STEP 2006 Solutions Thread

SimonM

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Daniel Freedman

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STEP III 2006, Question 2

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i) $\displaystyle I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos^2{\theta}}{1 - \sin{\theta} \sin{2 \alpha}} \ \mathrm{d} \theta$

Let $\theta = - \psi \implies \mathrm{d} \theta = - \mathrm{d}\psi$

$\\ \displaystyle \therefore I = - \int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} \frac{\cos^2{{(-\psi)}}}{1 - \sin{{(-\psi)}} \sin{2 \alpha}} \ \mathrm{d} \psi = \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\cos^2{{(\psi)}}}{1 + \sin{(\psi)} \sin{2 \alpha}} \ \mathrm{d} \psi = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos^2{\theta}}{1 + \sin{\theta} \sin{2 \alpha}} \ \mathrm{d} \theta$ , as theta and psi are just dummy variables, and sine is odd and cos is even.

$\\ \displaystyle \therefore 2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos^2{\theta}}{1 - \sin{\theta} \sin{2 \alpha}} + \frac{\cos^2{\theta}}{1 + \sin{\theta} \sin{2 \alpha}} \ \mathrm{d} \theta \\ \\ \\ = \int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} \frac{ 2 \cos^2{\theta}}{1 - \sin^2{\theta}\sin^2{2\alpha}} \ \mathrm{d} \theta = \int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} \frac{2}{\sec^2{\theta} - \tan^2{\theta}\sin^2{2\alpha}} \ \mathrm{d} \theta \\ \\ \\ = \int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} \frac{2}{1 + \tan^2{\theta} - \tan^2{\theta}(1 - \cos^2{2\alpha})} \mathrm{d}\theta \\ \\ \\ = \int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} \frac{2}{1+\tan^2{\theta}\cos^2{2\alpha}} \mathrm{d} \theta$

as required.

ii) $\displaystyle J = \int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} \frac{\sec^2{\theta}}{1+\tan^2{\theta}\cos^2{2 \alpha}} \mathrm{d} \theta = \sec^2{2 \alpha} \int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} \frac{\sec^2{\theta}}{\sec^2{2 \alpha} + \tan^2{\theta}} \mathrm{d} \theta$

Let $u = \tan{\theta} \implies \mathrm{d}\theta = \frac{\mathrm{d}u}{\sec^2{\theta}}$

$\\ \displaystyle \therefore J = \sec^2{2 \alpha} \int_{-\infty}^{\infty} \frac{1}{\sec^2{2 \alpha} + u^2} \mathrm{d}u = \sec^2{2 \alpha} \big[ \cos{2\alpha} \arctan{(u\cos{2\alpha})} \big]_{-\infty}^{\infty} \\ \\ \\ = \sec{2\alpha} \big[ \frac{\pi}{2} - - \frac{\pi}{2} \big] = \pi \sec{2\alpha}$

as $\cos{2\alpha} > 0$ for $0 < \alpha < \frac{\pi}{4}$

iii)

$\\ \displaystyle I\sin^2{2\alpha} + J\cos^2{2\alpha} = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2{2\alpha} + \cos^2{2\alpha}\sec^2{\theta}}{1 + \tan^2{\theta}\cos^2{2\alpha}} \mathrm{d} \theta \\ \\ \\ = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2{2\alpha} + \cos^2{2\alpha}\sec^2{\theta}}{1 + (\sec^2{\theta}-1)\cos^2{2\alpha}} \mathrm{d} \theta \\ \\ \\ = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2{2\alpha} + \cos^2{2\alpha}\sec^2{\theta}}{\sin^2{2\alpha} + \cos^2{2\alpha}\sec^2{\theta}} \mathrm{d}\theta = [\theta]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \pi$

$\\ \therefore I\sin^2{2\alpha} + \pi \sec{2\alpha}\cos^2{2\alpha} = \pi \\ \\ = \pi ( 1 - \cos{2\alpha}) \\ \\ = \pi ( 1 - (1-2\sin^2{2\alpha})) \\ \\ = \pi(2\sin^2{2\alpha})$

$\\ \therefore I = \frac{\pi(2\sin^2{2\alpha})}{\sin^2{2\alpha}} = \frac{\pi(2\sin^2{2\alpha})}{4\sin^2{\alpha}\cos^2{\alpha}} = \frac{1}{2}\pi \sec^2{\alpha}$

as required.

iv) If $\frac{\pi}{4} < \alpha < \frac{\pi}{2}$ , then $\cos{2\alpha} < 0$ , which will change the value of J to $-\pi \sec{2\alpha}$

$\\ \therefore I\sin^2{2\alpha} - \pi \sec{2\alpha} \cos^2{2\alpha} = \pi \\ \\ \implies I \sin^2{2\alpha} = \pi(2\cos^2{\alpha}) \\ \\ \implies I = \frac{ \pi(2\cos^2{\alpha}) }{4 \sin^2{\alpha} \cos^2{\alpha}} = \frac{1}{2} \pi \mbox{cosec}^2{\alpha}$

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Daniel Freedman

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STEP III 2006, Question 3

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i) tan is an odd function ie tan(-x) = - tan(x) and therefore can have no even powers of x in its series expansion

Prove $\cot{x} - \tan{x} = 2\cot{2x}$

$\mbox{LHS} = \frac{1}{\tan{x}} - \frac{\tan{x}}{1} = \frac{1 - \tan^2{x}}{\tan{x}} = 2 \left( \frac{1-\tan^2{x}}{2\tan{x}} \right) = 2 \left( \frac{1}{\frac{2\tan{x}}{1-\tan^2{x}}} \right) = \frac{2}{\tan{2x}} = 2\cot{2x} = \mbox{RHS}$

$\displaystyle \therefore \frac{1}{x} + \sum_{n=0}^{\infty} b_n x^n - \sum_{n=0}^{\infty} a_n x^n = 2 \left( \frac{1}{2x} + \sum_{n=0}^{\infty} b_n (2x)^n \right) \implies \sum_{n=0}^{\infty} \left( b_n x^n - a_n x^n \right) = 2^{n+1} \sum_{n=0}^{\infty} b_n x^n \implies b_n - a_n = 2^{n+1}b_n \ (x \neq 0 ) \implies a_n = b_n (1 - 2^{n+1})$

as required.

ii)

$\frac{1}{\tan{x}} + \frac{\tan{x}}{1} = \frac{1 + \tan^2{x}}{\tan{x}} = 2 \left( \frac{1+\tan^2{x}}{2\tan{x}} \right) = 2 \left( \frac{1}{\frac{2\tan{x}}{1+\tan^2{x}}} \right) = \frac{2}{\sin{2x}} = 2\mbox{cosec}{2x}$

$\therefore \cot{x} + \tan{x} = 2\mbox{cosec}{2x}$

$\\ \displaystyle \implies \frac{1}{x} + \sum_{n=0}^{\infty} b_n x^n + \sum_{n=0}^{\infty} a_n x^n = 2 \left( \frac{1}{2x} + \sum_{n=0}^{\infty} c_n (2x)^n \right) \implies b_n + a_n = 2^{n+1}c_n \ \ (x \neq 0) \implies c_n = b_n (2^{-n} - 1)$

as required.

iii)

$\displaystyle \left( 1 + x \sum_{n=0}^{\infty} b_n x^n \right)^2 + x^2 = \left( 1 + x \sum_{n=0}^{\infty} c_n x^n \right)^2 \Leftrightarrow (x \cot{x})^2 + x^2 = (x \mbox{cosec}{x})^2 \Leftrightarrow x^2 ( \cot^2{x} + 1 ) = x^2 \mbox{cosec}^2{x} \Leftrightarrow \cot^2{x} + 1 = \mbox{cosec}^2{x} \ (x \neq 0 ) \Leftrightarrow \cos^2{x} + \sin^2{x} = 1$

which is true by Pythagoras'.

$\\ \displaystyle \therefore \displaystyle \left( 1 + x \sum_{n=0}^{\infty} b_n x^n \right)^2 + x^2 = \left( 1 + x \sum_{n=0}^{\infty} c_n x^n \right)^2 \implies \left( 1 + x \sum_{n=0}^{\infty} b_n x^n \right)\left( 1 + x \sum_{n=0}^{\infty} b_n x^n \right) = \left( 1 + x \sum_{n=0}^{\infty} c_n x^n - x \right) \left( 1 + x \sum_{n=0}^{\infty} c_n x^n + x \right) \implies \left( 1 + x ( b_0 + b_1x + b_2x^2 + ... ) \right)^2 = (1 + x(c_0 + c_1x + c_2x^2 + ... ) - x)(1 + x(c_0 + c_1x + c_2x^2 + ... ) + x) \implies \left( 1 + x ( b_0 + b_1x + b_2x^2 + ... ) \right)^2= (1 + x((c_0 - 1) + c_1x + c_2x^2 + ... ))(1 + x((c_0+1) + c_1x + c_2x^2 + ... ) )$

EDIT: There was no point in factorising this way. I did the question whilst typing it up, and it seemed an obvious thing to do.

Equating coefficients gives:

$\\ x: \ \ 2b_0 = c_0 + 1 + c_0 - 1 \implies b_0 = c_0 = 0 \implies a_0 = 0$

$\\ x^2 : \ \ 2b_1 + b_0^2 = 2c_1 + (c_0+1)(c_0-1) \implies 2b_1 = -b_1 - 1 \implies 3b_1 = -1 \implies b_1 = -\frac{1}{3} \implies a_1 = (1 - 2^2)(- \frac{1}{3}) = 1$

as required.

$\\ x^4: \ \ 2b_3 + 2b_0 b_2 + b_1^2 = 2c_3 + (c_0 - 1)c_2 + c_1^2 + (c_0 + 1)c_2 \implies 2b_3 + \frac{1}{9} = 2c_3 + \frac{1}{36} \implies 2b_3 + \frac{1}{12} = 2(2^{-3}-1)b_3 \implies b_3 = -\frac{1}{45} \implies a_3 = (1-2^4)(-\frac{1}{45}) = \frac{1}{3}$

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Daniel Freedman

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STEP III 2006, Question 1

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$y = \frac{2x(x^2-5)}{x^2 - 4} = \frac{2x(x+\sqrt{5})(x-\sqrt{5})}{(x+2)(x-2)}$

Therefore the curve has vertical asymptotes with equations

and

.

The curve has roots $x = - \sqrt{5}, \ 0, \ \sqrt{5}$ .

As $x \to \pm \infty, \ y \to 2x$ , so the curve has an oblique asymptote with equation

.

As $x \to 0, \ y \to \frac{-10x}{-4} = \frac{5}{2}x$ . This is the equation of the tangent to the curve at the origin (can be verified by differentiating)

[SKETCH ATTACHED]

i)

$3x(x^2-6) = (x^2-4)(x+3) \implies \frac{2x(x^2-5)}{x^2 - 4} = \frac{2}{3}x + 2$

[SKETCH ATTACHED]

The graph shows that the two curves have three points of intersection, which means the above equation has three real roots

ii)

$4x(x^2-5) = (x^2 - 4)(5x - 2) \implies \frac{2x(x^2-5)}{x^2 - 4} = \frac{5}{2}x - 1$

[SKETCH ATTACHED]

The graph shows the two curves only have one point of intersection (using the fact that the line has the same gradient as the tangent to the curve at the origin). This means the above equation has one real root.

iii)

$\\ 4x^2(x^2-5)^2 = (x^2 - 4)^2 (x^2+1) \implies \left(\frac{2x(x^2-5)}{x^2 - 4} \right)^2 = x^2 + 1 \implies \frac{2x(x^2-5)}{x^2 - 4} = \pm \sqrt{x^2+1}$

[SKETCH ATTACHED]

The graph shows the curves have six points of intersection, which means the above equation has six real roots (the curves don't meet again because there are only two solutions of $\sqrt{x^2+1} = 2x$

Attached files

graph 1.GIF (9.4 KB)
graph 2.GIF (10.2 KB)
graph 3.GIF (10.7 KB)
graph 4.GIF (10.7 KB)

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Oh I Really Don't Care

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STEP III 8

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$\displaystyle \delta(1\times1) = \delta(1) + \delta(1)$ by property four. It follows $\delta(1) = 0$ By property three we see that $\delta \lambda = 0$ for any lambda in R

$\displaystyle \delta(x^2) = \delta(x \times x) = x\delta(x) + x\delta(x) = 2x$

$\displaystyle \delta(x\timesx^2) = x\delta(x^2) + x^2\delta(x) = 3x^2$

We introduce some notation and an operation we wish to show is equivalent to the operation delta. We define the operation upon a polynomial f(x) as follows; $\displaystyle \frac{d}{dx} f(x) = \lim_{dx \to 0} \frac{f(x+dx) - f(x)}{dx}$

We will now prove property 1

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$\displaystyle \frac{d}{dx}(x^n) = \lim_{dx \to 0}\frac{ (x+dx)^n - x^n}{dx} = \lim_{dx \to 0} \frac{x^n + nx^{n-1}dx - x^n + \phi}{dx} = (*)$ phi represents terms in the binomial expansion whose power of dx is greater then or equal to 2. It follow that (*) is equivalent to;

$\displaystyle \lim_{dx \to 0} \left(nx^{n-1} + \frac{\phi}{dx} \right) = nx^{n-1}$

Setting n = 0 we have $\frac{d}{dx}(x^0) = 0 x^{-1} = 0$

We now prove property 2 is met by this operation

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$\displaystyle \frac{d}{dx}\left(f(x) + g(x)\right)$

$\displaystyle = \lim_{dx \to 0} \frac{ f(x+dx) + g(x+dx) - f(x) - g(x)}{dx}$

$\displaystyle = \lim_{dx \to 0} \frac{ f(x+dx)-f(x)}{dx} + \lim_{dx \to 0} \frac{ g(x+dx)-g(x)}{dx}$

$\displaystyle = \frac{d}{dx} f(x) + \frac{d}{dx} g(x)$

We prove property three is met by the above operation.

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Let g(x) = $\lambda f(x)$

$\displaystyle \frac{d}{dx}g(x) = \lim_{dx \to 0} \frac{g(x+dx)-g(x)}{dx}$

$\displaystyle = \lim_{dx \to 0} \frac{\lambda f(x+dx) - \lambda f(x)}{dx}$

$\displaystyle = \lambda\left(\lim_{dx \to 0} \frac{f(x+dx) - f(x)}{dx}\right)$

$\displaystyle = \lambda\left(\frac{d}{dx} f(x) \right)$

We now prove the final property;

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$\displaystyle \frac{d}{dx} f(x) g(x)$

$\displaystyle= \lim_{dx \to 0} \frac{ f(x+dx)g(x+dx) - f(x)g(x)}{dx}$

$\displaystyle = \lim_{dx \to 0} \frac{ f(x+dx)g(x+dx) + g(x)f(x+dx) - g(x)f(x+dx) - f(x)g(x) }{dx}$

$\displaystyle = \lim_{dx \to 0} \left( g(x) \frac{f(x+dx)-f(x)}{dx} + f(x+dx) \frac{ g(x+dx)-g(x)}{dx}\right)$

We now need to show that delta (the operation given in the question) returns nx^{n-1} after being applied to x^n.

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We proceed by induction.

Assume that for some k in N that delta(x^k) = kx^(k-1). Assume that the result holds for k + 1.

$\displaystyle \delta(x^{k+1}) = \delta(x^k \times x) = x\delta(x^k) + x^k\delta(x) = kx^{k-1} \times x + x^k = (k+1)x^k$

P(k) implies P(k+1) and we know that delta(1) = 0 and delta(x) = 1. As a set which contains 0 and the successor function follows by mathematical induction, the set contains N. So delta(f(x)) is equivalent to d/dx (f(x)) if f(x) is a polynomial in x.

As required

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Glutamic Acid

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I/8:

(i) Let the base be OAC, so area = 1/2ca. Height is b, so the area is b*1/3*(1/2ca) = abc/6.

(ii) $\cos \theta$ is the angle between BC and AC. $BC = \left( \begin{array}{c} 0 \\ -b \\ 0 \end{array} \right) + \left( \begin{array}{c} 0 \\ 0 \\ c \end{array} \right) = \left( \begin{array}{c} 0 \\ -b \\ c \end{array} \right)$

$AC = \left( \begin{array}{c} -a \\ 0 \\ 0 \end{array} \right) + \left( \begin{array}{c} 0 \\ 0 \\ c \end{array} \right) = \left( \begin{array}{c} -a \\ 0 \\ c \end{array} \right)$

$\left( \begin{array}{c} 0 \\ -b \\ c \end{array} \right) . \left( \begin{array}{c} -a \\ 0 \\ c \end{array} \right) = \sqrt{a^2 + c^2} \sqrt{b^2 + c^2} \cos \theta \Rightarrow \cos \theta = \dfrac{c^2}{\sqrt{(a^2 + c^2)(b^2 + c^2)}}$

Area of triangle = " $\dfrac{1}{2}ab \sin \theta$ ". Note that $\sin^2 \theta = 1 - \cos^2 \theta$ , so $\sin^2 \theta = 1 - \dfrac{c^4}{(a^2 + c^2)(b^2 + c^2)} = \dfrac{a^2b^2 + a^2c^2 + b^2c^2}{(a^2 + c^2)(b^2 + c^2)}$

$\text{area} = \dfrac{1}{2} \sqrt{a^2 + c^2} \sqrt{b^2 + c^2} \dfrac{\sqrt{a^2b^2 + b^2c^2 +a^2c^2}}{\sqrt{(a^2 + c^2)(b^2 + c^2)}} = \dfrac{1}{2}\sqrt{a^2b^2 + b^2c^2 +a^2c^2}$

The area will clearly be equal to abc/6, only that the height of the new triangle is d.

$\dfrac{1}{6}abc = \dfrac{1}{6}(a^2b^2 + b^2c^2 +a^2c^2)^{1/2}d \Rightarrow a^2b^2c^2 = d^2(a^2b^2 + b^2c^2 +a^2c^2)$

$\Rightarrow \dfrac{1}{d^2} = \dfrac{a^2b^2 + b^2c^2 +a^2c^2}{a^2b^2c^2} \Rightarrow \dfrac{1}{d^2} = \dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2}$

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Daniel Freedman

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(Original post by DeanK22)
STEP III 8

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$\displaystyle \delta(1\times1) = \delta(1) + \delta(1)$ by property four. It follows $\delta(1) = 0$ By property three we see that $\delta \lambda = 0$ for any lambda in R

$\displaystyle \delta(x^2) = \delta(x \times x) = x\delta(x) + x\delta(x) = 2x$

$\displaystyle \delta(x\timesx^2) = x\delta(x^2) + x^2\delta(x) = 3x^2$

We introduce some notation and an operation we wish to show is equivalent to the operation delta. We define the operation upon a polynomial f(x) as follows; $\displaystyle \frac{d}{dx} f(x) = \lim_{dx \to 0} \frac{f(x+dx) - f(x)}{dx}$

We will now prove property 1

Spoiler:

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$\displaystyle \frac{d}{dx}(x^n) = \lim_{dx \to 0}\frac{ (x+dx)^n - x^n}{dx} = \lim_{dx \to 0} \frac{x^n + nx^{n-1}dx - x^n + \phi}{dx} = (*)$ phi represents terms in the binomial expansion whose power of dx is greater then or equal to 2. It follow that (*) is equivalent to;

$\displaystyle \lim_{dx \to 0} \left(nx^{n-1} + \frac{\phi}{dx} \right) = nx^{n-1}$

Setting n = 0 we have $\frac{d}{dx}(x^0) = 0 x^{-1} = 0$

We now prove property 2 is met by this operation

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$\displaystyle \frac{d}{dx}\left(f(x) + g(x)\right)$

$\displaystyle = \lim_{dx \to 0} \frac{ f(x+dx) + g(x+dx) - f(x) - g(x)}{dx}$

$\displaystyle = \lim_{dx \to 0} \frac{ f(x+dx)-f(x)}{dx} + \lim_{dx \to 0} \frac{ g(x+dx)-g(x)}{dx}$

$\displaystyle = \frac{d}{dx} f(x) + \frac{d}{dx} g(x)$

We prove property three is met by the above operation.

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Let g(x) = $\lambda f(x)$

$\displaystyle \frac{d}{dx}g(x) = \lim_{dx \to 0} \frac{g(x+dx)-g(x)}{dx}$

$\displaystyle = \lim_{dx \to 0} \frac{\lambda f(x+dx) - \lambda f(x)}{dx}$

$\displaystyle = \lambda\left(\lim_{dx \to 0} \frac{f(x+dx) - f(x)}{dx}\right)$

$\displaystyle = \lambda\left(\frac{d}{dx} f(x) \right)$

We now prove the final property;

Spoiler:

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$\displaystyle \frac{d}{dx} f(x) g(x)$

$\displaystyle= \lim_{dx \to 0} \frac{ f(x+dx)g(x+dx) - f(x)g(x)}{dx}$

$\displaystyle = \lim_{dx \to 0} \frac{ f(x+dx)g(x+dx) + g(x)f(x+dx) - g(x)f(x+dx) - f(x)g(x) }{dx}$

$\displaystyle = \lim_{dx \to 0} \left( g(x) \frac{f(x+dx)-f(x)}{dx} + f(x+dx) \frac{ g(x+dx)-g(x)}{dx}\right)$

We now need to show that delta (the operation given in the question) returns nx^{n-1} after being applied to x^n.

Spoiler:

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We proceed by induction.

Assume that for some k in N that delta(x^k) = kx^(k-1). Assume that the result holds for k + 1.

$\displaystyle \delta(x^{k+1}) = \delta(x^k \times x) = x\delta(x^k) + x^k\delta(x) = kx^{k-1} \times x + x^k = (k+1)x^k$

P(k) implies P(k+1) and we know that delta(1) = 0 and delta(x) = 1. As a set which contains 0 and the successor function follows by mathematical induction, the set contains N. So delta(f(x)) is equivalent to d/dx (f(x)) if f(x) is a polynomial in x.

As required

I'm pretty sure you don't need to prove that each of the four rules works for differentiation. You just need to prove, by induction as you have, that $\Delta(x^n) = nx^{n-1}$ , using the four rules. Then let $\displaystyle h(x) = \sum_{k=0}^n a_k x^k$ , and apply the operation to this function, making use of rules ii) and iii) to show that $\displaystyle \Delta(h(x)) = \sum_{k=0}^n k a_k x^{k-1}$ .

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Dadeyemi

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STEP III 2006, Question 4

initial part

let y = x
$\Rightarrow 2f(x) \equiv f(2x) \\ \Rightarrow 2f'(x)\equiv 2f'(2x) \\ \Rightarrow f''(x) \equiv 2f''(2x) \\ \Rightarrow f^{(n)} \equiv 2^{n-1}f^{(n)}(2x)$
Letting

we have

and $f^{n}(0) = 0$ for $n\neq 0$
so by Maclaurin $f(x) \equiv f'(0)x \equiv kx$ for some constant k

(i)

$G(x+y) \equiv ln(g(x+y)) \equiv ln(g(x)g(y)) \equiv ln(g(x)) + ln(g(y)) \equiv G(x)+G(y)$
so $G(x) \equiv f(x) = kx$
giving $g(x) = e^{kx}$

(ii)

$H(\ln (x+y)) \equiv h(xy) \equiv h(x)+h(y) \equiv H(\ln x)+G(\ln y)$
so $h(x) \equiv H(\ln x) \equiv f(\ln x) \equiv k\ln x$

(iii)

let $x = \tan \theta , y = \tan \phi$
$\displaystyle \Rightarrow t(\tan \theta) + t(\tan \phi) = t( \frac{\tan \theta +\tan \phi}{1- \tan \theta \tan \phi }) = t(\tan (\theta+ \phi))$
so $t (x) = t (\tan \theta) = f(\theta) = k\theta = k \arctan x$

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2006 I Question 13
Part (i)
The number of diamonds in one kilogram has a Poisson distribution of $D \sim \text{Poisson}(1)$ . We find that per 100 grams, $D \sim \text{Poisson}(\lambda = 0.1)$ . For 100T grams, $D_{T} \sim \text{Poisson}(\lambda = 0.1T)$ .

is the distribution of scores of the die.

Drawing 1, we have $P(T = 1 \cap D_{1}=0) = \frac{1}{6}(e^{-0.1})$ .
Similarly, $\displaystyle P((T=1 \cap D_{1})\cup(T=2\cap D_{2}=0)\cup\ldots\cup(T=6\cap D_{6}=0)) = \sum_{t=1}^6 \frac{1}{6}e^{-0.1t}=$
$\displaystyle = \frac{1}{6}e^{-0.1}\sum_{t=1}^6 e^{(-0.1)(t-1)} = \frac{e^{-0.1}}{6}\frac{1-e^{-0.6}}{1-e^{-0.1}}$ (recognising a finite geometric sum).

Part (i) - expectation
$\displaystyle E[D_T] = \sum_{t=1}^6 \frac{1}{6}E[D_t] = \frac{1}{6}\sum_{t=1}^6 0.1t = \frac{1}{6}(0.1+0.2+0.3+0.4+0.5+0.6) = \frac{2.1}{6} = \frac{0.7}{2} = 0.35$

Part (ii)
This time,

is a geometric distribution with $p = \frac{5}{6}$ . Thus, $P(T=t)=\frac{1}{6}\left(\frac{5}{6}\right)^{t-1}$ .

This time, our probability is:
$\displaystyle P((T=1 \cap D_{1}=0)\cup(T=2\cap D_{2}=0)\cup\ldots) = \sum_{t=1}^{\infty}P(T=t\cap D_{t}=0) =$
$\displaystyle = \sum_{t=1}^{\infty} \left(\frac{5}{6}\right)^{t-1}\frac{1}{6}e^{-0.1t} = \frac{e^{-0.1}}{6}\sum_{t=1}^{\infty} \left(e^{-0.1}\frac{5}{6}\right)^{t-1} =$
$\displaystyle = \frac{e^{-0.1}}{6}\frac{1}{1-\frac{5}{6}e^{-0.1}} = \frac{e^{-0.1}}{6-5e^{-0.1}}$ , as required.

Part (ii) - expectation
$\displaystyle E[D_T] = \sum_{t=1}^\infty E[D_t]P(T=t) = \sum_{t=1}^\infty 0.1t\frac{1}{6}\left(\frac{5}{6}\right)^{t-1} =\frac{1}{60}\sum_{t=1}^\infty t\left(\frac{5}{6}\right)^{t-1}$ .

Consider: $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} ((1-x)^{-1}) = (1-x)^{-2} = \frac{\mathrm{d}}{\mathrm{d}x} \sum_{t=0}^\infty x^t = \sum_{t=1}^\infty tx^{t-1}$ .

Thus, $\displaystyle E[D_T] = \frac{1}{60}\left(1-\frac{5}{6}\right)^{-2} = \frac{36}{60} = 0.6$ .

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Aurel-Aqua

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2006 I Question 14
Either I missed something crucial, or this question is incredibly easy. I'll write up the incredibly-easy interpretation of this question:

Part (i)
This is a geometric progression, where $\displaystyle P(\text{Red on rth attempt}) = pq^{r-1}$ , where $\displaystyle p=\frac{1}{n}$ , $\displaystyle q = \frac{n-1}{n}$ , so $\displaystyle P(\text{Red on rth attempt}) = \left(\frac{n-1}{n}\right)^{r-1}\frac{1}{n} = \frac{(n-1)^{r-1}}{n^r}$ .
Differentiating, we get: $\displaystyle \frac{\mathrm{d}P}{\mathrm{d}n} = \frac{n^r(r-1)(n-1)^{r-2} - rn^{r-1}(n-1)^{r-1}}{n^{2r}}$ . Equating to zero, $n^r(r-1)(n-1)^{r-2} = rn^{r-1}(n-1)^{r-1} \iff n(r-1) = r(n-1) \iff n(r-1-r) = -r \iff -n=-r \iff n = r$ . We know that this is the maximum as r tends to infinity, P tends to zero.

Part (ii)

$P(\text{Red on 1st}) = \frac{1}{n}$
$P(\text{Red on 2nd}) = \frac{n-1}{n}\cdot\frac{1}{n-1} = \frac{1}{n}$
$P(\text{Red on rth}) = \frac{n-1}{n}\cdot\frac{n-2}{n-1}\cdots\frac{n-(r-2)}{n-(r-3)}\cdot\frac{n-(r-1)}{n-(r-2)}\cdot\frac{1}{n-(r-1)}=\frac{1}{n}$ . Clearly maximum at n = 1.

Did I make a mistake here? It just looks too simple for a STEP question, I'm afraid.

Edit: see GHOSH-5's post here: http://www.thestudentroom.co.uk/show...3&postcount=41

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DFranklin

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(Original post by Daniel Freedman)
I'm pretty sure you don't need to prove that each of the four rules works for differentiation. You just need to prove, by induction as you have, that $\Delta(x^n) = nx^{n-1}$ , using the four rules. Then let $\displaystyle h(x) = \sum_{k=0}^n a_k x^k$ , and apply the operation to this function, making use of rules ii) and iii) to show that $\displaystyle \Delta(h(x)) = \sum_{k=0}^n k a_k x^{k-1}$ .

Actually, it's worse than that.

They've said "given that the operator $\Delta$ has these properties; show that...".

Dean has said "A particular operator $\delta$ has those properties, and for $\delta$ , we have ...."

But he hasn't shown that $\delta$ is the only possible operator with the properties.

This is one of those unfortunate cases where what he's done makes a reasonable amount of sense, but very definitely doesn't answer the actual question. I think you'd lose a ton of marks for this (even if the Examiner sympathised).

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Oh I Really Don't Care

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(Original post by DFranklin)
Actually, it's worse than that.

They've said "given that the operator $\Delta$ has these properties; show that...".

Dean has said "A particular operator $\delta$ has those properties, and for $\delta$ , we have ...."

But he hasn't shown that $\delta$ is the only possible operator with the properties.

This is one of those unfortunate cases where what he's done makes a reasonable amount of sense, but very definitely doesn't answer the actual question. I think you'd lose a ton of marks for this (even if the Examiner sympathised).

Thanks for pointing that out - certainly a mistake that will not be made this Summer (hopefully). Unfortunately I do not really know how to prove the Uniquness of this operator - could you please shed some light on to how you do that? Thanks.

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DFranklin

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Well, that's effectively what the question is asking you to show. I would:

(1) Prove by induction that $\Delta x^n = n x^{n-1}$ (using (i) and (iv)).
(2) Then use (iii) to prove $\Delta a_n x^n = n a_n x^{n-1}$
(3) Then use (ii) to prove $\Delta \sum_0^N a_n x^n = \sum_0^N n a_n x^{n-1}$ .

[Which effectively proves uniqueness of $\Delta$ for polynomials].

Then simply observe this is the same as the derivative in the case of polynomials.

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darkness9999

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I have different answers to STEP I Q 14 Im not to sure who is wrong...

Part (i)

Spoiler:

Show

From the diagram it is obvious that on the $(r-1)^{th}$ attempt you would have had to picked a blue one before the red one ...

So
$P(R=r)=\left(\frac{n}{n+1} \right)^{r-1} \times \frac{1}{n+1}$

$\Rightarrow \boxed{\frac{n^{r-1}}{(n+1)^r}}$

Now to find the maximum value of A we need to differentiate P(R=r) wtr n.

Let $f(x)= \frac{n^{r-1}}{(n+1)^r}$

Using the quatient rule:

$f‘(x)= \frac{(n+1)^r(r-1)n^{r-2}- n^{r-1}r(n+1)^{r-1}}{(n+1)^{2r}}$

$\Rightarrow n^{r-2}(n+1)^{r-2}\frac{(n+1)(r-1) - nr}{(n+1)^{2r}}$

$\Rightarrow n^{r-2}\frac{r-n-1}{(n+1)^{r+1}}$

Therefore it is maximised when $\boxed{n=r-1}$

Part (ii)

Spoiler:

Show

$P(R=1) = \frac{1}{n+1}$

$P(R=2) = \frac{n}{n+1}\times \frac{1}{n}$

$\Rightarrow \frac{1}{n+1}$

$P(R=3) = \frac{n}{n+1}\times \frac{n-1}{n}\times \frac{1}{n-1}$

$\Rightarrow \frac{1}{n+1}$

Hence
$\boxed{P(R=r)= \frac{1}{n+1}}$

This probability is decreasing as n increases therefore it is again maximised when $\boxed{n=r-1}$

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darkness9999

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STEP I: Q5

Part (i)

$\int \frac{3}{(x-4)\sqrt {2x+1}} \, dx$

Substitution:

$\frac{du}{dx} 2u = 2$

$u\times du = dx$

Therefore the integral becomes

$\int \frac{3u}{(x-4)\sqrt {2x+1}} \, du$

$\Rightarrow \int \frac{3u}{u \frac{u^2-9}{2}} \, du$

$\Rightarrow \int \frac{6}{(u-3)(u+3)} \, du$

Simplifying this further using partial fractions gives us

Spoiler:

Show

$\frac{6}{(u-3)(u+3)} \equiv \frac{A}{u-3} + \frac{B}{u+3}$

$u_3: 6= A6 \Rightarrow A=1$

$u_{-3}: 6= b(-6) \Rightarrow B=-1$

$\Rightarrow \frac{6}{(u-3)(u+3)} \Leftrightarrow \frac{1}{u-3} - \frac{1}{u+3}$

$\int \frac{1}{u-3} - \frac{1}{u+3} \, du$

$\Rightarrow \ln {u-3} - \ln {u+3} + C$

Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.

\Rightarrow \ln { \frac{u-3}{u+3} + C

If we now sub in the substitution we used we get

Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.

{\ln {\frac{\sqrt {2x+1}-3}{\sqrt {2x+1}+3}+C

Part (ii)

$\displaystyle\int^{\ln 8}_{\ln 3} \frac{2}{e^x\sqrt {e^x+1}} \, dx$

I used this substitution:

$\frac{du}{dx} 2u = e^x$

$\frac{2u\times du}{e^x} = dx$

So the integral becomes:

$\displaystyle\int^{3}_{2} \frac{4u}{e^{2x}\sqrt {e^x+1}} \, du$

$\Rightarrow \displaystyle\int^{3}_{2} \frac{4u}{(u^2-1)(u^2-1)u} \, du$

$\Rightarrow \displaystyle\int^{3}_{2} \frac{4}{(u+1)^2(u-1)^2} \, du$

Simplifying this further using partial fractions gives us

Spoiler:

Show

$\frac{4}{(u+1)^2(u-1)^2} \equiv \frac{A}{(u+1)^2} + \frac{B}{(u-1)^2} + \frac{C}{u-1} + \frac{D}{u+1}$

$u_{-1}: 4 = A4 \Rightarrow A=1$

$u_{1}: 4 = B4 \Rightarrow B=1$

$u_{0}: 2 = -C +D$

$u_{1}: 0 = -C -D$

$\Rightarrow C= -1 & D= 1$

$\Rightarrow \frac{4}{(u+1)^2(u-1)^2} \Leftrightarrow \frac{1}{(u+1)^2} + \frac{1}{(u-1)^2} - \frac{1}{u-1} + \frac{1}{u+1}$

$\displaystyle\int^{3}_{2} \frac{1}{(u+1)^2} + \frac{1}{(u-1)^2} - \frac{1}{u-1} + \frac{1}{u+1} \, du$

Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.

\Rightarrow \left[-\frac{1}{u+1} - \frac{1}{u-1} + \ln {\frac{u+1}{u-1} \right]^{3}_{2}

Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.

\Rightarrow \left( (-\frac{1}{4} -\frac{1}{2} + \ln {\frac{4}{2}) -(-\frac{1}{3} -1 + \ln 3) \right)

$\Rightarrow \boxed {\frac{7}{12} + \ln {\frac{2}{3}}}$

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Elongar

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STEP III 2006, Q5

We wish to show that $\alpha, \beta, \gamma$ form an equilateral iff

$\displaystyle \alpha^2 + \beta^2 + \gamma^2 - \alpha \beta - \beta \gamma - \gamma \alpha = 0$

Multiplying by two and factorising, we wish to show that,

$\displaystyle (\alpha - \beta)^2 + (\beta - \gamma)^2 + (\gamma - \alpha)^2 = 0$

holds for all equilateral triangles.

Note that $\alpha - \beta, \beta - \gamma, \gamma - \alpha$ are the sides of the triangle. If they are shown on an Argand diagram, the angle between any two consecutive sides is $\frac{\pi}{3}$ . Note also that the lengths of the sides are all of equal length, so the magnitudes of the complex numbers that represent them will also be equal.

Using this, we reconstruct the equation:

$\displaystyle \left(re^{i \theta}\right)^2 + \left(re^{i (\theta +\frac{\pi}{3})}\right)^2 + \left(re^{i (\theta +\frac{2\pi}{3})}\right)^2 = 0$

or, more simply,

$\displaystyle r^2 \left( e^{2 i \theta} + e^{2 i (\theta +\frac{\pi}{3})} + e^{2 i (\theta +\frac{2 \pi}{3})} \right) = 0$

$\displaystyle r^2 e^{2 i \theta} \left( 1 + e^{i \frac{\pi}{3}} + e^{i \frac{2 \pi}{3}} \right) = 0$

The bracket evaluates to 0, so we are done.

For the next part, we take the roots of the cubic to be $\alpha, \beta, \gamma$ , and expand, yielding Vieta's formula:

$\displaystyle z^3 - (\alpha + \beta + \gamma) z^2 + (\alpha \beta + \beta \gamma + \gamma \alpha) z - \alpha \beta \gamma = 0$

Algebra shows that:

$\displaystyle \alpha^2 + \beta^2 + \gamma^2 - \alpha \beta - \beta \gamma - \gamma \alpha = (\alpha + \beta + \gamma)^2 - 3 (\alpha \beta + \beta \gamma + \gamma \alpha) = a^2 - 3b = 0$

and the result follows immediately.

Now we write,

$p = Ae^{i \phi}$

The transformation

has the effect of rotating our equilateral triangle by the angle $\phi$ , magnifying it by a factor of

and translating it through the vector represented by

(would I need to show this?). Under these transformations, the equilateral triangle formed by the roots of the initial cubic remains an equilateral triangle.

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nuodai

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STEP I 2006 Question 2

Solution

The barn is square, so it doesn't matter which side the goat is on. Because the lengths are of side 2a and the rope is of length 4a, it makes sense that we should consider the case where the rope is between the middle of a side and the edge of the side

In this, let P be the position of the goat, O be the position where the rope is tethered, and ABCD be the barn, with O being along AB and A, B, C, D marking the corners of the barn, going clockwise. Bear in mind that this is much more easily explained with diagrams, so try drawing it if you don't follow.

Let's say that O is located along AB at distance x from A, where 0 < x < a (so that it is between the midpoint of AB and the point A)

The distance OA is

, so the distance OA is

and so on. This means that if the goat walks as far as it can clockwise, P will be located at a distance

from C along the line CD. So, the goat can make:
- A quarter-circle with radius CP = x
- A quarter-circle with radius BP = 2a + x
- A half-circle with radius OP = 4a
- A quarter-circle with radius AP = 4a - x
- A quarter-circle with radius DP = 2a - x

So the area would be:
$\newline \displaystyle \dfrac{\pi}{4}\Big[ x^2 + (2a+x)^2 + 2(4a)^2 + (4a - x)^2 + (2a - x)^2 \Big] \newline = \dfrac{\pi}{4} \Big[ x^2 + 4a^2 + 4ax + x^2 + 32a^2 + 16a^2 - 8ax + x^2 + 4a^2 - 4ax + x^2 \Big] \newline = \dfrac{\pi}{4} \Big[ 56a^2 - 8ax + 4x^2 \Big] = \pi(14 a^2 - 2ax + x^2)\newline = \pi([x - a]^2 + 13a^2)$

This therefore gives us the maximum and minimum values. The maximum value must occur when

since $0 \le x \le a$ , and the maximum value must occur when

, therefore $\boxed{ 13\pi a^2 \le A \le 14\pi a^2 }$

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Oh I Really Don't Care

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RE; Noudai

opting for the use of x as the distance from the side of the nearest corner of the barn and a diagram saves you considering the cases and gives the answer far quicker.

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nuodai

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(Original post by DeanK22)
RE; Noudai

opting for the use of x as the distance from the side of the nearest corner of the barn and a diagram saves you considering the cases and gives the answer far quicker.

I realised that a bit too late :p:

I've chopped it out of the answer though.

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qgujxj39

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STEP III 2006 Q7

Spoiler:

Show

Part i):

Using the quadratic formula we obtain

$u = \displaystyle\frac{-2\sinh x \pm \sqrt{4\sinh^2x + 4}}{2}$

$=-\sinh x \pm \cosh x$

If we let $u = \frac{dy}{dx}$ , we obtain the differential equation they're asking about, so:

$\frac{dy}{dx} = -\sinh x \pm \cosh x$

Using the condition that $\frac{dy}{dx} > 0$ at

, we see that the $\pm$ must be a +, hence

$\frac{dy}{dx} = \cosh x - \sinh x$

Integrating we obtain

$y = \sinh x - \cosh x + c$

and using y = 0 when x = 0 we find that c = 1.

Part ii):

Using the quadratic formula again to find dy/dx, we obtain (apologies for not writing it out in full):

$\frac{dy}{dx} = \displaystyle\frac{-1 \pm \cosh y}{\sinh y}$

$\implies \displaystyle\int\displaystyle\frac{\sinh y}{1 \pm \cosh y}\mathrm{d}y = \int - \mathrm{d}x$

We can integrate the left hand side using recognition: the top is the derivative of the bottom, so:

$\pm \ln (1 \pm \cosh y) = c - x$

Now, if the $\pm$ sign were a minus, then the condition x = 0 when y = 0 would be impossible, as this would mean that the arbitrary constant were infinite. We can therefore deduce that the $\pm$ must be a +. Therefore:

$\ln (1 + \cosh y) = c - x$

Using x = 0 when y = 0, we find that c = ln2. Therefore

$\ln (1 + \cosh y) = \ln 2 - x$

$\implies \cosh y = 2e^{-x} -1$ , which is the solution of the differential equation.

Writing the cosh in full exponentials and multiplying by 2 gives:

$e^y + e^{-y} = 4e^{-x} - 2$

To find the asymptotes, we consider two cases:

1) when $y \rightarrow \infty$ , we can disregard $e^{-y}$ and

because they are comparatively small, so $e^y \approx 4e^{-x}$ . Rearranging gives the equation of the asymptote, $y = \ln 4 - x$ .

1) when $y \rightarrow -\infty$ , we can disregard $e^{y}$ and

because they are comparatively small, so $e^{-y} \approx 4e^{-x}$ . Rearranging gives the equation of the other asymptote, $y = x - \ln 4$ .

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