(Updated as far as #111) SimonM - 26.04.2009

Done

STEP I:

1: Solution by SimonM

2: Solution by SimonM

3: Solution in Siklos booklet

4: Solution by brianeverit

5: Solution by Aurel-Aqua

6: Solution by SimonM

7: Solution by Aurel-Aqua

8: Solution by SimonM

9: Solution by lilman91

10: Solution by Aurel-Aqua

11: Solution by brianeverit

12: Solution by brianeverit

13: Solution by brianeverit

14: Solution by lilman91 and Aurel-Aqua

STEP II:

1: Solution by brianeverit

2: Solution by nota bene, Solution by Square

3: Solution by nota bene, #28, etc.

4: Solution by brianeverit

5: Solution by I hate maths

6: Solution by nota bene

7: Solution by nota bene

8: Solution by brianeverit

9: Solution by brianeverit

10: Solution by Aurel-Aqua

11: Solution by brianeverit

12: Solution by brianeverit

13: Solution by SimonM And Aurel-Aqua

14: Solution by SimonM

STEP III:

1: Solution by Zhen Lin (PDF)

2: Solution by Agrippa (PDF)

3: Solution by Zhen Lin (PDF)

4: Solution by Zhen Lin (PDF)

5: Solution by Zhen Lin (PDF)

6: Solution by squeezebox, #61

7: Solution by Zhen Lin (PDF), Post #52

8: Solution by squeezebox

9: Solution by brianeverit

10: Solution by squeezebox

11: Solution by brianeverit

12: Solution by Aurel-Aqua

13: Solution by toasted-lion

14: Solution by brianeverit

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

Done

STEP I:

1: Solution by SimonM

2: Solution by SimonM

3: Solution in Siklos booklet

4: Solution by brianeverit

5: Solution by Aurel-Aqua

6: Solution by SimonM

7: Solution by Aurel-Aqua

8: Solution by SimonM

9: Solution by lilman91

10: Solution by Aurel-Aqua

11: Solution by brianeverit

12: Solution by brianeverit

13: Solution by brianeverit

14: Solution by lilman91 and Aurel-Aqua

STEP II:

1: Solution by brianeverit

2: Solution by nota bene, Solution by Square

3: Solution by nota bene, #28, etc.

4: Solution by brianeverit

5: Solution by I hate maths

6: Solution by nota bene

7: Solution by nota bene

8: Solution by brianeverit

9: Solution by brianeverit

10: Solution by Aurel-Aqua

11: Solution by brianeverit

12: Solution by brianeverit

13: Solution by SimonM And Aurel-Aqua

14: Solution by SimonM

STEP III:

1: Solution by Zhen Lin (PDF)

2: Solution by Agrippa (PDF)

3: Solution by Zhen Lin (PDF)

4: Solution by Zhen Lin (PDF)

5: Solution by Zhen Lin (PDF)

6: Solution by squeezebox, #61

7: Solution by Zhen Lin (PDF), Post #52

8: Solution by squeezebox

9: Solution by brianeverit

10: Solution by squeezebox

11: Solution by brianeverit

12: Solution by Aurel-Aqua

13: Solution by toasted-lion

14: Solution by brianeverit

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

(edited 5 years ago)

Scroll to see replies

I just tried a few questions from paper III and it seemed relatively easy - I managed to complete 1, 3, 4, 5 and 7 as well as attempt (badly) 2 in about 2 hours. I was under the impression that the older papers were harder? Hmm...

I will post my solutions for 1, 3, 4, 5 and 7 shortly. In the meantime - does anybody have a hint for 2? There's obviously at least one root (x = 0), but I'm not sure I can see a way to count the non-trivial roots...

I will post my solutions for 1, 3, 4, 5 and 7 shortly. In the meantime - does anybody have a hint for 2? There's obviously at least one root (x = 0), but I'm not sure I can see a way to count the non-trivial roots...

Err... which paper?

Edit: ah, you edited. I'll have a look.

Edit: ah, you edited. I'll have a look.

In STEP questions like that, you can be almost certain that the second part relates to the first part in some way. In this case, put y = e^x - 1 - k tan^-1 x, and find dy/dx...

Aaah. I see. I will try that now.

In the meantime, these are my solutions for questions 1, 3, 4, 5, and 7 of paper III. Incidentally, one of my interview questions was to prove that there were only five regular polyhedra (last part of question 4), but by considering the sum of angles at each vertex instead... Much less long-winded, I think, even though it would be necessary to assume that all regular polyhedra are convex.

Also, I think this post needs updating... (There's also a thread for 2000 now.)

In the meantime, these are my solutions for questions 1, 3, 4, 5, and 7 of paper III. Incidentally, one of my interview questions was to prove that there were only five regular polyhedra (last part of question 4), but by considering the sum of angles at each vertex instead... Much less long-winded, I think, even though it would be necessary to assume that all regular polyhedra are convex.

Also, I think this post needs updating... (There's also a thread for 2000 now.)

Those all look fine to me. Funnily, I don't think I ever looked at this paper last year - I don't remember any of the (pure) questions at any rate! My feeling is those questions are relatively short, but also require a bit more familiarity with "non A-level maths" than the average. This is probably right up your street, so it worked out as a particularly easy paper for you.

I think I've got 8 and 6 (paper III) scribbled somewhere...I'll post them if I can find.

Those all look fine to me. Funnily, I don't think I ever looked at this paper last year - I don't remember any of the (pure) questions at any rate! My feeling is those questions are relatively short, but also require a bit more familiarity with "non A-level maths" than the average. This is probably right up your street, so it worked out as a particularly easy paper for you.

STEP III - Question 8

i)

$0 \leq x \leq 2\pi \Rightarrow n = 1$.

So the differential equation becomes:

$\frac{d^{2}y}{dx^{2}} + y = 0$, whose general solution is: $y(x) = A\cos(x) + B\sin(x)$.

Now, since when x=0, y=0 and $\frac{dy}{dx} = 1$, the particular solution is $y(x) =\sin(x)$ (*).

ii) Since $2\pi \leq x \leq 4\pi, \Rightarrow n = 2$.

So this time the differential equation is:

$\frac{d^{2}y}{dx^{2}} + 4y = 0$

general solution: $y(x) = C\cos(2x) + D\sin(2x)$ (**).

Now, since y and $\frac{dy}{dx}$ are continuous at $x=2\pi$, (**) has the same value as (*) at $x = 2\pi$ and also their derivatives take the same value. From this we can work out the values of C and D, (C=0 and D= $\frac{1}{2}$). Which gives us: $y(x) = \frac{1}{2}\sin(2x)$.

Using the same method in the previous part, we can deduce that $y(x) = \frac{1}{n}\sin(nx)$.

iii)

$\displaystyle\int^\infty_0 y^{2} \, dx = \displaystyle\int^{2\pi}_{0} y^{2} \, dx + \displaystyle\int^{4\pi}_{2\pi} y^{2} \, dx + \displaystyle\int^{6\pi}_{4\pi} y^{2} \, dx + ...........+ \displaystyle\int^{2r\pi}_{2(r-1)\pi} y^{2} \, dx + ....$

= $\displaystyle\int^{2\pi}_{0} \sin^{2}(x) \, dx + \displaystyle\int^{4\pi}_{2\pi} \frac{1}{2^{2}}\sin^{2}(x) \, dx + \displaystyle\int^{6\pi}_{4\pi} \frac{1}{3^{2}}\sin^{2}(x) \, dx + .....$

= $\displaystyle\int^{2\pi}_{0} (1-\cos(2x)) \, dx + \displaystyle\int^{4\pi}_{2\pi} \frac{1}{2^{2}}(1-\cos(4x)) \, dx + \displaystyle\int^{6\pi}_{4\pi} \frac{1}{3^{2}}(1-\cos(6x))\, dx + .....$

(using $\cos(2x) = 1 - 2\sin^{2}(x)$)

= $\frac{1}{2}[ 2\pi - 0] + \frac{1}{2^{2}}[\frac{1}{2}( 4\pi - 2\pi )] + \frac{1}{3^{2}}[\frac{1}{2}( 6\pi - 4\pi)] + ....$

= $\pi + \frac{\pi}{2^{2}} + \frac{\pi}{3^{2}} + ..... = \pi \displaystyle\sum_{n=1}^\infty \frac{1}{n^{2}}$

i)

$0 \leq x \leq 2\pi \Rightarrow n = 1$.

So the differential equation becomes:

$\frac{d^{2}y}{dx^{2}} + y = 0$, whose general solution is: $y(x) = A\cos(x) + B\sin(x)$.

Now, since when x=0, y=0 and $\frac{dy}{dx} = 1$, the particular solution is $y(x) =\sin(x)$ (*).

ii) Since $2\pi \leq x \leq 4\pi, \Rightarrow n = 2$.

So this time the differential equation is:

$\frac{d^{2}y}{dx^{2}} + 4y = 0$

general solution: $y(x) = C\cos(2x) + D\sin(2x)$ (**).

Now, since y and $\frac{dy}{dx}$ are continuous at $x=2\pi$, (**) has the same value as (*) at $x = 2\pi$ and also their derivatives take the same value. From this we can work out the values of C and D, (C=0 and D= $\frac{1}{2}$). Which gives us: $y(x) = \frac{1}{2}\sin(2x)$.

Using the same method in the previous part, we can deduce that $y(x) = \frac{1}{n}\sin(nx)$.

iii)

$\displaystyle\int^\infty_0 y^{2} \, dx = \displaystyle\int^{2\pi}_{0} y^{2} \, dx + \displaystyle\int^{4\pi}_{2\pi} y^{2} \, dx + \displaystyle\int^{6\pi}_{4\pi} y^{2} \, dx + ...........+ \displaystyle\int^{2r\pi}_{2(r-1)\pi} y^{2} \, dx + ....$

= $\displaystyle\int^{2\pi}_{0} \sin^{2}(x) \, dx + \displaystyle\int^{4\pi}_{2\pi} \frac{1}{2^{2}}\sin^{2}(x) \, dx + \displaystyle\int^{6\pi}_{4\pi} \frac{1}{3^{2}}\sin^{2}(x) \, dx + .....$

= $\displaystyle\int^{2\pi}_{0} (1-\cos(2x)) \, dx + \displaystyle\int^{4\pi}_{2\pi} \frac{1}{2^{2}}(1-\cos(4x)) \, dx + \displaystyle\int^{6\pi}_{4\pi} \frac{1}{3^{2}}(1-\cos(6x))\, dx + .....$

(using $\cos(2x) = 1 - 2\sin^{2}(x)$)

= $\frac{1}{2}[ 2\pi - 0] + \frac{1}{2^{2}}[\frac{1}{2}( 4\pi - 2\pi )] + \frac{1}{3^{2}}[\frac{1}{2}( 6\pi - 4\pi)] + ....$

= $\pi + \frac{\pi}{2^{2}} + \frac{\pi}{3^{2}} + ..... = \pi \displaystyle\sum_{n=1}^\infty \frac{1}{n^{2}}$

Zhen Lin

I just tried a few questions from paper III and it seemed relatively easy - I managed to complete 1, 3, 4, 5 and 7 as well as attempt (badly) 2 in about 2 hours. I was under the impression that the older papers were harder? Hmm...

I will post my solutions for 1, 3, 4, 5 and 7 shortly. In the meantime - does anybody have a hint for 2? There's obviously at least one root (x = 0), but I'm not sure I can see a way to count the non-trivial roots...

I will post my solutions for 1, 3, 4, 5 and 7 shortly. In the meantime - does anybody have a hint for 2? There's obviously at least one root (x = 0), but I'm not sure I can see a way to count the non-trivial roots...

I would find the stationary points by differentiating, then see where there is a +ve/-ve sign change between stationary points in the original equation, if that makes sense.

Zhen, for question 2, I found sketching the graph of f(x) makes it alot eaiser and clearer. (same for part ii) ).

Well, for the first part it's very easy to see the answer, even without sketching the graph - we know it's monotonic increasing from calculating the derivative. The second part is not so clear-cut because I can't tell whether $\left( e^x - 1 \right)$ is increasing/decreasing faster than $k \tan^{-1} x$. I tried generalebriety's suggestion and found a derivative that is similar to the first part, but then I got stuck because turning points are not necessarily the same as roots...

I dunno if what I did was correct, but I considered; g(x) = $\frac{e^{x} - 1}{tan^{-1}(x)}$. By playing around with it, I noticed that as x $\longrightarrow -\infty$, g(x) $\longrightarrow \frac{2}{\pi}$. If you want me to into anyfurther detail , I will.

Zhen Lin

Well, for the first part it's very easy to see the answer, even without sketching the graph - we know it's monotonic increasing from calculating the derivative. The second part is not so clear-cut because I can't tell whether $\left( e^x - 1 \right)$ is increasing/decreasing faster than $k \tan^{-1} x$. I tried generalebriety's suggestion and found a derivative that is similar to the first part, but then I got stuck because turning points are not necessarily the same as roots...

You know the roots must be between turning points, before the first turning point, or after the last one, so if there is a sign change between two turning points there must be one root between the turning points. It is quite easy to work out if there is another root before or after the last turning point.

II/7

This question seems too easy, so I've probably made a mistake:

$y=\frac{x}{\sqrt{x^2-2x+a}}$ Then $\frac{dy}{dx}= - \frac{x\frac{x-1}{\sqrt{x^2-2x+a}}-\sqrt{x^2-2x+a}}{x^2-2x+a}= - \frac{x(x-1)-(x^2-2x+a)}{(x^2-2x+a)^{\frac{3}{2}}}= - \frac{x-a}{(x^2-2x+a)^{\frac{3}{2}}} = \frac{a-x}{(x^2-2x+a)^{\frac{3}{2}}}$

Stationary points can occur when dy/dx=0 i.e. the numerator needs to be zero <=> 0=a-x <=>x=a Hence there is at most one stationary point. The derivative is undefined if $x^2-2x+a=0$ i.e. $x=\frac{2\pm\sqrt{4-4a}}{2}$ and for a>1 this has no (real) solutions and thus the function has more than zero stationary points. I.e. there is one stationary point, as desired.

i) a=2 f(0)=0 $f(2)=\sqrt{2}$ so stationary point at $(2,\sqrt{2})$ $f'(1.9)<0$and $f'(2.1)>0$ and thus the point is a maximum.

Also note $\displaystyle\lim_{x\to\pm\infty}[f(x)]=\pm1$

ii) a=1 f(0)=0 From above we have that there are no stationary points because the denominator of the function (and it's derivative) goes to 0 as a goes to 1 i.e. vertical asymptote at x=1 Same as in i) note that $\displaystyle\lim_{x\to\pm\infty}[f(x)]=\pm1$

Sketches see attached. (Very rough, would put much more effort in during a real exam)

This question seems too easy, so I've probably made a mistake:

$y=\frac{x}{\sqrt{x^2-2x+a}}$ Then $\frac{dy}{dx}= - \frac{x\frac{x-1}{\sqrt{x^2-2x+a}}-\sqrt{x^2-2x+a}}{x^2-2x+a}= - \frac{x(x-1)-(x^2-2x+a)}{(x^2-2x+a)^{\frac{3}{2}}}= - \frac{x-a}{(x^2-2x+a)^{\frac{3}{2}}} = \frac{a-x}{(x^2-2x+a)^{\frac{3}{2}}}$

Stationary points can occur when dy/dx=0 i.e. the numerator needs to be zero <=> 0=a-x <=>x=a Hence there is at most one stationary point. The derivative is undefined if $x^2-2x+a=0$ i.e. $x=\frac{2\pm\sqrt{4-4a}}{2}$ and for a>1 this has no (real) solutions and thus the function has more than zero stationary points. I.e. there is one stationary point, as desired.

i) a=2 f(0)=0 $f(2)=\sqrt{2}$ so stationary point at $(2,\sqrt{2})$ $f'(1.9)<0$and $f'(2.1)>0$ and thus the point is a maximum.

Also note $\displaystyle\lim_{x\to\pm\infty}[f(x)]=\pm1$

ii) a=1 f(0)=0 From above we have that there are no stationary points because the denominator of the function (and it's derivative) goes to 0 as a goes to 1 i.e. vertical asymptote at x=1 Same as in i) note that $\displaystyle\lim_{x\to\pm\infty}[f(x)]=\pm1$

Sketches see attached. (Very rough, would put much more effort in during a real exam)

- MAT 1996-2006 Solution Thread
- 2023 STEP 3 Math
- STEP 2016 Solutions
- STEP Maths I, II, III 1993 Solutions
- STEP maths I, II, III 1990 solutions
- STEP Maths I, II, III 1996 Solutions
- STEP maths I, II, III 1991 solutions
- STEP 2006 Solutions Thread
- Colours of Substances (IGCSE Chemistry Edexcel)
- Cambridge Math STEP 2
- Electrochemistry
- STEP 2005 Solutions Thread
- STEP I, II, III 2010 Solutions
- UKMT Senior challenge
- Concern regarding STEP 2023 preparation
- Physics/ Maths question help please
- Ideal and regular solutions question
- STEP result 2023
- How should I go about preparing for the STEP Maths exam?
- Alevel chemistry kw

Last reply 5 minutes ago

parents aren't letting me go on a uni residential - what do i do?Last reply 8 minutes ago

Zoom to Zoology? Uni open days, applications & more 2021-Present - a parents takeLast reply 11 minutes ago

Choosing Subjects for S5 to study Software engineering in the UkPosted 15 minutes ago

I am a pilot working towards a commercial licence. Ask me anything!Last reply 20 minutes ago

Turner and Townsend Degree Apprenticeships 2024Last reply 1 month ago

Edexcel GCSE Mathematics Paper 3 (1MA1 3) - 13th November 2023 [Exam Chat]Last reply 1 month ago

Edexcel A Level Mathematics Paper 1 (9MA0 01) - 6th June 2023 [Exam Chat]Maths Exams

2960

Last reply 1 month ago

AQA A Level Mathematics Paper 1 (7357/1) - 6th June 2023 [Exam Chat]Maths Exams

352

Last reply 1 month ago

Edexcel GCSE Mathematics Paper 2 (1MA1 2) - 10th November 2023 [Exam Chat]Last reply 1 month ago

Edexcel A-level Mathematics Paper 1 [6th June 2023] Unofficial MarkschemeMaths Exams

156

Last reply 2 months ago

Edexcel GCSE Mathematics Paper 1 (1MA1 1) - 8th November 2023 [Exam Chat]Last reply 3 months ago

Edexcel IGCSE Higher tier Mathematics A Paper 1 1H (4MA1) - 19th May 2023 [Exam Chat]Maths Exams

232

Last reply 3 months ago

AQA GCSE Mathematics Paper 3 (Higher) 8300/3H - 14th June 2023 [ Exam Chat]Maths Exams

247

Last reply 4 months ago

Edexcel GCSE Higher tier Maths Paper 1 1H (1MA1) - 19th May 2023 [Exam Chat]Maths Exams

970

Last reply 5 months ago

Edexcel GCSE higher tier Maths Paper 2 2H (1MA1) - 7th June 2023 [Exam Chat]Maths Exams

642

Last reply 5 months ago

Edexcel IGCSE Higher tier Mathematics A Paper 2 2H (4MA1) - 10th November 2023Last reply 7 months ago

OCR A-level Mathematics A Paper 3 (H240/03) - 20th June 2023 [Exam Chat]Maths Exams

169

Last reply 7 months ago

AQA Level 2 Further Maths 2023 Paper 1 (8365/1) & Paper 2 (8365/2) [Exam Chat]Maths Exams

489

Last reply 7 months ago

Edexcel A Level Mathematics Paper 3 (9MA0 03) - 20th June 2023 [Exam Chat]Maths Exams

1836

Last reply 7 months ago

AQA A Level Mathematics Paper 3 (7357/3) - 20th June 2023 [Exam Chat]Maths Exams

307

Last reply 7 months ago

OCR A-level Mathematics B Paper 3 (H640/03) - 20th June 2023 [Exam Chat]Maths Exams

132

Last reply 9 months ago

Edexcel A-level Further Mathematics Paper 1 (9FM0 01) - 25th May 2023 [Exam Chat]Maths Exams

902

Last reply 9 months ago

Edexcel A Level Mathematics Paper 2 (9MA0 02) - 13th June 2023 [Exam Chat]Maths Exams

2708

Last reply 1 month ago

Edexcel GCSE Mathematics Paper 3 (1MA1 3) - 13th November 2023 [Exam Chat]Last reply 1 month ago

Edexcel A Level Mathematics Paper 1 (9MA0 01) - 6th June 2023 [Exam Chat]Maths Exams

2960

Last reply 1 month ago

AQA A Level Mathematics Paper 1 (7357/1) - 6th June 2023 [Exam Chat]Maths Exams

352

Last reply 1 month ago

Edexcel GCSE Mathematics Paper 2 (1MA1 2) - 10th November 2023 [Exam Chat]Last reply 1 month ago

Edexcel A-level Mathematics Paper 1 [6th June 2023] Unofficial MarkschemeMaths Exams

156

Last reply 2 months ago

Edexcel GCSE Mathematics Paper 1 (1MA1 1) - 8th November 2023 [Exam Chat]Last reply 3 months ago

Edexcel IGCSE Higher tier Mathematics A Paper 1 1H (4MA1) - 19th May 2023 [Exam Chat]Maths Exams

232

Last reply 3 months ago

AQA GCSE Mathematics Paper 3 (Higher) 8300/3H - 14th June 2023 [ Exam Chat]Maths Exams

247

Last reply 4 months ago

Edexcel GCSE Higher tier Maths Paper 1 1H (1MA1) - 19th May 2023 [Exam Chat]Maths Exams

970

Last reply 5 months ago

Edexcel GCSE higher tier Maths Paper 2 2H (1MA1) - 7th June 2023 [Exam Chat]Maths Exams

642

Last reply 5 months ago

Edexcel IGCSE Higher tier Mathematics A Paper 2 2H (4MA1) - 10th November 2023Last reply 7 months ago

OCR A-level Mathematics A Paper 3 (H240/03) - 20th June 2023 [Exam Chat]Maths Exams

169

Last reply 7 months ago

AQA Level 2 Further Maths 2023 Paper 1 (8365/1) & Paper 2 (8365/2) [Exam Chat]Maths Exams

489

Last reply 7 months ago

Edexcel A Level Mathematics Paper 3 (9MA0 03) - 20th June 2023 [Exam Chat]Maths Exams

1836

Last reply 7 months ago

AQA A Level Mathematics Paper 3 (7357/3) - 20th June 2023 [Exam Chat]Maths Exams

307

Last reply 7 months ago

OCR A-level Mathematics B Paper 3 (H640/03) - 20th June 2023 [Exam Chat]Maths Exams

132

Last reply 9 months ago

Edexcel A-level Further Mathematics Paper 1 (9FM0 01) - 25th May 2023 [Exam Chat]Maths Exams

902

Last reply 9 months ago

Edexcel A Level Mathematics Paper 2 (9MA0 02) - 13th June 2023 [Exam Chat]Maths Exams

2708