STEP 2018 Solutions

Original post by Notnek

STEP I Q8 :

(i)

Spoiler

(ii)

Spoiler

(iii)

Spoiler

(iv)

Spoiler

(v)

Spoiler

You know what’s freakin awesome? We got an explicit solution of an otherwise unsolvable integral using functions that we defined ourselves without even knowing what the functions are. Also I’m not sure if it was *necessary* to give the second last integral in a form not involving inverse functions because the other part made you give the answer in s^-1 form

Reply 5

OP

Note that we have

\displaystyle f'(x) = \frac{(1-\log^2 x)(\log^3 x - 3\log^2 x-\log x-1)}{x^2 \log^2 x}

so that we can see that

f,f' = 0

when

\ln^2 x = 1

.

Let

u = \log t

, then for

x \in (0,1)

we have

Unparseable latex formula:

\displaystyle [br]\begin{align*}F(x) = \int_{-1}^{\log x} \frac{(1-u^2)^2}{u} \, \mathrm{d}u &= \left[\log |u| - u^2 + \frac{u^4}{4}\right]_{-1}^{\log x} \\ & = \log (-\log x) - \log^2 x + \frac{1}{4}\log^4 x + 1 - 1/4\end{align*}

For any

x > 0

with

x \neq 1

, we have

F(x) = \log |\log x| - \log^2 x + \frac{1}{4} \log^4 x + 1 - 1/4

i) So

F(x^{-1}) = \log |-\log x| - (-\log x)^2 + \frac{1}{4}(-\log x)^4 + 1 - 1/4 = F(x)

ii) Check on Desmos.

(edited 5 years ago)

Reply 6

OP

Original post by Student1256

You know what’s freakin awesome? We got an explicit solution of an otherwise unsolvable integral using functions that we defined ourselves without even knowing what the functions are.

It's just defining sine and cosine in one particular way and then doing the usual sine and cosine stuff.

Reply 7

Original post by Zacken

It's just defining sine and cosine in one particular way and then doing the usual sine and cosine stuff.

Yes when I was doing the exam I was like wow this really feels like trig. I’m just saying we didn’t need to know what the actual function is to use it to aid us in finding the integral

Reply 8

JeremyC

8

Original post by Zacken

It's just defining sine and cosine in one particular way and then doing the usual sine and cosine stuff.

That's what I thought when I first saw the question, but it's slightly different.

In the question,

s'(x) = c(x)^2

c'(x) = -s(x)^2

vs

s'(x) = c(x)

c'(x) = -s(x)

for the normal sine and cosine functions.

I agree with Student1256 - it's an awesome question. :biggrin:

(edited 5 years ago)

Reply 9

Notnek

21

Original post by Student1256

You know what’s freakin awesome? We got an explicit solution of an otherwise unsolvable integral using functions that we defined ourselves without even knowing what the functions are. Also I’m not sure if it was *necessary* to give the second last integral in a form not involving inverse functions because the other part made you give the answer in s^-1 form

I agree, it is a well put together question.

I suppose it's technically what they asked for but I'm unsure if it would get all the marks. @DFranklin what do you think?

(edited 5 years ago)

Reply 10

etothepiiplusone

15

I can do Q12, it'll be 30 minutes, watch this space!

EDIT: Q12 parts i) and ii) : (. I need to comne back to this tomorrow to add parts iii) and iv), because bizarrely I can't do the algebra without the time pressure, when I was able to do it with pressure in the exam ...

EDIT EDIT: Done @Zacken ! Although I will need to proofread this.

i)

Spoiler

ii)

Spoiler

iii)

Spoiler

iv)

Spoiler

(edited 5 years ago)

Reply 11

OP

Original post by JeremyC

That's what I thought when I first saw the question, but it's slightly different.

In the question,

s'(x) = c(x)^2

c'(x) = -s(x)^2

vs

s'(x) = c(x)

c'(x) = -s(x)

for the normal sign and cosine functions.

Ah right yeah, juggling typing and looking on a phone so missed that

Reply 12

Rohan77642

14

Original post by Zacken

Note that we have

\displaystyle f'(x) = \frac{(1-\log^2 x)(\log^3 x - 3\log^2 x-\log x-1)}{x^2 \log^2 x}

so that we can see that

f,f' = 0

when

\ln^2 x = 1

.

Let

u = \log t

, then for

x \in (0,1)

we have

\displaystyle

\begin{align*}F(x) = \int_{-1}^{\log x} \frac{(1-u^2)^2}{u} \, \mathrm{d}u &= \left[\log |u| - u^2 + \frac{u^4}{4}\right]_{-1}^{\log x} \\ & = \log (-\log x) - \log^2 x + \frac{1}{4}\log^4 x + 1 - 1/4\end{align*}

For any

x > 0

with

x \neq 1

, we have

F(x) = \log |\log x| - \log^2 x + \frac{1}{4} \log^4 x + 1 - 1/4

i) So

F(x^{-1}) = \log |-\log x| - (-\log x)^2 + \frac{1}{4}(-\log x)^4 + 1 - 1/4 = F(x)

ii) Check on Desmos.

How much do you (and other people) think I would get for the question if I did the parts highlighted in the solution?

Reply 13

OP

Original post by GcseLad-_-

@Zacken Opinions on difficulty of the paper?

Haven’t had a chance to look properly, but I’ve seen a few very easy/short questions on a short glance (Q4/8/etc...) but my opinion might be wrong nowadays, after two years of uni maths. Think it’s just a standard paper. Not too hard, not too easy.

(edited 5 years ago)

Reply 14

gnsbrg

7

STEP I Q5
I)

f(x)=k(x-1)(x-2)(x-3)(x-4)+1

where

k

is a non-zero real constant.

II)
From I) we know

P(x)=k(x-1)\ldots (x-N)+1

again non-zero

k

P(N+1)=k(N)\ldots 1+1=kN!+1

If

P(N+1)=1

,then

kN!=0

,which is impossible.

Given

P(N+1)=2

, then

kN!=1

,that is

k=\frac{1}{N!}

Therefore,

P(N+r)=\frac{1}{N!}(N+r-1)\ldots (r)+1=\frac{(N+r-1)!}{N!(r-1)!}+1={N+r-1\choose N}+1

Using this, it is plain that

r=2

verifies

P(N+r)=N+r

III)

Since the coefficient of

x^4

is 1,

k=1

[latex]S(x)=(x-a)(x-b)(x-c)(x-d)+2001

a) If there were one, we would have that

(e-a)(e-b)(e-c)(e-d)=17

This is impossible if

a,b,c,d

are distinct and integers as 17 is prime and three factors should equal 1.

b)

S(0)=2017

implies

abcd=16=2^4

Taking into account

Unparseable latex formula:

a&lt;b&lt;c&lt;d

, we find that there are 5 sets of integers

(a,b,c,d)=(-2,-1,2,4),(-4,-1,1,4),(-4,-2,1,2),(-2,-1,1,8),(-8,-1,1,2)

Thanks to @DFranklin for some corrections in this last bit.

(edited 5 years ago)

Reply 15

Original post by Notnek

I agree, it is a well put together question.

So you're saying you could leave the penultimate integral as s/c? I suppose it's technically what they asked for but I'm unsure if it would get all the marks. @DFranklin what do you think?

Previously it said show that the integral is equal to s^-1(u) +c and that’s it. So I’m pretty sure if you left the other integrals in this form it’d be okay. I’m not saying to leave it in x. You still have to substitute back u

Reply 16

Notnek

21

Original post by Student1256

Previously it said show that the integral is equal to s^-1(u) +c and that’s it. So I’m pretty sure if you left the other integrals in this form it’d be okay. I’m not saying to leave it in x. You still have to substitute back u

Can you post what you mean i.e. post an answer to the penultimate integral that you think would be okay?

Reply 17

JeremyC

8

Original post by Notnek

Can you post what you mean i.e. post an answer to the penultimate integral that you think would be okay?

If you write

x = s^{-1}(u)

then you could have:

\frac{s(x)}{c(x)} = \frac{u}{c(s^{-1}(u))}

This is what I answered (plus the constant) but I think your answer is probably nicer as it avoids the inverse function - although I think both should hopefully get the marks. :biggrin:

A similar approach can be used for the second part of (v).

Reply 18