STEP Maths I, II, III 2001 Solutions

SimonM

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SimonM

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Lets start on some of the more modern ones. I remember reading that one of the websites has removed some of these solutions

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Aurel-Aqua

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I Question 2

Part (i) $1+2x-x^2>\frac{2}{x}$ , $x\neq 0$ .

First it might help to find the points of equality: multiplying by x and rearranging, we get

, $x\neq 0$ . The roots are -1, 1 and 2. Now we must sketch the graph and notice that the inequality is satisfied for

(the zero due to the asymptote) and

.

Part (ii) $\sqrt{3x+10} > 2 + \sqrt{x+4}$ , $x \geq -\frac{10}{3}$ .

Because of the condition, squaring both sides still leaves an iff, giving: $3x + 10 > 4 + x + 4 + 4\sqrt{x+4}\implies 2x + 2 > 4\sqrt{x+4} \implies x + 1 > 2\sqrt{x+4}$ . Now by the condition, we can deduce that x > -1 because of the square root being always positive. Squaring again, $x^2+2x+1 > 4(x + 4) \implies x^2 - 2x - 15 > 0$ , giving

. From this, we can deduce that

and

, but our new condition,

only gives us

as the final solution.

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sonofdot

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STEP II 2001, Question 6

Part One

$\begin{array}{rl} \displaystyle\int_0^1 \frac{x^4}{1+x^2} \, dx & = \displaystyle\int_0^1 \frac{x^4 + x^2 - x^2}{1+x^2} \, dx \\ \br \\ & = \displaystyle\int_0^1 yx^2 - \frac{1+x^2 - 1}{1+x^2} \, dx \\ \br \\ & = \displaystyle\int_0^1 x^2 - 1 + \frac{1}{1+x^2} \, dx \\ \br \\ & = \displaystyle \left[ \frac{1}{3}x^3 - x + \arctan x \right]_0^1 \\ \br \\ & = \displaystyle\frac13 - 1 + \arctan 1 \\ \br \\ & = \displaystyle \boxed{\frac{\pi}{4} - \frac{2}{3}} \end{array}$

(i)

$\displaystyle\int_0^1 x^3 \arctan \left(\frac{1-x}{1+x} \right) \, dx$

$\displaystyle\frac{d}{dx} \left(\frac{1-x}{1+x} \right) = -\frac{2}{(1+x)^2}$

Let $\displaystyle u = \arctan \left(\frac{1-x}{1+x} \right)$
$\begin{array}{rl} \displaystyle\therefore \frac{du}{dx} & = \displaystyle - \frac{2}{(1+x)^2} \times \frac{1}{1+ (\frac{1-x}{1+x})^2} \\ \br \\ & \displaystyle = - \frac{2}{(1+x)^2} \times \frac{(1+x)^2}{(1+x)^2+(1-x)^2} \\ \br \\ & \displaystyle = - \frac{1}{1+x^2}\end{array}$

Let $\displaystyle \frac{dv}{dx} = x^3 \Leftarrow v = \frac14 x^4$

$\displaystyle\int_0^1 u\frac{dv}{dx} \, dx = \left[uv \right]_0^1 - \int_0^1 v \frac{du}{dx} \, dx$

$\begin{array}{rl} \displaystyle\therefore \int_0^1 x^3 \arctan \left(\frac{1-x}{1+x} \right) \, dx & = \displaystyle\left[ \frac14 x^4 \arctan \left(\frac{1-x}{1+x} \right) \right]_0^1 + \frac14 \int_0^1 \frac{x^4}{1+x^2} \\ \br \\ & \displaystyle = \frac14 \left( \frac{\pi}{4} - \frac{2}{3} \right) \\ \br \\ & \displaystyle = \boxed{\frac{\pi}{16} - \frac{1}{6}}\end{array}$

(ii)

$\displaystyle\int_0^1 \frac{(1-y)^3}{(1+y)^5} \arctan y \, dy$

Let $\displaystyle y = \frac{1-x}{1+x} \iff x = \frac{1-y}{1+y}$

$\displaystyle \frac{dx}{dy} = - \frac{2}{(1+y)^2} \Rightarrow dy = -\frac{1}{2}(1+y)^2 \, dx$

$\begin{array}{rl} \displaystyle\therefore \int_0^1 \frac{(1-y)^3}{(1+y)^5} \arctan y \, dy & = \displaystyle\int_1^0 \frac{x^3}{(1+y)^2} \arctan \left( \frac{1-x}{1+x} \right) \times -\frac{1}{2}(1+y)^2 \, dx \\ \br \\ & \displaystyle = \int_1^0 - \frac12 x^3 \arctan \left( \frac{1-x}{1+x} \right) \, dx \\ \br \\ & \displaystyle = \frac12 \int_0^1 x^3 \arctan \left(\frac{1-x}{1+x} \right) \, dx \\ \br \\ & \displaystyle = \frac12 \left( \frac{\pi}{16} - \frac{1}{6} \right) \\ \br \\ & \displaystyle = \boxed{\frac{\pi}{32} - \frac{1}{12}} \end{array}$

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ForGreatJustice

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STEP II Question 2

Sketch not included, but pretty simple (y=0 for 0<x<N, and y=1 for N<x<2N)

i)

for

can be split into 3 regions:

for

and

for

So by splitting the summation into these three ranges we get

$\displaystyle\sum_{k=1}^{2N} (-1)^{[k/2N]} k = \displaystyle\sum_{k=1}^{N-1} k - \displaystyle\sum_{k=N}^{2N-1} k + 2N$

$= \displaystyle\frac{1}{2}(N-1)(N) -(\displaystyle\frac{1}{2}(2N-1)(2N) -\displaystyle\frac{1}{2}(N-1)(N)) +2N$

$\displaystyle\sum_{k=1}^{2N} (-1)^{[k/2N]} k =2N-N^2$ as required

ii) $S_N= \displaystyle\sum_{k=1}^{2N} (-1)^{[k/2N]} 2^{-k}$

Splitting into 3 as before:

$S_N=\displaystyle\sum_{k=1}^{N-1} 2^{-k} - \displaystyle\sum_{k=N}^{2N-1} 2^{-k} + 2^{-2N}$

Using the formula for the sum of a geometric progression
$S_n=\displaystyle\frac{a(1-r^n)}{1-r}$ we get

$S_N=\displaystyle\frac{2^{-1}(1-(1/2)^{N-1})}{1-2^{-1}} +\displaystyle\frac{2^{-N}(1-(1/2)^{N})}{1-2^{-1}} + 2^{-2N}$

$S_N=1-(1/2)^{N-1}+2^{1-N}-2^{1-2N}+2^{-2N}$
Which could probably be made simpler, so now we can see as $N\to\infty$ then $S_N\to1$

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Aurel-Aqua

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I Question 2

,

. Well, we sketch the graph to show the two stationary points and crossing the x-axis at

,

. We can conclude the two stationary points exist, so there must be a real x-value for them to be found.

.
Since the two stationary points are distinct and real, by

, we get $(2(p+q+r))^2-4\cdot 3 \cdot (pq+qr+rp) > 0 \implies (p+q+r)^2 > 3(pq+qr+rp)$ .

Considering

, to have three roots, there must be a real solution to the quadratic, giving

. This implies that the two other roots are $\alpha = \frac{-g-\sqrt{g^2-2h}}{2}$ and $\beta = \frac{-g+\sqrt{g^2-2h}}{2}$ , and $\gamma = k$ . Letting $p = \alpha$ , $q = \beta$ , $r = \gamma$ in any order (since they can be changed around due to triple-symmetry ). Knowing that

and $pq+qr+rp = \alpha\beta + \alpha\gamma + \gamma\beta= h +\gamma(\alpha+\beta) = h + k(-g) = h - gk$ . Substituting into our previous result, we obtain: $(k-g)^2 > 3(h-gk) \implies (g-k)^2>3(h-gk)$ , as required. We can show that it is not necessary by letting

,

, $(1-1)^2>3(h-1)\implies h > 1 \implies 1^2 > 4h \text{ is not true}$ .

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Mark13

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STEP I - Q13

Given that plate is broken, the probability that it is broken by the mathematician is 0.25.

Therefore, the number of plates,

, broken by the mathematician can be represented by the following binomial distribution:

~

We need to find $P(X \geq 3)$ when n=5.

$P(X \geq 3) = P(X=3) + P(X=4) + P(X=5)$
$=\displaystyle \binom{5}{3}(0.25)^3(0.75)^2 + \displaystyle \binom{5}{4}(0.25)^4(0.75) + \displaystyle \binom{5}{5}(0.25)^5$

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qgujxj39

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I shall get going.

(Original post by SimonM)
Lets start on some of the more modern ones. I remember reading that one of the websites has removed some of these solutions

Meikleriggs only has solutions since 2004 now

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Glutamic Acid

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II/11:

Highest point of trajectory when v = 0, so 0 = v - gt and t = v/g. The horizontal distance travelled will be uv/g at this time, and the vertical will be $\dfrac{v^2}{2g}$ .

P will hit the ground when its vertical displacement is $-\dfrac{v^2}{2g}$ , so one must solve the quadratic equation $v_et - \dfrac{1}{2}gt^2 = -\dfrac{v^2}{2g}$ giving $t = \dfrac{v_e + \sqrt{v_e^2 + v^2}}{g}$ (using the quadratic formula). The horizontal distance travelled in this time will be $\dfrac{u(v_e + \sqrt{v_e^2 + v^2})}{g}$ .

Adding the distances: $R = \dfrac{uv}{g} + \dfrac{u(v_e + \sqrt{v_e^2 + v^2}}{g} \Rightarrow \dfrac{gR}{u} = v + v_e + \sqrt{v_e^2 + v^2}$ , as required.

If R > D then $v + v_e + \sqrt{v_e^2 + v^2} > \dfrac{Dg}{u} \Rightarrow \sqrt{v_e^2 + v^2} > \dfrac{Dg}{u} - (v + v_e) \Rightarrow v_e^2 + v^2 > \dfrac{D^2g^2}{u^2} - \dfrac{2Dg}{u}(v + v_e) + v^2 + 2vv_e + v_e^2$

$\Rightarrow \dfrac{2Dgv_e}{u} - 2vv_e > \dfrac{D^2g^2}{u^2} - \dfrac{2Dgv}{u} \Rightarrow v_e \left( \dfrac{2Dg}{u} - 2v \right) > \dfrac{Dg}{u} \left( \dfrac{Dg}{u} - 2v \right) \Rightarrow v_e > \dfrac{Dg}{2u} \left( \dfrac{\frac{Dg}{u} - 2v}{\frac{Dg}{u} - v} \right)$

$\Rightarrow v_e > \dfrac{Dg}{2u} \left( \dfrac{Dg - 2vu}{Dg - vu} \right)$ , as required.

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Aurel-Aqua

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III Question 1

$\displaystyle y = \ln\left(x+\sqrt{x^2+1}\right) \implies \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1+\frac{x}{\sqrt{x^2+1}}}{x+\sqrt{x^2+1}} =$
$\displaystyle = \frac{1}{\sqrt{x^2+1}}\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x} = \frac{1}{\sqrt{x^2+1}} \text{, given } \sqrt{x^2+1}+x \neq 0 \text{, which is clearly true}$

Case when

:
$(x^2+1)(-x(x^2+1)^{-3/2}) + x(x^2+1)^{-1/2} = x(x^2+1)^{-1/2}(-1+1) = 0$ .
Differentiating the given relation, we get:
$(x^2+1)y^{(n+3)}+(2+(2n+1))xy^{(n+2)} + (2n+1+n^2)y^{(n+1)} = 0 =$
$= (x^2+1)y^{((n+1)+2)} + (2(n+1)+1)xy^{(n+2)} + (n+1)^2y^{(n+1)} = 0$ .
Since it is true for n = 0, and it is also true for n + 1, by mathematical induction we conclude that it must also be true for n = 1, n = 2, and so forth for all positive values of n.

Substituting

into the relation, we get $y^{(n+2)}(0) = -n^2y^{(n)}(0)$ . Thus $y^{(0)}(0) = 0\implies y^{(2k)}(0) = 0$ , all even powers are zero. $y^{(1)}(0) = 1\implies y^{(3)}(0) = -1 \implies y^{(5)}(0) = 9 \implies y^{(7)}(0) = -225$ .

Maclaurin series with terms of even powers equal to zero is $y(x) \approx xy^{(1)}(0) + \frac{1}{3!}x^3y^{(3)}(0) + \frac{1}{5!}x^5y^{(5)}(0) + \frac{1}{7!}x^7y^{(7)}(0)$ , so $y(x)\approx x + \frac{-1}{3!}x^2 + \frac{9}{5!}x^5+\frac{-225}{7!}x^7 = x-\frac{1}{6}x^3 + \frac{3}{40}x^5 - \frac{5}{112}x^7$ .

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DFranklin

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(Original post by Mark13)
STEP I - Q13

Given that plate is broken, the probability that it is broken by the mathematician is 0.25.

Therefore, the number of plates,

, broken by the mathematician can be represented by the following binomial distribution:

It seems to me you are assuming independance between the plates - that is, knowing that a particular plate was broken by the mathematician doesn't affect the probability that another plate was broken by the mathematician. It's not at all obvious to me such an assumption is justified.

(It isn't justified for general distributions; imagine the case where each student either breaks 5 plates with probability k, or otherwise breaks no plates at all. Then if you know the mathematican broke a plate, you know he broke 5 of them, and you'd end up with an answer for 'broke > 3 plates' of 1/4).

On the other hand, it wouldn't surprise me if you were right, either. I confess I have a tendancy to overcomplicate these problems. You'd still need some kind of justification though (based on the distribution being Poisson).

I did try doing this the 'brute force' way by looking at all possibilities (e.g. 1st student breaks 0 plates, 2nd student 1 plate, 3rd student 1 plate, mathmo breaks 3 plates), and it feels a little tough for STEP I. Assuming I made no mistakes, I do get a different answer, however. I'll have another go when I'm not literally writing on the back of an envelope in the kitchen.

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SimonM

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Question 4, STEP I 2001

Spoiler:

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Consider $(\cos \theta + i \sin \theta)^3 = \cos^3 \theta + i \sin 3\theta = \cos^3 \theta + 3\cos^2 \theta \sin \theta i - 3\cos \theta \sin^2 \theta -3i\sin^3\theta$ by De Moivre's Theorem.

Therefore $\displaystyle \tan 3 \theta = \frac{\cos^2 \theta \sin \theta - 3\sin^3 \theta}{\cos^3 \theta -3\cos \theta \sin^2 \theta}$

Dividing top and bottom by $\cos^3 \theta$ yields:

$\displaystyle \tan 3 \theta = \frac{3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta}$

Since $\theta \in (0, \frac{\pi}{2})$ , $\sin \theta = \sqrt{1- \cos^2 \theta}$

Therefore $\tan \theta = \frac{\sqrt{1-\frac{4}{5}}}{\frac{2}{\sqrt{5}}} = \frac{1}{2}$

Therefore $\tan 3 \theta = \frac{3 \frac{1}{2} - \frac{1}{8}}{1-\frac{3}{4}} = \frac{11}{2}$

$\tan 3 \cos^{-1} x = \frac{11}{2}$

The range of $3 \cos^{-1} x$ is $[0,3\pi]$ . Considering that as the domain of $\tan \alpha$ , there are three places where $\tan \alpha = \frac{11}{2}$ . $\tan$ is periodic with period of $\pi$

Therefore $\cos^{-1} x = \cos^{-1}(2/\sqrt{5}) + \frac{n\pi}{3}$ n =0,1,2

So $\displaystyle x = \cos \left ( \cos^{-1}(2/\sqrt{5}) + \frac{n\pi}{3}\right ) = \boxed{\left \{ \frac{2}{\sqrt{5}}, \frac{1}{2\sqrt{5}}(2-\sqrt{3}), -\frac{1}{2\sqrt{5}}(2+\sqrt{3}) \right \} }$

The range of $\frac{1}{3} \tan ^{-1} x$ is $(-\frac{1}{3}, \frac{1}{3})$ . This gives cos a smaller range between $(-\frac{\pi}{2}, \frac{\pi}{2})$ , so there will be two solutions, on negative and one positive, that is

$\displaystyle \boxed{ \left \{ -\frac{11}{2}, \frac{11}{2} \right \}}$

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Glutamic Acid

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II/5:

dy/dx = 0 when x = 0, so there's a stationary point. $\dfrac{\text{d}^2y}{\text{d}x^2} = e^{-x^2}(2x^4 - 5x^2 + 1)$ , which equals 1 at x = 0, so it's a minimum point. Consider $\displaystyle \int x(1 - x^2)e^{-x^2} \, \text{d}x$ , letting u = x^2 gives du = 2x dx so it becomes
$\displaystyle \int \frac{1}{2}e^{-u} - \frac{1}{2}ue^{-u} \, \text{d}u$ . From inspection, $y = \frac{1}{2}ue^{-u} \Rightarrow y = \frac{1}{2}x^2e^{-x^2}$ . No constant as the curve passes through the origin. At x = 1, y = $\frac{1}{2}e^{-1}$ and $\dfrac{\text{d}^2y}{\text{d}x^2} = -2e^{-1}$ ; negative, so a maximum.

The other stationary point is $(-1, \dfrac{1}{2}e^{-1}$ as it's an even function (which is to be expected as the derivative is odd).

For C_2: $\dfrac{\text{d}^2y}{\text{d}x^2} = e^{-x^3}(3x^5 - 3x^3 - 3x^2 + 1)$ . At x = 0, this equals 1; so minimum. At x = 1, this equals $-2e^{-1}$ , negative, so maximum.

The derivatives of C_1 and C_2 are equal at x = 0 and x = 1. If the derivative of C_2 is greater than C_1 for 0 < x < 1 then it follows that $k > \frac{1}{2}e^{-1}$ . This relies on showing that $e^{-x^3} > e^{-x^2} \Leftrightarrow e^{x^2} > e^{x^3} \Leftrightarrow x^2 < x^3 \Leftrightarrow x^2 (x - 1) < 0$ for 0 < x < 1, which can be shown with a very simple sketch. As e^(x^2) and e^(x^3) < 1 for 0 < x < 1, when taking logarithms the inequality is reversed.

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qgujxj39

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STEP III 2001, QUESTION 3

Consider the equation

where b and c are real numbers.

(i) Show that the roots of the equation are real and positive if and only if

and $b^2 \geq 4c > 0$ , and sketch the region of the b-c plane in which these conditions hold.

Spoiler:

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let the roots be

then by the quadratic formula:
$x_1 = \frac{b + \sqrt{b^2 - 4c}}{2}$
$x_2 = \frac{b - \sqrt{b^2 - 4c}}{2}$
Clearly $x_1 \geq x_2$ , so both roots are positive iff

therefore
$b > \sqrt{b^2 - 4c}$ (1)

If $b \leq 0$ , then $\sqrt{b^2 - 4c} < 0$ , which cannot be true if the roots are real. Therefore

(first condition), and thus we can square both sides of equation (1):

or

(second condition)

Finally, $b^2 - 4c \geq 0$ (third condition), otherwise the roots would clearly be complex.

The three conditions obtained are equivalent to those stipulated in the question.

Pathetic sketch, where the curve drawn is the positive half of $c = \frac{b^2}{4}$ and the shaded area underneath (inclusive of the curve itself) is where the conditions hold:

(ii) Sketch the region of the b-c plane in which the roots are real and less than 1 in magnitude.

Spoiler:

Show

By assumption, we can take $x_1 \geq x_2$ , therefore these conditions become

, as these both imply the other inequalities involved. These lead to:

$\sqrt{b^2 - 4c} < 2 - b$
$\sqrt{b^2 - 4c} < b + 2$
The right hand sides of these must clearly be positive, therefore

. On this assumption we can square both inequalities, which (after some manipulation) become:

Plotting these on a graph, together with $b^2 - 4c \geq 0$ , we get something looking vaguely like:

(Apologies for any dodgy copypasting, notepad is a ***** to work with sometimes.)

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Glutamic Acid

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(Original post by DFranklin)
It seems to me you are assuming independance between the plates - that is, knowing that a particular plate was broken by the mathematician doesn't affect the probability that another plate was broken by the mathematician. It's not at all obvious to me such an assumption is justified.

(It isn't justified for general distributions; imagine the case where each student either breaks 5 plates with probability k, or otherwise breaks no plates at all. Then if you know the mathematican broke a plate, you know he broke 5 of them, and you'd end up with an answer for 'broke > 3 plates' of 1/4).

On the other hand, it wouldn't surprise me if you were right, either. I confess I have a tendancy to overcomplicate these problems. You'd still need some kind of justification though (based on the distribution being Poisson).

I did try doing this the 'brute force' way by looking at all possibilities (e.g. 1st student breaks 0 plates, 2nd student 1 plate, 3rd student 1 plate, mathmo breaks 3 plates), and it feels a little tough for STEP I. Assuming I made no mistakes, I do get a different answer, however. I'll have another go when I'm not literally writing on the back of an envelope in the kitchen.

Hmm, I got 53/512, which is rounded to 0.1035., using the brute force method.

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SimonM

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STEP I 2001, Question 5

Spoiler:

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(i) $\displaystyle \int_0^1 \frac{1}{(1+tx)^2} \, dx = \frac{1}{t} \int_0^1 \frac{t}{(1+tx)^2} \, dx = \frac{1}{t} \left [ - \frac{1}{1+tx} \right ]_0^1 = \frac{1}{t} \left (1 - \frac{1}{1+t} \right ) = \frac{1}{1+t}$

(ii) $\displaystyle \int_0^1 \frac{-2x}{(1+tx)^3} \, dx = \frac{1}{t} \int_0^1 \frac{-2tx}{(1+tx)^3} \, dx = \frac{1}{t} \int_0^1 \frac{-2(1+tx) + 2}{(1+tx)^3} \, dx =$
$\displaystyle \frac{-2}{t} \int_0^1 \frac{1}{(1+tx)^2} \, dx + \frac{1}{t^2} \int_0^1 \frac{2t}{(1+tx)^3} \, dx = \frac{-2}{t(1+t)} + \frac{1}{t^2} \left [ - \frac{1}{(1+tx)^2} \right ]_0^1 =$
$\displaystyle \frac{-2}{t(1+t)} + \frac{1}{t^2} \left ( 1 - \frac{1}{(1+t)^2} \right ) =$
$\displaystyle - \frac{1}{(1+t)^2}$

Conjecture

$\displaystyle \int_0^1 \frac{d}{dt} \frac{1}{(1+tx)^2} \, dx = \frac{d}{dt} \frac{1}{(1+t)}$

So,

$\displaystyle \int_0^1 \frac{6x^2}{(1+x)^4} \, dx = \frac{2}{(1+1)^3} = \frac{1}{4}$

Which happens to be true. (For more info, look into differentiation under the integral sign)

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vahik92

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(Original post by Aurel-Aqua)
I Question 2

Part (i) $1+2x-x^2>\frac{2}{x}$ , $x\neq 0$ .

First it might help to find the points of equality: multiplying by x and rearranging, we get

, $x\neq 0$ . The roots are -1, 1 and 2. Now we must sketch the graph and notice that the inequality is satisfied for

(the zero due to the asymptote) and

.

Part (ii) $\sqrt{3x+10} > 2 + \sqrt{x+4}$ , $x \geq -\frac{10}{3}$ .

Because of the condition, squaring both sides still leaves an iff, giving: $3x + 10 > 4 + x + 4 + 4\sqrt{x+4}\implies 2x + 2 > 4\sqrt{x+4} \implies x + 1 > 2\sqrt{x+4}$ . Now by the condition, we can deduce that x > -1 because of the square root being always positive. Squaring again, $x^2+2x+1 > 4(x + 4) \implies x^2 - 2x - 15 > 0$ , giving

. From this, we can deduce that

and

, but our new condition,

only gives us

as the final solution.

You're genius you know, I've got 98% in my C2 - but I was clueless in this question - you're really a maths genius

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DFranklin

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(Original post by Glutamic Acid)
Hmm, I got 53/512, which is rounded to 0.1035., using the brute force method.

Yeah, I'd actually expect Mark13's method to give the right answer, I just don't think it's justified. I was a little surprised to get a different answer doing it by hand, and I probably have made a mistake - that's why I said I'd need to check.

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Aurel-Aqua

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Can someone do 2001 III Q8?

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SimonM

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STEP I, Question 14

Spoiler:

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The probability that the best candidate is amongst the first n candidates, is the same as the probability that any candidate is amongst the first n candidates. This probability is $\frac{n}{N}$

Spoiler:

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The probability the best amongst the first n candidates is best or second best is the probability it is the best is there + probability that the second best is there - probability that both are there. This is

$2\frac{\binom{N-1}{n-1}}{\binom{N}{n}} - \frac{\binom{N-2}{n-2}}{\binom{N}{n}} = \frac{2 \frac{N-1}{n-1} - 1}{\frac{N(N-1)}{n(n-1)}} = \frac{(2 N - n-1) n}{N(N-1)}$

Let the candidates be

where ability at maths is measured by one's initial.

Possibilities for interviews are

So the probability that the best guy is there is $\frac{6}{12} = \frac{1}{2} = \frac{n}{N}$

The probability that one of the best two are there is $\frac{10}{12} = \frac{5}{6} = \frac{5\times2}{4 \times 3}$

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STEP Maths I, II, III 2001 Solutions

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