STEP maths I, II, III 1990 solutions

I am

Edit: Okay I'll start typing up II/1

Prove that both

x^4-2x^3+x^2

and

x^2-8x+17

are non-negative for all real x.
Factorise

x^4-2x^3+x^2

to

(x^2-1)^2

, which clearly is positive.
We can verify that

x^2-8x+17

is positive for all real x by rewriting it as

(x-4)^2+1

, which clearly is positive.

The equation

x^4-2x^3+x^2-8x+17=0

has clearly no real solutions for

x\le0

as we have earlier shown that

x^4-2x^3+x^2

is always positive, and -8(-x)+17 is always positive since two negative numbers multiplied is positive...

For the interval x>2 we can consider

x^4-2x^3

as we have earlier shown that

x^2-8x+17

always is positive for all real x. Factorise it to get

x^3(x-2)

. For positive x

x^3

is always positive, and for numbers over 2

(x-2)

is always postitive (and when x=2 the whole equation is still positive, as we have the x^2 etc. terms as well.)

Now consider the interval

0<x\le2

. Factorise

x^4-2x^3+x^2

to

(x^2-1)^2

, which clearly is positive. For 17-8x where x is less than 2 it is obvious that it will never be negative.

Prove that the equation

x^4-x^3+x^2-4x+4=0

has no real roots. First, consider the interval x<1, where we can use the factorisation into

4(1-x)+x^2((x-\frac{1}{2})^2+\frac{3}{4})

. Now all squared things are clearly positive, as is the constant, and for the chosen interval 4(1-x) is positive as well.
Consider x=1, here everything stays the same, except the 4(1-x) part that disappears, and the expression as a whole is still positive.
Now consider x>1 - factorise into

x^3(x-1)

(which is positive as the cube of a positive number is positive, and x-1 for x>1 is positive) and

(x-2)^2

, which obviously is positive as it is an even power.
Therefore , for all x the equation

x^4-x^3+x^2-4x+4=0

has no real solutions (as all y-values are above the x-axis).

Likely I have a few typos and reversed inequalities in there, point out if you see any :smile:

Reply 2

nota

Go open a thread now I know you want it

Not in public.
:redface:

Ahh...it's not copying the formatting ._.
I don't want to have to type out all those links again!

Reply 3

generalebriety

16

Rabite

Not in public.
:redface:

Ahh...it's not copying the formatting ._.
I don't want to have to type out all those links again!

Was that a hint? :wink:

Spoiler

Reply 4

I was just moaning without intent actually, as I do (I would fit right into HnR). Thanks! :wink:

Reply 5

III/9 Easy as

\pi e

tant = sinhu
v = sechu = 1/coshu = 1/sqrt(1+sinh^2u)
= 1/sqrt(1+tan^2t) = 1/sqrt(sec^2t) = 1/sect = cost

dt/du = (dv/du)/(dv/dt) = (-sechutanhu)/(-sint)
= sechu.sqrt(1-sech^2u)/sqrt(1-cos^2t)
= v.sqrt(1-v^2)/sqrt(1-v^2) = v

dv/du = -sechutanhu, -2dv/du = 2sechutanhu
= 2v.sqrt(1-v^2) = 2costsint = 2sin2t
d2v/du2 = d/du(-sechutanhu) = sechutanh^2u - sech^3u
-coshud2v/du2 = sech^2u - tanh^2u = v^2 - (1-v^2)
= cos^2t - sin^2t = cos2t

du/dt.d2v/dt2 = 1/v.(-cost) = -cost/cost = -1
dv/dt.d2u/dt2 = -sint.d/dt(1/v) = -sint.d/dt(sect) = -sint.sect.tant
= -tan^2t
(du/dt)^2 = 1/v^2 = 1/cos^2t = sec^2t = 1 + tan^2t
Sum of these parts = - 1 - tan^2t + 1 + tan^2t = 0

Reply 6

I don't want to type up the following questions at the moment, but I've done them if anyone needs hints:

III/1

Spoiler

III/2

Spoiler

III/5 (now written up, see post #11)

Spoiler

Rabite, you're missing a link to the 1991 thread :tongue:

Reply 7

insparato

Im having a go at II/8 atm

Reply 8

nota bene

I'll have a look at I/7.

edit: having looked more I think II/10 suits me, but I doubt I'll manage before I go to bed:/

Reply 9

insparato

Im tired. The integrals in the last two parts of II/8 are screaming at me. But i cant think straight when im tired so ill try crack the last two parts(basically one part to be honest) tomorrow :smile:

.

Reply 10

III/5

\begin{pmatrix} n \\ r \end{pmatrix} + \begin{pmatrix} n \\ r-1 \end{pmatrix} = \frac{n!}{r!(n-r)!} + \frac{n!}{(r-1)!(n-r+1)!} = \frac{n!(n-r+1)}{r!(n-r+1)!} + \frac{n!r}{r!(n-r+1)!}

= \frac{n!(n-r+1) + n!r}{r!(n-r+1)!} = \frac{n!(n+1)}{r!(n-r+1)!} = \frac{(n+1)!}{r!(n+1-r)!} = \begin{pmatrix} n+1 \\ r \end{pmatrix}

By induction. Case when n=1 is clearly just the product rule:

(uv)^{(1)} = u^{(1)}v + uv^{(1)}

Assume true for n=k, then differentiate again to get:

(uv)^{(k+1)} = u^{(k+1)}v + u^{(k)}v^{(1)} + \begin{pmatrix} k \\ 1 \end{pmatrix}u^{(k)}v^{(1)} + \begin{pmatrix} k \\ 1 \end{pmatrix}u^{(k-1)}v^{(2)} + \begin{pmatrix} k \\ 2 \end{pmatrix}u^{(k-1)}v^{(2)} + ... + uv^{(k+1)}

= u^{(k+1)}v + (\begin{pmatrix} k \\ 0 \end{pmatrix}+\begin{pmatrix} k \\ 1 \end{pmatrix}) u^{(k)}v^{(1)} + (\begin{pmatrix} k \\ 1 \end{pmatrix}+ \begin{pmatrix} k \\ 2 \end{pmatrix})u^{(k-1)}v^{(2)} + ... + uv^{(k+1)}

= u^{(k+1)}v + \begin{pmatrix} k+1 \\ 1 \end{pmatrix}u^{(k)}v^{(1)} + \begin{pmatrix} k+1 \\ 2 \end{pmatrix}u^{(k-1)}v^{(2)} + ... + uv^{(k+1)}

Which is the proposition with n=(k+1), therefore by induction it is true for all natural n.

y=\arcsin x => \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} => \frac{d^2y}{dx^2} = \frac{x}{(1-x^2)^\frac{3}{2}}

Therefore, the proposition for n=0 reduces to:

\frac{x}{\sqrt{1-x^2}} - \frac{x}{\sqrt{1-x^2}} = 0

which is obviously true.
Now assume that the proposition is true for n=k and differentiate:

(1-x^2)y^{(k+3)} - 2xy^{(k+2)} - (2k+1)xy^{(k+2)} - (2k+1)y^{(k+1)} - k^2y^{(k+1)} = 0

(1-x^2)y^{(k+3)} - (2(k+1)+1)xy^{(k+2)} - (k+1)^2y^{(k+1)} = 0

Which is the proposition for n=k+1. Therefore, by induction, the proposition is true for all non-negative integer n.

I hate latex, it took me probably 3 times as long to type this up as to actually do it on paper.

Reply 11

II/16 is awesome by the way if anyone wants to attempt it :biggrin:

Reply 12

Aw. *pat pat* (at Insparato)
I tried III/4 - think I got it, but I'm a bit sleepy for LaTeX...

Reply 13

I've been doing III/6, I'm OK down to the part where it says 'state the area of the region bounded by E' (thanks to the parameterisation method that I picked up after I think Rabite used it), and I was wondering whether there was an easy way to work that out based on the original curve, whose area is obviously pi, and the transformation.

Reply 14

DFranklin

18

insparato

Im tired. The integrals in the last two parts of II/8 are screaming at me. But i cant think straight when im tired so ill try crack the last two parts(basically one part to be honest) tomorrow :smile:

.

I got this out, but boy did I find it tricky:

Spoiler

Reply 15

I knew there was something funny about that!
It's an ellipse, I'm quite sure. I think you're meant to work out that

"An ellipse centered at the origin can be viewed as the image of the unit circle under a linear map associated with a symmetric matrix A = PDPt, D being a diagonal matrix with the eigenvalues of A, both of which are real positive, along the main diagonal, and P being a real unitary matrix having as columns the eigenvectors of A."
(From Wikipedia)

And the enclosed by an ellipse is pi*ab where a and b are the 'radii' along the y axis and x axis.

Reply 16

insparato

I got the first two parts out. For the 3rd

Im stuck with

2 \displaystyle [y(t)]^2 = \int [y(t)]^4 + 2 \int [x(t)]^2y(t)y'(t)

The thing is for

\int 2[x(t)]^2y(t)y'(t)

could you just do

u = blah

du = blah

dv = x(t)y(t)

v = \int x(t)y(t) = 0

Therefore

\int [x(t)]^2y(t)y'(t)

= blah(0) - \int blah(0) = 0

Hmm i havent looked at the spoiler because this very frustrating. Even if i get it out, i doesn't make me a better mathematician for being able to do it. There seems to be loads of ways you can do by parts with these integrals all leading on to different integrals. It just seems thats its luck you pick the right combination of u = .. and dv/dx = ... and maybe a bit of experience in these matters.

Reply 17

DFranklin

18

Speleo

I've been doing III/6, I'm OK down to the part where it says 'state the area of the region bounded by E' (thanks to the parameterisation method that I picked up after I think Rabite used it), and I was wondering whether there was an easy way to work that out based on the original curve, whose area is obviously pi, and the transformation.

You just need to multiply by the determinant

=\frac{4}{25}(9 \times 6 - (-2) \times (-2)) = 8

. (If the determinant had turned out to be -ve you would need to take the modulus too).

Reply 18