Farhan.Hanif93
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There weren't many solutions floating around so I thought why not? Papers are linked in the appropriate headings.

1992 Paper I:
A1: i - ii - iii - iv - v - vi - vii - viii - ix - x (Solutions by EEngWillow [i,iii], El Moro [iv])
B2 -
B3 - Solution by Farhan.Hanif93
B4 -
B5 -
B6 -
C7 - Solution by Farhan.Hanif93
C8 - Solution by Farhan.Hanif93
C9 - Solution by Farhan.Hanif93
C10 -
C11 -


1993 Paper I:
A1: i - ii - iii - iv - v - vi - vii - viii - ix - x (Solutions by Femto [i,vii,x], EEngWillow [v,vi], El Moro [ii,iii])
B2 - Solution by El Moro
B3 -
B4 - Solution by Femto
B5 -
B6 -
C7 - Solution by electriic_ink
C8 - Solution by electriic_ink
C9 - Solution by electriic_ink
C10 -
C11 -

1993 Paper II:
A1 -
A2 -
A3 -
A4 -
B5 - Solution by Farhan.Hanif93
B6 -
B7 -
B8 -
C9 -
C10 -
C11 -
C12 -
D13 -
D14 -
D15 -
D16 -

1994 Paper I:
A1: i - ii - iii - iv - v - vi - vii - viii - ix - x (Solutions by ben-smith)
B2 - Solution by ben-smith
B3 - Solution by ben-smith
B4 - Solution by ben-smith
B5 - Solution by ben-smith
B6 - Solution by desy.kris
C7 -
C8 -
C9 -
C10 -
C11 -

1995 Paper I:
A1: i - ii - iii - iv - v - vi - vii - viii - ix - x (Solutions by fGDu [i,ii,iii], dnumberwang [i,ii,iii,iv])
B2 -
B3 -
B4 -
B5 - Solution by twig
B6 -
C7 -
C8 - Solution by dnumberwang
C9 -
C10 -
C11 -

1995 Paper II:
A1 -
A2 - Solution by EEngWillow
A3 -
A4 -
B5 -
B6 -
B7 -
B8 -
C9 -
C10 -
C11 -
C12 -
D13 -
D14 -
D15 - Solution by EEngWillow
D16 -
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EEngWillow
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#2
(Original post by Farhan.Hanif93)
There weren't many solutions floating around so I thought why not? Papers are linked in the appropriate headings.

1992 Paper I:
Solutions
A1: i - ii - iii - iv - v - vi - vii - viii - ix - x
B2 -
B3 -
B4 -
B5 -
B6 -
C7 -
C8 -
C9 -
C10 -
C11 -


1993 Paper I:
Solutions
A1: i - ii - iii - iv - v - vi - vii - viii - ix - x
B2 -
B3 -
B4 -
B5 -
B6 -
C7 -
C8 -
C9 -
C10 -
C11 -


1993 Paper II:
Solutions
A1 -
A2 -
A3 -
A4 -
B5 -
B6 -
B7 -
B8 -
C9 -
C10 -
C11 -
C12 -
D13 -
D14 -
D15 -
D16 -


1994 Paper I:
Solutions
A1: i - ii - iii - iv - v - vi - vii - viii - ix - x
B2 -
B3 -
B4 -
B5 -
B6 -
C7 -
C8 -
C9 -
C10 -
C11 -


1995 Paper I:
Solutions
A1: i - ii - iii - iv - v - vi - vii - viii - ix - x
B2 -
B3 -
B4 -
B5 -
B6 -
C7 -
C8 -
C9 -
C10 -
C11 -


1995 Paper II:
Solutions
A1 -
A2 -
A3 -
A4 -
B5 -
B6 -
B7 -
B8 -
C9 -
C10 -
C11 -
C12 -
D13 -
D14 -
D15 -
D16 -
Unless you feel like printing these off and running them down to Manor Royal before I leave work in about 15 minutes, I'll have to look at them when I get home. File-sharing sites are blocked here
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Farhan.Hanif93
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1992 I - C8
(a)
I_1 = \displaystyle\int_0^{\frac{\pi}{  4}} \dfrac{2\sin ^2x}{2\sin x \cos x} dx = \displaystyle\int_0^{\frac{\pi}{  4}} \tan x dx = [-\ln (\cos x)]^{\frac{\pi}{4}}_0 = \boxed{\dfrac{1}{2}\ln 2}


(b)
Note that I_{2r+1}-I_{2r-1} = \displaystyle\int_0^{\frac{\pi}{  4}} \dfrac{\cos (4rx - 2x) - \cos (4rx+2x)}{\sin 2x} dx

=\displaystyle\int_0^{\frac{\pi}  {4}} \dfrac{2\sin 4rx\sin 2x}{\sin 2x} dx

=\left[-\dfrac{\cos 4rx}{2r}\right]^{\frac{\pi}{4}}_0

=\dfrac{1-cos (\pi r x)}{2r}

Depending on whether r is even or odd, \cos (\pi rx) = (-1)^r

Hence \boxed{I_{2r+1}-I_{2r-1} = \dfrac{1-(-1)^r}{2r}}, as required.

Note that I_{2r}-I_{2r-2} = \displaystyle\int^{\frac{\pi}{4}  }_0 \dfrac{\cos ((4r-2)x - 2x) - \cos ((4r-2)x + 2x)}{\sin 2x} dx

=\displaystyle\int^{\frac{\pi}{4  }}_0 \dfrac{2\sin (4r-2)x \sin 2x}{\sin 2x} dx

=\left[-\dfrac{1}{2r-1}\cos (4r-2)x \right]^{\frac{\pi}{4}}_0

\implies \boxed{I_{2r} - I_{2r-2} =\dfrac{1}{2r-1}}, as required.


(c)
I_8 = \dfrac{1}{7} + I_6 = \dfrac{1}{7} + \dfrac{1}{5}  + \dfrac{1}{3} + I_2 = \dfrac{1}{7} + \dfrac{1}{5}  + \dfrac{1}{3}  +1 = \boxed{\dfrac{176}{105}}

I_9 = I_7 = \dfrac{1}{3} + I_5 = \dfrac{1}{3} + I_3 = \dfrac{1}{3} + 1 + I_1 = \boxed{\dfrac{4}{3}+ \dfrac{1}{2}\ln 2}

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Farhan.Hanif93
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(Original post by EEngWillow)
Unless you feel like printing these off and running them down to Manor Royal before I leave work in about 15 minutes, I'll have to look at them when I get home. File-sharing sites are blocked here
No chance. :p:

You'll have to try them when you get back unfortunately.
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EEngWillow
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(Original post by Farhan.Hanif93)
No chance. :p:

You'll have to try them when you get back unfortunately.
Thought as much. Oh well, guess I'll have to read the newest Dexter novel on the bus home instead. What. A. Shame. :P
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Farhan.Hanif93
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1992 I - B3
(a)
\boxed{\cos 2\theta = 2\cos ^2 \theta -1}


(b)
\cos \theta > 0 over \left(0,\dfrac{\pi}{2}\right)


(c)
Note that \cos \left(\dfrac{\pi}{4}\right) = \cos \left(2\times \dfrac{\pi}{8}\right) = 2\cos ^2 \left(\dfrac{\pi}{8}\right) - 1
\implies \cos \left(\dfrac{\pi}{8}\right) = \sqrt{\dfrac{\frac{1}{\sqrt 2} + 1}{2}} (positive root since \cos (\frac{\pi}{8})>0)
\implies \boxed{2\cos \left(\dfrac{\pi}{8}\right) = \sqrt{2+\sqrt 2}}}, as required.


(d)
Note that \cos 4\theta =0 when \theta = \dfrac{\pi}{8}. Let c=\cos \theta and note that \cos 4\theta = 2\cos ^2(2\theta) - 1 = 8c^4-8c^2+1

Since this quartic has a root c=\cos (\frac{\pi}{8}), we can introduce a new variable x=2c which will yield a polynomial with one of it's roots as 2\cos (\frac{\pi}{8}), namely:
\boxed{x^4-4x^2+2=0}


(e)
Note that x^4-4x^2 +2 =0 \implies (x^2-(\sqrt 2+2))(x^2+(\sqrt 2 -2))=0 \implies \boxed{x=\pm \sqrt{2+\sqrt 2}, \pm \sqrt{2-\sqrt 2}}


(f)
Note that \cos 4\phi = 0 \iff  \boxed{\phi = (2n+1)\dfrac{\pi}{8}: n\in \mathbb{Z}}, which is also the solution set for f(x)=0.


(g)
Taking the principle values only, f(x)=0 \implies x= 2\cos \left(\dfrac{\pi}{8}\right) or 2\cos \left(\dfrac{4\pi}{8}\right) or 2\cos \left(\dfrac{5\pi}{8}\right) or 2\cos \left(\dfrac{7\pi}{8}\right).

Given that 2\cos \left(\dfrac{\pi}{8}\right) > 2\cos \left(\dfrac{3\pi}{8}\right)>2\c  os \left(\dfrac{5\pi}{8}\right) > 2\cos \left(\dfrac{7\pi}{8}\right), it follows that \boxed{\sqrt{2+\sqrt 2} = 2\cos \left(\dfrac{\pi}{8}\right), \sqrt{2-\sqrt 2} = 2\cos \left(\dfrac{3\pi}{8}\right), -\sqrt{2-\sqrt 2} = 2\cos \left(\dfrac{5\pi}{8}\right), -\sqrt{2+\sqrt 2} = 2\cos \left(\dfrac{7\pi}{8}\right)}

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Dirac Spinor
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1994 Paper 1 Section A
i)
Spoiler:
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d/dx[sin(cos(x^2))]=-sin(x^2)cos(cos(x^2))2x

ii)
Spoiler:
Show


 \dfrac{1}{(x+1)(x^2+3)} =\frac{a}{x+1}+\frac{bx+c}{x^2+3  }=\dfrac{bx^2+bx+cx+c+ax^2+3a}{(  x+1)(x^2+3)}

a+b=0

b+c=0

c+3a=1

\Rightarrow a=1/4,b=-1/4,c=1/4

\therefore \frac{1}{(x+1)(x^2+3)}=\frac{1}{  4(x+1)}+\frac{1-x}{4(x^2+3)}

iii)
Spoiler:
Show


\displaystyle \int^e_1(1+x)lnxdx=\displaystyle \int^e_1D[x+x^2/2]lnxdx

=[lnx(x+x^2/2)]^e_1-\displaystyle \int^e_1(x+x^2/2)1/xdx

=e+e^2/2-\displaystyle \int^e_11+x/2dx=e+e^2/2-[x+x^2/4]^e_1

=(e^2+5)/4

iv)
Spoiler:
Show


\dfrac{1}{4n^2-1}=\frac{1}{2}(\frac{1}{2n-1}-\frac{1}{2n+1})

\therefore 1/3+1/(15)+1/(35)+...+1/(4n^2-1)=\frac{1}{2}(1-\frac{1}{2n+1})

=\dfrac{n}{2n+1}

v)
Spoiler:
Show


Using the change of base law:

xyz=log_ab*log_bc*log_ca=log_ab*  \frac{log_ac}{log_ab}*\frac{log_  aa}{log_ac}=log_aa=1

vi)
Spoiler:
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y=x^3-3x+1 \Rightarrow y'3x^2-3

y'=0 \Rightarrow x=\pm 1

y(\pm 1)=-1,3
Checking end points:


y(-3/2)=17/8

y(5/2)=73/8

\therefore Max=73/8, min=-1

vii)
Spoiler:
Show


\mathbf{p}=k\begin{pmatrix} 3 \\ -1 \end{pmatrix}, \mathbf{q}=\begin{pmatrix} q_1 \\ q_2 \end{pmatrix}

\mathbf{q}.\mathbf{a}=0 \Rightarrow 3q_1-q_2=0

\mathbf{b}=\mathbf{p}+\mathbf{q} \Rightarrow \begin{pmatrix} 1 \\ -3 \end{pmatrix}=\begin{pmatrix} 3k+q_1 \\ -k+q_2 \end{pmatrix}

\Rightarrow  q_2=-3+k \Rightarrow q_1=\frac{k-3}{3}

\therefore 1=3k+\frac{k-3}{3} \Rightarrow k=0.6 \Rightarrow q_1=-0.8 \Rightarrow q_2=-2.4

viii)
Spoiler:
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x^2-2x+y^2+4y-4=0

\Rightarrow (x-1)^2-1+(y+2)^2-4-4=0

\Rightarrow (x-1)^2+(y+2)^2=9
So centre (1,-2) and radius 3

ix)
Spoiler:
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d=\sqrt{(t^2)^2+(t-3)^2}=\sqrt{t^4+t^2-6t+9}

\therefore D[d]=1/2(t^4+t^2-6t+9)^{-1/2}(4t^3+2t-6)=1/2(t^4+t^2-6t+9)^{-1/2}(t-1)(2t^2+2t+3)
So minimum distance occurs at t=1 which means the minimum distance is \sqrt5

x)
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considery=k \in \mathbb{R}

(k-a)x^2+k\Rightarrow x^2=\frac{-k}{k-a}
Which has no roots for k greater than or equal to a. This means that we have a horizontal asymptote y=a. The graph looks like y=x^2 close to the origin but, as x tends to infinity or -ve infinity, y tends to a
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EEngWillow
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Sadly run out of time midquestion for now, but some of them are done:

1992 Paper 1, question 1.

i)
Spoiler:
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\dfrac{d}{dx} sin(asin^{-1}x) use chain rule with u = asin^{-1}x

\dfrac{du}{dx} = \dfrac{a}{\sqrt{1-x^2}} and \dfrac{dy}{du} = cosu

\dfrac{dy}{dx} = \dfrac{dy}{du} \dfrac{du}{dx} = \dfrac{acos(asin^{-1}x)}{\sqrt{1-x^2}}


iii)
Spoiler:
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\displaystyle\int_0^4 \dfrac{dx}{1 + \sqrt{x}} let u = 1 + \sqrt{x}

\displaystyle\int_0^4 \dfrac{dx}{1+\sqrt{x}} = 2\displaystyle\int_1^3 \dfrac{u-1}{u}du = 2\displaystyle\int_1^3 1 - \frac{1}{u}du = 2\left[u - ln|u|\right]^3_1

2\left[u - ln|u|\right]^3_1 = 2\left[(3 - ln(3)) - (1 - ln(1))\right] = 4 - ln(9)


Think those are right; I'll look at the rest in a bit.
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electriic_ink
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Contributing 1993 I C7

part 0

Let I := \displaystyle \int_0^a f(t) \  \mathrm{d}t = \displaystyle \int_0^{a/2} f(t) \ \mathrm{d}t + \displaystyle \int_{a/2}^a f(t) \ \mathrm{d}t

Using the linear substitution  t = a - u, \displaystyle \int_{a/2}^a f(t) \ \mathrm{d}t becomes \displaystyle \int_0^{a/2} f(a-u) \ \mathrm{d}u

So I = \displaystyle \int_0^{a/2} f(t) \ \mathrm{d}t +  \displaystyle \int_0^{a/2} f(a-u) \ \mathrm{d}u = \displaystyle \int_0^{a/2} f(x)  +  f(a-x) \ \mathrm{d}x


part 1

Define  J := \displaystyle \int_0^{\pi} x \sin ^3 x \ \mathrm{d}x = \displaystyle \int_0^{\pi/2} x \sin ^3 x + (\pi - x) \sin ^3 (\pi -x) \ \mathrm{d}x

Because \sin (\pi - x) = \sin x, J=\displaystyle \pi \int_0^{\pi/2}  \sin ^3 x \ \mathrm{d}x

De Moivre's Theorem shows us that  4 \sin^3 x = 3 \sin x - \sin 3x. So  J = \dfrac{\pi}{4}\displaystyle \int_0^{\pi/2} 3 \sin x - \sin 3x \ \mathrm{dx} = \dfrac{\pi}{4}\left (3 - \dfrac{1}{3} \right ) = \dfrac{2\pi}{3}


part 2

This is a nightmare. Let K be the integral given in the question. By the result in part b):

 K = \pi \displaystyle \int_{0}^{\pi/2} \dfrac{\sin x}{\sqrt{4 - \cos ^2 x}} \ \mathrm{d} x.

Substitute  u = \cos x to get:

 K = \pi \displaystyle \int_{0}^{1} \dfrac{1}{\sqrt{4 - u^2}} \ \mathrm{d} u.

And then  u = 2 \sin t to get:

 K = \pi \displaystyle \int_{0}^{\pi/6} 1 \ \mathrm{d} t.


Hopefully these are all right, although please correct if not.
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electriic_ink
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Contributing 1993 I C8

part a

\tan ^{93}x
2x \tan ^{93}(x^2)
2x \tan ^{93}(x^2) - \tan ^{93} x

(Generally, \dfrac{\mathrm{d}}{\mathrm{d} x} \displaystyle \int_{g(x)}^{h(x)} f(t) \ \mathrm{d}t = h'(x)f(h(x)) - g'(x)f(g(x)))


part b

\ln (\ln x) = \displaystyle \int_{1}^{\ln x} \dfrac{1}{t} \ \mathrm{d} t

So \dfrac{\mathrm{d}}{\mathrm{d}x} \ln (\ln x) = \dfrac{\mathrm{d}}{\mathrm{d}x} \displaystyle \int_{1}^{\ln x} \dfrac{1}{t} \ \mathrm{d} t = \dfrac{1}{x \ln x}


For the parts question, pick v' = \dfrac{1}{t} and u=\ln (\ln t) to get:

\displaystyle \int_{2}^{x} \dfrac{\ln (\ln t)}{t} \ \mathrm{d} t = \left [\ln(t) \ln (\ln t) - \ln t \right ]^x_2 = (\ln x)(\ln (\ln x) - 1) -  (\ln 2)(\ln (\ln 2) - 1)
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dnumberwang
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1995 I C8

a
Same as STEP II 2006, Q4.

\displaystyle\int_0^{\pi} x f(\sin x) \ dx

\displaystyle y = \pi - x \implies \frac{dy}{dx} = -1 \implies dy = -dx

\displaystyle\int_0^{\pi} x f(\sin x) \ dx = \int_0^{\pi} (\pi - y) f(\sin (\pi-y)) \ dy


\sin x = \sin (\pi - x)

\displaystyle \int_0^{\pi} (\pi - y) f(\sin (\pi-y)) \ dy = \int_0^{\pi} \pi f(\sin y) \ dy - \int_0^{\pi} y f(\sin y) \ dy

rearranging gives
\displaystyle\int_0^{\pi} x f(\sin x) \ dx + \int_0^{\pi} y f(\sin y) \ dy = \pi \int_0^{\pi}  f(\sin y) \ dy



\displaystyle\int_0^{\pi} x f(\sin x) \ dx = \int_0^{\pi} y f(\sin y) \ dy

 \displaystyle 2\int_0^{\pi} x f(\sin x) \ dx = \pi \int_0^{\pi}  f(\sin x) \ dx


b
Using the first part,  \displaystyle \int_0^{\pi} x \sin^3 x \ dx = \frac{\pi}{2} \int_0^{\pi} \sin^3 x \ dx

\displaystyle\frac{\pi}{2} \int_0^{\pi} \sin^3 x \ dx = \frac{\pi}{2} \int_0^{\pi} \sin x  \times \sin^2 x \ dx =  \frac{\pi}{2} \int_0^{\pi} \sin x (1 - \cos^2 x) \ dx





= \displaystyle \frac{\pi}{2} \int_b^a \sin x (1 - \cos^2 x) \ \frac{dx}{du} \ du

u = \cos x \implies \displaystyle \frac{du}{dx} = - \sin x \implies \frac{dx}{du} = \frac {-1}{\sin x}

\displaystyle \frac{\pi}{2} \int_b^a \sin x (1 - \cos^2 x) \ \frac{dx}{du} \ du = \frac{\pi}{2} \int_{-1}^{1} 1 - u^2 \ du = \left[u- \frac{u^3}{3}\right]_{-1}^1 = \frac{2}{3} \pi
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fGDu
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1995 I

Please correct me if these are wrong. I have done up to B4 so I'll edit my post when I have time.

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(i)

\frac{2a}{x^2-a^2}=\frac{A}{x-a}+\frac{B}{x+a}



B( x - a)+A( x + a) = 2a



Let x = a



A(2a) = 2a



A = 1 



Let x = -a



B(-2a) = 2a



B = -1


(ii)

 a = \begin{pmatrix} -1\\2\end{pmatrix}



b = \begin{pmatrix} 5\\5\end{pmatrix}



b=ta+p

\begin{pmatrix} 5\\5\end{pmatrix} = t\begin{pmatrix} -1\\2\end{pmatrix}+p





Let p = \begin{pmatrix} p_1\\p_2\end{pmatrix}


p is perpendicular to a, so

a\bullet p=0



\begin{pmatrix} -1\\2\end{pmatrix}\begin{pmatrix} p_1\\p_2\end{pmatrix}=0



-1(p_1)+2(p_2)=0



-p_1+2p_2=0



p_1=2p_2



if p_1=2, p_2=1


p = x\begin{pmatrix} 2\\1\end{pmatrix}



\begin{pmatrix} 5\\5\end{pmatrix} = t\begin{pmatrix} -1\\2\end{pmatrix} + x\begin{pmatrix} 2\\1\end{pmatrix}

Equating components, we have
5 = -t + 2x

t = 2t + x



15 = 5x



x = 3



t = 1



so p = 3\begin{pmatrix} 2\\1\end{pmatrix}

(iii)






2\lvert x \rvert+\lvert x-3 \rvert\geq6

when

x\leq0



\lvert x \rvert = -x

\lvert x-3 \rvert = -x+3



-2x-x+3\geq6

-3x\geq3

x\leq-1

when

0 < x < 3



\lvert x \rvert = x

\lvert x-3 \rvert = -x+3



2x-x+3\geq6

x+3\geq6

x\geq3

when
x\geq3



\lvert x \rvert = x

\lvert x-3 \rvert = x-3



2x+x-3\geq6

3x\geq9

x\geq3

so either x\leq-1
or x\geq3


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Farhan.Hanif93
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(Original post by fGDu)
...
They're fine. Although I'd appreciate it if you posted your answers to the longer questions in separate posts. That way, if anyone came looking for them in future, they wouldn't have to scroll through hundreds of other solutions first. :p: It's fine to keep all of your section A questions in one post, though.

Been a bit busy today, will probably get some up tomorrow.
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fGDu
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(Original post by Farhan.Hanif93)
They're fine. Although I'd appreciate it if you posted your answers to the longer questions in separate posts. That way, if anyone came looking for them in future, they wouldn't have to scroll through hundreds of other solutions first. :p: It's fine to keep all of your section A questions in one post, though.

Been a bit busy today, will probably get some up tomorrow.
I will do, I've had a big malfunction with the modulus sign in (iii) so I'll try to sort that out.
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dnumberwang
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1995 I, A1

EDIT: spent too long faffing about with latex

(i)
\displaystyle \frac{2a}{x^2-a^2} = \frac{2a}{(x+a)(x-a)} = \frac{m}{x+a} + \frac{n}{x-a}
m (x-a) + n (x+a) = 2a
(m+n)x + (n-m)a = 2a
n = 1, m = -1
\displaystyle \frac{2a}{x^2-a^2} = \frac{1}{x-a} - \frac{1}{x+a}


(ii)

\begin{pmatrix} 5 \\ 5 \end{pmatrix} = t \begin{pmatrix} -1 \\ 2 \end{pmatrix} + \begin{pmatrix} x \\ y \end{pmatrix}

p perpendicular to a, so -x + 2y = 0, x = 2y

\begin{pmatrix} 5 \\ 5 \end{pmatrix} = \begin{pmatrix} -t \\ 2t \end{pmatrix} + \begin{pmatrix} 2y \\ y \end{pmatrix}

2y - t = 5 and 2y + t = 5

t = 1 and y = 3, p = \begin{pmatrix} 6 \\ 3 \end{pmatrix}


(iii)
When x \geq 3, f(x) = 2x + x - 3 = 3x + 3
f(x) \geq 6 for x \geq 3

When 0 < x < 3, f(x) = 2x + 3 - x = x + 3
f(x) \not\geq 6 between 0 < x < 3

When x \leq 0, f(x) = -2x + 3 - x = 3 - 3x
f(x) \geq 6 when x \leq -1

2|x| + |x-3| \geq 6 when x \geq 3 or x \leq -1


(iv)
as n is even: \displaystyle\sum_{r=1}^{n} r (-1)^r  = -1 + 2 - 3 + 4 - \cdots - (n-1) + n

= (2-1) + (4-3) + \cdots + [n-(n-1)] = 1 +1 +1 + \cdots =  \displaystyle \frac{n}{2}
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Farhan.Hanif93
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(Original post by fGDu)
I will do, I've had a big malfunction with the modulus sign in (iii) so I'll try to sort that out.
What LaTeX tags are you using? I think TSR's native one will respond slightly better.
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fGDu
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Dw, Dnumberwang has much neater solutions than mine so use his.
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Farhan.Hanif93
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#18
1992 I - C7
Note that z=\dfrac{y-x}{y+x} \implies \dfrac{dz}{dx} = \dfrac{(y+x)(\frac{dy}{dx}-1) - (y-x)(\frac{dy}{dx}+1)}{(y+x)^2} = \boxed{\dfrac{2}{(x+y)^2}\left(x  \dfrac{dy}{dx}-y\right)}, as required.

For the next part, note that f(x)\left(x\dfrac{dy}{dx}-y\right) = y^2-x^2 \implies f(x)\times \dfrac{2}{(x+y)^2}\left(x\dfrac{  dy}{dx}-y\right) = \dfrac{2(y-x)}{y+x}

Let z=\dfrac{y-x}{y+x} \implies \dfrac{dz}{dx} = \dfrac{2}{(x+y)^2}\left(x\dfrac{  dy}{dx}-y\right). Hence our DE becomes:
\boxed{f(x)\dfrac{dz}{dx} = 2z}, as required.

Considering (x^3+x^2+x+1)\left(x\dfrac{dy}{d  x}-y\right)=(x-1)(y^2-x^2) (\star).

\implies \dfrac{(x+1)(x^2+1)}{x-1} \times \dfrac{2}{(y+x)^2}\left(x\dfrac{  dy}{dx}-y\right) = \dfrac{2(y-x)}{y+x}

By the above result, setting z=\dfrac{y-x}{y+x} gives:
(\star): \dfrac{(x+1)(x^2+1)}{x-1}\dfrac{dz}{dx} = 2z

\implies \displaystyle\int \dfrac{1}{2z}dz = \displaystyle\int \dfrac{x-1}{(x+1)(x^2+1)}dx

Consider the values of the constants A, B and C such that:
 \dfrac{x-1}{(x+1)(x^2+1)} \equiv \dfrac{A}{x+1}+\dfrac{Bx+C}{x^2+  1}

\implies x-1 = A(x^2+1)+(Bx+C)(x+1)

Setting x=-1 \implies  A=-1
Setting x=0 \implies C=0
Setting x=1 \implies B=1

Hence (\star): \dfrac{1}{2}\ln |z| = \displaystyle\int \left(\dfrac{x}{x^2+1} - \dfrac{1}{x+1}\right) dx = \dfrac{1}{2}\ln (x^2+1) - \ln |x+1| + C

\implies \ln z = \ln \dfrac{k^2(x^2+1)}{(x+1)^2}} where C=\ln k

\implies \dfrac{y-x}{y+x} = \dfrac{k^2(x^2+1)}{(x+1)^2}}

y(1)=3 \implies  k=1

\implies y\left(1- \dfrac{(x^2+1)}{(x+1)^2}\right) = x\left(1 + \dfrac{(x^2+1)}{(x+1)^2}\right)

\implies \boxed{y=x^2+x+1}
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Farhan.Hanif93
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#19
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#19
1992 I - C9
First part
Let I = \dfrac{1}{2}\displaystyle\int^{n  +1}_n (x-n)(n+1-x)f''(x)dx

Consider I for integration by parts with u=\dfrac{1}{2} (x-n)(n+1-x) \implies \dfrac{du}{dx} = n-x + \dfrac{1}{2}, \dfrac{dv}{dx} = f''(x) \implies v=f'(x):
\implies I=-\displaystyle\int^{n+1}_n (n-x +\frac{1}{2})f'(x)dx

Consider I for IBP again with u=-n+x -\dfrac{1}{2} \implies \dfrac{du}{dx}=1, \dfrac{dv}{dx}=f'(x) \implies v= f(x):
\implies  \boxed{\dfrac{1}{2}\displaystyle  \int^{n+1}_n (x-n)(n+1-x)f''(x)dx = \dfrac{1}{2}[f(n)+f(n+1)] - \displaystyle\int_n^{n+1}f(x)dx} as required.

Second part
Note that f''(x)\geq 0 and (x-n)(n+1-x)\geq 0 \iff (x-n)(n+1-x)f''(x)\geq 0. From a sketch of any function that is non-negative over the interval of integration, it follows that I\geq 0.

Also, f''(x)\geq 0 and  (x-n)(n+1-x)f''(x) \leq \dfrac{1}{4} \iff I \leq \dfrac{1}{2}\displaystyle\int ^{n+1}_n \dfrac{1}{4}f''(x)dx = \dfrac{1}{8}[f'(n+1)-f'(n)]

It follows that \boxed{0\leq \dfrac{1}{2}[f(n)+f(n+1)] - \displaystyle\int_n^{n+1}f(x)dx \leq \dfrac{1}{8}[f'(n+1)-f'(n)]} (\star), as was to be shown.

Third part
Note that the inequality (\star) yields:

\displaystyle\sum_{n=1}^{N-1} 0 \leq \dfrac{1}{2}\displaystyle\sum_{n  =1}^{N-1} [f(n)+f(n+1)] - \displaystyle\sum_{n=1}^{N-1}\displaystyle\int_n^{n+1}f(x)d  x  \leq \dfrac{1}{8}\displaystyle\sum_{n  =1}^{N-1}[f'(n+1)-f'(n)]

Note that the LHS sum is clearly zero and

\dfrac{1}{2}\displaystyle\sum_{n  =1}^{N-1} [f(n)+f(n+1)] = \dfrac{1}{2}[f(1)+2f(2)+2f(3)+...+2f(N-1) + f(N)]

= \dfrac{1}{2}[2f(1)+2f(2)+...+2f(N) - f(1) - f(N)]

= \dfrac{1}{2}\displaystyle\sum_{n  =1}^{N}[2f(n)] - \dfrac{1}{2}[f(1)+f(N)]

Also, \displaystyle\sum_{n=1}^{N-1} \displaystyle\int_n^{n+1}f(x)dx =  \displaystyle\int_1^{2}f(x)dx + \displaystyle\int_2^{3}f(x)dx +...+ \displaystyle\int_{N-2}^{N-1}f(x)dx + \displaystyle\int_{N-1}^{N}f(x)dx
=\displaystyle\int_1^{N}f(x)dx

Finally, the sum on the RHS telescopes to f'(N)-f'(1).

It follows that \boxed{0\leq \displaystyle\sum_{1}^N f(n) - \dfrac{1}{2}[f(N)+f(1)] - \displaystyle\int_1^{N}f(x)dx \leq \dfrac{1}{8}[f'(N)-f'(1)]}, as required.

Final part
Let f(x)=\dfrac{1}{x}. Note that f''(x)=\dfrac{2}{x^3}\geq 0; \forall x \in \mathbb{N}. Hence it follows that f(x)=\dfrac{1}{x} can be plugged into our inequality from part 3 to give:

0\leq \displaystyle\sum_{1}^N \dfrac{1}{n} - \dfrac{1}{2}\left(\dfrac{1}{N}+1  \right) - [\ln x]^N_1 \leq \dfrac{1}{8}\left(1-\dfrac{1}{N^2}\right)

\iff \boxed{0 \leq \displaystyle\sum_{1}^N \dfrac{1}{n} - \ln N \leq \dfrac{5}{8} + \dfrac{1}{2N}-\dfrac{1}{8N^2}}, as desired.

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Farhan.Hanif93
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#20
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#20
1993 II - B5
Note that the stone descends a vertical height of OA=l\sin \alpha. Let v be the stone's speed at B. By the conservation of energy, \dfrac{1}{2}mv^2=mgl\sin \alpha
\boxed{\implies v=\sqrt{2gl\sin \alpha}}, as required.

Let t_1 be the time is takes for the stone to slide from A to B. Using s=\left(\dfrac{u+v}{2}\right)t with u=0, v=(2gl\sin \alpha)^{\frac{1}{2}}, s=l, t=t_1:

\boxed{t_1= 2\left(\dfrac{l}{2g\sin \alpha}\right)^{\frac{1}{2}}}

Let t_2 represent the time it takes for the stone to go from B to C such that the total time taken for the stone to go from A to C, T, is given by T=t_1+t_2.

Note that the horizontal component of the stone's velocity is given by (2gl\sin \alpha)^{\frac{1}{2}}\cos \alpha. Also, BC=l(1-\cos \alpha), hence:

v=\dfrac{s}{t} \implies t_2 =  \dfrac{l(1-\cos \alpha)}{(2gl\sin \alpha)^{\frac{1}{2}}\cos \alpha} = \left(\dfrac{l}{2g\sin \alpha}\right)^{\frac{1}{2}} \left(\dfrac{1}{ \cos \alpha} -1\right)

\implies T= \left(\dfrac{l}{2g\sin \alpha}\right)^{\frac{1}{2}} \left(2+\dfrac{1}{\cos \alpha} -1\right) = \boxed{\left(\dfrac{l}{2g\sin \alpha}\right)^{\frac{1}{2}} \left(1+\dfrac{1}{\cos \alpha}\right)}, as required.

By the product rule, \dfrac{dT}{d\alpha} =-\dfrac{1}{2}(\cos \alpha + 1)\sqrt{\dfrac{l}{2g\sin ^3\alpha}} + \sqrt{\dfrac{l}{2g\sin \alpha}}(\sec \alpha \tan \alpha)

\dfrac{dT}{d\alpha} =0 \implies \sec \alpha \tan \alpha - \dfrac{\cos \alpha + 1}{2\sin \alpha}=0

\implies 2\sin ^2 \alpha - \cos ^3 \alpha - \cos ^2 \alpha =0

\implies \cos ^3\alpha +3\cos ^2\alpha - 2=0

\implies (\cos \alpha +1)(\cos ^2 \alpha + 2\cos \alpha -2)=0

\implies \cos \alpha = -1 or \cos \alpha = -1\pm \sqrt 3.

Note that \alpha is acute, hence \cos \alpha \not=0 and 0< \cos \alpha < 1 \implies \cos \alpha \not = -1-\sqrt 3.

It follows that T takes an extreme value when \boxed{\cos \alpha = \sqrt 3 -1}, as required.

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