, so we're left with a load of positive terms. This is clearly not equal to zero (>2).


f'(x)


 \displaystyle f(x) = \frac{(1+x)^m - (1-x)^m}{(1+x)^m + (1-x)^m}



Expanding the numerator, terms with like expressions in powers of m and m-1 cancel, leaving:

 \displaystyle = \frac{2m(1+x)^m(1-x)^{m-1} + 2m(1-x)^m(1+x)^{m-1}}{[(1+x)^m + (1-x)^m]^2}

 \displaystyle = \frac{2m(1+x)^{m-1}(1-x)^{m-1}[(1+x)+(1-x)]}{[(1+x)^m + (1-x)^m]^2}

 \displaystyle f'(x) = \frac{4m(1+x)^{m-1}(1-x)^{m-1}}{[(1+x)^m + (1-x)^m]^2}


Graphs


 \displaystyle f(x) = \frac{(1+x)^5 - (1-x)^5}{(1+x)^5 + (1-x)^5}

 \displaystyle f'(x) = \frac{20(1+x)^{4}(1-x)^{4}}{[(1+x)^5 + (1-x)^5]^2}

Turning points can only be at x = ±1 since the rest of f'(x) is never 0.

At x = 1, f(x) = 1
At x = -1, f(x) = -1

At 0, f(x) = 0

f'(x) is positive everywhere, so the turning points must be points of inflexion.

As x tends to infinity, so do(1+x)⁵ and -(1-x)⁵, so f(x) tends to infinity.

As x tends to minus infinity, so do (1+x)⁵ and -(1-x)⁵, so f(x) tends to minus infinity.

After sketching f(x), 1/f(x) can be sketched by inspection.

f(x) and 1/f(x)
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STEP I question 4

(Not 100% confident, since I always make mistakes with intervals)

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let theta = x for notational purposes


\displaystyle \frac{sinx+1}{cosx}\leq1 \iff \frac{sinx+1}{cosx} - 1\leq0 \iff \frac{sinx-cosx+1}{cosx}\leq0 \iff \frac{\sqrt{2}sin(x-\frac{\pi}{4})+1}{cosx} \leq 0

Numerator changes sign when \displaystyle sin\left(x-\frac{\pi}{4}\right)= -\frac{1}{ \sqrt{2}}

for  \displaystyle -\frac{\pi}{4} \leq x -\frac{\pi}{4} < \frac{7 \pi}{8}

 \implies x-\frac{\pi}{4} = \frac{5 \pi}{4} or  x- \frac{\pi}{4} = -\frac{\pi}{4}

 \implies Numerator changes sign when  x= \frac{3 \pi}{2} or  x= 0

Denominator changes sign when  x=\frac{\pi}{2} or  x= \frac{3\pi}{2}

Therefore we must consider certain intervals:

Let  \displaystyle f(x) = \frac{\sqrt{2}sin(x-\frac{\pi}{4})+1}{cosx}

 0 < x < \frac{\pi}{2} \implies f(x)>0

 \frac{\pi}{2}< x < \frac{3 \pi}{2} \implies f(x)<0

 \frac{3 \pi}{2}< x < 2 \pi \implies f(x)<0

Note:  f(0) = 0

Therefore when considering the interval  0 \leq x < 2 \pi

\displaystyle \frac{sinx+1}{cosx}\leq1 \iff  \boxed{x= 0} or  \boxed{\frac{\pi}{2}< x < \frac{3 \pi}{2}} or  \boxed{\frac{3 \pi}{2}< x < 2 \pi}

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SimonM
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You could show

\displaystyle \frac{1+\sin x}{\cos x} = \cot \left ( \frac{\pi}{4} - \frac{x}{2} \right ) = \tan \left ( \frac{x}{2} - \frac{\pi}{4} \right )
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III/8

Differentiation


 \displaystyle (y+2x)^3(y=4x)=c

 \displaystyle 3(\frac{dy}{dx} + 2)(y + 2x)^2(y-4x) + (\frac{dy}{dx} - 4)(y+2x)^3 = 0

 \displaystyle 3\frac{dy}{dx}(y+2x)^2(y-4x) + \frac{dy}{dx}(y+2x)^3 = 4(y+2x)^3 - 6(y+2x)^2(y-4x)

 \displaystyle \frac{dy}{dx}[3(y+2x)^2(y-4x)+(y+2x)^3] = 4(y+2x)^3 - 6(y+2x)^2(y-4x)

 \displaystyle \frac{dy}{dx} = \frac{4(y+2x)^3 - 6(y+2x)^2(y-4x)}{3(y+2x)^2(y-4x)+(y+2x)^3}

 \displaystyle = \frac{4(y+2x) - 6(y-4x)}{3(y-4x)+(y+2x)}

 \displaystyle = \frac{4y + 8x - 6y + 24x}{3y - 12x + y + 2x}

 \displaystyle = \frac{32x - 2y}{4y - 10x} = \frac{16x - y}{2y - 5x}


General form of derivative


 \displaystyle \frac{d}{dx} (y+ax)^n(y+bx)

 \displaystyle = n (\frac{dy}{dx} + a)(y+ax)^{n-1}(y+bx) + (\frac{dy}{dx} + b)(y+ax)^n

 \displaystyle = \frac{dy}{dx} [n(y+ax)^{n-1}(y+bx)+(y+ax)^n] = -an(y+ax)^{n-1}(y+bx)-b(y+ax)^n

 \displaystyle \frac{dy}{dx} = \frac{-an(y+bx) - b(y +ax)}{n(y+bx)+(y+ax)}

 \displaystyle \frac{dy}{dx} = \frac{-(an+b)y - ab(n+1)x}{(n+1)y + (nb+a)x}


Solution of Differential Equation


 \displaystyle  \frac{dy}{dx} = \frac{10x - 4y}{3x - y}

Trying to solve this using dy/dx as above proved fruitless. I divided by -(n+1) to get the y in the denominator as it appears in the question:

 \displaystyle \frac{abx + \frac{an+b}{n+1}y}{-\frac{nb+a}{n+1} - y} = \frac{10x - 4y}{3x - y}

So we have three equations to solve for three unknowns:

 \displaystyle ab = 10

 \displaystyle \frac{an+b}{n+1} = -4

 \displaystyle -\frac{nb+a}{n+1} = 3

So:

 \displaystyle an + b = -4n - 4 \Rightarrow n(a+4) = -b - 4

 \displaystyle nb + a = -3n - 3 \Rightarrow n(b+3) = -a - 3

Dividing these gives  \displaystyle \frac{a+4}{b+3} = {b+4}{a+3}

Multiplying by (b+3)(a+3) and expanding:

 \displaystyle a^2 + 7a + 12 = b^2 + 7b + 12

Obviously a contender solution here is  a = b = \pm \sqrt{10} (from ab = 10), but this doesn't work when trying to find n in our original equations. So we must press on, and subtract 12 from each side.

 \displaystyle a^2 - b^2 + 7(a - b) = 0

 \displaystyle (a - b)(a + b) + 7(a-b) = 0

I think it's safe to divide by (a - b) here, since we've discounted a = b as a solution.

 \displaystyle a + b + 7 = 0 \ , b = \frac{10}{a}

 a^2 + 7a + 10 = 0

 \displaystyle (a+5)(a+2) = 0  \Rightarrow (a, \ b) = (-5, \ -2), (-2, \ -5)

Taking  (a, \ b) = (-5, \ -2) gives  n = 2 so a general solution to the differential equation is:

 \displaystyle (y - 5x)^2(y-2x) + C

Taking  (a, \ b) = (-2, \ -5) gives  n = \frac{1}{2} so another solution is:

 \displaystyle (y - 2x)^{\frac{1}{2}}(y-5x) + C
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(Original post by SimonM)
You could show

\displaystyle \frac{1+\sin x}{\cos x} = \cot \left ( \frac{\pi}{4} - \frac{x}{2} \right ) = \tan \left ( \frac{x}{2} - \frac{\pi}{4} \right )
I would've never through of that :o:
But does my solution check out?

PS: I think you missed post #5
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STEP II Q3

First Part:
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Let x be an irrational number.

Suppose (in hope of a contradiction) that \sqrt[3]{x} = \frac{p}{q} where p,q are integers

 \displaystyle \implies x = \frac{p^3}{q^3}

But x is irrational so this is a contradiction therefore our original assumption was false. Hence the cube root of x is irrational if x is irrational.


Second part:

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Suppose statement is true for n=k.
 \displaystyle U_{k+1} = 5^{\frac{1}{3^{k+1}}} = \left( 5^{\frac{1}{3^{k}}}\right)^{\fra  c{1}{3}}

By the inductive hypothesis  5^{\frac{1}{3^{k}}} is irrational and it follow from our result in the first part of the question that the cube root of this must be irrational therefore it must be true for n=k+1.

The case where n=1 is true (from the question) therefore the result is true for all positive integral value of n.


Last part:

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Define a sequence  A_n = m+U_n-1 of which all the terms are clearly irrational since U_n is irrational for all n and a rational number + irrational number is always irrational.

As  n \to \infty
 \frac{1}{3^{n}} \to 0
 \implies U_n \to 1 As n increases without bound.

Therefore the sequence  A_n converges to m

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I did III/Q4 yesterday, I'll type it up soon.
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Question 5, STEP I, 2003
(i)
 (2x+\frac{1}{x^2})^6 = (2x)^6 + 6(2x)^5(\frac{1}{x^2}) + 15(2x)^4(\frac{1}{x^2})^2 +\cdots

 = 64x^6 + \cdots + 15\times 2^4 \times x^4 \times \frac{1}{x^4} + \cdots


 = 64x^6 + \cdots + \boxed{240} + \cdots \ \ \ \square



 (ax^3 + \frac{b}{x^2})^{5n} = (ax^3)^{5n} + 5n(ax^3)^{5n-1} (\frac{b}{x^2}) + \cdots + \binom{5n}{r} \cdot a^{5n-r} \cdot b^{r} \cdot x^{5(3n-r)} + \cdots

and we require  5(3n-r) = 0

 \implies r = 3n

therefore, the term independent of x is given by:

 \boxed{\binom{5n}{3n} \cdot a^{2n} \cdot b^{3n}}
(ii)
 f(x) = (x^6 + 3x^5)^{\frac{1}{2}}

 f(x) = x^3(1+\frac{3}{x})^{\frac{1}{2}} \iff |x| > 3

 f(x) = x^3 \left [ 1 + \cdots + \dfrac{(\frac{1}{2})(\frac{-1}{2})(\frac{-3}{2})}{3!} (\frac{3}{x})^3 + \cdots \right ]

 f(x) = x^3 \left [ 1 + \cdots + \frac{27}{16x^3} + \cdots \right ]

therefore the term independent of x is  \boxed{\frac{27}{16}} \ \ \ \square

 f(x) = x^3 (1+\frac{3}{x})^{\frac{1}{2}}

 \iff f(x) = x^3 (\frac{3}{x})^{\frac{1}{2}} (1+\frac{x}{3})^{\frac{1}{2}}  \iff |x| < 3

 \iff f(x) = x^{\frac{5}{2}} \sqrt{3} (1+\frac{x}{3})^{\frac{1}{2}}

 \iff f(x) = x^{\frac{5}{2}} \sqrt{3} \left [ 1 + \frac{x}{6} -\frac{x^2}{72} + \cdots \right ]

 \iff f(x) = x^{\frac{5}{2}} \sqrt{3} + \frac{x^{\frac{7}{2}} \sqrt{3}}{6} + \cdots

and as the powers of x are increasing, beginning at  x^{\frac{5}{2}} , there will be no term independent of x.
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STEP I question 3

Let x= theta for notational purposes
(i)
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 \displaystyle 2sin\left(\frac{x}{2}\right) = sinx \iff 2sin\left(\frac{x}{2}\right) = 2sin\left(\frac{x}{2}\right)cos\  left(\frac{x}{2} \right)
 \displaystyle \iff 2sin\left(\frac{x}{2}\right) \left[1-cos\left(\frac{x}{2}\right) \right] = 0

 \iff sin\left(\frac{x}{2}\right)=0 or  cos\left(\frac{x}{2}\right)=1

Now:
 sin\left(\frac{x}{2}\right) = 0 \iff \frac{x}{2} = n\pi
 cos\left(\frac{x}{2}\right) = 1 \iff \frac{x}{2} = 2n\pi

Where n is an integer.

But  \{2n\pi:n \in \mathbb{Z}\} \subset \{n\pi:n \in \mathbb{Z}\}

Therefore  \displaystyle 2sin\left(\frac{x}{2}\right) = sinx \iff sin\left(\frac{x}{2}\right) =0


(ii)

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Let  tan\left(\frac{x}{2}\right) = t

 \displaystyle 2t = \frac{2t}{1-t^2} \iff t^3 = 0 \iff t=0 \iff sin\left(\frac{x}{2}\right)=0 \iff \frac{x}{2} = n\pi
\iff x = 2n\pi

Where n is an integer.


(iii)

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 2cos\left(\frac{x}{2}\right) = cosx \iff 2cos\left(\frac{x}{2}\right) = 2cos^2\left(\frac{x}{2}\right)-1
 \iff 2cos^2\left(\frac{x}{2}\right)-2cos\left(\frac{x}{2}\right)-1 =0

 \displaystyle \iff cos\left(\frac{x}{2}\right) = \frac{2 \pm  \sqrt{12}}{4} = \frac{1 \pm  \sqrt{3}}{2}

But  \displaystyle \left|\frac{1 + \sqrt{3}}{2} \right|>1
Hence no solutions arise from this.

Also,  \displaystyle \frac{1 - \sqrt{3}}{2} < 0

So we use  \displaystyle \frac{\sqrt{3}-1}{2} to find the other angles.

Cosx is -ve for  \displaystyle \frac{\pi}{2}&lt;x&lt;\frac{3 \pi}{2} when considering 0<x<2pi

Therefore the angles that we want are:

 \displaystyle \frac{x}{2} = (2n+1)\pi \pm cos^{-1}\left(\frac{\sqrt{3} -1}{2} \right)

 \implies \displaystyle x = (4n+2)\pi \pm 2cos^{-1}\left(\frac{\sqrt{3} -1}{2} \right)

And this is equal to  \boxed{\displaystyle (4n+2)\pi \pm 2 \phi \right)}

As required.

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Question 7, STEP I, 2003
First Part
 10k-k^2 = 25-(k-5)^2 \implies 10k-k^2 \leq 25 with equality if k = 5.

 k(10-k) \geq 0 for k being an integer between 0 and 9 inclusive

therefore  0 \leq 10k-k^2 \leq 25 \ \ \ \square

 k(k+1)(k-1) is merely a product of three consecutive integers, and we know that exactly one integer will be a multiple of 3, so we know that this expression will have a factor of 3, and so is divisible by 3.
Second Part
 N = 100a + 10b + c \implies S = a + b^2 + c^3 for  1 \leq a \leq 9 and  0 \leq b, c \leq 9

 100a + 10b + c = a + b^2 + c^3

 \iff 99a +10b-b^2 = c(c+1)(c-1)

Now as 99a and c(c+1)(c-1) are multiples of 3, therefore b(10-b) must also be a multiple of 3, so we require b = {1, 3, 4, 6, 7, 9}.
We also can see than c(c+1)(c-1) must be greater than or equal to 99, which implies that  c^3-c must be greater than 99.
This implies that c = {5, 6, 7, 8, 9}.

By rewriting the equation:

 99a = c(c+1)(c-1) - (10b-b^2)

We can write down all possible values of the LHS, as we just look for the multiples of 99, with respect to the range of a:
99a = {99, 198, 297, 396, 495, 594, 693, 792, 891}

We can also write down all possible values of c(c+1)(c-1), with respect to our possible values of c:
 5^3 - 5 = 120
 6^3 - 6 = 210
 7^3 - 7 = 336
 8^3 - 8 = 504
 9^3 - 9 = 720

We can also write down all the possible values of b(10-b) with respect to our set of b:
b(10-b) = {9, 21, 24}

Notice that, going back to the equation 99a = c(c+1)(c-1) - b(10-b), the maximum of the RHS is 720 - 9 = 711, so we can reject the two higher values of 99a, ie: 99a = {99, 198, 297, 396, 495, 594, 693}

Now by looking at cases of c(c+1)(c-1):

c(c+1)(c-1) = 120. We see that b(10-b) = 21 provides us with a solution, ie: c = 5, b = {7, 3} a = 1

c(c+1)(c-1) = 210, and by checking the set of b(10-b), subtracting them from 210, we do not get a multiple of 99, so there is no solution for c = 6

c(c+1)(c-1) = 336, and once more no solution, with the possible values of b(10-b), so no solution for c = 7

c(c+1)(c-1) = 504, and by checking the set of b(10-b), we see that b(10-b) = 9 provides us with a solution, ie: c = 8, b = {1, 9}, a = 5

c(c+1)(c-1) = 720, and once more, no solution, so no solution for c = 9.

Therefore our solutions are 135, 175, 518, 598.

I have a feeling I've made a mistake somewhere, and I've missed a solution.
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Unbounded
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Question 9, STEP I, 2003
First Part
Considering the situation vertically, and using s=ut+\frac{1}{2}at^2 , and denoting t to be the time when the particle reaches P.

we find that:  h = Vt\sin \theta -\frac{g}{2}t^2

And considering it horizontally:

d = Vt\cos \theta

 \iff t = \dfrac{d}{V\cos \theta}

substituting this into the first equation gives:

 h = d\tan \theta -\dfrac{gd^2}{2V^2\cos^2 \theta}

 \iff \dfrac{2V^2h\cos^2 \theta}{gd^2} = \dfrac{2V^2 \tan \theta \cos^2 \theta}{gd} - 1

 \iff \dfrac{2kh}{d}\cos^2 \theta = 2k\tan \theta \cos^2 \theta - 1

 \iff 1 - 2kT\cos^2 \theta + \dfrac{2kh}{d}\cos^2 \theta = 0

 \iff \sin^2 \theta + \cos^2 \theta - 2kT\cos^2 \theta + \dfrac{2kh}{d}\cos^2 \theta = 0

 \iff \dfrac{\sin^2 \theta}{\cos^2 \theta} + 1 - 2kT + \dfrac{2kh}{d} = 0

 \iff \boxed{T^2 - 2kT + \dfrac{2kh}{d} + 1 = 0} \ \ \ \square
Second Part
Considering  T^2 - 2kT + \frac{2kh}{d} + 1 = 0 as a quadratic in T, there exist two distinct solutions (and correspondingly two distinct angles) if and only if the discriminant is greater than zero.

 \iff (-2k)^2 -4\left(\frac{2kh}{d} + 1\right) &gt; 0

 \iff k^2 - \frac{2kh}{d} &gt; 1

 \iff \left(k-\frac{h}{d}\right)^2 - \frac{h^2}{d^2} &gt; 1

 \iff \left(k-\frac{h}{d}\right)^2 &gt; \frac{h^2+d^2}{d^2}

 \iff k-\frac{h}{d} &gt; \frac{\pm\sqrt{h^2+d^2}}{d}

And as  \frac{\sqrt{h^2+d^2}}{d} &gt; \frac{-\sqrt{h^2+d^2}}{d}

Then  k-\frac{h}{d} &gt; \frac{\sqrt{h^2+d^2}}{d}

 \iff \boxed{kd &gt; h + \sqrt{h^2+d^2}} \ \ \ \square
Third Part
Looking back at the quadratic in T:  T^2 - 2kT + \frac{2kh}{d} + 1 = 0

Let it factorise into two linear factors: T^2 - 2kT + \frac{2kh}{d} + 1 = (\tan \alpha-m)(\tan \beta-n) = 0

ie.  \tan \alpha = m and  \tan \beta = n

By expanding the brackets, we see that  mn = \frac{2kh}{d} + 1 and also that  m+n = 2k

Now looking at the equation  \alpha + \beta = \pi - \mathrm{arctan} \frac{d}{h}

 \iff \tan (\alpha + \beta) = \tan (\pi + \mathrm{arctan} \frac{d}{h})

 \iff \dfrac{\tan \alpha + \tan \beta}{1-\tan \alpha \tan \beta} = \dfrac{\tan \pi - \frac{d}{h}}{1- \frac{d}{h}\tan \pi}

 \iff \dfrac{m+n}{1-mn} = \dfrac{0-\frac{d}{h}}{1-0}

 \iff \dfrac{2k}{1-\frac{2kh}{d} - 1} = \dfrac{-d}{h}

 \iff \dfrac{1}{-\frac{h}{d}} = \dfrac{-d}{h} which is necessarily true.

So the statement  \alpha + \beta = \pi - \mathrm{arctan} \frac{d}{h} is true.
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Question 8, STEP I, 2003
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Let the B = yV. the rate at which B increases with respect to time is:

 \dfrac{\mathrm{d}B}{\mathrm{d}t} = V\dfrac{\mathrm{d}y}{\mathrm{d}t  }

From the question, we see that:

 V \dfrac{\mathrm{d}y}{\mathrm{d}t} = kVxy

 \iff \dfrac{\mathrm{d}y}{\mathrm{d}t} = ky(1-y)

 \iff \dfrac{1}{y(1-y)} \ \mathrm{d}y = k \ \mathrm{d}t

 \iff \displaystyle\int \dfrac{1}{y(1-y)} \ \mathrm{d}y = \displaystyle\int k \ \mathrm{d}t

 \iff \displaystyle\int \left(\dfrac{1}{y} + \dfrac{1}{1-y}\right) \ \mathrm{d}y = kt + C where C is some constant

 \iff \ln y - \ln (1-y) =  kt + C

 \iff \ln \left(\dfrac{y}{1-y}\right) = kt + C

 \iff \dfrac{y}{1-y} = De^{kt} where  D = e^C

 \iff y = De^{kt} - yDe^{kt}

 \iff y(1 + De^{kt}) = De^{kt}

 \iff \boxed{y = \dfrac{De^{kt}}{1+De^{kt}}} \ \ \ \square


When A completely turns into B, y = 1.

 \iff 1 = \dfrac{De^{kt}}{1+De^{kt}}

 \iff 1 + De^{kt} = De^{kt}

 \iff 1 = 0 and we have a contradiction and can conclude that A never completely turns into B.

I have a very strong feeling I've missed something, in the question
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qgujxj39
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STEP III 2003, Q4

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\frac{dy}{dt} = 1 + 3t^2, \frac{dx}{dt} = 2t, therefore \frac{dy}{dx} = \frac{1 + 3t^2}{2t}

Using y - y_1 = m(x - x_1), we obtain the equation of the tangent:

y - t(1 + t^2) = \frac{1 + 3t^2}{2t}(x - t^2)

This meets the curve again when x = T^2, y = T(1 + T^2). Substituting this in and rearranging to find a quadratic in T^2, we obtain:

(2t)T^3 - (1 + 3t^2)T^2 + (2t)T + (t^4 - t^2) = 0

By the factor theorem, we find that T = t is a root. We can factorise and divide by (T - t) (because T \neq t) to obtain

(2t)T^2 - (1 + t^2)T + (t - t^3) = .

Using the quadratic formula, we obtain

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

One of these gives T = t, which we can disregard.

Therefore, T = \frac{1 - t^2}{2t} as required.



The double angle formula for tangent is:

\tan 2\theta = \displaystyle\frac{2\tan \theta}{1 - \tan^2 \theta}

Therefore, \cot 2\theta = \displaystyle\frac{1 - \tan^2 \theta}{2\tan \theta}

So, if t_i = \tan \theta, then t_{i + 1} = \cot 2\theta

Therefore, if t_0 = \tan (\frac{7\pi}{18}), then
t_1 = \cot (\frac{7\pi}{9})

We can also apply the formula similarly to the cotangent function, so

t_2 = \tan (\frac{14\pi}{9})
t_3 = \cot (\frac{28\pi}{9})

Using the identities \cot(\theta + \pi) = \cot \theta and \cot(\frac{\pi}{2} - \theta) = \tan \theta, we get

t_3 = \tan (\frac{7\pi}{18}) = t_0 as required.

Another value of t_0 which satisfies the required properties is \tan(\frac{\pi}{9}).

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Question 10, STEP I, 2003
First Part
Because the lamina is uniform, we can model masses by areas.

The area of the rectangle ABCD is pq. It's centre of mass is at the point  (\frac{p}{2} , \frac{q}{2})

The ara of the triangle ABX is qr/2 and it's centre of mass is  (\frac{r}{3}, \frac{2q}{3} )

Therefore we have:

 pq\displaystyle\binom{\frac{p}{2  }}{\frac{q}{2}} - \frac{qr}{2} \displaystyle\binom{\frac{r}{3}}  {\frac{2q}{3}} = q(p-\frac{r}{2}) \displaystyle\binom{a}{b}

Equating the coordinates, we get:

 \frac{p^2q}{2} - \frac{qr^2}{6} = aq(p-\frac{r}{2})

 \iff 3p^2 - r^2 = 3a(2p-r)

 \iff \boxed{a = \dfrac{3p^2-r^2}{3(2p-r)}}

And also:
 \frac{pq^2}{2} - \frac{q^2r}{3} = bq(p-\frac{r}{2})

 \iff 3pq-2qr = 3b(2p-r)

 \iff \boxed{b = \dfrac{q(3p-2r)}{3(2p-r)}}
Second Part
 \tan 45 = \dfrac{b}{a}

 \iff 1 = \dfrac{q(3p-2r)}{3p^2-r^2}

 \iff r^2 -2rq - 3p^2 +3pq = 0

 \iff r = \dfrac{2q\pm \sqrt{4q^2-4(3p(q-p))}}{2}

 \iff r = q \pm \sqrt{q^2 - 3pq + p^2}

If  r = q + \sqrt{q^2 - 3pq + p^2} , then r would be greater than q, which causes X not to lie in the closed interval [B,C], so the area cut away from the lamina would not be a triangle, however we are told it is a triangle, therefore r < q.

 \implies \boxed{r = q - \sqrt{q^2 - 3pq + p^2}} \ \ \ \square
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STEP III 2003 Q10

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n.b. I don't know how to LaTex the dots denoting differentiation w.r.t to time, so I'm just using x for displacement, v for speed and a for acceleration.

a = kxv
\rightarrow v\frac{dv}{dx} = kxv
\rightarrow \frac{dv}{dx} = kx

Integrating gives
v = \frac{1}{2}kx^2 + c
and using the initial conditions gives
v = \frac{1}{2}kx^2 + \frac{1}{2}kd^2

Separating variables gives

\displaystyle\int\frac{\mathrm{d  }x}{x^2 + d^2} = \displaystyle\int\frac{k}{2}\mat  hrm{d}t

\rightarrow \frac{1}{d} \arctan(\frac{x}{d} = \frac{1}{2}kt + c

Using the initial conditions gives c = \frac{1}{d}\arctan 1 = \frac{\pi}{4d}

Therefore, upon rearranging a bit, we get

x = d\tan(\frac{1}{2}dkt + \frac{pi}{4})

As t \rightarrow \frac{\pi}{2dk}, the argument of tangent \rightarrow \frac{\pi}{2} and hence x \rightarrow \infty as required.



In this second case, our first integration combined with the initial conditions gives
v = \frac{1}{2}kx^2 + U - \frac{1}{2}kd^2
=\frac{1}{2}k(x^2 - (d^2 - \frac{2U}{k})

Now, for simplicity, let a^2 = d^2 - \frac{2U}{k}.

Separating variables we obtain

\displaystyle\int\frac{\mathrm{d  }x}{x^2 - a^2} = \displaystyle\int\frac{k}{2}\mat  hrm{d}t

\rightarrow \frac{1}{2a}\ln(\displaystyle\fr  ac{x-a}{x+a}) = \frac{1}{2}kt + c
which becomes
\rightarrow \frac{1}{2a}\ln(\displaystyle\fr  ac{x-a}{x+a}) = \frac{1}{2}kt + \frac{1}{2a}\ln(\displaystyle\fr  ac{d-a}{d+a})

Rearranging a lot, we get the rather nasty answer:

Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
x = \sqrt{d^2 - \frac{2U}{k}} \displaystyle\frac{1 + \frac{d - \sqrt{d^2 - \frac{2U}{k}}}}{d + \sqrt{d^2 - \frac{2U}{k}}} e^{\sqrt{d^2 - \frac{2U}{k}}kt}{1 - \frac{d - \sqrt{d^2 - \frac{2U}{k}}}{d + \sqrt{d^2 - \frac{2U}{k}}} e^{\sqrt{d^2 - \frac{2U}{k}}kt}


As x \rightarrow \infty, the fraction \rightarrow -1

Therefore x \rightarrow -\sqrt{d^2 - \frac{2U}{k}}.

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