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STEP I Q2

STEP I Q2



I will update my post with the last bit ...!
Reply 2
STEP I 2008 Question 2

Solution



EDIT: Beaten to it! Well we did it different ways so I'll put them both up :p:
STEP III Question 1

Using the suggested hint;

(x+y)3=(ax+by)(x+y)=(ax2+by2)+xy(a+b)=(1+5xy)5 \displaystyle \frac{(x+y)}{3} = (ax+by)(x+y) = \left(ax^2 + by^2\right) + xy(a+b) = \frac{(1+5xy)}{5}

Doing the exact idea again;

(x+y)5=(ax2+by2)(x+y)=(ax3+by3)+xy(ax+by)=(3+7xy)21 \displaystyle \frac{(x+y)}{5} = (ax^2+by^2)(x+y) = \left(ax^3+by^3\right) + xy(ax + by) = \frac{(3+7xy)}{21}

Dividing the first equation obtained by the following one results in [after rearranging];

xy=335 xy = \frac{3}{35} along with x+y=67 x+y = \frac{6}{7}

(xα)(xβ)=x2(α+β)x+αβ (x - \alpha )(x - \beta) = x^2 - (\alpha + \beta)x + \alpha \beta

So x and y are the roots of the quadratic;

p26p7+335=0 p^2 - \frac{6p}{7} + \frac{3}{35} = 0

It follows;

x=37±23035  and  y=3723035 \displaystyle x = \frac{3}{7} \pm \frac{2\sqrt{30}}{35} \; and \; y= \frac{3}{7} \mp \frac{2\sqrt{30}}{35}

Subsituting these into equation (i) and equation (ii) gives two simulateneous eqnts in a and b that are too tedious for me to solve without being in an exam.
STEP III, Question 2

This is a classic example of a telescoping series and the summation is as follows:

r=0n((r+1)krk)=(n+1)k \displaystyle \sum_{r=0}^{n} \left((r+1)^k - r^k \right) = (n+1)^k

We then proceed by binomial expansion;

r=0n((r+1)krk)=(n+1)k=kr=0nrk1+(k2)r=0nrk2++r=0n1 \displaystyle \sum_{r=0}^{n} \left((r+1)^k - r^k \right) = (n+1)^k = k\sum_{r=0}^{n}r^{k-1} + \binom{k}{2}\sum_{r=0}^{n} r^{k-2} + \dots + \sum_{r=0}^{n} 1

Using the notation;

Sk(n)=r=0nrk \displaystyle S_k(n) = \sum_{r=0}^{n}r^k it follows that;

kSk1(n)=(n+1)k(n+1)(k2)Sk2(n)(kk1)S1(n) \displaystyle kS_{k-1}(n) = (n+1)^k - (n+1) - \binom{k}{2} S_{k-2}(n) - \dots - \binom{k}{k-1} S_1(n)

Applying the above with k = 4,5 to find S_3 and S_4 respectively quickly works.

(ii) DFranklin solutions;

Claim: S_k(n) is a poly of order k+1 for all k.

Proof: Use strong induction on k. Suppose true for k < N.

Then by (*), SN(n)=(n+1)N+1+S_N(n) = (n+1)^{N+1} + "the sum of a load of polynomials that are of order < N+1 (by induction hypothesis)". Since the sum of an N+1 degree poly and a load of polys of degree < N+1 is a poly of degree N+1, S_N(n) is a N+1 degree poly. Since S(0) = n is a degree 1 poly, the result follows by induction.

For the other bits, I would go:

Write S_N(n) = a_0 +a_1 n + a_2 n^2 +... + a_N n^N + a_{N+1} n^{N+1}

Since S_N(0) = 0 (sum of no terms is zero) we must have a_0 = 0 so the constant term must be zero.
Since S_N(1) = 1 (sum from 1 to 1 of 1^N is 1) we must have a_0 + a_1+...+a_N+a_{N+1} = 1.
Reply 5
STEP III 2008 Question 6

Solution



In other news I found this paper pretty tough. I think this was the only full solution I managed when I did it as a mock :<
STEP III, Question 4

(i) Letting y = x so I don't refer to a y or x by mistake the problem will be restated as;

show that

x > tanh(x/2) if x > 0

Let y1=x  and  y2=tanh(x2) y_1 = x \; and \; y_2 = tanh(\frac{x}{2}) when x = 0, y_1 == y_2 = 0

We then consider the derivitive and see that (y2)<1 (y_2)' < 1 when x > 0 and the result follows.

archosh(z) > (z-1)/root(z^2-1) Let z = cosh(x) the result trivially follows from the above after rewriting (x^2-1) as (x-1)(x+1) and remembering the identity linking cosh(x) with sinh^2(x/2)

(ii) We need to that if in an interval [a,b] that f(x) => g(x) it follows that;

abf(x)dxabg(x)dx \displaystyle \int_{a}^{b} f(x) dx \ge \int_{a}^{b} g(x) dx

Using IBP and splitting the integral we see that;

1xarcosh(x)dx1xx1x21dx \displaystyle \int_1^x arcosh(x) dx \ge \int_1^x \frac{x-1}{\sqrt{x^2-1}} dx

and the result follows.

Integrating (ii) wrt x gives the required result.
Reply 7
STEP I Question 3

Introduction

Part (i)



Part (ii) coming when I can be bothered with tedium.
STEP III, question 8

1 + 2P = 0 , <---> p = -1/2

S(1+px) = (x/3) + (x*x/6) + (x*x*x/12) + ... - (x*x/6) - (x*x*x/12) - .... = x/3

It follows

S(1x2)=x3    S=2x3(2x) \displaystyle S_{\infty}(1-\frac{x}{2}) = \frac{x}{3} \iff S_{\infty} = \frac{2x}{3(2-x)} (as clearly we know that x =/= 2 .

We can do a similar thing by considering pxn+12n+1 \frac{px^{n+1}}{2^{n+1}} and it follows:

Sn+1=(1xn+12n+1)2x3(2x) \displaystyle S_{n+1} = \left(1 - \frac{x^{n+1}}{2^{n+1}} \right) \frac{2x}{3(2-x)}

(ii) Do the same and we obtain: q = 1 , p = -2.5

T(1+px+qx2)=2+3x    T=4+6x(2x)(12x) \displaystyle T_{\infty} (1+px+qx^2) = 2 + 3x \iff T_{\infty} = \frac{4+6x}{(2-x)(1-2x)}

We then proceed by partial fractions:

4+6x2(1x2)(12x)=A1x2+B12x \displaystyle \frac{4+6x}{2(1-\frac{x}{2})(1-2x)} = \frac{A}{1-\frac{x}{2}} + \frac{B}{1-2x}

B = 14/3 and A = -8/3

Then proceed as above;

Tn+1=143(1(2x)n+112x)43(1(x2)n+12x) \displaystyle T_{n+1} = \frac{14}{3} \left(\frac{1-(2x)^{n+1}}{1-2x}\right) - \frac{4}{3} \left(\frac{1-(\frac{x}{2})^{n+1}}{2-x}\right)
Reply 9
DeanK22
(ii) To me this is blatantly obviously true. It has basically been shown by the working above. I do not know what the eaminer wants from this - just look at the above epression? I guess you could talk about the relationship between integration and summation? For a STEP III I am not going to tell you that saying "well it is kin of obvious" is the thing to do but this seems like the case here.
I think you'd be expected to do a proof by induction for the first bit.

e.g. Claim: S_k(n) is a poly of order k+1 for all k.

Proof: Use strong induction on k. Suppose true for k < N.

Then by (*), SN(n)=(n+1)N+1+S_N(n) = (n+1)^{N+1} + "the sum of a load of polynomials that are of order < N+1 (by induction hypothesis)". Since the sum of an N+1 degree poly and a load of polys of degree < N+1 is a poly of degree N+1, S_N(n) is a N+1 degree poly. Since S(0) = n is a degree 1 poly, the result follows by induction.

For the other bits, I would go:

Write S_N(n) = a_0 +a_1 n + a_2 n^2 +... + a_N n^N + a_{N+1} n^{N+1}

Since S_N(0) = 0 (sum of no terms is zero) we must have a_0 = 0 so the constant term must be zero.
Since S_N(1) = 1 (sum from 1 to 1 of 1^N is 1) we must have a_0 + a_1+...+a_N+a_{N+1} = 1.

To be honest, your "it is kind of obvious" comment looks dubious here. There are some pretty clear steps that you need to go through. And waffling about links between integration and summation indicates to me that you are missing the point of the question.
Reply 10
nuodai do you think you could use the same format for the first post as the other years (ie using "Solution by name" as the hyperlink. The only exception to this is 2007 which I intend to fix in the near future)
DFranklin

For the other bits, I would go:

Write S_N(n) = a_0 +a_1 n + a_2 n^2 +... + a_N n^N + a_{N+1} n^{N+1}

Since S_N(0) = 0 (sum of no terms is zero) we must have a_0 = 0 so the constant term must be zero.
Since S_N(1) = 1 (sum from 1 to 1 of 1^N is 1) we must have a_0 + a_1+...+a_N+a_{N+1} = 1.


:o: I just read the part about proving it would be a polynomial of degree k+1 - these other bits look abit more involved. Do you think it would be neccessary to prove the bit about the degree of the polnomial or would it be OK to say that from the previous working it can be seen?
Reply 12
SimonM
nuodai do you think you could use the same format for the first post as the other years (ie using "Solution by name" as the hyperlink. The only exception to this is 2007 which I intend to fix in the near future)

Consider it done.
DeanK22
:o: I just read the part about proving it would be a polynomial of degree k+1 - these other bits look abit more involved. Do you think it would be neccessary to prove the bit about the degree of the polnomial or would it be OK to say that from the previous working it can be seen?
No, it wouldn't be OK.
II/7:.

(i) y=u(1+x2)1/2dydx=dudx(1+x2)1/2+xu(1+x2)1/2y = u(1+x^2)^{1/2} \Rightarrow \dfrac{\text{d}y}{\text{d}x} = \dfrac{\text{d}u}{\text{d}x}(1+x^2)^{1/2} + xu(1+x^2)^{-1/2}

(1+x2)1/2u[dudx(1+x2)1/2+xu(1+x2)1/2]=xu(1+x2)1/2+x1+x2\Rightarrow \dfrac{(1+x^2)^{-1/2}}{u} \left[ \dfrac{\text{d}u}{\text{d}x}(1+x^2)^{1/2} + xu(1+x^2)^{-1/2} \right] = xu(1+x^2)^{1/2} + \dfrac{x}{1+x^2}

1u2dudx=x(1+x2)1/21u=13(1+x2)3/2+c\Rightarrow \dfrac{1}{u^2}\dfrac{\text{d}u}{\text{d}x} = x(1+x^2)^{1/2} \Rightarrow \dfrac{-1}{u} = \dfrac{1}{3}(1+x^2)^{3/2} + c when x = 0, y = 1 so u = 1 so c = -4/3.

Solution: (1+x2)1/2y=13(1+x2)3/243- \dfrac{(1+x^2)^{1/2}}{y} = \dfrac{1}{3}(1+x^2)^{3/2} - \dfrac{4}{3}.

(ii) Let y=u(1+x3)1/3dydx=dudx(1+x3)1/3+ux2(1+x3)2/3y = u(1+x^3)^{1/3} \Rightarrow \dfrac{\text{d}y}{\text{d}x} = \dfrac{\text{d}u}{\text{d}x}(1+x^3)^{1/3} + ux^2(1+x^3)^{-2/3}

(1+x3)1/3u[dudx(1+x3)1/3+ux2(1+x2)2/3]=ux2(1+x2)1/3+x21+x3\Rightarrow \dfrac{(1+x^3)^{-1/3}}{u} \left[ \dfrac{\text{d}u}{\text{d}x} (1+x^3)^{1/3} + ux^2(1+x^2)^{-2/3} \right] = ux^2(1+x^2)^{1/3} + \dfrac{x^2}{1 + x^3}

1u2dudx=x2(1+x3)1/31u=14(1+x3)4/3+c\Rightarrow \dfrac{1}{u^2} \dfrac{\text{d}u}{\text{d}x} = x^2(1+x^3)^{1/3} \Rightarrow \dfrac{-1}{u} = \dfrac{1}{4}(1+x^3)^{4/3} + c
u = 1, again, so c = - 5/4.

Solution: (1+x3)1/3y=14(1+x3)4/354- \dfrac{(1+x^3)^{1/3}}{y} = \dfrac{1}{4}(1+x^3)^{4/3} - \dfrac{5}{4}.

(iii) I conjecture the solution to be (1+xn)1/ny=1n+1(1+xn)1+1/nn+2n+1- \dfrac{(1+x^n)^{1/n}}{y} = \dfrac{1}{n+1}(1+x^n)^{1 + 1/n} - \dfrac{n+2}{n+1}. To prove it, one would make the substitution y = u(1+x^n)^(1/n).
II/5:

0π/2sin2x1+sin2xdx=0π/22cosxsinx1+sin2xdx\displaystyle \int_0^{\pi/2} \frac{\sin 2x}{1 + \sin^2 x} \, \text{d}x = \int_0^{\pi/2} \frac{2 \cos x \sin x}{1 + \sin^2 x} \, \text{d}x Letting u = sin x, this becomes
012u1+u2du=[ln(1+u2)]01=ln2\displaystyle \int_0^1 \frac{2u}{1 + u^2} \, \text{d}u = \left[ \ln(1 + u^2) \right]^1_0 = \ln 2.

0π/2sinx1+sin2xdx=0π/2sinxcos2x2dx\displaystyle \int_0^{\pi/2} \frac{\sin x}{1 + \sin^2 x} \, \text{d}x = \int_0^{\pi/2} \frac{- \sin x}{\cos^2 x - 2} \, \text{d}x u = cos x, this becomes
10duu22=122[lnx2x+2]10=122ln2+121\displaystyle \int_1^0 \frac{\text{d}u}{u^2 -2} = \frac{1}{2 \sqrt{2}} \left[ \ln \left| \frac{x - \sqrt{2}}{x + \sqrt{2}} \right| \right]^0_1 = \frac{1}{2 \sqrt{2}} \ln \frac{\sqrt{2} + 1}{\sqrt{2} - 1}

(1+2)5=1+52+20+202+20+42=41+292<99292<582<22<4(1 + \sqrt{2})^5 = 1 + 5 \sqrt{2} + 20 + 20 \sqrt{2} + 20 + 4 \sqrt{2} = 41 + 29 \sqrt{2} < 99 \Leftrightarrow 29 \sqrt{2} < 58 \Leftrightarrow \sqrt{2} < 2 \Leftrightarrow 2 < 4.
2>1.42>1.96\sqrt{2} > 1.4 \Leftrightarrow 2 > 1.96

As 22>21.42^{\sqrt{2}} > 2^{1.4} (exponential functions are strictly increasing), sufficient to show that 27/5>1+227>(1+2)52^{7/5} > 1 + \sqrt{2} \Leftrightarrow 2^7 > (1 + \sqrt{2})^5, and 128 > 99 is enough to show this holds.

Suppose ln2>122ln2+12122ln2>ln(1+2)22ln2>ln(2+1)ln22>ln(1+2)22>2+1\ln 2 > \dfrac{1}{2 \sqrt{2}} \ln \dfrac{\sqrt{2} + 1}{\sqrt{2} - 1} \Leftrightarrow 2 \sqrt{2} \ln 2 > \ln (1 + \sqrt{2})^2 \Leftrightarrow \sqrt{2} \ln 2 > \ln (\sqrt{2} + 1) \Leftrightarrow \ln 2^{\sqrt{2}} > \ln(1 + \sqrt{2}) \Leftrightarrow 2^{\sqrt{2}} > \sqrt{2} + 1, previously established, so the first integral is greater. And we are done, forever.
II/9:

h=40tsinα1210t2-h = 40t \sin \alpha - \frac{1}{2} \cdot 10t^2 and 20=40tcosα20 = 40t \cos \alpha therefore h=40sinα2cosα54sec2αh=20tanα54(tan2α+1)-h = \dfrac{40 \sin \alpha}{2 \cos \alpha} - \dfrac{5}{4} \sec^2 \alpha \Rightarrow -h = 20 \tan \alpha - \dfrac{5}{4}(\tan^2 \alpha + 1)
tanα=20±4005(5/4h)5/2 \Rightarrow \tan \alpha = \dfrac{20 \pm \sqrt{400 - 5(5/4 -h)}}{5/2}.
The larger root means the ball will loop up in the air for ages and ages and ages like some sort of stupid duck so we choose the smaller root.
As t = sec a / 2, and cos a approx 1, t is approx 1/2.

(ii) h > 5/4.

(iii)tanα=20400+25/45/2=2020(1+1/64)1/25/22020(1+1/128)5/2=1/16\tan \alpha = \dfrac{20 - \sqrt{400 + 25/4}}{5/2} = \dfrac{20 - 20(1 + 1/64)^{1/2}}{5/2} \approx \dfrac{20 - 20(1 + 1/128)}{5/2} = -1/16 (first term binomial expansion). Using the tan a = a approximation and that 1 radian is 57 degres, alpha is approximately 57/16 degrees.
STEP III 2008, Question 5

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STEP III 2008, Question 3

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STEP III 2008, Question 10

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