STEP maths I, II, III 1992 solutions

(Updated as far as post #159.) SimonM - 23.03.2009

Solutions for earlier STEP papers are not available on the internet, so we're making our own. Please submit any solution to any problem which is currently unsolved (red above); if you see any mistakes in solutions posted, please point them out. :smile:

There are still a lot of questions to do in the 1993 thread; don't forget to check there too! :biggrin:

Several of the other threads still have one or two unsolved questions too - links are at the bottom of this post.

N.B. the "usual" past paper website has the papers mislabelled; please check the front of the paper, not the filename, to see which paper you're doing. (Mathematics = I, further mathematics A = II, further mathematics B = III.)

STEP I:
1: Solution by Lusus Naturae
2: Solution by nota bene / DFranklin
3: Solution by Speleo
4: Solution by Speleo
5: Solution by ad absurdum and DFranklin
6: Solution by brianeverit
7: Solution by DeathAwaitsU, insparato and DFranklin
8: Solution by insparato and DFranklin
9: Solution by *bobo*
10: Solution by *bobo*
11: Solution by *bobo*
12: Solution by brianeverit
13: Solution by *bobo*
14: Solution by brianeverit
15: Solution by TheDuck
16: Solution by generalebriety

STEP II:
1: Solution by Speleo
2: Solution by generalebriety
3: Solution by generalebriety
4: Solution by generalebriety
5: Solution by brianeverit
6: Solution by khaixiang
7: Solution by ad absurdum
8: Solution by Lusus Naturae
9: Solution by generalebriety and mikelbird
10: Solution by generalebriety and khaixiang
11: Solution by *bobo*
12: Solution by brianeverit
13: Solution by *bobo*
14: Solution by generalebriety
15: Solution by ben-smith
16: Solution by brianeverit

STEP III:
1: Solution by khaixiang
2: Solution by nota bene
3: Solution by khaixiang
4: Solution by SimonM
5: Solution by SimonM
6: Solution by khaixiang
7: Solution by generalebriety
8: Solution by khaixiang
9: (Incomplete) Solution by generalebriety Full solution by mikelbird
10: Solution by Glutamic Acid
11: Solution by *bobo*
12: Solution by *bobo*
13: Solution by *bobo*
14: Solution by brianeverit
15: Solution by ben-smith
16: Solution by ben-smith

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 14 years ago)

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Reply 1

generalebriety

Oh, and by the way - I'm doing II Q4. Back off. :p:

$\displaystyle \text{Consider }\lozenge (\lambda\mu ) = \lambda\lozenge\mu + \mu\lozenge\lambda \text{ [a]} \\[br]\text{But } \lozenge (\lambda\mu ) = \lambda\lozenge\mu \text{ [d]} \\[br]\therefore \lambda\lozenge\mu + \mu\lozenge\lambda = \lambda\lozenge\mu \Rightarrow \mu\lozenge\lambda = 0 \Rightarrow \lozenge\lambda = 0.\\ \\[br]\lozenge x^2 = x\lozenge x + x\lozenge x \text{ [a]} \\[br]= x.1 + x.1 \text{ [c]} \\[br]= 2x.\\[br]\lozenge x^3 = x^2 \lozenge x + x \lozenge x^2 \text{ [a]} \\[br]= x^2.1 + x.2x \text{ [c]} \\[br]= 3x^2. \\ \\[br][br]\text{Proposition: } \lozenge x^n = nx^{n-1}. \text{ Proceed by induction; assume true for } n \leq k. \\[br]\lozenge x^k = kx^{k-1} \\[br]\lozenge x^{k+1} = \lozenge (x^k.x) = x^k\lozenge x + x\lozenge x^k \text{ [a]}[br]= x^k.1 + x.kx^{k-1}\\[br]= (k+1)x^k \text{as required}. \\[br]\therefore \lozenge x^n = nx^{n-1}. \\ \\[br]h(x) = k_0 + k_1x + k_2x^2 + k_3x^3 + \dots + k_rx^r + \dots\\[br]\lozenge h(x) = \lozenge (k_0 + k_1x + k_2x^2 + k_3x^3 + \dots + k_rx^r + \dots ) \\[br]\lozenge h(x) = \lozenge (k_0) + \lozenge (k_1x) + \lozenge (k_2x^2) + \lozenge (k_3x^3) + \dots + \lozenge (k_rx^r) + \dots \text{ [b]} \\[br]\lozenge h(x) = 0 + k_1\lozenge (x) + k_2\lozenge (x^2) + k_3\lozenge (x^3) + \dots + k_r\lozenge (x^r) + \dots \text{ [d]} \\[br]\lozenge h(x) = k_1(1) + k_2(2x) + k_3(3x^2) + \dots + k_r(rx^{r-1}) + \dots \\[br]\therefore \lozenge h(x) = \frac{\text{d}}{\text{d}x} h(x).[br]$

Reply 2

Speleo

II/1

Someone might want to check v.) and vi.), v.) because it seems so easy, vi.) because I'm not certain I can neglect the terms I did... (limit's right though).

Think I'll attempt Lusus' challenge now, I expect to fail miserably :tongue:

92ii1d.JPG58.2KB

Reply 3

Swayum

I'll give STEP I q7 a shot.

$g(0) = b$
and $g(1) = a+b$
$g(n) = an + b$

$ng(1) = an + bn$
$g(n) = ng(1) - bn + b = ng(1) - ng(0) + g(0)$
$g(n) = ng(1) - (n-1)g(0)$

As n, g(0) and g(1) are integers, g(n) must be an integer too.

$f(-1) = a - b + c$
and $f(0) = c$
and $f(1) = a + b + c$
and $f(n) = an^2 + bn + c$

$f(-1) + f(1) = 2a + 2c$
$\frac{f(-1) + f(1)}{2} - c = a = \frac{f(-1) + f(1)}{2} - f(0)$

So $f(n) = n^2(\frac{f(-1) + f(1)}{2} - f(0)) + bn + c$

$Let I = a = \frac{f(-1) + f(1)}{2} - f(0)$

(I is the integer form of a)

$nf(1) = an + bn + cn$
$nf(1) - In - cn = bn$
$nf(1) - In - nf(0) = bn$

We've now got integer forms of an^2, bn and c so f(n) can be expressed as an integer:

$f(n) = an^2 + bn + c$
$f(n) = In^2 + nf(1) - In - nf(0) + c$
$f(n) = In^2 + nf(1) - In - nf(0) + f(0)$

All of which are integers. A more complete f(n):

$f(n) = n^2(\frac{f(-1) + f(1)}{2} - f(0)) + nf(1) - n(\frac{f(-1) + f(1)}{2} - f(0)) - nf(0) + f(0)$

$f(n) = n^2(\frac{f(-1) + f(1)}{2} - f(0)) + nf(1) - n(\frac{f(-1) + f(1)}{2} - f(0)) - f(0)(n - 1)$

And by the by, in case someone wants a justification of why $\frac{f(-1) + f(1)}{2}$ is an integer:

$f(-1) + f(1) = 2a + 2c$
$f(-1) + f(1) = 2(a+c)$

The left hand side is an integer. This implies that the right hand side is an integer. As I can take out a factor of 2 out of the right hand side, it implies that the right hand side is an even integer. This then implies that the left hand side is an even integer. And so the left hand side is divisible by 2.

Don't have enough time for the last bit.

Reply 4

insparato

Ive done most of STEP II Q2 and Done STEP II Q8

Reply 5

nota bene

I'll try III/2 (Matrices) I think I'm able to do it but I'm not 100% sure

EDIT:
$A^2=(I+aJ)^2=I^2+2aJ+a^2J^2$

$J^2=J\timesJ=\begin{pmatrix} 1&1 \\ 1&1\end{pmatrix} \begin{pmatrix} 1&1 \\ 1&1\end{pmatrix}=\begin{pmatrix} 1\times1+1\times1&1\times1+1\times1 \\ 1\times1+1\times1&1\times1+1\times1 \end{pmatrix}=\begin{pmatrix} 2&2 \\ 2&2\end{pmatrix}=2J$
Furthermore an identity matrix I^n=I (If David reads this he'll probably tell me to prove it :biggrin:

Honestly I don't know how much of this we are allowed to assume, either way I'll leave it out for the time being). Also multiplying by an Identity matrix does not change anything (could possibly also require a proof, but I really don't see how we are supposed to prove all these things in 30 minutes (although the proofs are not hard)).
Therefore $A^2=(I+aJ)^2=I+2aJ+a^2J^2=I+2aJ+2a^2J=I+\frac{1}{2}((1+2a)^2-1)J$
As desired...
$A^3=(I+aJ)^3=I+3aJ+3a^2J^2+a^3J^3$ Using $J^2=2J$ and $J^3=2J^2$ (for confirmation:
$J^3=J\timesJ^2=\begin{pmatrix} 1&1 \\ 1&1\end{pmatrix} \begin{pmatrix} 2&2 \\ 2&2\end{pmatrix}=\begin{pmatrix} 1\times2+1\times2&1\times2+1\times2 \\ 1\times2+1\times2&1\times2+1\times2 \end{pmatrix}=\begin{pmatrix} 4&4 \\ 4&4\end{pmatrix}=2J^2$ )

Then we have $A^3=(I+aJ)^3=I+3aJ+3a^2J^2+a^3J^3=I+3aJ+6a^2J+4a^3J=I+\frac{1}{2}((1+2a)^3-1)J$
As desired...

$A^4=(I+aJ)^4=I+4aJ+6a^2J^2+4a^3J^3+a^4J^4$ Let's show J^4=2J^3:
$J^4=J\timesJ^3=\begin{pmatrix} 1&1 \\ 1&1\end{pmatrix} \begin{pmatrix} 4&4 \\ 4&4\end{pmatrix}=\begin{pmatrix} 1\times4+1\times4&1\times4+1\times4 \\ 1\times4+1\times4&1\times4+1\times4 \end{pmatrix}=\begin{pmatrix} 8&8 \\ 8&8\end{pmatrix}=2J^3$
Using the previously proven information it can now be rewritten as:
$A^4=(I+aJ)^4=I+4aJ+12a^2J+16a^3J+8a^4=I+\frac{1}{2}((1+2a)^4-1)J$

Moving on I will first prove $J^n=2^{n-1}J$ The case n=2 is already proven earlier. Therefore we assume the statement to be true for all n=k where $n\geq2$ .
Multiply by J we have $J\times2^{k-1}J=\begin{pmatrix} 1&1 \\ 1&1\end{pmatrix} \begin{pmatrix} 2^{k-1}&2^{k-1} \\ 2^{k-1}&2^{k-1}\end{pmatrix}=\begin{pmatrix} 1\times2^{k-1}+1\times2^{k-1}&1\times2^{k-1}+1\times2^{k-1} \\ 1\times2^{k-1}+1\times2^{k-1}&1\times2^{k-1}+1\times2^{k-1} \end{pmatrix}=\begin{pmatrix} 2^k&2^k \\ 2^k&2^k\end{pmatrix}=2^kJ$
Q.E.D. (P(k)->P(k+1) by mathematical induction)

Now we move on to proving the statement that $A^n=(I+aJ)^n=I+\frac{1}{2}((1+2a)^n-1)J$ The case n=1 is already proven. Assume true for n=k, and multiply by (I+aJ) (and using a bit of information we know is true).
$(I+aJ)(I+\frac{1}{2}((1+2a)^k)-1)J)=I+\frac{1}{2}((1+2a)^k-1)J+aJ+2a\frac{1}{2}((1+2a)^k-1)J=I+(\frac{1}{2}(1+2a)^k-\frac{1}{2}+a+a(1+2a)^k-a)J=I+\frac{1}{2}((1+2a)(1+2a)^k-1)=I+\frac{1}{2}((1+2a)^{k+1}-1)J$
As desired; therefore by the principle of mathematical induction we have proven that P(k)->P(k+1) Q.E.D.

Now I hope I have not made too many typos or other stupid things...

Reply 6

Kolya

STEP II Q8

i)
$\displaystyle \int \frac{x}{(x-1)(x^2-1)} \ dx$

But $x^2-1 = (x-1)(x+1)$

$\displaystyle \therefore \int \frac{x}{(x-1)^2(x+1)} \ dx$

We will now sort this out into partial fractions:

$\frac{x}{(x-1)^2(x+1)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+1}$
$x = A(x-1)(x+1) + B(x+1) + C(x-1)^2$

Setting $x = 1$ gives us

$B = \frac{1}{2}$

Setting $x = 2$ gives us $6A + 2C = 1$ (*)

Setting $x = 3$ gives us $8A + 4C = 1$ (#)

Solving (*) and (#) by the elimination method gives us:

$A = \frac{1}{4} \\ C = -\frac{1}{4}$

Therefore we now have to solve

$\displaystyle \int \frac{1}{4(x-1)} + \frac{1}{2(x-1)^2} - \frac{1}{4(x+1)} \ dx$

Taking out a constant gives

$\displaystyle \frac{1}{4} \int \frac{1}{(x-1)} + \frac{2}{(x-1)^2} - \frac{1}{(x+1)} \ dx = \\ \frac{1}{4} \int \frac{1}{(x-1)} + 2(x-1)^{-2} - \frac{1}{(x+1)} \ dx$

Which integrates to

$\frac{1}{4}(ln(x-1) - ln(x+1) - \frac{2}{x-1}) + c$

ii.
$\displaystyle \int \frac{1}{3cosx+4sinx} \ dx$

Let $3cosx+4sinx = Rsin(x+ \alpha) \\ = Rsinxcos\alpha + Rcosxsin \alpha \ \Rightarrow R = 5 , \ \alpha = arctan(\frac{3}{4})$

So now we are integrating:

$\displaystyle \int \frac{1}{5sin(x+arctan(\frac{3}{4})} \ dx = \frac{1}{5} \int csc(x+arctan(\frac{3}{4}) \ dx \\ = -\frac{1}{5}ln|csc(x+arctan(\frac{3}{4})) + cot(x+arctan(\frac{3}{4}))| + c$

iii.
$\displaystyle \int \frac{1}{shx} \ dx = \int \frac{2}{e^x-e^{-x}}$

Let $u = e^x \Leftrightarrow \frac{dx}{du} = \frac{1}{u}$

So:

$\displaystyle \int {2}{u-\frac{1}{u}} \cdot \frac{1}{u} \ du = int\frac{2}{u^2-1} \ du$

Using the cover rule for partial fractions gives

$\displaystyle \int \frac{1}{u-1} - \frac{1}{u+1} \ dx = ln(u-1) - ln(u+1) + C = ln|(\frac{e^x-1}{e^x+1})| + c$

Done.

Reply 7

insparato

STEP II
Q8 ii) Turn the denominator into 5cos (x - arctan3/4) and integrates easily

Q8 iii) put it in to exponentials and use the substitution u = e^x i get ln |(e^x -1)/(e^x+1)| + C

Reply 8

generalebriety

insparato

STEP II
Q8 ii) Turn the denominator into 5cos (x - arctan3/4) and integrates easily

Q8 iii) put it in to exponentials and use the substitution u = e^x i get ln |(e^x -1)/(e^x+1)| + C

Do it.

Reply 9

insparato

I did on paper but lusus started typing it up :p:

Reply 10

generalebriety

Might as well do II/2.

$\displaystyle y = x\frac{\text{d}y}{\text{d}x} - \cosh\bigg(\frac{\text{d}y}{\text{d}x}\bigg)\; (*)\\[br]\frac{\text{d}y}{\text{d}x} = x\frac{\text{d}^2y}{\text{d}x^2} + \frac{\text{d}y}{\text{d}x} - \frac{\text{d}^2y}{\text{d}x^2}\sinh \bigg(\frac{\text{d}y}{\text{d}x}\bigg)\\[br]0 = \frac{\text{d}^2y}{\text{d}x^2} \bigg( x - \sinh \bigg(\frac{\text{d}y}{\text{d}x}\bigg)\bigg) \\[br]\therefore \frac{\text{d}^2y}{\text{d}x^2} = 0 \Rightarrow y = Ax + B\\[br]\text{or } x - \sinh \bigg(\frac{\text{d}y}{\text{d}x}\bigg) = 0 \Rightarrow \sinh^{-1} x = \frac{\text{d}y}{\text{d}x} \\[br]\Rightarrow y = \int \sinh^{-1} \text{d}x = x \sinh^{-1} x - \int\frac{x}{\sqrt{1 + x^2}} \text{d}x = x \sinh^{-1} x - (1+x^2)^{1/2} + C.$

$\displaystyle (1)\;\; y = Ax+B: \frac{\text{d} y}{\text{d} x} = A. \\[br](*): Ax + B = Ax - \cosh A \\[br]B = -\cosh A \\[br]\therefore y = Ax - \cosh A.$

$\displaystyle (2)\;\; y = x \sinh^{-1} x - (1+x^2)^{1/2} + C:\\[br]\frac{\text{d}y}{\text{d}x} = \sinh^{-1} x + \frac{x}{\sqrt{1+x^2}} - \frac{x}{\sqrt{1+x^2}} = \sinh^{-1} x.\\[br](*): x\sinh^{-1} x - (1+x^2)^{1/2} + C = x\sinh^{-1} x - \cosh (\sinh^{-1} x).\\[br]- (1+x^2)^{1/2} + C = - \cosh (\sinh^{-1} x).\\[br]u = \sinh^{-1} x \Rightarrow \sinh u = x \Rightarrow \cosh u = \sqrt{1+x^2} \\[br]\Rightarrow \cosh (\sinh^{-1} x) = (1+x^2)^{1/2}.\\[br](*): - (1-x^2)^{1/2} + C = - (1+x^2)^{1/2} \\[br]C = 0\\[br]\therefore y = x \sinh^{-1} x - (1+x^2)^{1/2} .$

Reply 11

insparato

Just need to finish off the general solution on Q2 im having alittle bit of difficulty at the near the end. You might as well finish it off :p:

Reply 12

Speleo

My question is done up there ^

Also, good job on the clickable links to each question :smile:

Reply 13

generalebriety

I/3.

(i)
$\displaystyle \int_{-\pi}^\pi |\sin x| \text{d}x = 2\int_0^\pi \sin x \text{d}x = -2[\cos x]_0^\pi = 4.$

(ii)
$\displaystyle \int_{-\pi}^\pi \sin |x| \text{d}x = 2\int_0^\pi \sin x \text{d}x = 4.$

(iii)
$\displaystyle \int_{-\pi}^\pi x \sin x \text{d}x = [-x \cos x]_{-\pi}^\pi + \int_{-\pi}^\pi \cos x \text{d}x = 2\pi + [\sin x]_{-\pi}^{pi} = 2\pi.$

Edit: ah, to hell with it. I'll change the first post, Speleo corrected me on everything. :biggrin:

Left the stuff above for easy LaTeXing.

Reply 14

Kolya

II/8 is now complete.

(Aaron, do I have any corrections to make, or do you agree?)

Reply 15

insparato

That looks fine to me Lusus.

Reply 16

generalebriety

*beautifies first post*

I should've done art instead... :p:

Reply 17

Speleo

Just did I/3, got a different answer for every part :rofl:

Pretty sure I'm right, too :wink:

92i3ad.JPG47.5KB

92i3bd.JPG16.3KB

Reply 18

generalebriety

Speleo

Just did I/3, got a different answer for every part :rofl:

Pretty sure I'm right, too :wink:

****'s sake. :bawling:

I never could do trig. And I'm tired and I didn't have breakfast today and the moon's in Jupiter and I don't like the colour of what you're wearing.

Cheers. :p: