STEP Maths I, II, III 1988 solutions Watch

maltodextrin
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(Original post by kabbers)
III/1:

Sketch y = \frac{x^2 e^{-x}}{x+1}

I'm not going to bother posting most of the differentiating legwork, there arent too many tricks, its just an algebra bash.

Differentiating (quotient rule is your friend), we get

\frac{dy}{dx} = e^{-x}\frac{x(2-x^2)}{(x+1)^2}

So we have turning points at x = 0, and x = +\sqrt{2}, -\sqrt{2}


Differentiating again, we get

\frac{d^2 y}{dx^2} = e^{-x}\frac{(2-3x^2)(x+1)^2 - 2(x+1)(2x-x^3) - (2x - x^3)(x+1)^2}{(x+1)^4}

Substituting in values of x for the turning points, we find that x = 0 is a minimum, and the other two maximums.


Noticing the denominator of y, we find that there is an asymptote at x = -1.

And because of the exponential properties of e^-x, y tends to zero as x \to +\infty

y tends to -\infty as x \to -\infty since the numerator remains positive and given the exponenential properties of e^x much greater than the denominator, while the denominator becomes negative.


So the graph will look http://www.thestudentroom.co.uk/show...8&postcount=14



Prove \displaystyle 0 < \int^{\infty}_0 \frac{x^2 e^{-x}}{1+x} dx < 1

First note that \frac{x^2}{x+1} = \frac{x^2+x}{x+1} - \frac{x}{x+1} = x - \frac{x}{x+1} = x - (\frac{x+1}{x+1} - \frac{1}{x+1}) = x - 1 + \frac{1}{x+1}

Hence we may split the integral into \displaystyle\int^{\infty}_0 (x - 1)e^{-x} dx + \int^{\infty}_0 \frac{1}{x+1}e^{-x} dx

Consider \displaystyle\int^{\infty}_0 (x - 1)e^{-x} dx

= \displaystyle\int^{\infty}_0 xe^{-x} dx - \int^{\infty}_0 e^{-x} dx

= \displaystyle([-xe^{-x}]^{\infty}_0 + \int^{\infty}_0 e^{-x} dx - \int^{\infty}_0 e^{-x} dx

= 0


Now consider \displaystyle\int^{\infty}_0 \frac{1}{x+1}e^{-x} dx

I posit the inequality \frac{1}{x+1}e^{-x} < e^{-x} for x > 0

\frac{1}{x+1} < 1
1 < (x+1)
0 < x

So our inequality holds.

So:

\displaystyle\int^{\infty}_0 \frac{1}{x+1}e^{-x} dx < \int^{\infty}_0 e^{-x} dx

\displaystyle\int^{\infty}_0 \frac{1}{x+1}e^{-x} dx < [-e^{-x}]^{\infty}_0

\displaystyle\int^{\infty}_0 \frac{1}{x+1}e^{-x} dx < 1

Thus \displaystyle\int^{\infty}_0 (x - 1)e^{-x} dx + \int^{\infty}_0 \frac{1}{x+1}e^{-x} dx < 1

And therefore, \int^{\infty}_0 \frac{x^2 e^{-x}}{1+x} dx < 1


Now notice that the graph of y = \frac{x^2 e^{-x}}{1+x} is greater than 0 for x > 0, and hence so is the infinite integral.

So, \displaystyle 0 < \int^{\infty}_0 \frac{x^2 e^{-x}}{1+x} dx < 1

please point out any mistakes
I'm probably wrong but to show that the integral lies between 0 and 1 would it not be sufficient to say that

(x^2)(e^-x)/(x + 1) < (x^2)(e^-x)/x = x(e^-x) for 0 < x < infinity

then integrate x(e^-x), which equals 1. Seems a lot simpler?
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DFranklin
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Yeah, that's fine.

While we're nitpicking - taking the 2nd derivative to deduce the nature of the turning points is a lot more work than simply sketching x(2-x^2) to argue whether the derivative is increasing or decreasing.
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brianeverit
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1988 Paper 2 number 4
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1988 Paper 2 number 8
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1988 Paper 2 numbers 10,11 & 12
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1988 Paper 2 numbers 13 & 14
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1988 Paper 2 numbers 15 & 16
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SimonM
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Question 8, STEP III

Spoiler:
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\displaystyle \frac{dy}{dx} = \frac{dy}{dt} / \frac{dx}{dt} = \frac{2a}{2at} = \frac{1}{t}

Therefore the equation of the gradient and normal respectively are:

\displaystyle \frac{y-2at}{x-at^2} = \frac{1}{t} and \displaystyle \frac{y-2at}{x-at^2} = -t

For the triangle formed by the tangents, the vertices will be the intersection of the tangents. By symmetry we need only compute this for one vertex:

\displaystyle \begin{cases} y = \frac{1}{t_1} x+ at_1 \\ y = \frac{1}{t_2} x + at_2 \end{cases} \Rightarrow x \left ( \frac{1}{t_1} - \frac{1}{t_2} \right ) = a(t_2 - t_1) \Rightarrow x = at_1t_2, y = a(t_1+t_2)

Therefore the area of the triangle of tangents is:

\displaystyle A_1 = \frac{1}{2} \begin{vmatrix} at_1t_2 & a(t_1+t_2) & 1 \\ at_2t_3 & a(t_2+t_3) & 1 \\ at_3t_1 & a(t_3+t_1) & 1 \end{vmatrix}

For the triangle formed by normals, the vertices will be the intersection of the normals. Again by symmetry we need only consider one pair:

\displaystyle \begin{cases} y = -t_1 x + at_1^3 + 2at_1 \\ y = -t_2 x + at_2^3 + 2at_2 \end{cases} \Rightarrow x(t_1 - t_2) = a(t_1^3 - t_2^3)+2a(t_1-t_2) \\ \Rightarrow x = a(t_1^2+t_1t_2 + t_2^2 + 2), y = -at_1t_2(t_1+t_2)

This gives us an area

\displaystyle A_2 = \begin{vmatrix} a(t_1^2+t_1t_2 + t_2^2 + 2) & -at_1t_2(t_1+t_2) & 1 \\

a(t_2^2+t_2t_3 + t_3^2 + 2) & -at_2t_3(t_2+t_3) & 1 \\

a(t_3^2+t_3t_1 + t_1^2 + 2) & -at_3t_1(t_3+t_1) & 1 \end{vmatrix}

\displaystyle A_1 = \frac{1}{2} \begin{vmatrix} at_1t_2 & a(t_1+t_2) & 1 \\ at_2t_3 & a(t_2+t_3) & 1 \\ at_3t_1 & a(t_3+t_1) & 1 \end{vmatrix} = \frac{1}{2} \begin{vmatrix} at_1t_2 & a(t_1+t_2) & 1 \\ at_2(t_3-t_1) & a(t_3-t_1) & 0 \\ at_1(t_3-t_2) & a(t_3-t_2) & 0 \end{vmatrix} \\ = \frac{a^2}{2}|(t_3-t_1)(t_3-t_2)(t_2-t_1)|

\displaystyle A_2 = \begin{vmatrix} a(t_1^2+t_1t_2 + t_2^2 + 2) & -at_1t_2(t_1+t_2) & 1 \\

a(t_2^2+t_2t_3 + t_3^2 + 2) & -at_2t_3(t_2+t_3) & 1 \\

a(t_3^2+t_3t_1 + t_1^2 + 2) & -at_3t_1(t_3+t_1) & 1 \end{vmatrix} \\ =\begin{vmatrix} a(t_1^2+t_1t_2 + t_2^2 + 2) & -at_1t_2(t_1+t_2) & 1 \\

a(t_2(t_3-t_1) + t_3^2-t_1^2 ) & -at_2(t_1-t_3)(t_1+t_2+t_3) & 0 \\

a(t_3^2-t_2^2+t_1(t_3-t_2)) & -at_1(t_2-t_3)(t_1+t_2+t_3) & 0 \end{vmatrix} \\

=\\ \begin{vmatrix} a(t_1^2+t_1t_2 + t_2^2 + 2) & -at_1t_2(t_1+t_2) & 1 \\

a(t_3-t_1)(t_1+t_2+t_3) & -at_2(t_1-t_3)(t_1+t_2+t_3) & 0 \\

a(t_3-t_2)(t_1+t_2+t_3) & -at_1(t_2-t_3)(t_1+t_2+t_3) & 0 \end{vmatrix} \\

= \frac{a^2}{2} |(t_3-t_1)(t_3-t_2)(t_2-t_1)(t_1+t_2+t_3)^2| = (t_1+t_2+t_3)^2 A_1

As required.

If they are congruent, the area of the triangle will be zero: t_1+t_2+t_3 = 0
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r2enigma
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Can somebody please post the solution for STEP III Q3?
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hannah.j
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(Original post by SimonM)
Done
Am I missing something or is the solution, well, missing?
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Mensorah
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Where can i get STEP 2 solution 2
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Dirac Spinor
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(Original post by SimonM)
....
STEP III Q10
Consider the positions of the dogs as time goes on. They will form a square of decreasing side length that rotates about the centre. If we take the position of one of the dogs to be (x,y) from the centre, then we see by symmetry that the position of one the next dogs along is (-y,x). We can also see that the tangent at any moment goes through the position of the next dog along, so:
\frac{dy}{dx}=\frac{y-x}{x-(-y)}=\frac{\frac{y}{x}-1}{1+\frac{y}{x}}
let:
u=\frac{y}{x} \Rightarrow \frac{dy}{dx}=\frac{du}{dx}x+u

\therefore \frac{du}{dx}x+u=\frac{u-1}{1+u}

\Rightarrow \frac{du}{dx}x=\frac{-u^2-1}{1+u}

\Rightarrow \frac{dx}{du} \frac{1}{x}=\frac{1+u}{-u^2-1}

\Rightarrow  \frac{1}{x}dx=\frac{1+u}{-u^2-1}du
let \theta be the angle between the x axis and the straight line to the point (x,y). Since u=\frac{y}{x} and, by trigonometry, \frac{y}{x}=Tan\theta \Rightarrow u=Tan\theta. So, substituting into the RHS and integrating we get:
lnx+c_1=-ln(sec\theta)-\theta+c_2.
By trigonometry, again, x=rcos\theta where r is the distance from the origin to the point (x,y) \Rightarrow ln(rcos\theta)+ln(sec\theta)=-\theta +c \Rightarrow ln(r)=-\theta +c \Rightarrow r=e^{-\theta}e^c
e^c is constant so r=e^{-\theta}\lambda for some constant \lambda as required.
To generalise to n-dogs held at the vertices of an n-gon it must be noted that the next dog along will have the coordinates of the first dog bur rotated \frac{2\pi}{n} about the centre of the field i.e:
\begin{pmatrix} cos(\frac{2\pi}{n}) & -sin(\frac{2\pi}{n})\\sin(\frac{2  \pi}{n}) & cos(\frac{2\pi}{n}) \end{pmatrix}\begin{pmatrix}x\\y  \end{pmatrix}=\begin{pmatrix}xco  s\frac{2\pi}{n}-ysin\frac{2\pi}{n}\\xsin\frac{2 \pi}{n}+ycos\frac{2\pi}{n} \end{pmatrix}
So, like the previous part,:
\frac{dy}{dx}=\frac{y-xsin\frac{2 \pi}{n}-ycos\frac{2 \pi}{n}}{x-xcos\frac{2 \pi}{n}+ysin\frac{2 \pi}{n}}=\frac{yx^{-1}-sin\frac{2 \pi}{n}-yx^{-1}cos\frac{2 \pi}{n}}{1-cos\frac{2 \pi}{n}+yx^{-1}sin\frac{2 \pi}{n}}
let:
u=\frac{y}{x} \Rightarrow \frac{dy}{dx}=\frac{du}{dx}x+u

\therefore \frac{du}{dx}x=\frac{-sin\frac{2 \pi}{n}-u^2sin\frac{2 \pi}{n}}{1-cos\frac{2 \pi}{n}+usin\frac{2 \pi}{n}}

\Rightarrow \frac{dx}{du}\frac{1}{x}=-cosec(\frac{2 \pi}{n})\frac{1-cos\frac{2 \pi}{n}+usin\frac{2 \pi}{n}}{u^2+1}
let:
u=Tan\theta \Rightarrow \frac{du}{d\theta}=sec^2\theta

\Rightarrow \frac{dx}{du}\frac{1}{x}=-cosec(\frac{2 \pi}{n})\frac{1-cos\frac{2 \pi}{n}+tan\theta sin\frac{2 \pi}{n}}{sec^2\theta}

\Rightarrow \frac{1}{x}dx=(-cosec(\frac{2 \pi}{n})+cot\frac{2 \pi}{n}-Tan\theta)d\theta
Integrating both sides we get:
lnx+c=(-cosec(\frac{2 \pi}{n})+cot\frac{2 \pi}{n})\theta-ln\sec\theta +c
And now, just like in the previous part:
ln(rcos\theta)+ln(sec\theta)+c=(-cosec(\frac{2 \pi}{n})+cot\frac{2 \pi}{n})\theta+c

\Rightarrow r=e^{-\theta(cosec(\frac{2 \pi}{n})-cot\frac{2 \pi}{n})}e^c= e^{-\theta (\frac {1-cos\frac{2 \pi}{n}}{sin\frac{2 \pi}{n}})}\lambda=e^{-\theta Tan(\frac{\pi}{n})}\lambda for some constant \lambda. Letting n=4 we can check and see that it gives the same answer as the first part.
EDIT: Just realised that the final result simplified to the much nicer form that it is currently in.
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Dzwx777
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Where can I find the step past papers before 1997? I have been looking for them for days, but all the links do not work now. Can you help me? thanks a lot!
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Dirac Spinor
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(Original post by Dzwx777)
Where can I find the step past papers before 1997? I have been looking for them for days, but all the links do not work now. Can you help me? thanks a lot!
http://www.mathshelper.co.uk/oxb.htm
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mikelbird
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(Original post by insparato)
III/3

 |z - i| = 2 is the circle on the complex plane with radius 2 and centre (0,i)

so x^2 + (y - 1)^2 = 2

and therefore in paramatrised form

 x = \sqrt2 cos\theta

 y  = 1 + \sqrt2 sin \theta

I've omitted sketches however I have described them in sufficient detail to make an easy sketch if one chooses.

(i)

 w = \frac{z + i}{z - i}

 wz - wi = z + i

 wz - z = i + wi

 z(w - 1) = i(w + 1)

 z = \frac{i(w+1)}{w-1}

 z - i = \frac{i(w +1)}{w - 1} - i = \frac{i(w + 1) - i(w-1)}{w - 1}

 = \frac{1 + i}{w - 1}

 |z - i| = \left|\frac{1+i}{w-1}\right|

 2 = \frac{|1 + i|}{|w - 1|}

 |w - 1| = \frac{\sqrt2}{2}

Okay so its a circle of radius  \frac{\sqrt2}{2} and the centre is (1,0) [/latex]

(ii)

If z = x + iy and is real then y must = 0

so

 w = u + iv = \frac{x + i(y + 1)}{x + i(y - 1)} = \frac{x + i}{x - i}

 (u + iv)(x - i) = x + i

 ux - ui + vxi + v = x + i

 ux + v + i(-u + vx - 1) = x

-u + vx - 1 must = 0 (this is important bit on)

so

 ux + v = x

 x(1 - u) = v

 x = \frac{v}{1 - u}

Now from before

 -u + vx - 1 = 0

 vx = u + 1

 x = \frac{u+1}{v}

so

 \frac{v}{1 - u} = \frac{u+1}{v}

 v^2 = (u + 1)(1 - u)

 u^2 + v^2 = 1

So in the W plane, when z is real, you get a circle centre(0,0) with radius 1

(iii)

if z = x + iy and z is imaginary then x = 0

 u + vi = \frac{x + i(y + 1)}{x + i(y - 1)}

 u + vi = \frac{i(y + 1)}{i(y - 1)}

 (u + vi)(iy - i) = yi + i

 uyi - ui - vy + v = yi + i

 yi - uyi = -ui - vy + v - i

 y(i-ui) = i(-1 -u) - vy + v

 -vy + v = 0

 y = \frac{i(-1-u)}{i(1 - u)}

 y = \frac{-1 -u}{1 - u}

 1 = \frac{-1 - u}{1 - u}

 -1 - u = 1 -u

 0 = 2

Wahey! wtf ? I'm sure i got something different when jotting it down, something that made sense! hmm have i gone wrong somewhere? I think I'm doing the right method.
erm....I have something different....
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brianeverit
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1988 STEP III number 3.
I'm afraid I disagree with the solutions published so far. Here is my solution.

 |z-i|=2 \Rightarrow |x+(y-1)i|=2 \text{ where }z=x+iy
 \Rightarrow x^2+(y-1)^2=4 \text{ so parametric equations are }x=2\cos\theta, y=2\sin\theta+1
 (i)  w=\dfrac{z+i}{z-i} \Rightarrow z=\left( \dfrac{w+1}{w-1} \right)i \text{ and }|z-i|=2 \Rightarrow \left|\dfrac{w+1}{w-1}i-i\right|=2 \Rightarrow left| \dfrac{2i}{w-1} \right|=2
 \Rightarrow |2i|=2|w-1| \text{ i.e. }|w-1|=1 \text{ so it is a circle, centre }w=1 \text{ radius }1
 (ii) z \text{ real } \Rightarrow \text {Im}\left( \dfrac{(u+1)i-v}{(u-1)i-v} \right)= \text{ Im}\left( \dfrac{((u+1)i-v)((u-1)i-v)}{(u-1)^2+v^2} \right)= -\dfrac{2uv}{(u-1)^2+v^2}=0
 \text{hence, }2uv=0 \text{  i.e.  It is the two axes}
 (iii) z\text{ imaginary }\Rightarrow \text{Re} \left( \dfrac{w+1}{w-=1} \right)i=0 \text{ i.e. } \dfrac {v^2-(u^2-1)}{(u-1)^2+v^2}=0 \Rightarrow v^2-u^2+1=0 \text{  i.e. A hyperbola}
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brianeverit
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1988 STEP I question 1 Alternative Solution

 \text{h}(x)=\dfrac{\ln x}{x} \Rightarrow \text{h}'(x)=\dfrac{1- \ln x}{x^2}=0 \text{ when } \ln x=1\Rightarrow x=\text{e}
 \text{for small }h, \text{h}'(1-h)&gt;0 \text{ while h}'(1+h)&lt;0 \text { so a maximum point at }(\text{e,e}^{-1}
 x=1 \Rightarrow \text{h}(x)=0
 \text {h}(x) \rightarrow 0 \text{ as }x \rightarrow \infty
 \text {h}(x) \rightarrow -\infty \text{ as }x \rightarrow 0

n^m=m^n \Rightarrow m \ln n=n \ln m \Rightarrow \dfrac{\ln n}{n}=\dfrac{\ln m}{m}
 \text{Clearly, }m \text{ and }n\text{ must lie on either side of the turning point}
\text{ so the only possible values for the smaller of }m \text{ and }n \text{ are } 1 \text{ and }2
\text{ there is obviously no other integer with }1^m=m^1 \text{ but } 2^4=4^2
 \text{ so only pair of  integers is }2 \text { and }4
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brianeverit
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1988 STEP I question 2 Alternative solution

 \text{g}(x)= \text{ f}(x)+\dfrac{1}{\text{f}(x)} \Rightarrow \text{ g}'(x)=\text{ f}'(x) -\dfrac{\text{f}'(x)}{(\text{f}(x  ))^2}
 \text{and g}''(x)=\text{ f}''(x)- \dfrac{(\text{f}(x))^2\text{f}''  (x)-\text{f}'(x).2\text{f}(x)\text{f  }'(x)}{(\text{f}(x))^4}
\text{i.e. g}''(x)=\text{ f}''(x)- \dfrac{\text{f}''(x)}{(\text{f}(  x))^2}+ \dfrac{2(\text{f}'(x))^2}{(\text  {f}(x))^3}
 \text{f}(x)=4+ \cos2x+2 \sin x \Rightarrow \text{f}'(x)=2\cosx-2 \sin2x \text{ and f}''(x)=-2 \sin x-4\cos2x
\text{so g}'(x)=2\cos x-2\sin x-\dfrac{2\cos x-2\sin 2x}{(4+\cos 2x+2\sin x)^2}
 \text{g}'(x)=0 \text{ when }}2\cos x-2\sin 2x=0 \text{ or }(4+\cos 2x+2\sin x)^2=1
 4+\cos 2x+2\sin 2x&gt;1 \text{ for all }x \text{ so solutions are given by }2 \cos x-2 \sin 2x=0
\Rightarrow \cos x-2\sin x\cos x=0 \Rightarrow \cos x=0 \text{ or }\sin x =\dfrac{1}{2} \text{ so }x=\dfrac{\pi}{6},\dfrac{\pi}{2}  ,\dfrac{5\pi}{6} \text{ or }\dfrac{3\pi}{2}
\text{g}(x) \text{ is a continuous function since }4+\cos 2x+2\sin x \text{ cannot be zero }
\text{ so we need only consider one of the turning points }
x=\dfrac{\pi}{2} \Rightarrow \text{ f}(x)=5,\text{ f}'(x)=0\text{ and f}''(x)=2 \Rightarrow \text{ g}''(x)&gt;0
\text{so a minimum at }x=\dfrac{\pi}{2}
 \text{ g}\left(\dfrac{\pi}{6}\right)= \dfrac{11}{2}+ \dfrac{2}{11}= \dfrac{125}{22}, \text{ g}\left(\dfrac{\pi}{2}\right)=5+ \dfrac{1}{5}= \dfrac{26}{5}, \text{ g} \left( \dfrac{5\pi}{6} \right)= \dfrac{11}{2}+ \dfrac{2}{11}= \dfrac{125}{22} \text{ and g} \left( \dfrac{3\pi}{2} \rihjt)=1+1=2
 \text{so stationary points are maxima at } \left( \dfrac{\pi}{6}, \dfrac{125}{22} \right) \text{ and }\left( \dfrac{5\pi}{6}, \dfrac{125}{22} \right) \text{ and minima at }\left( \dfrac{\pi}{2}, \dfrac{26}{5}  \right) \text{ and } \left( \dfrac{3\piu}{2}, 2\right)
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brianeverit
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1988 STEP I question 7 Alternative solution

 \text{f}(x)=ax^2+bx+c \Rightarrow \text{ f}'(x)=2ax+b
 \text{f}(1)(x+ \frac{1}{2})+\text{ f}(-1)(x- \frac{1}{2})-2\text{ f}(0)x=(a+b+c)(x+ \frac{1}{2})+(a-b+c)(x- \frac{1}{2})-2cx
=2ax+2cx+b-2cx=2ax+b=\text{ f}'(x)
 \text{Since f}'(x) \text{ is linear it must attain it's maximum value at either }x=1 \text{ or }x=-1
\text{f}'(-1)=-\frac{1}{2}\text{f}(1)- \frac{3}{2}\text{f}(-1)-2\text{f}(0) \text{ and f}'(1)= \frac{3}{2}\text{f}(1)+ \frac{1}{2}\text{f}(-1)-2\text{f}(0)
 \text{f}(-1),\tewxt{ f}(0) \text{ and f}(1) \text{ are all between }\pm{1} \Rightarrow |\text{f}'(x)| \leq4
 \text{Consider }2x^2-1 \text{ then }-1 \leq \text{ f}(x0) \leq1 \text{ i.e. }\text{f}(x)| \leq1 \text{ for all }x \text { with }|x| \leq1
 \text{and f}'(x)=4x \text{ lies between }\pm4 \text{ so this function satisfies the requirements}
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brianeverit
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(Original post by nota bene)
I/9

i) \displaystyle\int_1^e \frac{\log x}{x^2} dx Recall that \int \log x dx= x(\log (x)-1)+c, now we can use this to integrate by parts.
I=\displaystyle\int \frac{\log x}{x^2}dx=\frac{1}{x^2}x(\log(x)-1)-\displaystyle\int-2\frac{x(\log(x)-1)}{x^3}dx
-I=\frac{1}{x}(\log(x)-1)+2\frac{1}{x}=\frac{\log x +1}{x}
Putting in the limits gives -I=\frac{2}{e}-1= So I=1-\frac{2}{e}

ii) \displaystyle\int \frac{\cos(x)}{\sin(x)\sqrt{1+\s  in(x)}}dx Let u^2=1+\sin(x)\Rightarrow 2u\frac{du}{dx}=\cos(x)\Leftrigh  tarrow dx=\frac{2u}{\cos(x)}du
So the integral becomes \displastyle\int \frac{2}{\sin x}du=\displaystyle\int\frac{2}{u  ^2-1}du=\displaystyle\int\frac{1}{u-1}du-\displaystyle\int\frac{1}{u+1}du  =\log|u-1|-\log|u+1|+c=\log|\frac{u-1}{u+1}|+c
Going back to x we have \log|\frac{u-1}{u+1}|+c=\log|\frac{\sqrt{1+\s  in x}-1}{\sqrt{1+\sin x}+1}|+c
Alternative solution for part (b)

 \int \dfrac { \cos x}{ \sin x \sqrt{1+ \sin x}}dx
\text{ so putting }\sin x= \tan^2 \theta \text{ so } \cos x \dfrac {dx}{d \theta}=2 \tan x\sec^2 x
 \text{integral becomes } \int \dfrac{2\tan \theta \sec^2 \theta d\theta}{\tan^2 \theta \sec\theta }=\int \dfrac{2\sec\theta}{\tan\theta}d  \theta=\int 2\csc\theta d\theta=-\ln{|\csc\theta+\cot\theta|}+c
 \tan\theta=\sqrt{x} \Rightarrow \cot\theta=\dfrac{1}{\sqrt{x}},\  sin\theta= \dfrac{\sqrt{x}}{\sqrt{1+x}} \text{ so }\csc\theta=\dfrac{\sqrt{1+x}}{ \sqrt{x}}
\text{Hence }\int \dfrac{ \cos x}{ \sin x \sqrt{1+\sin x}}dx=-\ln A \left| \dfrac{1+ \sqrt{1+x}}{\sqrt{x}} \right|=\ln A\left| \dfrac{\sqrt x}{1+\sqrt{1+x}} \right|
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