STEP Maths I, II, III 1989 solutions

OP

I am no expert at STEP, so don't hesitate to correct me, it is more than likely made that I have made a mistake. :redface:

STEP I - Question 8

Using good old de Moivre's theorem;

\cos(4\theta) = \Re( \cos(\theta) + i\sin(\theta))^{4} = \Re(\cos^{4}(\theta) + 4i\cos^{3}(\theta)\sin(\theta) - 6\cos^{2}(\theta)\sin^{2}(\theta) - 4i\sin^{3}(\theta)\cos(\theta) + \sin^{4}(\theta) ) = cos^{4}(\theta) - 6\cos^{2}(\theta)\sin^{2}(\theta) = 8\cos^{4}(\theta) - 8\cos^{2}(\theta) + 1

(using

\cos^{2}(\theta) + \sin^{2}(\theta) \equiv 1

and similary;

[br]\cos(6\theta) = \Re( \cos(\theta) + i\sin(\theta))^{6} = \Re(\cos^{6}(\theta) + 6i\cos^{5}(\theta)\sin(\theta) - 15\cos^{4}(\theta)\sin^{2}(\theta) - 20i\sin^{3}(\theta)\cos^{3}(\theta) + 15\sin^{4}(\theta)\cos^{2}(\theta) + 6i\cos(\theta)\sin^{5}(\theta) - \sin^{6}(\theta) ) = 32\cos^{6}(\theta) - 48\cos^{4}(\theta) +18\cos^{2}(\theta) - 1

Now, consider:

\frac{1}{2}\cos(6\theta) - \frac{1}{2}cos(4\theta)

\frac{1}{2}\cos(6\theta) - \frac{1}{2}cos(4\theta) = 16c^{6} - 28c^{4} + 13c^{2} -1

Where c is

\cos(\theta)

.

Now let

x = cos(\theta)

,

so:

16x^{6} - 28x^{4} + 13x^{2} -1 = 16c^{6} - 28c^{4} + 13c^{2} -1 = \frac{1}{2}\cos(6\theta) - \frac{1}{2}cos(4\theta) = 0[br]

\Rightarrow \cos(6\theta) = \cos(4\theta)

\Rightarrow 6\theta = 2n\pi \pm 4\theta

Which gives;

\theta = \frac{n\pi}{5}

(

n= 1,2,3,4

).
These values of

\theta

give distinct values of

\cos(\theta)

or

\theta = n\pi

(

n= 0,1

)

Hence the roots of the equation are:

x = \cos(0), \cos(\pi), \cos(\frac{\pi}{5}), \cos(\frac{2\pi}{5}), \cos(\frac{3\pi}{5})

and

\cos(\frac{4\pi}{5})

.

Reply 2

OP

STEP III - Question 10

Lets assume that the result is true for n = k;

\displaystyle\sum_{r=1}^k r(r+1)(r+2)(r+3)(r+4) = \frac{1}{6}k(k+1)(k+2)(k+3)(k+4)(k+5)

(*)

[br]\Rightarrow \displaystyle\sum_{r=1}^{k+1} r(r+1)(r+2)(r+3)(r+4) = \frac{1}{6}k(k+1)(k+2)(k+3)(k+4)(k+5) + (k+1)(k+2)(k+3)(k+4)(k+5)

\Rightarrow \displaystyle\sum_{r=1}^{k+1} r(r+1)(r+2)(r+3)(r+4) = \frac{1}{6}(k+1)(k+2)(k+3)(k+4)(k+5)(k+6)

This is the same as (*) except k+1 replaces k. Hence if the result is true for n=k, its true for n= k+1.

When n=1,

LHS of (*) = 1x2x3x4x5 = 120
RHS of (*) = (1/6)x1x2x3x4x5x6 = 120.

So (*) is true for n=1.
Hence, by induction;

\displaystyle\sum_{r=1}^n r(r+1)(r+2)(r+3)(r+4) = \frac{1}{6}n(n+1)(n+2)(n+3)(n+4)(n+5)

\forall n \geq 1

.

Since;

r^{5} < r(r+1)(r+2)(r+3)(r+4)

\Rightarrow \displaystyle\sum_{r=1}^n r^{5} < \displaystyle\sum_{r=1}^n r(r+1)(r+2)(r+3)(r+4)

\Rightarrow \displaystyle\sum_{r=1}^n r^{5} < \frac{1}{6}n(n+1)(n+2)(n+3)(n+4)(n+5)

.

Using (*);

\displaystyle\sum_{r=0}^{n-1} r(r-1)(r-2)(r-3)(r-4) = \frac{1}{6}n(n-1)(n-2)(n-3)(n-4)(n-5)

r(r-1)(r-2)(r-3)(r-4) < r^{5}

\Rightarrow \displaystyle\sum_{r=0}^{n-1} r^{5} > \displaystyle\sum_{r=0}^{n-1} r(r-1)(r-2)(r-3)(r-4)

\Rightarrow \displaystyle\sum_{r=0}^{n-1} r^{5} > \frac{1}{6}(n-5)(n-4)(n-3)(n-2)(n-1)n

In this case,

f(x) = x^{5}

From the previous parts, we know that:

\displaystyle\sum_{r=0}^{n-1} \frac{a^{6}}{n^{6}}r^{5} > \frac{a^{6}}{6n^{6}}(n-5)(n-4)(n-3)(n-2)(n-1)n

and clearly, also:

\displaystyle\sum_{r=0}^{n-1} \frac{a^{6}}{n^{6}}r^{5} < \frac{a^{6}}{6n^{6}}n(n+1)(n+2)(n+3)(n+4)(n+5)

as

n\longrightarrow \infty

,

\frac{a^{6}}{6n^{6}}(n-5)(n-4)(n-3)(n-2)(n-1)n

and

\frac{a^{6}}{6n^{6}}n(n+1)(n+2)(n+3)(n+4)(n+5) \longrightarrow \frac{a^{6}}{6}

So,

\lim_{n\to \infty} \displaystyle\sum_{r=0}^{n-1} \frac{a^{6}}{n^{6}}r^{5} = \frac{a^{6}}{6}

Using a similar arguement, we can show that:

\lim_{n\to \infty} \displaystyle\sum_{r=1}^{n} \frac{a^{6}}{n^{6}}r^{5} = \frac{a^{6}}{6}

We have shown that the limits exist and are equal to

\frac{a^{6}}{6}

.

Hence;

\displaystyle\int^a_0 x^{5} \, \mathrm{d}x = \frac{a^{6}}{6}

Reply 3

kabbers

4

STEP I, Q2

The For

x > 0

, find

\displaystyle\int x\ln x dx

Using integration by parts (

\displaystyle\int u'v dx = uv - \int uv' dx

):

\displaystyle\int x\ln x dx = \frac{1}{2}x^2 \cdot \ln x - \int \frac{1}{2}x^2 \cdot \frac{1}{x} dx

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

\displaystyle = \frac{1}{2}(x^2 \cdot \ln x - \frac{1}{2} x^2) + C

\displaystyle = \frac{1}{4}x^2(2 \ln x - 1) + C

By approximating the area corresponding to

\int^1_0 x\ln(1/x) dx

by n rectangles of equal width and with their top right hand vertices on the curve

y = x\ln(1/x)

, show that, as

n \to \infty

,

\displaystyle\frac{1}{2}(1 + \frac{1}{n})\ln n - \frac{1}{n^2}[ln(\frac{n!}{0!}) + ln(\frac{n!}{1!}) + ln(\frac{n!}{2!}) + ... ln(\frac{n!}{(n-1)!})] \to \frac{1}{4}

(*)

Firstly, note that

\displaystyle \int^1_0 x\ln(1/x) dx

\displaystyle= \int^1_0 x(-\ln(x)) dx

\displaystyle= -\int^1_0 x\ln(x) dx

\displaystyle= -[\frac{1}{4}x^2(2 \ln x - 1)]^1_0

(by the above indefinite integral)

\displaystyle= \lim_{t \to 0} -[\frac{1}{4}x^2(2 \ln x - 1)]^1_t

\displaystyle= \lim_{t \to 0} -\frac{1}{4}(1^2(\ln 1 - 1) - t^2(2 \ln t - 1))

\displaystyle= \lim_{t \to 0} \frac{1}{4}(1 + 2t^2\ln t - t^2))

\displaystyle= \lim_{t \to 0} \frac{1}{4}(1 + 2t(t\ln t) - t^2))

\displaystyle= \frac{1}{4}

The sum of the areas of our rectangles is going to be

\displaystyle (**) \sum^n_{r=1} \frac{1}{n} (\frac{r}{n}\ln(\frac{1}{r/n}))

\displaystyle = \sum^n_{r=1} \frac{1}{n} (\frac{r}{n}\ln(\frac{n}{r}))

\displaystyle = \frac{1}{n^2} \sum^n_{r=1} r\ln(\frac{n}{r})

\displaystyle = \frac{1}{n^2} (1\ln(\frac{n}{1}) + 2\ln(\frac{n}{2}) + 3\ln(\frac{n}{3}) + ... + n\ln(\frac{n}{n}))

\displaystyle = \frac{1}{n^2} \ln((\frac{n}{1})(\frac{n}{2})^2(\frac{n}{3})^3 \cdot \cdot \cdot (\frac{n}{n})^n)

\displaystyle = \frac{1}{n^2} \ln(\frac{n\cdot n^2\cdot n^3\cdot\cdot\cdot n^n}{1\cdot 2^2\cdot 3^3\cdot\cdot\cdot n^n})

\displaystyle = \frac{1}{n^2} \ln(\frac{n^{1+2+3+4+...+n}}{1\cdot 2^2\cdot 3^3\cdot\cdot\cdot n^n})

\displaystyle = \frac{1}{n^2} (ln(n^{1+2+3+4+...+n})-\ln(1\cdot 2^2\cdot 3^3\cdot\cdot\cdot n^n))

\displaystyle = \frac{1}{n^2} (ln(n^{\frac{1}{2}n(n+1)})-\ln(1\cdot 2^2\cdot 3^3\cdot\cdot\cdot n^n))

\displaystyle = \frac{1}{n^2} ({\frac{1}{2}n(n+1)}ln(n)-\ln(1\cdot 2^2\cdot 3^3\cdot\cdot\cdot n^n))

\displaystyle = \frac{1}{n^2}{\frac{1}{2}n(n+1)}ln(n) - \frac{1}{n^2} \ln(1\cdot 2^2\cdot 3^3\cdot\cdot\cdot n^n)

\displaystyle = \frac{1}{2}(1+\frac{1}{n})ln(n) - \frac{1}{n^2} \ln(1\cdot 2^2\cdot 3^3\cdot\cdot\cdot n^n)

Claim:

1\cdot 2^2\cdot 3^3\cdot\cdot\cdot k^k = \frac{k!}{0!}\cdot\frac{k!}{1!}\cdot\frac{k!}{2!} \cdot\cdot\cdot\frac{k!}{k!}

So for k + 1, we must prove that

\frac{k!}{0!}\cdot\frac{k!}{1!}\cdot\frac{k!}{2!} \cdot \cdot \cdot \frac{(k+1)!}{(k+1)!}

is the equation we arrive at.

Basis case (let k=1):

1^1 = \frac{1!}{0!}\cdot\frac{1!}{1!} = 1

Inductive step:

1\cdot 2^2 \cdot 3^3 \cdot \cdot \cdot k^k\cdot (k+1)^{k+1} = (k+1)^{k+1}(\frac{k!}{0!} \cdot \frac {k!}{1!} \cdot \frac{k!}{2!} \cdot \cdot \cdot \frac{k!}{k!})

= ((k+1)\frac{k!}{0!}(k+1) \cdot \frac{k!}{1!}(k+1) \cdot \frac{k!}{2!} \cdot\cdot\cdot (k+1)\frac{k!}{k!})

= (\frac{(k+1)!}{0!}\cdot\frac{(k+1)!}{1!}\cdot\frac{(k+1)!}{2!}\cdot\cdot\cdot\frac{(k+1)!}{k!})

= (\frac{(k+1)!}{0!}\cdot\frac{(k+1)!}{1!}\cdot\frac{(k+1)!}{2!}\cdot\cdot\cdot\frac{(k+1)!}{k!}\cdot \frac{(k+1)!}{(k+1)!})

So,

\displaystyle (**) = \frac{1}{2}(1+\frac{1}{n})ln(n) - \frac{1}{n^2} \ln(\frac{n!}{0!}\cdot\frac{n!}{1!}\cdot\frac{n!}{2!}\cdot\cdot\cdot\frac{n!}{n!})

\displaystyle = \frac{1}{2}(1+\frac{1}{n})ln(n) - \frac{1}{n^2} (\ln(\frac{n!}{0!}) + \ln(\frac{n!}{1!}) + \ln(\frac{n!}{2!}) + ... + \ln(\frac{n!}{(n-1)!}) + \ln(\frac{n!}{n!}))

\displaystyle = \frac{1}{2}(1+\frac{1}{n})ln(n) - \frac{1}{n^2} (\ln(\frac{n!}{0!}) + \ln(\frac{n!}{1!}) + \ln(\frac{n!}{2!}) + ... + \ln(\frac{n!}{(n-1)!}))

As

n \to \infty

, our sum tends to the integral

(*)

. So,

\displaystyle \frac{1}{2}(1+\frac{1}{n})ln(n) - \frac{1}{n^2} (\ln(\frac{n!}{0!}) + \ln(\frac{n!}{1!}) + \ln(\frac{n!}{2!}) + ... + \ln(\frac{n!}{(n-1)!})) \to \frac{1}{4}

is this ok? have i missed anything out or not explained each step in enough detail?

Reply 4

Question 4, STEP I

With six points, each point is attached to five other points, so at least three lines radiating from A must be of the same colour (gold, say). The points which are joined to A must then be joined to each other by silver lines, or an entirely gold triangle would be made. However, then a triangle with entirely silver edges can be made from these three points.

Diagram showing five points is attached.

fivepoints.jpg4.9KB

Reply 5

Question 5, STEP I

\displaystyle (1+x)^{n} = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^{2} + \cdots + \binom{n}{n}x^{n}

a)
Let x = 1.

\displaystyle 2^{n} = \binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \cdots + \binom{n}{n}

Let x = -1

\displaystyle 0 = \binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \cdots + \binom{n}{n}

As n is even. So

\displaystyle \binom{n}{0} + \binom{n}{2} + \cdots + \binom{n}{n} = \binom{n}{1} + \binom{n}{3} + \cdots + \binom{n}{n-1} = 2^{n-1}

As their sum is

\displaystyle 2^{n}

b) Suppose that

\displaystyle \binom{k}{1} + 2\binom{k}{2} + \cdots + k\binom{k}{k} = k2^{k-1}

Then

\displaystyle \binom{k}{0} + 2\binom{k}{1} + 3\binom{k}{2} + \cdots + (k+1)\binom{k}{k} = k2^{k-1} + 2^{k}

Note that

\displaystyle \binom{k}{r} + \binom{k}{r-1} = \binom{k+1}{r}

and that

\displaystyle \binom{k}{k} = \binom{k+1}{k+1}

So, upon adding these two expressions, we get

\displaystyle \binom{k+1}{1} + 2\binom{k+1}{2} + \cdots + (k+1)\binom{k+1}{k+1} = k2^{k-1} + k2^{k-1} + 2^{k} = (k+1)2^{k}

Also,

\displaystyle \binom{1}{1} = 1 = 1 \times 2^{0}

So true by induction.

\displaystyle \displaystyle \sum_{r=0}^{n} \left(r + (-1)^{r}\right) \binom{n}{r} = \left(\binom{n}{1} + 2\binom{n}{2} + \cdots + n \binom{n}{n}\right) + \left(\binom{n}{0} - \binom{n}{1} + \cdots + (-1)^{n}\binom{n}{n}\right)

\displaystyle = n2^{n-1}

Reply 6

OP

STEP II Q1

[br]\cos(3\theta) = \cos(2\theta + \theta) = \cos(2\theta)\cos(\theta) - \sin(2\theta)\sin(\theta) = (2cos^{2}(\theta) - 1)\cos(\theta) - 2\sin(\theta)\cos(\theta)sin(\theta) = 2cos^{3}(\theta) - \cos(\theta) - 2(1-\cos^{2}(\theta))\cos(\theta) = 4cos^{3}(\theta) - 3\cos(\theta).[br]

Substituting x= y-a into the equation:

24x^{3} - 72x^{2} + 66x -19 = 24(y-a)^{3} - 72(y-a)^{2} + 66(y-a) -19 = 0

Notice that the reduced form shown in the question has no

z^{2}

term, so we need to find the value of a which will get rid of the

y^{2}

term.

Expanding out and collectiing terms we get:

24y^{3} - 72(a+1)y^{2} + (72a^{2} + 144a +66)y - 24a^{3} - 72a^{2} -66a -19 = 0

So the value of a to get rid of the

y^{2}

is -1.

So we now have:

24y^{3} - 72(a+1)y^{2} + (72a^{2} + 144a +66)y - 24a^{3} - 72a^{2} -66a -19 = 24t^{3} - 6y -1 = 0

\Rightarrow 24y^{3} - 6y = 1

(*)

Substituting y=z/b into (*);

24(\frac{z}{b})^{3} - 6(\frac{z}{b}) = 1

dividing both sides by 2 and multiplying by

b

;

12\frac{z^{3}}{b^{2}} - 3z = \frac{b}{2}

.

And so to make this equal to;

4z^{3}- 3z = \frac{b}{2}

we make b =

\pm \sqrt3

.

Lets take b =

\sqrt3

and let z =

\cos(\theta)

so;

4z^{3}- 3z = 4\cos^{3}(\theta) - 3\cos(\theta) = \cos(3\theta) = \frac{\sqrt3}{2}

\Rightarrow 3\theta = \frac{pi}{6}, \frac{11\pi}{6}

and

\frac{13\pi}{6}

(these give distinct values of

cos\theta

, and hence, three distinct solutions.)

we know that; x = y - a = (z/b) - a.

Hence the solutions of the equation are:

x = \frac{\cos(\frac{\pi}{18})}{\sqrt3} + 1 , \frac{\cos(\frac{11\pi}{18})}{\sqrt3} +1

and

\frac{\cos(\frac{13\pi}{18})}{\sqrt3} + 1

Reply 7

Glutamic Acid

17

Question 11, STEP I.

Vertical component =

v \sin \theta

Using s = ut + 1/2at^2.

-h = v \sin \theta - 1/2gt^2

1/2gt^2 - v \sin \theta - h = 0

Quadratic in t. Ignoring the negative root.

t = \dfrac{v \sin \theta + \sqrt{v^2 \sin^2 \theta + 2hg}}{g}

R = horizontal component x time.
Horizontal component =

v \cos \theta

R = v \cos \theta (\dfrac{v \sin \theta + \sqrt{v^2 \sin^2 \theta + 2hg}}{g}

R = \dfrac{v^2 \sin \theta \cos \theta + v \cos \theta \sqrt{v^2 \sin^2 \theta + 2hg}}{g}

Multiply both numerator and denominator by 2.

R = \dfrac{2v^2 \sin \theta \cos \theta + 2v \cos \theta \sqrt{v^2 \sin^2 \theta + 2hg}}{2g}

Using

\sin (2\theta) = 2 \sin \theta \cos \theta

We can simplify it.

R = \dfrac{v^2 \sin 2\theta + 2v \cos \theta \sqrt{v^2 \sin^2 \theta + 2hg}}{2g}

Splitting up Mr. Big Fraction.

R = \dfrac{v^2 \sin 2\theta}{2g} + \dfrac{2v \cos \theta \sqrt{v^2 \sin^2 \theta + 2hg}}{2g}

Ignore the first fraction for the moment, we'll work on the second one.

\dfrac{2v \cos \theta \sqrt{v^2 \sin^2 \theta (1 + \dfrac{2hg}{v^2 \sin^2 \theta})}}{2g}

Unparseable latex formula:

\dfrac{2v^2 \cos \theta \sin \theta}{2g} \times \sqrt{1 + \dfrac{2hg}{v^2 \sin^2 \theta}

We can use our old friend Mr. Double Angle Formula to simplify it:

\dfrac{v^2 \sin 2\theta}{2g} \sqrt{1 + \dfrac{2hg}{v^2 \sin^2 \theta}}

The whole fraction looks like:

Unparseable latex formula:

R = (\dfrac{v^2\sin 2\theta}{2g}) + (\dfrac{v^2 \sin 2\theta}{2g}) \sqrt{1 + \dfrac{2hg}{v^2 \sin^2 \theta}

And lo and behold, there's a common factor to both fractions. Let's take it out:

R = \dfrac{v^2}{2g}\sin 2\theta(1 + \sqrt{1 + \dfrac{2hg}{v^2 \sin^2 \theta}})

And that's what we're looking for.
*I'll finish off the last small bit in a few minutes*

Reply 8

Swayum

18

STEP I Question 6

y = f(x)

dy/dx = f'(x)

The normal to the curve will therefore have gradient -1/f'(x). The equation of the normal would be

y - f(x) = \frac{-1}{f'(x)}(x' - x)

(where x' is the general x coordinate of the normal).

At the point Q, x' = 0 as it cuts the y axis.

y - f(x) = \frac{x}{f'(x)}

y = \frac{x}{f'(x)} + f(x)

The distance PQ can be worked out using Pythagoras' theorem.

PQ^2 = (f(x) - (\frac{x}{f'(x)} + f(x)))^2 + x^2

= \frac{x^2}{f'(x)^2} + x^2

It's given that

PQ^2 = e^{x^2} + x^2

So

\frac{x^2}{f'(x)^2} + x^2 = e^{x^2} + x^2

\frac{x^2}{f'(x)^2} = e^{x^2}

\frac{x^2}{e^{x^2}} = f'(x)^2

\frac{x}{e^{0.5x^2}} = f'(x)

\int 1 \mathrm{d}y = \int \frac{x}{e^{0.5x^2}} \mathrm{d}x

y = \int xe^{-0.5x^2} \mathrm{d}x

y = -e^{-0.5x^2} + c

(use a substitution of u = x^2 if you can't see why)

-2 = -1 + c

c = -1

y = -e^{-0.5x^2} - 1

Reply 9

STEP III, Q8

\frac{\mathrm{d}^{2}x}{\mathrm{d}t^{2}} = 4\frac{\mathrm{d}x}{\mathrm{d}t} - 4\frac{\mathrm{d}y}{\mathrm{d}t}

\frac{\mathrm{d}^{2}x}{\mathrm{d}t^{2}} - 4\frac{\mathrm{d}x}{\mathrm{d}t} + 4x = 48e^{2t} + 48e^{-2t}

Let

u = \frac{\mathrm{d}x}{\mathrm{d}t} - 2x

Then

\frac{\mathrm{d}u}{\mathrm{d}t} - 2u = 48e^{2t} + 48e^{-2t}

Using an integrating factor,

e^{-2t}

, we get

ue^{-2t} = \int 48 + 48e^{-4t} \; \mathrm{d}t = 48t - 12e^{-4t} + c

u = e^{2t}(48t + c) - 12e^{-2t}

\frac{\mathrm{d}x}{\mathrm{d}t} - 2x = e^{2t}(48t + c) - 12e^{-2t}

xe^{-2t} = \int 48t + c - 12e^{-4t} \; \mathrm{d}t = 24t^{2} + ct + 3e^{-4t} + d

x = e^{2t}(24t^{2} + ct + d) + 3e^{-2t}

Applying boundary conditions:

0 = d + 3 \Rightarrow d = -3

\frac{\mathrm{d}x}{\mathrm{d}t} = 2e^{2t}(24t^{2} + ct + d) + e^{2t}(48t + c) - 6e^{-2t} = 4(x-y)

2d + c - 6 = 0 \Rightarrow c = 6 - 2d = 6 + 6 = 12

x = e^{2t}(24t^{2} + 12t - 3) + 3e^{-2t}

4y = 4x - \frac{\mathrm{d}x}{\mathrm{d}t} = e^{2t}(48t^{2} - 4t - 18) + 18e^{-2t} = 4te^{2t}(12t - 1) - 36\sinh 2t

y = te^{2t}(12t - 1) - 9\sinh 2t

---

I think that I shall restrict myself to a maximum of three solutions per day (at least at the start), so that everyone has a chance to answer some if they want.

Reply 10

kabbers

4

Hang on a sec, mine was mislabelled, should be Step I not step II.. sorry about that!

edit: thanks.. now typing up II/2 :p:

Reply 11

kabbers

4

STEP II/2

\displaystyle \tan x = \sum^{\infty}_0 a_n x^n

\displaystyle x \cot x = 1 + \sum^{\infty}_1 b_n x^n

\displaystyle (*) \cot x = \frac{1}{x} + \sum^{\infty}_1 b_n x^{n-1}

\displaystyle = \frac{1}{x} + \sum^{\infty}_0 b_{n+1} x^n

\displaystyle 2 \cot 2x = \cot x - \tan x

Therefore, by (*),

\displaystyle 2 \cot 2x = \frac{1}{x} + \sum^{\infty}_0 b_{n+1} x^n - \sum^{\infty}_0 a_n x^n

\displaystyle = \frac{1}{x} + \sum^{\infty}_0 (b_{n+1} - a_n) x^n

we can also use (*) to express 2cot2x as

\displaystyle 2(\frac{1}{2x} + \sum^{\infty}_0 b_{n+1} (2x)^n) = \frac{1}{x} + 2\sum^{\infty}_0 b_{n+1} 2^n x^n

Therefore,

\displaystyle \frac{1}{x} + 2\sum^{\infty}_0 b_{n+1} 2^n x^n = \frac{1}{x} + \sum^{\infty}_0 (b_{n+1} - a_n) x^n

\displaystyle \sum^{\infty}_0 b_{n+1} 2^{n+1} x^n = \sum^{\infty}_0 (b_{n+1} - a_n) x^n

we can compare the coefficients of the respective summations and say

\displaystyle b_{n+1} 2^{n+1} = b_{n+1} - a_n

\displaystyle a_n = b_{n+1} - b_{n+1} 2^{n+1}

\displaystyle a_{n-1} = b_n(1 - 2^n)

for the second part, note that

\displaystyle x \csc x = 1 + \sum^{\infty}_1 c_n x^n

\displaystyle \csc x = \frac{1}{x} + \sum^{\infty}_1 c_n x^{n-1}

\displaystyle = \frac{1}{x} + \sum^{\infty}_0 c_{n+1} x^n

we have to find an identity for 2csc2x

\displaystyle 2 \csc 2x = \frac{2}{\sin 2x} = \frac{2}{2 \sin x \cos x} = \frac{1}{\sin x \cos x}

(using partial fractions)

\displaystyle = \frac{\sin^2 x + cos^2 x}{\sin x \cos x} = \frac{\cos x}{\sin x} + \frac{\sin x}{\cos x} = \cot x + \tan x

Also,

\displaystyle 2 \csc 2x = 2(\frac{1}{2x} + \sum^{\infty}_0 c_{n+1} (2x)^n)

\displaystyle = 2(\frac{1}{2x} + \sum^{\infty}_0 c_{n+1} 2^n x^n)

\displaystyle = \frac{1}{x} + \sum^{\infty}_0 c_{n+1} 2^{n+1} x^n

\displaystyle \cot x + \tan x = \frac{1}{x} + \sum^{\infty}_0 b_{n+1} x^n + \sum^{\infty}_0 a_n x^n

\displaystyle 2 \csc 2x = \frac{1}{x} + \sum^{\infty}_0 b_{n+1} x^n + \sum^{\infty}_0 a_n x^n

2 \csc 2x = \cot x + \tan x

, so:

\displaystyle \frac{1}{x} + \sum^{\infty}_0 c_{n+1} 2^{n+1} x^n = \frac{1}{x} + \sum^{\infty}_0 b_{n+1} x^n + \sum^{\infty}_0 a_n x^n

\displaystyle \sum^{\infty}_0 c_{n+1} 2^{n+1} x^n = \sum^{\infty}_0 b_{n+1} x^n + \sum^{\infty}_0 a_n x^n

\displaystyle \sum^{\infty}_0 c_{n+1} 2^{n+1} x^n = \sum^{\infty}_0 (b_{n+1} + a_n) x^n

Comparing coefficients of the polynomials (in x) of the expansion,

\displaystyle c_{n+1} 2^{n+1} = b_{n+1} + a_n

Rearranging and substituting in the earlier definition of

a_{n-1} = b_n(1-2^n) \Longrightarrow \frac{a_{n-1}}{1-2^n} = b_n

,

\displaystyle c_{n+1} 2^{n+1} = \frac{a_{n}}{1-2^{n+1}} + a_n

\displaystyle c_{n+1} = (\frac{1}{1 - 2^{n+1}}+1)(\frac{1}{2^{n+1}})a_n

\displaystyle c_{n+1} = (\frac{2 - 2^{n+1}}{1 - 2^{n+1}})(\frac{1}{2^{n+1}})a_n

\displaystyle c_{n+1} = (\frac{1 - 2^n}{1 - 2^{n+1}})(\frac{1}{2^n})a_n

\displaystyle c_{n+1} = (\frac{2^n - 1}{2^{n+1} - 1})(\frac{1}{2^n})a_n

\displaystyle c_{n+1} = (\frac{2^{n-1} - 1}{2^{n} - 1})(\frac{1}{2^{n-1}})a_{n-1}

any corrections welcome !

Reply 12

*bobo*

8

STEP II, (11) attached

attachments 2&3 are in the wrong order

step1989q11.doc553.0KB

step1989q11p3.doc210.4KB

step1989q11p2.doc162.3KB

Reply 13

*bobo*

8

Step III (2)

step1989q2p1.doc116.2KB

step1989q2p2.doc80.9KB

step1989q2p3.doc73.2KB

Reply 14

*bobo*

8

STEP II (6) attached

step1989q6p1.doc81.4KB

step1989q62.doc97.3KB

Reply 15

OP

STEP II Question 3

2(u + iv) = e^{x+iy} - e^{-(x+iy)} = 2\sinh(x + iy) = 2\sinh(x)\cosh(iy) + 2\cosh(x)\sinh(yi) = 2\sinh(x)\cos(y) + 2i\sin(y)\cosh(x)

Equating real and imaginary parts:

u = \sinh(x)\cos(y)

(*)

v = \sin(y)\cosh(x)

(**)

if x=a, then:

u = \sinh(a)\cos(y)

(1)

v = \sin(y)\cosh(a)

(2)

Now, subbing (1) and (2) into:

\cos^{2}y + \sin^{2}y = 1

We get:

(\frac{u}{\sinh(a)})^{2} + (\frac{v}{\cosh(a)})^{2} = 1 (\triangle)

As required.

Substituting y=b into (*) and (**):

u = \sinh(x)\cos(b)

(3)

v = \cosh(x)\sin(b)

(4).

This time, using

\cosh^{2}x - \sinh^{2}x = 1

, we end up with:

(\frac{v}{\sin(b)})^{2} - (\frac{u}{\cos(b)})^{2} = 1 (\heartsuit)

.

Since

0 < \sin(b) < 1

and

\cosh(x) \geq 1

,

\Rightarrow v>0

. Hence, (

\heartsuit

) corresponds to a single hyeprbola branch, which is above the line v=0.

One point of intersection is:

u = \sinh(a)\cos(b)

and

v = \sin(b)\cosh(a)

.

Differentiating

\triangle

implicitly we get:

\frac{dv}{du} = (\frac{u}{v})\coth^{2}(a)

.

Which is equal to:

-\cot(b)\coth(a)

at

u = \sinh(a)\cos(b)

and

v = \sin(b)\cosh(a)

.

Differentiating

\heartsuit

implicitly we get:

\frac{dv}{du} = (\frac{u}{v})\tan^{2}(b)

.

Which is,

\tanh(a)\tan(b)

when

u = \sinh(a)\cos(b)

and

v = \sin(b)\cosh(a)

.

\tanh(a)\tan(b)

X

-\cot(b)\coth(a) = -1

Hence the curves intersect and right angles at this point.

Regarding the sketch, I think the elipse becomes a vertical straight line from v = -1 to v = 1, and the hyperbola becomes the line v = 1.

Reply 16

STEP 1, Q3

\textbf{x} = x\textbf{a}, \; \textbf{y} = (1-x)\textbf{b}

\textbf{c} = \frac{1}{3}(2\textbf{a} + \textbf{b}), \textbf{d} = \frac{1}{3}(\textbf{a} + 2\textbf{b})

(By the ratio theorem.)

\overrightarrow{CY} = (\frac{2}{3} - x)\textbf{b} - \frac{2}{3}\textbf{a}

\overrightarrow{DX} = (x - \frac{1}{3})\textbf{a} - \frac{2}{3}\textbf{b}

These vectors are perpendicular, so the scalar product is zero.

(\frac{2}{9} + x - x^{2})\textbf{a}.\textbf{b} - \frac{2}{9} = 0

(Note that the scalar product is commutative and distributive.)

\textbf{a}.\textbf{b} = \frac{-2}{9x^{2} - 9x - 2}

Let

f(x) = \frac{-2}{9x^{2} - 9x - 2}, \; 0 \leq x \leq 1

f'(x) = \frac{2(18x - 9)}{(9x^{2}-9x-2)^{2}}

Minimum and maximum values occur at the endpoint of a range or at a turning point. There is a turning point when x = 1/2.

f(0) = 1, \; f(\frac{1}{2}) = \frac{8}{17}, \; f(1) = 1

So

\frac{8}{17} \leq \textbf{a}.\textbf{b} \leq 1

Note that

\textbf{a}.\textbf{b} = |a||b|\cos\theta = \cos\theta

Where

\theta = \angle AOB, \; 0 < \theta < \pi

The maximum value is

\theta

occurs when

\cos\theta = \frac{8}{17}

Suppose that

\lambda \overrightarrow{CY} = \overrightarrow{CE},\; \mu \overrightarrow{DX} = \overrightarrow{DE}

Then

\lambda (\frac{1}{6} \textbf{b} - \frac{2}{3} \textbf{a}) = \frac{1}{3} ( \textbf{b} - \textbf{a} ) + \mu ( \frac{1}{6} \textbf{a} - \frac{2}{3} \textbf{b} )

So

\frac{1}{6} \lambda = \frac{1}{3} - \frac{2}{3}\mu

-\frac{2}{3}\lambda = -\frac{1}{3} + \frac{1}{6}\mu

\lambda = \mu = \frac{2}{5}

\textbf{e} = \textbf{c} + \frac{2}{5}\overrightarrow{CY} = \frac{2}{5}\textbf{a} + \frac{2}{5}\textbf{b}

By the cosine rule,

AB^{2} = \frac{18}{17}

Let F be the midpoint of AB. By Pythagoras,

OF^{2} = 1 - \frac{9}{34} = \frac{25}{34}, \; OF = \frac{5}{\sqrt{34}}

OE = \frac{3}{5} \times \frac{5}{\sqrt{34}} = \frac{3}{\sqrt{34}}

Reply 17

OP

*bump* (just to remind people that this thread is still here :biggrin:

) - I have solutions that I can type up if no one else wants to, but I really don't want to take over the thread..

Reply 18

OP

STEP I Q7 graph.doc200.7KB

STEP I - Question 7

*Graph Attached*

If we work out the area of a quarter of the rectangle, which is in the top right hand bit of the axes, and maximise this, it make calculation soooo much eaiser. (Since we can forget about modulus signs and y being negative) :

Let the area of the rectangle be

A

.

Base of smaller rectangle = 1-x
height =

\sqrt(1-(1-x)) = \sqrt x

so

\frac{1}{4}A = (1-x)\sqrt x

which we want to maximise.

\frac{d\frac{1}{4}A}{dx} = \frac{1-3x}{2\sqrt x}

Which is zero when

x = \frac{1}{3}

Finding the second derivative and subsituting x =

\frac{1}{3}

in shows

\frac{1}{4}A

is a maximum at this point.

So

A_{max} = 4 \times [(1-\frac{1}{3})\frac{1}{\sqrt 3}] = \frac{8}{\sqrt 27}

.

________

We can do the second bit in the same way, since its a closed curve that is symmetrical in both axes.

Let the area of the second rectangle be

B

.
Working in the top right quadrant again;
Base = 1-x
height =

(1-(1-x))^{\frac{n}{2m}} = x^{\frac{n}{2m}}

.

So

\frac{1}{4}B = x^{\frac{n}{2m}}(1-x)

.

\frac{d\frac{1}{4}B}{dx} = x^{\frac{n}{2m} -1}(\frac{n}{2m} - x(1 + \frac{n}{2m}))

Which is zero when x = 0 or x =

\frac{n}{2m + n}

.

Clearly the area is a minimum when x= 0, and so is a maximum when x =

\frac{n}{2m + n}

.

\Rightarrow B_{max} = 4 \times [(\frac{n}{2m+n})^{\frac{n}{2m}}(1 - \frac{n}{2m +n})] = 4(\frac{n}{2m + n})^{\frac{n}{2m}}(\frac{2m}{2m+n})

.

Reply 19

nota bene