STEP maths I, II, III 1991 solutions

I'm attempting I/3 I'll see if I sort it out...

Okay, I'll post what I have now (probably should scan a sketch tomorrow as well), but it is not finished, and likely contains a few mistakes

Spoiler

I'm tired of this one, I'll take it up tomorrow again and try to sort it out.

Reply 3

Doing III/1.

Edit: seems Speleo's doing this in another thread. The cheek! :p:

III/2:

Unparseable latex formula:

\displaystyle[br]\text{Similarity } \Leftrightarrow \\[br]P_1P_2 : P_1P_3 = Q_1Q_2 : Q_1Q_3 \text{ and } \angle P_2P_1P_3 = \angle \Q_2Q_1Q_3 \\[br]\Leftrightarrow \biggl\lvert \frac{z_1-z_2}{z_1-z_3} \biggr\rvert = \biggl\lvert \frac{w_1-w_2}{w_1-w_3} \biggr\rvert \text{ and } \arg \bigg(\frac{z_1-z_2}{z_1-z_3}\bigg) = \arg \bigg(\frac{w_1-w_2}{w_1-w_3}\bigg) \\[br]\Leftrightarrow \frac{z_1-z_2}{z_1-z_3} = \frac{w_1-w_2}{w_1-w_3} \\[br]\Leftrightarrow (z_1-z_2)(w_1-w_3) = (w_1-w_2)(z_1-z_3) \\[br]\Leftrightarrow z_1w_2 - z_1w_3 - z_2w_1 + z_2w_3 + z_3w_1 - z_3w_2 = 0 \\[br]\Leftrightarrow \det \begin{pmatrix} z_1&z_2&z_3 \\ w_1&w_2&w_3 \\ 1&1&1 \end{pmatrix} = 0.

(i)

Unparseable latex formula:

\displaystyle[br]\frac{z_1-z_2}{z_1-z_3} = \frac{z^2_1-z^2_2}{z^2_1-z^2_3}\\[br]\Rightarrow \frac{z^2_1 - z^2_3}{z_1 - z_3} = \frac{z^2_1 - z^2_2}{z_1 - z_2} \\[br]\Rightarrow z_1 + z_3 = z_1 + z_2 \\[br]\Rightarrow z_2 = z_3 \text{ which is a contradiction. Hence triangles are not similar}.

(ii)

Unparseable latex formula:

\displaystyle[br]\text{Similarity } \Leftrightarrow \frac{z^3_1 - z^3_3}{z_1 - z_3} = \frac{z^3_1 - z^3_2}{z_1 - z_2} \\[br]\Leftrightarrow z^2_1 + z_1z_3 + z^2_3 = z^2_1 + z_1z_2 + z^2_2 \\[br]\Leftrightarrow z_1(z_3-z_2) = -(z_2+z_3)(z_3-z_2) \\[br]\Leftrightarrow z_1+z_2+z_3 = 0 \Leftrightarrow \text{ centroid is origin}.

(iii)

Unparseable latex formula:

\displaystyle[br]\text{Equilateral } \Leftrightarrow \triangle P_1P_2P_3 \text{ similar to } P_2P_3P_1 \\[br]\Leftrightarrow \frac{z_1-z_2}{z_1-z_3} = \frac{z_2-z_3}{z_2-z_1} \\[br]\Leftrightarrow 2z_1z_2 - z^2_1 - z^2_2 = z^2_3 + z_1z_2 - z_1z_3 - z_2z_3 \\[br]\text{and result follows}.

Reply 4

Speleo

7

It's yours if you want it, I'm not going to be able to contribute properly for a day or two. I'll finish it if it's still left after that.

Reply 5

Speleo

It's yours if you want it, I'm not going to be able to contribute properly for a day or two. I'll finish it if it's still left after that.

Might as well give it a go then. :smile:

III/1:

(a)

Unparseable latex formula:

\displaystyle[br]\frac{6}{r(r+1)(r+3)} = \frac{2}{r} - \frac{3}{r+1} + \frac{1}{r+3} \text{ (partial fractions)} \\[br]\therefore \sum_1^n \frac{6}{r(r+1)(r+3)} = \sum_1^n \bigg[ \frac{2}{r} - \frac{3}{r+1} + \frac{1}{r+3} \bigg] \\[br]= \bigg[ \frac{2}{1} + \frac{2}{2}+ \frac{2}{3} + \dots + \frac{2}{n} \bigg] - \bigg[ \frac{3}{2} + \frac{3}{3} + \frac{3}{4} + \dots + \frac{3}{n+1} \bigg] + \bigg[ \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \dots + \frac{1}{n+3} \bigg] \\[br]= \frac{2}{1} + \frac{2}{2} + \frac{2}{3} - \frac{3}{2} - \frac{3}{3} - \frac{3}{n+1} + \frac{1}{n+1} + \frac{1}{n+2} + \frac{1}{n+3} \\[br]= \frac{7}{6} - \frac{2}{n+1} + \frac{1}{n+2} + \frac{1}{n+3}.

(b)

Unparseable latex formula:

\displaystyle \ln (1 + x + x^2 + x^3) = \ln (1 - x^4) - \ln (1 - x) \\[br]= x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^3}{3} + \frac{x^3}{3} + \dots - x^4 - \frac{x^8}{2} - \frac{x^{12}}{3} - \dots \\[br]= x + \frac{x^2}{2} + \frac{x^3}{3} - \frac{3x^4}{4} + \dots + \frac{x^{4r-3}}{4r-3} + \frac{x^{4r-2}}{4r-2} + \frac{x^{4r-1}}{4r-1} - \frac{3x^{4r}}{4r} + \dots

(c)

Unparseable latex formula:

\displaystyle[br]f(x) = e^{x\ln (x+1)} = e^y = 1 + y + \frac{y^2}{2!} + \frac{y^3}{3!} + \frac{y^4}{4!} + \dots \\[br]y = x\ln (x+1) = x\Big[ x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots\Big] = x^2 - \frac{x^3}{2} + \frac{x^4}{3} + \dots \\[br]\therefore e^{x\ln (x+1)} = 1 + \Big[ x^2 - \frac{x^3}{2} + \frac{x^4}{3}\Big] + \frac{[x^2 + \dots]^2}{2!} + \frac{[x^2 + \dots]^3}{3!} + \dots \\[br]= 1 + x^2 - \frac{1}{2}x^3 + \frac{5}{6}x^4 + \dots

Reply 6

generalebriety

Would appreciate if someone could check (b) for me, I'm not sure it's right...

I'm working it through now; firstly, where did you get the 1's from? surely they shouldn't be there? (or am I being stupid? anyway those doesn't affect the result)

I think the (4r-1)x^(4r) seems wrong I'm trying to work out what it should be...

edit: having looked at the first four terms where both series get involved I'm inclined to believe it shall plainly be

\frac{3x^{4r}}{4r}

Reply 7

nota bene

I'm working it through now; firstly, where did you get the 1's from? surely they shouldn't be there? (or am I being stupid? anyway those doesn't affect the result)

I think the (4r-1)x^(4r) seems wrong I'm trying to work out what it should be...

edit: having looked at the first four terms where both series get involved I'm inclined to believe it shall plainly be

\frac{3x^{4r}}{4r}

Oh - of course the 1s shouldn't be there, sorry. I'm crap at the Maclaurin series.

I think you're right about the other bit too, but I shall mostly take it on trust... need to do some French literature work, I shouldn't be doing maths. :biggrin:

Cheers.

Reply 8

generalebriety

I think you're right about the other bit too, but I shall mostly take it on trust... need to do some French literature work, I shouldn't be doing maths. :biggrin:

Hehe I recognise doing maths instead of what we are supposed to do :ninja:

.

About the question, I looked at more terms and as the difference between the two series always is a multiple of 4 I'm quite 100% sure I am right about that 3 now :smile:

.

In fact I think all questions on power series, maclaurin and taylor seem very accessible and fairly easy on these old papers (what the hell I've been able to do them!).

Reply 9

nota bene

Hehe I recognise doing maths instead of what we are supposed to do :ninja:

.

About the question, I looked at more terms and as the difference between the two series always is a multiple of 4 I'm quite 100% sure I am right about that 3 now :smile:

.

In fact I think all questions on power series, maclaurin and taylor seem very accessible and fairly easy on these old papers (what the hell I've been able to do them!).

Yeah, but, I haven't. :hmmm:

Reply 10

DFranklin

18

nota bene

I'm working it through now; firstly, where did you get the 1's from? surely they shouldn't be there? (or am I being stupid? anyway those doesn't affect the result)

I think the (4r-1)x^(4r) seems wrong I'm trying to work out what it should be...

edit: having looked at the first four terms where both series get involved I'm inclined to believe it shall plainly be

\frac{3x^{4r}}{4r}

The 2nd line of your answer to (b) looks crazy; did you just forget to put in all the denominators in the log expansions? (i.e. a LaTeX error rather than a maths one).

Reply 11

DFranklin

The 2nd line of your answer to (b) looks crazy; did you just forget to put in all the denominators in the log expansions? (i.e. a LaTeX error rather than a maths one).

Did you mean to quote me there?

I assume you didn't, and yes there should be denominators in general's expression I think someone was lazy latexing :p:

.

Reply 12

justinsh

4

STEP I question 1

First part:

\sin^2\theta+sin^2\phi+sin^2\psi+2\sin\theta\sin \phi \sin\psi\\

=\sin\theta[\cos(\phi+\psi)+2\sin\phi\sin\psi]+\sin^2\phi+\sin^2\psi\\

=\sin\theta[\cos\phi\cos\psi+\sin\phi\sin\psi]+\sin^2\phi+\sin^2\psi\\

=\cos(\phi+\psi)\cos(\phi-\psi)+\sin^2\phi+\sin^2\psi\\

=\cos^2\phi\cos^2\psi-\sin^2\phi\sin^2\psi+\sin^2\phi+\sin^2\psi\\

=(1-\sin^2\phi)(1-\sin^2\psi)-\sin^2\phi\sin^2\psi+\sin^2\phi+\sin^2\psi\\

=1

Second part:

Letting

\theta=\phi=\pi/5

and

\psi=\pi/10

, using the first part we have:

2\sin^2\frac{\pi}{5}+\sin^2\frac{\pi}{10}+2\sin^2 \frac{\pi}{5}\sin\frac{\pi}{10}-1=0\\

\Rightarrow8\sin^2\frac{\pi}{10}\cos^2\frac{\pi}{10}+\sin^2\frac{\pi}{10}+8\sin^3\frac{\pi}{10}\cos^2\frac{\pi}{10}-1=0\\

\Rightarrow8\sin^2\frac{\pi}{10}-8\sin^4\frac{\pi}{10}+\sin^2\frac{\pi}{10}+8\sin^3\frac{\pi}{10}-8\sin^5\frac{\pi}{10}-1=0\\

\Rightarrow(8\sin^3\frac{\pi}{10}+\sin^2\frac{\pi}{10}-1)+\sin^2\frac{\pi}{10}(1-8\sin^2\frac{\pi}{10}-8\sin^3\frac{\pi}{10})=0

\Rightarrow(8\sin^3\frac{\pi}{10}+8\sin^2\frac{\pi}{10}-1)\cos^2\frac{\pi}{10}=0\\

\Rightarrow8\sin^3\frac{\pi}{10}+8\sin^2\frac{\pi}{10}-1=0

Reply 13

DFranklin

The 2nd line of your answer to (b) looks crazy; did you just forget to put in all the denominators in the log expansions? (i.e. a LaTeX error rather than a maths one).

I think it was more a fatigue error than either - I certainly seem to have got the rest of it wrong from there on.

Reply 14

nota bene

Did you mean to quote me there?

I assume you didn't, and yes there should be denominators in general's expression I think someone was lazy latexing :p:

.

No, I just can't do the Maclaurin series. At all. I always make stupid mistakes and I know none of them off the top of my head. :s-smilie:

Reply 15

Typing up I/6

i)

g(x)=\frac{2x^2+3}{x^4+4}

gives

g'(x)=\frac{-4x^5-12x^3+16x}{(x^4+4)^2}

Since the denominator is non-negative, for the derivative to =0 the numerator must be zero. Factorizing renders

4x(4-3x^2-x^4)

x=0 is a trivial solution, putting u=x^2 we can solve the quadratic that is left, giving

u_1=-4 \text{, and, } u_2=1

we can reject -4 as that gives imaginary solutions for x and we are left with x^2=1 i.e.

x=\pm1

. This agrees with the reasoning in the question.

"Hence the stationary values are given by x=0, g(x)=3/4 and

x=\pm1

, g(x)=1" True.

"Since 3/4<1, there is a maxima at

x=\pm1

and a minimum at x=0"
The statement happens to be true, as we shall notice, but the argument is invalid as we cannot know if any of the points is a point of inflection. Therefore we test the points by looking at the value of the second derivative.

g''(x)=\displaystyle\frac{(x^4+4)^2(-20x^4-36x^2+16)-(-4x^5-12x^3+16x)(4x^3)}{(x^4+4)^4}

At x=0 g''(x)>0 which indicates a minimum, and at both x=1 and x=-1 g''(x)<0 which indicates a maximum.

"Thus we must have

\frac{3}{4} \le g(x) \le1

for all x"
This is not true as we know nothing of what happens when

x<-1 \text{, or,} x>1

. We can easily prove the statement wrong by e.g. considering x=5 which gives g(x)=0.08...

ii) Their first statement that

\displaystyle\int(1-x)^{-3}dx=-3(1-x)^{-4}

is false as they are differentiating (missing out a sign change as well) instead of integrating. It shall be:

\displaystyle\int(1-x)^{-3}dx=\frac{1}{2}(1-x)^{-2}

Evaluating the integral

\displaystyle\int_{-1}^3(1-x)^{-3}

means we have

[\frac{1}{2(1-x)^2}]_{-1}^3

\frac{1}{8}-\frac{1}{8}=0

as proposed (but with faulty argument as they differentiated...)

Reply 16

generalebriety

No, I just can't do the Maclaurin series. At all. I always make stupid mistakes and I know none of them off the top of my head. :s-smilie:

Well, what is the problem :s-smilie:

? I got the same answer as you (when including the fractions half-way :p:

).

I'm probably missing something obvious, must be the lack of sleep...

Reply 17

nota bene

Well, what is the problem :s-smilie:

? I got the same answer as you (when including the fractions half-way :p:

).

I'm probably missing something obvious, must be the lack of sleep...

Um, because I copied your answer without thinking when you corrected me. Because I hate the Maclaurin series. :wink:

There's no problem. I just do stupid things like ln(1+x) = 1 when x=0. Or I differentiate wrongly (especially with complicated functions). Or my arithmetic goes haywire. And I can't remember any of them off the top of my head. :s-smilie:

Reply 18