Papers found here. (Credit to Tarquin Digby for those [and Charles98 for his photos])

STEP II:

1: Solution by mikelbird

2: Solution by LightBlueSoldier

3: Solution by Stray

4: Solution by Khallil

5: Solution by Blazy

6: Solution by DFranklin

7: Solution by cliverlong

8: Solution by LightBlueSoldier, Solution by Stray

9: Solution by Brammer

10: Solution by JosephML

11: Solution by Brammer

12: Solution by shamika

13: Solution by DJMayes

STEP III:

1: Solution by jtSketchy

2: Solution by Brammer

3: Solution by edrraa, Solution by mikelbird

4: Solution by Farhan.Hanif93

5: Solution by metaltron

6: Solution by ctrls

7: Solution by Farhan.Hanif93

8: Solution by DFranklin

9: Solution by JosephML

10: Solution by JosephML

11: Solution by Brammer

12: Solution by DJMayes

13: Solution by DJMayes

STEP II:

1: Solution by mikelbird

2: Solution by LightBlueSoldier

3: Solution by Stray

4: Solution by Khallil

5: Solution by Blazy

6: Solution by DFranklin

7: Solution by cliverlong

8: Solution by LightBlueSoldier, Solution by Stray

9: Solution by Brammer

10: Solution by JosephML

11: Solution by Brammer

12: Solution by shamika

13: Solution by DJMayes

STEP III:

1: Solution by jtSketchy

2: Solution by Brammer

3: Solution by edrraa, Solution by mikelbird

4: Solution by Farhan.Hanif93

5: Solution by metaltron

6: Solution by ctrls

7: Solution by Farhan.Hanif93

8: Solution by DFranklin

9: Solution by JosephML

10: Solution by JosephML

11: Solution by Brammer

12: Solution by DJMayes

13: Solution by DJMayes

(edited 8 years ago)

Scroll to see replies

Some photos: II 2:

And II 8

Posted from TSR Mobile

And II 8

Posted from TSR Mobile

II/4: I'll edit the last part into this post in a short while.

Spoiler

(edited 4 years ago)

Original post by JosephML

III/9

.

.

I only got up as far as the result involving uksin - how many marks do you think I'd have got?

Original post by JosephML

III/9

I know it's trivial but didn't you also have to show that when t=0, v=u (initial conditions)?

Original post by Kasel

I know it's trivial but didn't you also have to show that when t=0, v=u (initial conditions)?

Ah, you're right. I missed that bit. I'll correct the solution, thanks.

Original post by CD315

I only got up as far as the result involving uksin - how many marks do you think I'd have got?

I should suspect somewhere between 8 and 12 marks.

Original post by Khallil

II/4

Do you think we would have had to prove arctan(x) + arctan(1/x)=pi/2?

There are a few short proofs but I forgot to include any in my solution.

STEP III, Q4

It's worth pointing out that a very similar question appeared in a IB tripos exam for the variational principles course.

(i)

(ii)

It's worth pointing out that a very similar question appeared in a IB tripos exam for the variational principles course.

(edited 10 years ago)

Original post by Kasel

...

I did but I left it out of the solution above for brevity. The one I wrote in the paper was very similar to this, but it looks really forced with regards to the 1-1 in the denominator.

$\begin{aligned} \phi = \arctan b + \arctan \frac{1}{b} \implies \tan \phi & = \dfrac{\tan \left( \arctan b \right) + \tan \left( \arctan \frac{1}{b} \right)}{1 - \tan \left( \arctan b \right)\tan \left( \arctan \frac{1}{b} \right)} \\ & = \dfrac{b + \frac{1}{b}}{1 - 1} \end{aligned}$

$\therefore \ \tan \phi = \displaystyle \lim_{n \to 0} \left( \dfrac{b + \frac{1}{b}}{n} \right) \implies \phi = \arctan \left( \lim_{n \to 0} \left( \dfrac{b + \frac{1}{b}}{n} \right) \right) = \dfrac{\pi}{2}$

(edited 10 years ago)

Original post by Khallil

...

To avoid limits you can say let y=arctan(x), then tan(y)=x

and

cot(y)=(1/x)=tan(pi/2 - y)

which gives arctan(x)+arctan(1/x)= y+pi/2-y=pi/2

Original post by Kasel

...

Ah, of course!

STEP III - Q12

Spoiler

(edited 10 years ago)

Original post by Stray

STEP II - Q8 (LaTeX)

$\{n, a, b, c_r\} \mathbb{Z}+$

$c_r, (0 \leq r \leq n) = \texttt{coefficient of } x^r \texttt{ in } (a + bx)^n$

$c_r = ^n\mathrm{C}_r \cdot a^{n-r} b^r= \frac{n!}{r!(n-r)!} \cdot a^{n-r} b^r$

$c_m \geq c_r \texttt{ for } 0 \leq r \leq n$

If $c_m$ is a maximum, it must be $\geq c_{m+1}$ and $\geq c_{m-1}$

Taking first, $c_{m+1}$ :

$\frac{n!a^{n-(m+1)} b^{(m+1)}}{(m+1)!(n-(m+1))!} \leq \frac{n!a^{n-m} b^m}{m!(n-m)!}$

Divide both sides by the common factor

$\frac{n!a^{n-(m+1)}b^m}{m!(n-(m+1))!}$

Giving

$\frac{b}{m+1} \leq \frac{a}{n-m}$

$bn-bm \leq am+a$

$bn-a \leq am+bm$

$\frac{bn-a}{a+b} \leq m$

This is almost the LHS of what we are required to show, just needs a little fiddling

$\frac{bn+b-(a+b)}{a+b} \leq m$

$\frac{b(n+1)}{a+b} -1 \leq m (**)$

Now repeat the process for $c_{m-1}$:

$\frac{n!a^{n-(m-1)} b^{(m-1)}}{(m-1)!(n-(m-1))!} \leq \frac{n!a^{n-m} b^m}{m!(n-m)!}$

Divide both sides by the common factor

$\frac{n!a^{n-m}b^{m-1}}{(m-1)!(n-m)!}$

Giving

$\frac{a}{n-(m-1)} \leq \frac{b}{m}$

$am \leq bn + b - bm$

$m(a+b) \leq b(n+1)$

$m \leq \frac{b(n+1)}{a+b}$

Which combines with (**) to give the required solution

$\frac{b(n+1)}{a+b} -1 \leq m \leq \frac{b(n+1)}{a+b}$

Considering the possible values of $m$

$\texttt{Let } y=\frac{b(n+1)}{a+b}$

$y-1 \leq m \leq y$

Either $y$ and $y-1$ are both integers, or $y$ and $y-1$ are non-integer values either side of an integer, so if $m$ is an integer, it is either $y$ and $y-1$ (2 values) or it lies between them (1 value).

For the next parts, we are asked to find either the value of $m$ where there is 1 value, or the larger value if there are 2. This means we can consider the function $G(n,a,b)$ to produce the Floor of the upper limit - so if the upper limit evaluates to 8.9 then the value is 8, but if it is 9.0 it is 9, etc. The Floor symbol is $\lfloor$.

$G(n,a,b) = \lfloor{\frac{b(n+1)}{a+b}}$

i)

$G(9,2,3) = \lfloor{\frac{3(9+1)}{2+3}} = \lfloor\frac{30}{5} = 6$

ii)

$G(2k,a,a) = \lfloor{\frac{a(2k+1)}{a+a}} = \lfloor\frac{2k+1}{2} = \lfloor{(k + 0.5)} = k$

$G(2k-1,a,a) = \lfloor{\frac{a(2k-1+1)}{a+a}} = \lfloor\frac{2k}{2} = k$

iii)

If we fix $n$ and $b$, the only part of $G$ that can alter is the denominator containing $a$, so for large $G$ we want small $a$, so $G_{max}$ is when $a=1$. (Zero is not a positive integer).

iv)

For fixed $a=1$ and (unknown) fixed $n$, the part of the function varying is:

$\frac{b}{b+1} (***)$

This tends to 1 as b tends to infinity, but will never actually reach 1, so $(***)(n+1)$ tends towards but never reaches $n+1$ so for fixed $n$

$G(n,1,b)_{max} =n \texttt{ when } b = \infty$

---

This was a nice question once you get into it, but I can see that the layout of the question might make it look more complex than it actually was.

$\{n, a, b, c_r\} \mathbb{Z}+$

$c_r, (0 \leq r \leq n) = \texttt{coefficient of } x^r \texttt{ in } (a + bx)^n$

$c_r = ^n\mathrm{C}_r \cdot a^{n-r} b^r= \frac{n!}{r!(n-r)!} \cdot a^{n-r} b^r$

$c_m \geq c_r \texttt{ for } 0 \leq r \leq n$

If $c_m$ is a maximum, it must be $\geq c_{m+1}$ and $\geq c_{m-1}$

Taking first, $c_{m+1}$ :

$\frac{n!a^{n-(m+1)} b^{(m+1)}}{(m+1)!(n-(m+1))!} \leq \frac{n!a^{n-m} b^m}{m!(n-m)!}$

Divide both sides by the common factor

$\frac{n!a^{n-(m+1)}b^m}{m!(n-(m+1))!}$

Giving

$\frac{b}{m+1} \leq \frac{a}{n-m}$

$bn-bm \leq am+a$

$bn-a \leq am+bm$

$\frac{bn-a}{a+b} \leq m$

This is almost the LHS of what we are required to show, just needs a little fiddling

$\frac{bn+b-(a+b)}{a+b} \leq m$

$\frac{b(n+1)}{a+b} -1 \leq m (**)$

Now repeat the process for $c_{m-1}$:

$\frac{n!a^{n-(m-1)} b^{(m-1)}}{(m-1)!(n-(m-1))!} \leq \frac{n!a^{n-m} b^m}{m!(n-m)!}$

Divide both sides by the common factor

$\frac{n!a^{n-m}b^{m-1}}{(m-1)!(n-m)!}$

Giving

$\frac{a}{n-(m-1)} \leq \frac{b}{m}$

$am \leq bn + b - bm$

$m(a+b) \leq b(n+1)$

$m \leq \frac{b(n+1)}{a+b}$

Which combines with (**) to give the required solution

$\frac{b(n+1)}{a+b} -1 \leq m \leq \frac{b(n+1)}{a+b}$

Considering the possible values of $m$

$\texttt{Let } y=\frac{b(n+1)}{a+b}$

$y-1 \leq m \leq y$

Either $y$ and $y-1$ are both integers, or $y$ and $y-1$ are non-integer values either side of an integer, so if $m$ is an integer, it is either $y$ and $y-1$ (2 values) or it lies between them (1 value).

For the next parts, we are asked to find either the value of $m$ where there is 1 value, or the larger value if there are 2. This means we can consider the function $G(n,a,b)$ to produce the Floor of the upper limit - so if the upper limit evaluates to 8.9 then the value is 8, but if it is 9.0 it is 9, etc. The Floor symbol is $\lfloor$.

$G(n,a,b) = \lfloor{\frac{b(n+1)}{a+b}}$

i)

Unparseable latex formula:

G(9,1,3) = \lfloor{\frac{3(9+1)}{1+3}} = \lfloor\frac{30}{4} = \floor{7.5} = 7

$G(9,2,3) = \lfloor{\frac{3(9+1)}{2+3}} = \lfloor\frac{30}{5} = 6$

ii)

$G(2k,a,a) = \lfloor{\frac{a(2k+1)}{a+a}} = \lfloor\frac{2k+1}{2} = \lfloor{(k + 0.5)} = k$

$G(2k-1,a,a) = \lfloor{\frac{a(2k-1+1)}{a+a}} = \lfloor\frac{2k}{2} = k$

iii)

If we fix $n$ and $b$, the only part of $G$ that can alter is the denominator containing $a$, so for large $G$ we want small $a$, so $G_{max}$ is when $a=1$. (Zero is not a positive integer).

iv)

For fixed $a=1$ and (unknown) fixed $n$, the part of the function varying is:

$\frac{b}{b+1} (***)$

This tends to 1 as b tends to infinity, but will never actually reach 1, so $(***)(n+1)$ tends towards but never reaches $n+1$ so for fixed $n$

$G(n,1,b)_{max} =n \texttt{ when } b = \infty$

---

This was a nice question once you get into it, but I can see that the layout of the question might make it look more complex than it actually was.

The last part is incorrect. In particular the value of infinity does not exist.

Posted from TSR Mobile

- STEP 2016 Solutions
- Hello! Join my STEP study
- Ial chemistry unit 5 edexcel
- How should I go about preparing for the STEP Maths exam?
- MAT 1996-2006 Solution Thread
- STEP Maths I, II, III 2001 Solutions
- STEP Maths I, II, III 1993 Solutions
- STEP maths I, II, III 1990 solutions
- STEP maths I, II, III 1991 solutions
- STEP 2006 Solutions Thread
- STEP 2
- Difficulty of Various Mathematics Entrance Exams (MAT, STEP, Gaokao, JEE, Suneung)
- STEP Maths I, II, III 1988 solutions
- why are the arrows this way around in this galvanic cell?
- Computer Science at Imperial College London.
- Ideal and regular solutions question
- STEP Maths I,II,III 1987 Solutions
- Physics/ Maths question help please
- STEP Maths I, II, III 1989 solutions
- STEP-II for 2025 entry

Latest

Trending

Last reply 3 weeks ago

Edexcel A Level Mathematics Paper 3 (9MA0 03) - 20th June 2024 [Exam Chat]Maths Exams

386

1104

Last reply 4 weeks ago

Edexcel GCSE Mathematics Paper 1 Higher (1MA1 1H) - 16th May 2024 [Exam Chat]Maths Exams

419

1460

Last reply 1 month ago

AQA Level 2 Further Maths 2024 Paper 2 (8365/2) - 19th June [Exam Chat]Maths Exams

94

461

Last reply 1 month ago

Edexcel A-level Further Mathematics Paper 1 (9FM0 01) - 22nd May 2024 [Exam Chat]Maths Exams

180

615

Last reply 1 month ago

AQA A-level Mathematics Paper 3 (7357/3) - 20th June 2024 [Exam Chat]Maths Exams

140

627

Last reply 1 month ago

Edexcel A Level Mathematics Paper 2 (9MA0 02) - 11th June 2024 [Exam Chat]Maths Exams

498

1842

Last reply 1 month ago

AQA A-level Mathematics Paper 2 (7357/2) - 11th June 2024 [Exam Chat]Maths Exams

171

545

Last reply 3 months ago

AQA GCSE Mathematics Paper 1 Foundation (8300/1F) - 16th May 2024 [Exam Chat]Maths Exams

45

63

Last reply 3 months ago

Edexcel GCSE Mathematics Paper 3 Higher (1MA1 3H) - 10th June 2024 [Exam Chat]Maths Exams

265

1273

Last reply 3 months ago

Edexcel GCSE Statistics Paper 2 Higher Tier (1ST0 2H) - 17th June 2024 [Exam Chat]Maths Exams

23

30

Last reply 3 months ago

Edexcel GCSE Mathematics Paper 1 Foundation (1MA1 1F) - 16th May 2024 [Exam Chat]Maths Exams

42

47

Last reply 3 months ago

AQA GCSE Mathematics Paper 2 Higher (8300/2H) - 3rd June 2024 [Exam Chat]Maths Exams

136

414

Last reply 3 months ago

OCR GCSE Mathematics Paper 4 Higher (J560/04) - 16th May 2024 [Exam Chat]Maths Exams

87

197

Last reply 3 months ago

Edexcel A-level Further Mathematics Paper 2 (9FM0 02) - 3rd June 2024 [Exam Chat]Maths Exams

132

297

Last reply 3 months ago

Edexcel A Level Further Mathematics Paper 3D (9FM0 3D) - 21st June 2024 [Exam Chat]Maths Exams

53

121

Last reply 3 months ago

AQA Level 2 Further Maths 2024 Paper 1 (8365/1) - 11th June [Exam Chat]Maths Exams

168

709

Last reply 3 months ago

OCR A Level Mathematics A Paper 2 (H240/02) - 11th June 2024 [Exam Chat]Maths Exams

109

222

Last reply 3 months ago

Edexcel A Level Mathematics Paper 1 (9MA0 01) - 4th June 2024 [Exam Chat]Maths Exams

553

1452

Last reply 3 months ago

Edexcel GCSE Mathematics Paper 2 Higher (1MA1 2H) - 3rd June 2024 [Exam Chat]Maths Exams

298

1257