STEP maths I, II, III 1990 solutions

the last part can be derived naturally by written k as a function of p，and we find the range when p<0

Original post by SimonM

STEP II Question 7

f'(t)+f(t)-kf(t-1)=0

If there is a solution of the form

f(t) = Ae^{pt}

then

Ape^{pt} + Ae^{pt} -kAe^{p(t-1)} = 0

That is

Unparseable latex formula:

pe^{pt} +e^{pt} = ke^{p(t-1}}

Or

\boxed{p+1 = ke^{-p}}

as required.

If

k > 0

there will be one solution. (I'm not going to stick my graphs up, (no scanner)) (However, visualise the graphs

y=x+1

and

y = ke^{-x}

If our two graphs are tangential, there will be one solution (this is also the critical point between two solutions and no solutions for

k<0

So

\frac{d}{dx} (x+1) = 1

and

\frac{d}{dx} (ke^{-x}) = -ke^{-x}

That gives us

1 = -ke^{-x} \Rightarrow x = \ln (-k)

Substituting back in gives us

\ln (-k)+1 = ke^{-\ln (-k)} \Rightarrow \ln(-k)+1 = -1 \Rightarrow \boxed{k = -e^{-2}}

So we have

0 solutions if

k < -e^{-2}

1 solution if

k = -e^{-2}

or

k >0

2 solutions if

0>k>-e^{-2}

The final question asks us to find where the solutions are negative.

Consider the critical values when x = 0 is a solution. This gives us

k = 1

When

k<0

,

ke^{-x} < 0

so for this to equal

x+1

, we must have

x<-1

, which gives a negative p

When

k>0

, if

ke^{-x}<x+1

, consider at

x=0

, then the solution will be negative, that is

k<1

So this gives negative p for

\boxed{k<1}

Reply 182

EXM, why in the first part, m is neglected?
I think the better way is to suppose ka-c/ka-b= ka-c^2/ka-b^2=ka-c^3/ka-b^3
thus we can have one variable, and by substitute finally we have (a-b)(a-c)=0 and that is an contradiction.

Original post by Simba

III/2:

i) OA = m(a^3 i + a^2 j + ak)

BC = b^3 i + b^2 j + bk + n[(c^3 - b^3)i + (c^2 - b^2)j + (c - b)k]

For an intersection:

b^3 + n(c^3 - b^3) = a^3. --- (1)
b^2 + n(c^2 - b^2) = a^2. --- (2)
b + n(c - b) = a. --- (3)

Sub (3) in (2):

b^2 + n(c^2 - b^2) = [b + n(c - b)]^2 = b^2 + 2bn(c - b) + n^2.(c - b)^2.

c^2 - b^2 = 2b(c - b) + n(c - b)^2.

(c + b)(c - b) - 2b(c - b) - n(c - b)^2 = 0.

(c - b)[c + b - 2b - n(c - b)] = 0.

(c - b)[c - b - n(c - b)] = 0.

(c - b)(c - b)(1 - n) = 0.

But c =/= b, so n = 1 (if they intersect).

Sub in (3):

b + c - b = a.

c = a.

But c =/= a, so the lines don't intersect.

QED.

ii) cos(AOB) = a.b/|a||b| = [(ab)^3 + (ab)^2 + ab]/root(a^6 + a^4 + a^2).root(b^6 + b^4 + b^2)

= [(ab)^3 + (ab)^2 + ab]/ab.root[(a^4 + a^2 + 1)(b^4 + b^2 + 1)]

= [(ab)^2 + ab + 1]/root[(a^4 + a^2 + 1)(b^4 + b^2 + 1)].

= 3/root(a^4.b^4 + a^4.b^2 + a^4 + a^2.b^4 + a^2.b^2 + a^2 + b^4 + b^2 + 1)

= 3/root(3 + a^2 + a^4 + b^2 + a^2 + b^4 + b^2)

= 3/root(a^4 + b^4 + 2a^2 + 2b^2 + 3)

Since a and b are variable, we can make a (or b) as large as we like, so the denominator can approach infinity. So cos(AOB) can be made arbitrarily close to 0. Values of a and b very close to 1 (but not equal to 1 since a and b are distinct) give cos(AOB) arbitrarily close to 1. It is not possible for a sum of squares (+3) to be negative, so cos(AOB) > 0, outruling pi/2 < AOB < pi.

Hence 0 < cos(AOB) < 1, and so 0 < AOB < pi/2.

QED.

Reply 183

we can continue the method of the complex numbers in part 3, quite easy
let z=cosθ+isinθ, and let w=secθ
LHS=Re(1+zw+z^2w^2+...)=Re (1-(zw)^n)/1-zw, which we can calculated easily

Original post by SimonM

Question 4

Unparseable latex formula:

\begin{array} {l}[br]\displaystyle \cos \alpha + \cos ( \alpha + 2 \beta ) + \ldots + \cos ( \alpha + 2 n \beta ) \\[br]\displaystyle = \rm Re \, \left ( e^{i\alpha} + e^{i(\alpha+2\beta)} + \ldots + e^{i(\alpha+2n\beta)} \right ) \\[br]\displaystyle = \rm Re \, \left ( e^{i\alpha} ( 1+e^{i2\beta}+\ldots+e^{i2n\beta}) \right ) \\[br]\displaystyle = \rm Re \, \left ( e^{i\alpha} \left (\frac{e^{i2(n+1)\beta} - 1}{e^{i2\beta}-1} \right ) \right ) \\[br]\displaystyle = \rm Re \, \left ( e^{i\alpha} \frac{e^{i(n+1)\beta}}{e^{i\beta}} \frac{e^{i(n+1)\beta}-e^{-i(n+1)\beta}}{e^{i\beta}-e^{-i\beta}} \right ) \\[br]\displaystyle = \rm Re \, \left ( e^{i (\alpha + n\beta)} \frac{\sin(n+1)\beta}{\sin \beta} \right ) \\[br]\displaystyle = \frac{\sin (n+1)\beta \cos (\alpha + n \beta)}{\sin \beta} \end{array}

Unparseable latex formula:

\begin{array}{l}[br]\displaystyle \cos \alpha + \binom{n}{1} \cos ( \alpha + 2\beta ) + \ldots + \binom{n}{r} \cos ( \alpha + 2r \beta ) + \ldots + \cos ( \alpha + 2n \beta ) \\[br]\displaystyle = \rm Re \, \left ( e^{i\alpha} \left (1+e^{i2\beta} \right )^n \right ) \\[br]\displaystyle = \rm Re \, \left ( e^{i \alpha} \left ( 2 \cos \beta e^{i \beta} \right )^n \right ) \\[br]\displaystyle = \rm Re \, \left ( e^{i( \alpha + n \beta )} 2^n cos ^n \beta \right ) \\[br]\displaystyle = 2^n \cos^n \beta \cos ( \alpha + n \beta) [br]\end{array}[br]

For our final one, we proceed by induction.

Theorem

\displaystyle 1+\cos \theta \sec \theta + \ldots \cos r \theta \sec^r \theta + \ldots + \cos n \theta \sec ^n \theta = \frac{\sin (n+1)\theta \sec^n \theta}{\sin \theta}

Proof (by Induction)

Base Case

When

n=0

LHS = 1

\displaystyle RHS = \frac{\sin \theta \times 1}{\sin \theta} = 1 = LHS

as required

Inductive Hypothesis

Assume that, for all

k\leq n

,

\displaystyle 1+\cos \theta \sec \theta + \ldots \cos r \theta \sec^r \theta + \ldots + \cos k \theta \sec ^k \theta = \frac{\sin (k+1)\theta \sec^k \theta}{\sin \theta}

Inductive Step

If

\displaystyle 1+\cos \theta \sec \theta + \ldots + \cos r \theta \sec^r \theta + \ldots + \cos n \theta \sec ^n \theta = \frac{\sin (n+1)\theta \sec^n \theta}{\sin \theta}

Then by adding

\cos (n+1)\theta \sec^{n+1} \theta

to each side we get

Unparseable latex formula:

\begin{array}{l} \displaystyle 1+\cos \theta \sec \theta + \ldots + \cos n \theta \sec ^n \theta +\cos (n+1)\theta \sec^{n+1} \theta \\ [br]\displaystyle = \frac{\sin (n+1)\theta \sec^n \theta}{\sin \theta} + \cos (n+1)\theta \sec^{n+1} \theta \\[br]= \displaystyle \frac{\sin(n+1)\theta \sec^n \theta + \sin \theta \cos (n+1) \theta \sec^{n+1}}{\sin \theta} \\[br]= \displaystyle \frac{\sec^{n+1} \theta}{\sin \theta} \left ( \sin(n+1)\theta \cos \theta + \sin \theta \cos (n+1) \theta \right ) \\[br]= \displaystyle \frac{\sec^{n+1} \theta}{\sin \theta} \sin(n+2) \end{array}

Therefore, if our inductive hypothesis is true for n, it is true for n+1. Since it is true for 0, by the principle of mathematical induction, it is true for all nonnegative integers

Reply 184

i find it pretty tricky to do the last part of Ⅱ13
we just need“the length of smaller 2sides of the triangle is larger the largest” to get what we want
and it‘s the key of this problem，rather than the angle---i think

Reply 185

Pade

2

Original post by Simba

II/4:

Line segments:

1: 1 line segment.
2: 4 line segments.
3: 9 line segments.
4: 16 line segments.

So it is reasonable to suppose that n lines are split into n^2 line segments.

Each new line that we introduce splits every existing line into one more piece, and is itself intersected into as many parts as there are lines after it has been added.

So the number of line segments for n lines is 1 + (1 + 2) + (2 + 3) + ... + (n - 1 + n) = 1 + 3 + 5 + ... + 2n - 1.

\sum_{k=1}^n (2k - 1) = n(n+1) - n = n^2

QED.

Planes:

Each new line cuts the existing n - 1 lines in n - 1 places, creating n new regions.

One region exists to begin with.

So the number of planes for n lines is 1 + 1 + 2 + 3 + 4 + ... + n

1 +

\sum_{k=1}^n k = n(n+1)/2 = (n^2+n)/2 = (n^2 + n + 2)/2

as required.

QED.

Alternatively, for the second part, could we not have used Euler's formula for planar graphs and with some modifications arrive at the result?

Reply 186

Original post by Pade

Alternatively, for the second part, could we not have used Euler's formula for planar graphs and with some modifications arrive at the result?

Good idea. However,
(i)Euler's formula for planar graphs should be carefully used with a clear statement
(ii)Normally, if you use this formula in STEP without a detailed explanation, you'll lose some marks in STEP. It's better to use the basic methods to figure out STEP questions.

Reply 187

Pade

2

Original post by cxs

Good idea. However,
(i)Euler's formula for planar graphs should be carefully used with a clear statement
(ii)Normally, if you use this formula in STEP without a detailed explanation, you'll lose some marks in STEP. It's better to use the basic methods to figure out STEP questions.

Ok, thanks.

Reply 188

In here, there is a nice parametric solution for STEP III, Q2.

Reply 189

Billord

6

Original post by Dystopia

Attempt at STEP III, Q8.

Spoiler

It might help you to use your initial conditions as soon as you get any kind of constant. I did that and got something slightly nicer to look at. I don't know latex to forgive my presentation:
((x^2-2x+2)e^x-(56x-94)e^2)/(1-x^3)
However I am extremely prone to making mistakes so this could too be wrong

Reply 190

richmaths

13

Original post by SimonM

STEP I, Question 7

\dfrac{dy}{dx} +Py=Q

Let

y = uv

So

\displaystyle uv' + vu' + Puv = Q \Rightarrow v' + v \left (\frac{u'}{u} + P \right) = \frac{Q}{u}

Therefore, we can express

v'

in terms of

Q, u, x

if

P+\dfrac{u'}{u} = 0

That is

\boxed{u'+Pu = 0}

.

Solving

\displaystyle \frac{dy}{dx} - \frac{2y}{x+1} = (x+1)^{\frac{5}{2}}

Let

y = uv

, to find a suitable u, solve

u'+Pu =0

, that is

\frac{du}{dx} -\frac{2u}{x+1} = 0

Some separating variables gives

u = A(x+1)^2

We must now solve

v' = \frac{(x+1)^{\frac{5}{2}}}{A(x+1)^2}

, that is

v' = \frac{1}{A}(x+1)^{\frac{1}{2}}

, so

v = \frac{2}{3A}(x+1)^{\frac{3}{2}}+D

Therefore

y = uv = \frac{2}{3}(x+1)^{\frac{7}{2}}+F(x+1)^2

y(0) = \frac{2}{3}+G = 1 \Rightarrow G = \frac{1}{3}

Therefore

\boxed{y = \frac{1}{3} \Big (2(x+1)^{\frac{7}{2}}+(x+1)^{2}\Big)}

Valid for all

x>-1

(I think)

why would u'/u + P = 0?

Reply 191

siyuan04

3

Original post by Dystopia

Attempt at STEP III, Q8.

Spoiler

I know this is 14 years into the future, but I can confirm that you have indeed made an error. The right hand side of the original question was (x^3+3x)e^x, not (x^3+3x^2)e^x.

Reply 192

RuoyanC

Original post by SimonM

STEP II Question 7

f'(t)+f(t)-kf(t-1)=0

If there is a solution of the form

f(t) = Ae^{pt}

then

Ape^{pt} + Ae^{pt} -kAe^{p(t-1)} = 0

That is

Unparseable latex formula:

pe^{pt} +e^{pt} = ke^{p(t-1}}

Or

\boxed{p+1 = ke^{-p}}

as required.

If

k > 0

there will be one solution. (I'm not going to stick my graphs up, (no scanner)) (However, visualise the graphs

y=x+1

and

y = ke^{-x}

If our two graphs are tangential, there will be one solution (this is also the critical point between two solutions and no solutions for

k<0

So

\frac{d}{dx} (x+1) = 1

and

\frac{d}{dx} (ke^{-x}) = -ke^{-x}

That gives us

1 = -ke^{-x} \Rightarrow x = \ln (-k)

Substituting back in gives us

\ln (-k)+1 = ke^{-\ln (-k)} \Rightarrow \ln(-k)+1 = -1 \Rightarrow \boxed{k = -e^{-2}}

So we have

0 solutions if

k < -e^{-2}

1 solution if

k = -e^{-2}

or

k >0

2 solutions if

0>k>-e^{-2}

The final question asks us to find where the solutions are negative.

Consider the critical values when x = 0 is a solution. This gives us

k = 1

When

k<0

,

ke^{-x} < 0

so for this to equal

x+1

, we must have

x<-1

, which gives a negative p

When

k>0

, if

ke^{-x}<x+1

, consider at

x=0

, then the solution will be negative, that is

k<1

So this gives negative p for

\boxed{k<1}

I think for the last question you forgot k
must be greater than -e^(-2) as if x is less than it there's no such solution.

Reply 193

Original post by brianeverit

1990 Paper 1 numbers 11 & 13

I get a more involved solution for no 13 . First part is correct since cylinder is held still so can be ignored in the dynamics. But for the second part need to take into consideration the normal reaction of the rail against the cylinder in resolving the forces. This gives the normal reaction against the horizontal plane to be (sqrt(3)(m+M)g)/(sqrt(3) + /mu). Then using moments we get /theta is at most arcsin((Sqrt(3)*/mu*(m+M))/((sqrt(3) + /mu)m)). Substituting the values given this calculates theta to be 0.64 rads or 36.7 deg

(edited 10 months ago)

Reply 194

Original post by siyuan04

I know this is 14 years into the future, but I can confirm that you have indeed made an error. The right hand side of the original question was (x^3+3x)e^x, not (x^3+3x^2)e^x.

Actually, no. The original paper states (x^3+3x^2)e^x. For some reason there is a misprint in the STEP database rendering of this question.

Reply 195

Original post by Speleo

III/9 Easy as

\pi e

tant = sinhu
v = sechu = 1/coshu = 1/sqrt(1+sinh^2u)
= 1/sqrt(1+tan^2t) = 1/sqrt(sec^2t) = 1/sect = cost

dt/du = (dv/du)/(dv/dt) = (-sechutanhu)/(-sint)
= sechu.sqrt(1-sech^2u)/sqrt(1-cos^2t)
= v.sqrt(1-v^2)/sqrt(1-v^2) = v

dv/du = -sechutanhu, -2dv/du = 2sechutanhu
= 2v.sqrt(1-v^2) = 2costsint = 2sin2t
d2v/du2 = d/du(-sechutanhu) = sechutanh^2u - sech^3u
-coshud2v/du2 = sech^2u - tanh^2u = v^2 - (1-v^2)
= cos^2t - sin^2t = cos2t

du/dt.d2v/dt2 = 1/v.(-cost) = -cost/cost = -1
dv/dt.d2u/dt2 = -sint.d/dt(1/v) = -sint.d/dt(sect) = -sint.sect.tant
= -tan^2t
(du/dt)^2 = 1/v^2 = 1/cos^2t = sec^2t = 1 + tan^2t
Sum of these parts = - 1 - tan^2t + 1 + tan^2t = 0

16 years later and noting that the first two terms in part (iv} relate to the product rule makes this part very straightforward, Given previous parts, not sure the examiner intended people to see that trick?

Reply 196

Original post by brianeverit

1990 Paper III number 11

I get a slightly different answer here - (17*sqrt(2)/6)ga. Think OP made a minor arithmetic error. Also find it interesting that the question states 'find in terms of a, M and g' given that M has no involvement in the final answer.

Reply 197