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STEP maths I, II, III 1991 solutions

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Reply 40
Avatar for gyrase
gyrase
3
II/1, Let t = lambda, alpha = z

h(x) = ax^2+bx+c = a(x^2+bx/a+c/a) = a((x+b/2a)^2-(b/2a)^2+c/a), so we require (b/2a)^2 = c/a, so b^2 = 4ac, i.e.discriminant = 0

3x^2+4x+t(x^2-2) = (3+t)x^2+4x-2t where a=3+t, b=4,c=-2t and we require that 16 = -8t(3+t), 0 = 3t+t^2+2, t= -1, -2.

(1) f(x)-g(x) = 2(x+1)^2
(2) f(x)-2g(x) = (x+2)^2

f(x) = 2x(1)-(2) = 4(x+1)^2 - (x+2)^2
g(x) = (1)-(2) = 2(x+1)^2 - (x+2)^2, hence A = 4, B = -1, C = 2, D = -1, m = 1, n = 2.

3x^2+4x+t(x^2+z) = (3+t)x^2+4x+zt, so a=3+t,b=4, c=zt, we require b^2 = 4ac, so 16 = 4(3+t)zt, 0 = 3zt+zt^2-4, a=z,b=3z,c=-4 then 9z^2 = -16z, z = -16/9.
Reply 41
Avatar for Rabite
Rabite
8
Ooh, very neat solution.

Has anyone done Q8 in II? I get 2ln{cosh½x}, but it doesn't seem to work when I put it back...
Reply 42
Avatar for khaixiang
khaixiang
5
Rabite
Ooh, very neat solution.

Has anyone done Q8 in II? I get 2ln{cosh½x}, but it doesn't seem to work when I put it back...


Yeah, that's the answer I got, double-checked it with Maple just now. :smile: But I really can't be bothered to substitute in, maybe you plug it back wrongly?
Reply 43
Avatar for gyrase
gyrase
3
Rabite
Ooh, very neat solution.

Has anyone done Q8 in II? I get 2ln{cosh½x}, but it doesn't seem to work when I put it back...

I'm not too sure, I breifly did it and got u = e^-x and -(e^x+e^-x)
Reply 44
Avatar for khaixiang
khaixiang
5
Decota
I'm not too sure, I breifly did it and got u = e^-x and -(e^x+e^-x)


That's the first part, Rabite is talking about the second part of the question. :smile:
Reply 45
Avatar for gyrase
gyrase
3
khaixiang
That's the first part, Rabite is talking about the second part of the question. :smile:

oh i see. 2ln(cosh½x) is indeed correct.
Reply 46
Avatar for Rabite
Rabite
8
Well I'll type it up :p:

u2+2usinh⁡x−1=0 u^2 + 2u \sinh x - 1 = 0
The quadratic formula involves fractions, and I hate fractions, so let's complete the square:
(u+sinh⁡x)2−1−sinh⁡2x=0 (u+\sinh x)^2 - 1 - \sinh ^2 x=0

u+sinh⁡x=±cosh⁡xu+\sinh x= ± \cosh x

u=cosh⁡x−sinh⁡x u = \cosh x -\sinh x Or −(cosh⁡x+sinh⁡x) -(\cosh x+\sinh x)

coshx-sinhx is uh...e^{-x} and -(coshx+sinhx) is -e^x.
y`>0 at x=0 so we have to choose the + solution.

dydx=e−x \frac{dy}{dx} = e^{-x}
Integrate and find the +C to get
y=1−e−x y = 1-e^{-x}.

For the next bit, let u = dy/dx for a start:
u2sinh⁡x+2u−sinh⁡x=0 u^2 \sinh x + 2u - \sinh x =0
u2+2ucosechx−1=0 u^2 + 2u cosech x - 1 =0
(u+cosechx)2−1−cosech2x=0 (u+cosech x)^2 - 1 - cosech^2 x =0
u+cosechx=±coth⁡x u+cosech x = ±\coth x
u=coth⁡x−cosechx u = \coth x - cosech x Or −(coth⁡x+cosechx) -(\coth x + cosech x) .

coth⁡x−cosechx=cosh⁡x−1sinh⁡x \coth x - cosech x = \frac{\cosh x - 1}{\sinh x}
=tanh⁡½x = \tanh{½x} (Consider double "angle" formulae)
But u = dy/dx:
dydx=tanh⁡½x \frac{dy}{dx} = \tanh{½x}
Tanh = sinh / cosh, which is a derivative over its function:
y=2ln⁡cosh⁡½x+C y = 2 \ln \cosh ½ x +C

The solution passes through (0,0); so C =0.

(Taking the negative solution when we square rooted earlier results in something that can't pass through (0,0))

[edit] Is there no /cosech in Latex?
Reply 47
Avatar for khaixiang
khaixiang
5
Rabite


[edit] Is there no /cosech in Latex?


TSR's LaTeX is messed up then, the proper code is \csch
Nice work btw, though "1" and "coth(x)" is missing somewhere in the middle. All the STEP III questions are so unbearably long and tedious :frown: Good thing the style was revamped.
Reply 48
Avatar for Rabite
Rabite
8
I tried \csch too :frown:

Anyway. I think I've found the missing buggers, but maybe I've still missed one...
I kept doing stuff like \sinhx which just didn't come out as anything.

Thanks for that :p:
Speleo

*bawls*:frown: Matrices are easy:/


Speleo
Also, does anyone have the '05 papers? (II and III '06 are available if you go back a couple hundred pages to last June in this forum).

Drop me a pm with your email, I think I have II and III from 2005, but lack I from 2005 and I from 2004, all others I think I have.
Reply 50
Avatar for Rabite
Rabite
8
nota bene
*bawls* Matrices are easy:/

*fills a 4x4 Matrix's elements with 1=0 and drops it on nota*

Maybe they just forgot to put it in the syllabus >>;;
They're so tedious that they didn't deserve their own section or something.
nota bene
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

Some observations I made was that if we let the first complex number have a length a the second will have a+ad etc. as long as we are in the first quadrant. This gives zn+1=zn+adn−1z_{n+1}=z_n+ad^{n-1}Maybe I'm missing something, but I don't see how this works; you seem to have gone directly from talking about the lengths of the vectors to their actual (complex) values. In particular, I don't understand how you're taking into account the value of theta.

For what it's worth, I think the correct formula for the first bit is:

Spoiler

Rabite
II, 2:
I'm not entirely sure how to finish this one off, so someone else do that.We find the equations for the asymptotes meet at (1,-1), and your guess that the axes are at +/-45 degrees seems a good one.

So we suspect that the axes have equations x+y=1, x-y =2. You could probably stop there, but partly to reassure myself, and partly for the hell of it, we can go further:

We look for α,β\alpha, \beta such that α(x−y−2)2−β(x+y)2\alpha(x-y-2)^2-\beta(x+y)^2 resembles the given equation.

Easiest is to look first at the coefficients of x and y, which tells us that α=4\alpha = 4, and then the coefficient of x^2 tells us β=1\beta = 1.

Expanding out, we find 4(x−y−2)2−(x+y)2=3x2−10xy+3y2−16x+16y+164(x-y-2)^2-(x+y)^2 = 3x^2-10xy+3y^2-16x+16y+16 and so using the given equation for the curve, we obtain (x+y)2−4(x−y−2)2=1(x+y)^2 - 4(x-y-2)^2 = 1.

So we have explicitly obtained the equation of a hyperbola in terms of our new axes, confirming our guess.
Someone let me know when stuff's finished. :p:
DFranklin
Maybe I'm missing something, but I don't see how this works; you seem to have gone directly from talking about the lengths of the vectors to their actual (complex) values. In particular, I don't understand how you're taking into account the value of theta.

For what it's worth, I think the correct formula for the first bit is:

Spoiler


Yes, I have rncis(nθ)r^ncis(n\theta) towards the end, but it needs to be mentioned in the beginning. In general that whole post is totally messed up and should not be looked at:p:.

Someone else feel free to do this question, or I'll try to get it correct later.

generalebriety: I think my solution to I/6 is okay unless I've made a stupid mistake somewhere. (added Decota's comment, just to be on the safe side; I had seen it was not differentiated correctly, but not bothered to mention it as it was totally wrong either way.) Also take away this question as 'partially solved' until I fix it; it is horrible at the moment!
Reply 55
Avatar for justinsh
justinsh
4
STEP I question 8

(a) Let y=π−x⇒dy=−dxy=\pi-x \Rightarrow dy=-dx.

I=∫0πxf(sin⁡x)dx\displaystyle I=\int_0^\pi xf(\sin x)dx\\
=−∫π0(π−y)f(sin⁡y)dy\displaystyle=-\int_\pi^0 (\pi-y) f(\sin y)dy\\
=π∫0πf(sin⁡x)dx−∫0πxf(sin⁡x)dx\displaystyle=\pi\int_0^\pi f(\sin x)dx-\int_0^\pi xf(\sin x)dx\\
=π∫0π/2f(sin⁡x)dx+π∫π/2πf(sin⁡x)dx−I\displaystyle=\pi\int_0^{\pi/2} f(\sin x)dx+\pi\int_{\pi/2}^\pi f(\sin x)dx-I\\
=π∫0π/2f(sin⁡x)dx−π∫π/20f(sin⁡y)dy−I\displaystyle=\pi\int_0^{\pi/2} f(\sin x)dx-\pi\int_{\pi/2}^0 f(\sin y)dy-I\\
=π∫0π/2f(sin⁡x)dx+π∫0π/2f(sin⁡x)dx−I\displaystyle=\pi\int_0^{\pi/2} f(\sin x)dx+\pi\int_0^{\pi/2} f(\sin x)dx-I\\
=2π∫0π/2f(sin⁡x)dx−I\displaystyle=2\pi\int_0^{\pi/2} f(\sin x)dx-I\\
∴I=π∫0π/2f(sin⁡x)dx\displaystyle\therefore I=\pi\int_0^{\pi/2} f(\sin x)dx

A shorter alternative due to DFranklin:

I=∫0π/2xf(sin⁡x)dx+∫π/2πxf(sin⁡x)dx\displaystyle I = \int_0^{\pi/2} x f(\sin x)dx + \int_{\pi/2}^\pi x f(\sin x)dx
=∫0π/2xf(sin⁡x)dx−∫π/20(π−y)f(sin⁡(π−y))dy\displaystyle = \int_0^{\pi/2} x f(\sin x)dx -\int_{\pi/2}^0 (\pi-y)f(\sin(\pi-y))dy (by substitution)
=∫0π/2(x+(π−x))f(sin⁡x)dx\displaystyle = \int_0^{\pi/2} (x+(\pi-x)) f(\sin x)dx (recombining)
=π∫0π/2f(sin⁡x)dx\displaystyle = \pi \int_0^{\pi/2} f(\sin x)dx

Using this last result:

J=∫0πx2+sin⁡xdx\displaystyle J=\int_0^\pi \frac{x}{2+\sin x}dx\\
=π∫0π/212+sin⁡xdx \displaystyle =\pi\int_0^{\pi/2} \frac{1}{2+\sin x}dx\\

Let
Unparseable latex formula:

\displaystylet=\tan\frac{x}{2} \Rightarrow dt=\frac{1}{2}(1+\tan^2\frac{x}{2})dx=(1+t^2)dx

. Also, sin⁡x=2t1+t2\displaystyle\sin x=\frac{2t}{1+t^2}. Hence:

J=π∫0112+2t1+t22dtt2+1\displaystyle J=\pi\int_0^1 \frac{1}{2+\frac{2t}{1+t^2}}\frac{2dt}{t^2+1}\\
=π∫01dtt2+t+1\displaystyle=\pi\int_0^1 \frac{dt}{t^2+t+1}
=π∫01dt(t+12)2+34\displaystyle=\pi\int_0^1 \frac{dt}{(t+\frac{1}{2})^2+\frac{3}{4}}
=π∫01dt(2t3+13)2+1\displaystyle=\pi\int_0^1 \frac{dt}{(\frac{2t}{\sqrt{3}}+\frac{1}{\sqrt{3}})^2+1}
=π23∫1/33/3duu2+1\displaystyle=\pi\frac{2}{\sqrt{3}}\int_{1/\sqrt{3}}^{3/\sqrt{3}} \frac{du}{u^2+1}
=π23(π3−π6)\displaystyle=\pi\frac{2}{\sqrt{3}}\left(\frac{\pi}{3}-\frac{\pi}{6}\right)
=3π29\displaystyle=\frac{\sqrt{3}\pi^2}{9}

(b) Let u=(sin⁡−1t)2⇒du=2sin⁡−1t1−t2\displaystyle u=(\sin^{-1}t)^2 \Rightarrow du=\frac{2\sin^{-1}t}{\sqrt{1-t^2}}

Boundaries: (sin⁡−10)2=0,(sin⁡−11)2=(π2)2=π24\displaystyle(\sin^{-1}0)^2=0, (\sin^{-1}1)^2=\left(\frac{\pi}{2}\right)^2=\frac{\pi^2}{4}

∫01(sin⁡−1t)cos⁡[(sin⁡−1t)2]1−t2dt\displaystyle\int_0^1\frac{(\sin^{-1}t)\cos[(\sin^{-1}t)^2]}{\sqrt{1-t^2}}dt\\
=12∫0π2/4cos⁡udu=12sin⁡π24\displaystyle=\frac{1}{2}\int_0^{\pi^2/4}\cos u du=\frac{1}{2}\sin\frac{\pi^2}{4}
justinsh
My part (a) is rather long, is there a shorter way?
I=∫0π/2xf(sin⁡x) dx+∫π/2πxf(sin⁡x) dxI = \int_0^{\pi/2} x f(\sin x) \,dx + \int_{\pi/2}^\pi x f(\sin x) \,dx. Sub y=π−xy = \pi-x in the 2nd integral to get

I=∫0π/2xf(sin⁡x) dx−∫π/20(π−y)f(sin⁡(π−y)) dyI = \int_0^{\pi/2} x f(\sin x) \,dx -\int_{\pi/2}^0 (\pi-y)f(\sin(\pi-y))\,dy. Recombine the two integrals to get

I=∫0π/2(x+(π−x))f(sin⁡x) dxI = \int_0^{\pi/2} (x+(\pi-x)) f(\sin x) \, dx

I=π∫0π/2f(sin⁡x) dxI = \pi \int_0^{\pi/2} f(\sin x) \, dx
Reply 57
Avatar for justinsh
justinsh
4
STEP I questions 4

Please tell me what you think, I'm not certain of what I did.
2.jpeg345.0KB
1.jpeg267.8KB
Reply 58
Avatar for gyrase
gyrase
3
*Bump*

I will rep anyone who posts a solution for II/4 in its entirety. (the trigonometry one)

kthnx.
Reply 59
Avatar for Rabite
Rabite
8
Hmm...
I can do the first bit and the last bit.

But not the middle bit (unless we resort to complex numbers, but that's foul play right?)

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