STEP Maths I, II, III 1989 solutions

STEP I (Mathematics)

STEP II (F.Maths A)
1: Solution by squeezebox
2: Solution by kabbers
3: Solution by squeezebox
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5: Solution by Dystopia
6: Solution by *bobo*
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11:Solution by *bobo*
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14:Solution by squeezebox
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STEP III (F.Maths B)
1: Solution by squeezebox
2: Solution by *bobo*
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8: Solution by Dystopia
9: Solution by squeezebox
10: Solution by squeezebox
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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

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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.

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 $\cos(\frac{4\pi}{5})$.
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;

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

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

$\displaystyle = n2^{n-1}$
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 $\frac{\cos(\frac{13\pi}{18})}{\sqrt3} + 1$
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*
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)

$-2 = -1 + c$

$c = -1$

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