STEP Maths II, III 2009 Solutions

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Zacken
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#1
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#1
(Updated as far as #57)

STEP II:
1: Solution by Zacken
2: Solution by IrrationalRoot
3: Soluions by Zacken
4: Solution by Zacken
5: Solution by Zacken
6: Solution by Zacken
7: Solution by B_9710
8: Solution by Farhan.Hanif93
9: Solution by A-cute-Angle
10: Solution by Zacken
11: Solution by The-Spartan
12: Solution by Farhan.Hanif3
13: Solution by Farhan.Hanif3

STEP III:
1: Solution by IrrationalRoot
2: Solution by Farhan.Hanif93
3: Solution by Zacken
4: Solution by EricPiphany
5: Solution by IrrationalRoot
6: Solution by Farhan.Hanif93
7: Solution by Farhan.Hanif93
8: Solution by IrrationalRoot
9: Solution by Farhan.Hanif93
10: Solution by Farhan.Hanif93
11: Solution by Farhan.Hanif93
12: Solution by Zacken
13: Solution by Farhan.Hanif3

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Zacken
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STEP II, Question 1

First part

We note that since x^4 + y^4 = u is unchanged if x \mapsto -x or y \mapsto -y as well as y \mapsto \pm x then the equation of symmetries are the axes (i.e: x=0 and y=0) and the lines y=\pm x.

Since xy = v is unchanged under y \mapsto \pm x then the equation of symmetry is y=\pm x


Second part
A quick sketch:
Spoiler:
Show
Image
proves useful.

We can immediately identify A as (\alpha, \beta), B as (\beta, \alpha), C as (-\alpha, -\beta) and D as (-\beta, -\alpha).


Third part
A quick check of the gradients reveals that m(AB) = \frac{\beta - \alpha}{\alpha - \beta} = -1, and likewise m(BC) = 1, m(CD) = -1 and m(DA) = 1. Hence ABCD forms a rectangle.

Furthermore, the lengths are:

\displaystyle 

\begin{equation*}|AB|^2 = |CD|^2 = 2(\alpha - \beta) \Rightarrow |AB| = |CD| = \sqrt{2}(\alpha - \beta)\end{equation*} (since \alpha - \beta).

Also:

\displaystyle

\begin{equation*} |AD|^2 = |BC|^2 = 2(\alpha + \beta)^2 \Rightarrow |AD| = |BC| = \sqrt{2}(\alpha + \beta)\end{equation*}.

The area is then \Delta = 2(\alpha - \beta)(\alpha + \beta) = 2(\alpha^2 - \beta^2).

Remember that \alpha, \beta lie on both curves, hence: \alpha^4 + \beta^4 = u and \alpha \beta = v. So that:

\displaystyle 

\begin{equation*} (\alpha^2 - \beta^2)^2 = \alpha^4 + \beta^4 - 2\alpha^2 \beta^2 = u - 2v^2 \end{equation*}.

Hence, the area is:

\displaystyle 

\begin{equation*}\Delta = 2\sqrt{u - 2v^2} \end{equation*}

and verifying this in the given case u = 81 and v=4 yields \Delta = 2\sqrt{49} = 2 \times 7 = 14 as required.
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Zacken
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STEP II, Question 4

First part
Write p(x) -1 = (x-1)^5 g_0(x) for some fourth degree polynomial g_0(x). Then p(x) = (x-1)^5g(x) + 1 \Rightarrow p(1) = 1.


Second part
Using the product rule, we have:

\displaystyle 

\begin{equation*}p'(x) = (x-1)^5g'_0(x) + 5(x-1)^4g_0(x) = (x-1)^4(g'_0(x)(x-1) + 5g_0(x)) \end{equaiton*}

So p'(x) is clearly divisible by (x-1)^4.


Third part
Write p(x) = (x+1)^5g_1(x) + 1 for some fourth degree polynomial g_1(x).

Then p'(x) = (x+1)^4(g'_1(x)(x+1) + 5g_1(x)) in a similar fashion as (ii). Also note that p(-1) = -1

But since p(x)[ is a 9th degree polynomial, then p'(x) is an 8th degree polynomial.

However, we have that (x-1)^4 | p'(x) and (x+1)^4 | p'(x).

So it must be the case that p'(x) = k(x-1)^4(x+1)^4 = k(x^2 - 1)^4

From this, we can expand: p'(x) = k(x^8 - 4x^6  + 6x^4 - 4x^2 + 1), so that, if we integrate:

\displaystyle 

\begin{equation*}p(x) = k\left(\frac{x^9}{9} - \frac{4x^7}{7} + \frac{6x^5}{5} - \frac{4x^3}{3} + x\right) + c\end{equation*}

Now, using p(\pm 1) = \pm 1 gives us: \displaystyle 1 = k\left(\frac{128}{315}\right) + c and \displaystyle -1 = -k\left(\frac{128}{315}\right) + c

Adding both these equations yields 0 = c and hence k = \frac{315}{128} so that:

\displaystyle 

\begin{equation*}p(x) = \frac{315}{128}\left(\frac{x^9}{9} - \frac{4x^7}{7} + \frac{6x^5}{5} - \frac{4x^3}{3} + x\right)\end{equation*}
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Zacken
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STEP II, Q5:

(i)
(\sqrt{x-1} +1)^2 = x-1 + 2\sqrt{x-1} + 1 = x + 2\sqrt{x-1} as expected.

Also, note that (\sqrt{x-1} -1)^2 = x-1 - 2\sqrt{x-1} + 1 = x - 2\sqrt{x-1}.

We can then state that \sqrt{x + 2\sqrt{x-1}} = |\sqrt{x-1} + 1| and likewise \sqrt{x - 2\sqrt{x-1}} = |\sqrt{x-1} - 1|

This means that our integral is nothing more than:

\displaystyle 

\begin{equation*}\int_5^{10} \frac{|\sqrt{x-1} + 1| + |\sqrt{x-1}-1|}{\sqrt{x-1}} \, \mathrm{d}x = \int_5^10 \frac{2\sqrt{x-1}}{\sqrt{x-1}} \, \mathrm{d}x = \int_5^{10} 2 \, \mathrm{d}x = 10

Where we take the positive root of the square root since it is positive in the interval [5, 10] that we are integrating over.


(ii)
Since there is a sign change in the integrand at x=2 we'll need to split the integral into two regions and flip the sign of the first region. That is:

\displaystyle

\begin{equation*} \int_{5/4}^{2} \frac{|\sqrt{x-1} -1 |}{\sqrt{x-1}} \, \mathrm{d}x + \int_{2}^{10} \frac{|\sqrt{x-1} -1 |}{\sqrt{x-1}} \, \mathrm{d}x = -\int_{5/4}^{2} \frac{\sqrt{x-1} -1 }{\sqrt{x-1}} \, \mathrm{d}x + \int_{2}^{10} \frac{\sqrt{x-1} -1 }{\sqrt{x-1}} \, \mathrm{d}x \end{equation*}.

Integrating \displaystyle \int (x-1)^{-1/2} \, \mathrm{d}x = 2\sqrt{x-1}, we get:

\displaystyle 

\begin{equation*}-[x]_{5/4}^2 + [x]_2^{10} - 2[\sqrt{x-1}]_{5/4}^{2} + 2[\sqrt{x-1}]_{2}^{10} = -2 + \frac{5}{4} + 10 - 2 + 2 - 1 - 6 + 2 = 4 \, \frac{1}{4}\end{equation*}


(iii)


Note that (\sqrt{x+1} - 1)^2 = x+1 - 2\sqrt{x+1} + 1 = x - 2\sqrt{x+1} + 2.

Hence \sqrt{x - 2\sqrt{x+1} + 2} = |\sqrt{x+1} - 1|.

Since the entirety of our integrand is always positive for all valid x, and using the difference of two squares on the denominator:

\displaystyle

\begin{equation*} \int_{5/4}^{10} \frac{\sqrt{x-1} + 1 + \sqrt{x+1} - 1}{\sqrt{x-1}\sqrt{x+1}}  = \int_{5/4}^{10} (x+1)^{-1/2} + (x-1)^{-1/2} \, \mathrm{d}x \end{equation*}

We finally end up with:

\displaystyle

\begin{equation*}2[\sqrt{x+1} + \sqrt{x-1}]_{5/4}^{10} = 2\left(\sqrt{11} + 3 -2\right) = 2(1 + \sqrt{11})\end{equation*}

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IrrationalRoot
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STEP III, Question 1

p and q
Let m_{AB} denote the gradient of any line AB.

We have m_{SV}=\dfrac{nv-ms}{v-s} and m_{UT}=\dfrac{nu-mt}{u-t}.

Therefore the equation of SV is

y-ms=m_{SV} (x-s) \Rightarrow y=ms+\dfrac{nv-ms}{v-s} (x-s),

so

0=ms+\dfrac{nv-ms}{v-s} (p-s) 

\Rightarrow p=s-ms\left(\dfrac{v-s}{nv-ms}\right)

=\dfrac{snv-ms^2-ms(v-s)}{nv-ms}

=\dfrac{msv-ms^2+ms^2-snv}{ms-nv}=\dfrac{(m-n)sv}{ms-nv},

as required.

To get q, we note that we can simply replace v with u and s with t so that q=\dfrac{(m-n)tu}{mt-nu}.


st and s+t
If S and T lie on the circle x^2+(y-c)^2=r^2, then s^2+(ms-c)^2=r^2 and t^2+(mt-c)^2=r^2.

Therefore a quadratic equation satisfied by both s and t is a^2+(ma-c)^2=r^2, that is, (m^2+1)a^2-2mca+(c^2-r^2)=0.

Considering the sum and product of the roots gives st=\dfrac{c^2-r^2}{m^2+1} and s+t=\dfrac{2mc}{m^2+1}.


p+q=0
We now also have uv=\dfrac{c^2-r^2}{n^2+1} and u+v=\dfrac{2nc}{n^2+1}. Hence, \dfrac{st}{uv}=\dfrac{n^2+1}{m^2+1} and \dfrac{s+t}{u+v}=\dfrac{2mc}{m^2+1} \times  \dfrac{n^2+1}{2nc}=\dfrac{m(n^2+1)}{n(m^2+1)}.

Therefore p+q=\dfrac{(m-n)sv}{ms-nv} + \dfrac{(m-n)tu}{mt-nu}=(m-n)\left(\dfrac{sv}{ms-nv}+\dfrac{tu}{mt-nu}\right), so it suffices to show that

\dfrac{sv}{ms-nv}+\dfrac{tu}{mt-nu}=0 and so it suffices to show that

sv(mt-nu)=tu(ms-nv)=0 \Leftrightarrow mvst-nusv+must-nutv=0 \Leftrightarrow mvst+must=nusv+nutv \Leftrightarrow mst(u+v)=nuv(s+t) \Leftrightarrow \dfrac{m}{n}\left(\dfrac{st}{uv}\right)=\dfrac{s+t}{u+v} \Leftrightarrow \dfrac{m}{n}\left(\dfrac{n^2+1}{m^2+1}\right)=\dfrac{m(n^2+1)}{n(m^2+1)}

which is true, so we conclude that p+q=0, as required.
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Krollo
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#6
(Original post by Zacken)
(Updated as far as #5)

STEP II:
1: Solution by Zacken
2:
3:
4: Solution by Zacken
5: Solution by Zacken
6:
7:
8:
9:
10:
11:
12:
13:

STEP III:
1: Solution by IrrationalRoot
2:
3:
4:
5:
6:
7:
8:
9:
10:
11:
12:
13:

Solutions written by TSR members:
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I'll take ii q11 for now. Will probably focus on the applied seeing as no-one else seems to like it :-P

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IrrationalRoot
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STEP III, Question 5

yz+zx+xy=\dfrac{(x+y+z)^2-(x^2+y^2+z^2)}{2}=\dfrac{(1)^2-2}{2}=-\dfrac{1}{2}
as required.

x^2y+x^2z+y^2z+y^2x+z^2x+z^2y=x^2(y+z)+y^2(z+x)+z^2(x+y)= x^2(1-x)+y^2(1-y)+z^2(1-z)=x^2+y^2+z^2-(x^3+y^3+z^3)=2-3=-1
as required.

-\frac{1}{2}=-\frac{1}{2} \times 1=(yz+zx+xy)(x+y+z)=xyz+y^2z+yz^2+x^2z+xyz+xz^2+x^2y+xy^2+xyz=3xyz+x^2y+x^2z+y^2z+y^2x+z^2x+z^2y=3xyz-1



\Rightarrow xyz=\dfrac{1-\frac{1}{2}}{3}=\dfrac{1}{6}
as required.

We require

S_{n+1}=aS_n+bS_{n-1}+cS_{n-2} 



\Leftrightarrow x^{n+1}+y^{n+1}+z^{n+1}=ax^n+ay^n+az^n+bx^{n-1}+by^{n-1}+bz^{n-1}+cx^{n-2}+cy^{n-2}+cz^{n-2} 



\Leftrightarrow x^{n-2}(x^3-ax^2-bx-c)+y^{n-2}(y^3-ay^2-by-c)+z^{n-2}(z^3-az^2-bz-c)=0.

It therefore suffices to find a, b and c such that the cubic m^3-am^2-bm-c=0 for m=x,y,z.

Therefore we can take a=x+y+z=1, b=-(yz+zx+xy)=\dfrac{1}{2} and c=xyz=\dfrac{1}{6}.
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EricPiphany
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#8
Can I reserve STEP III Q4?
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IrrationalRoot
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#9
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#9
(Original post by EricPiphany)
Can I reserve STEP III Q4?
I was gonna do that one but I'm pretty new to Latex and so I'd be terribly slow with all those integrals and stuff so go ahead haha.

I'd like to reserve II Q2 (i.e. I'm doing it now) .

EDIT: Meant to say Q2 sorry.
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EricPiphany
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STEP III Q4

(i).
Laplace transform of \displaystyle e^{-bt}f(t) is
 \displaystyle \int_0^\infty e^{-st}\left(e^{-bt}f(t)\right) \,\mathrm{d}t = \int_0^\infty e^{-(s+b)t}f(t)\right) \,\mathrm{d}t = F(s+b).

(ii).
Laplace transform of \displaystyle f(at) is
 \displaystyle \int_0^\infty e^{-st}f(at) \,\mathrm{d}t , making the substitution  \displaystyle u = at, \mathrm{d}u = a \mathrm{d}t,
 \displaystyle a^{-1} \int_0^\infty e^{-\frac{s}{a}u}f(u) \,\mathrm{d}u =  \displaystyle a^{-1} \int_0^\infty e^{-\frac{s}{a}t}f(t) \,\mathrm{d}t = a^{-1}F\left(\frac{s}{a}\right).

(iii).
Laplace transform of  \displaystyle f'(t) is
 \displaystyle \int_0^\infty e^{-st}f'(t) \,\mathrm{d}t = \left[e^{-st}f(t)\right]_0^\infty + s\int_0^\infty e^{-st}f(t) \,\mathrm{d}t = -f(0)+sF(s), using integration by parts.

(iv).
Laplace transform of  \displaystyle \sin(t) is
 \displaystyle \int_0^\infty e^{-st}\sin(t) \,\mathrm{d}t = \left[-e^{-st}\cos(t)\right]_0^\infty -s \int_0^\infty e^{-st}\cos(t) \,\mathrm{d}t =
\displaystyle =1 - s \left[e^{-st}\sin(t)\right]_0^\infty -s^2 \int_0^\infty e^{-st}\sin(t) \,\mathrm{d}t = 1- s^2 \int_0^\infty e^{-st}\sin(t) \,\mathrm{d}t.
So, \displaystyle \mathrm{I} = 1-s^2 \mathrm{I}, so \displaystyle \mathrm{I}=\frac{1}{1+s^2}.

Now, \displaystyle \cos(t) is the derivative of \displaystyle \sin(t), so by part (iii), Laplace transform of  \displaystyle \cos(t) is \dfrac{s}{1+s^2}.
And by part (ii), Laplace transform of \displaystyle \cos(qt) is \displaystyle q^{-1} \left(\frac{s/q}{1+(s/q)^2}\right) = \frac{s}{s^2+q^2}.
Thus, by part (i), Laplace transform of \displaystyle e^{-pt}\cos(qt) is \displaystyle \frac{s+p}{q^2+(s+p)^2}.

What a mess, feel free to edit and repost.
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IrrationalRoot
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#11
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#11
(Original post by StrangeBanana)
Doing STEP III, Q5
Done a while ago!!
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username1258398
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(Original post by IrrationalRoot)
Done a while ago!!
hmm

I'm tired

EDIT:
I'll come back tomorrow and see if any are still available
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IrrationalRoot
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STEP II, Question 2

(i)
\dfrac{dy}{dx}=a^{\sin(\pi e^x)} \ln a \times \dfrac{d}{dx} \sin(\pi e^x)=a^{\sin(\pi e^x)} \ln a \times \cos(\pi e^x) \times \pi e^x.

\dfrac{dy}{dx}=0 

\Leftrightarrow \cos(\pi e^x)=0 (a^{\sin(\pi e^x)}, \ln a, \pi e^x \not= 0)

\Leftrightarrow \pi e^x=\dfrac{(2k+1)\pi}{2}, k\in\mathbb{Z}

\Leftrightarrow e^x=\dfrac{2k+1}{2} 

\Leftrightarrow x=\ln\frac{1}{2}, \ln\frac{3}{2}, \ln\frac{5}{2}, \ldots .

Therefore y=a^{\sin(\frac{\pi}{2})}, a^{\sin(\frac{3\pi}{2})}, a^{\sin(\frac{5\pi}{2})}, \ldots=a,\frac{1}{a},a,\frac{1}{a}, \ldots.

Therefore coordinates of stationary points are (\ln\frac{1}{2},a), (\ln\frac{3}{2},\frac{1}{a}), (\ln\frac{5}{2},a), (\ln\frac{7}{2}, \frac{1}{a}), \ldots.


(ii)
(ii) \ln y=\sin(\pi e^x)\lna, so for small x,

\ln y \approx \sin(\pi (1+x)) \ln a = \sin(\pi + \pi x) \ln a =-\sin(\pi x) \ln a \approx -\pi x \ln a (x is small)

\Rightarrow y \approx e^{-\pi x \ln a} \approx 1-\pi x \ln a

as required.


(iii)
(iii) The graph should look sinusoidal, with maxima y=a at x=\ln\dfrac{1}{2}, \ln\dfrac{5}{2}, \ln\dfrac{9}{2}, \ldots and minima y=\frac{1}{a} at \ln\dfrac{3}{2}, \ln\dfrac{7}{2}, \ln\dfrac{11}{2}, \ldots.

These extrema should become closer together as x increases. Note that for x<\ln\dfrac{1}{2}, y is strictly increasing and as x \rightarrow -\infty, y \rightarrow 1^+.


(iv)
(iv) The maxima occur at x=\ln\left(\dfrac{4k-3}{2}\right) for k=1,2,3, \ldots Therefore the required area is approximately

\dfrac{1}{2}\left(a+\dfrac{1}{a}\right)\left(\ln\left(\dfrac{4k-1}{2}\right)-\ln\left(\dfrac{4k-3}{2}\right)\right)

+\dfrac{1}{2}\left(a+\dfrac{1}{a}\right)\left(\ln \left(\dfrac{4k+1}{2}\right) - \ln \left(\dfrac{4k-1}{2}\right)\right)



=\left(\dfrac{a^2+1}{2a}\right) \left(\ln\left(\dfrac{4k+1}{2} \right)-\ln\left(\dfrac{4k-3}{2}\right)\right)



=\left(\dfrac{a^2+1}{2a}\right) \ln\left(\dfrac{4k+1}{4k-3}\right)



=\left(\dfrac{a^2+1}{2a}\right) \ln\left(\dfrac{4k-3}{4k-3}+\dfrac{4}{4k-3}\right)



=\left(\dfrac{a^2+1}{2a}\right) \ln\left(1+\left(k-\dfrac{3}{4}\right)^{-1}\right)
as required.
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EricPiphany
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(Original post by Zacken)
STEP II, Question 4

So it must be the case thatp'(x) = k(x-1)^4(x+1)^4 = k(x^2-1)^4
Just a small typo
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Zacken
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#15
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#15
(Original post by IrrationalRoot)
...
Would have been simpler to note that \sin (\pi e^x) is a bounded function, so that a^{\sin (\pi e^x)} is maximum when \sin (\pi e^x) is 1, giving rise to a^1 and \sin (\pi e^x) = -1 giving rise to a^{-1}.
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Zacken
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(Original post by EricPiphany)
Just a small typo
I'm being incredibly thick, but I can't see it?
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EricPiphany
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#17
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#17
(Original post by Zacken)
I'm being incredibly thick, but I can't see it?
coz I've tried to correct it
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Zacken
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#18
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#18
(Original post by EricPiphany)
coz I've tried to correct it
Ahhh, yeah lol, dumb mistake that one. Thanks!
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Zacken
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#19
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#19
STEP II, Question 3:

\displaystyle \tan \left(\frac{\pi}{4} - \frac{x}{2}\right) = \frac{1 - \tan \frac{x}{2}}{1 + \tan \frac{x}{2}} = \frac{\cos \frac{x}{2} - \sin \frac{x}{2}}{\sin \frac{x}{2} + \cos \frac{x}{2}} = \frac{1 - 2\sin \frac{x}{2}\cos \frac{x}{2}}{\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}} = \frac{1 - \sin x}{\cos x}.

Which simplifies down to \sec x - \tan x.

(i)
Set x = \frac{\pi}{4} to get \displaystyle \tan \frac{\pi}{8} = \sec \frac{\pi}{4} - \tan \frac{\pi}{4} = \sqrt{2} - 1.

Hence, using the fact that:

\displaystyle 

\begin{equation*}\tan \frac{11\pi}{24} = \tan \left(\frac{\pi}{3} + \frac{\pi}{8}\right) = \frac{\tan \frac{\pi}{3} + \tan \frac{\pi}{8}}{1 - \tan \frac{\pi}{3} \tan \frac{\pi}{8}} = \frac{\sqrt{3} + \sqrt{2} - 1}{\sqrt{3} - \sqrt{6} + 1} \end{equation*}

as required.



(ii)


\displaystyle

\begin{equation*}\frac{\sqrt{3} + \sqrt{2} - 1}{1 - \sqrt{6} + \sqrt{3}}  \times \frac{1 + \sqrt{3} + \sqrt{6}}{1 + \sqrt{3} + \sqrt{6}} = \frac{1 + 2\sqrt{2} + \sqrt{3}}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3}+1} = 2 + \sqrt{2} + \sqrt{3} + \sqrt{6}\end{equation*}


(iii)


Set x = \frac{11\pi}{24}, then:
\displaystyle 

\begin{equation*}\tan \left(\frac{\pi}{48}\right) = \sec \frac{11\pi}{24} - \tan \frac{11\pi}{24}\end{equation*}

But, we know that:

\displaystyle

\begin{equation*} \sec^2 \frac{11\pi}{24} = 1 + \tan^2 \frac{11\pi}{24} = 1 + (2 + \sqrt{2} + \sqrt{3} + \sqrt{6})^2 = 16 + 10\sqrt{2} + 8\sqrt{3} + 6\sqrt{6}\end{equation*}

Hence: \sec \frac{11\pi}{24} = \sqrt{16 + 10\sqrt{2} + 8\sqrt{3} + 6\sqrt{6}}.

Giving us:

\displaystyle 

\begin{equation*}\tan \frac{\pi}{48} = \sec \frac{11\pi}{24} - \tan \frac{11\pi}{24} = \sqrt{16 + 10\sqrt{2} + 8\sqrt{3} + 6\sqrt{6}} - 2 - \sqrt{2} - \sqrt{3} - \sqrt{6} \end{equation*}

as required.


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Farhan.Hanif93
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#20
Report 6 years ago
#20
STEP II Q8

Solution
Straightforward diagram with A, B and C non-collinear; P lying on the line AB between A and B; and Q lying on the line CA "beyond" A. Note:

\begin{array}{rcl}

CQ \times BP = AB \times AC &\iff& \mu |\mathbf{a}-\mathbf{c}| \times \lambda|\mathbf{a}-\mathbf{b}| = AB \times AC\\ 

&\iff& \mu \lambda (CA \times BA) = AB \times AC\\

&\iff& \boxed{\mu  =1/\lambda} \ (\star)

\end{array}

Furthermore, using (\star) and the definitions of \mathbf{p} and \mathbf{q}, observe that any point on the line PQ has position vector:

\begin{array}{rcl}

\tau \mathbf{q} + (1-\tau) \mathbf{p} &=& \tau [(1/\lambda)\mathbf{a} + (1-1/\lambda) \mathbf{c}] + (1-\tau) [\lambda \mathbf{a} + (1-\lambda) \mathbf{b}]\\

&=& [\tau / \lambda + (1-\tau) \lambda] \mathbf{a} +(1-\tau)(1-\lambda) \mathbf{b} + \tau(1-1/\lambda)\mathbf{c} \ (\star \star)

\end{array}

By taking \tau = \dfrac{\lambda}{\lambda-1} for any \lambda \in (0,1) (and verifying by direct substitution into (\star \star)), the coefficients of \mathbf{a}, \mathbf{b} and \mathbf{c} collapse to -1,1 and 1 respectively. That is, \mathbf{d}=-\mathbf{a}+\mathbf{b}+\mathbf{c} always lies on the line PQ, as required.

Finally, note that \mathbf{d}=-\mathbf{a}+\mathbf{b}+\mathbf{c} \Rightarrow \vec{AB} = -\vec{DC} so that the (opposite) sides AB and DC are parallel and of equal length. It follows that the quadrilateral ABDC is a parallelogram.
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