STEP I, II, III 2002 Solutions
Maths exam discussion - share revision tips in preparation for GCSE, A Level and other maths exams and discuss how they went afterwards.
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STEP I, II, III 2002 SolutionsSTEP I:
1: Solution by Unbounded
2: Solution by Unbounded
3: Solution by nota bene
4: Solution by SimonM
5: Solution by nota bene
6: Solution by Unbounded
7: Solution by sonofdot
8: Solution by Unbounded
9: Solution by cliverlong
10: Solution by Dadeyemi
11:Solution by Unbounded
12: Solution by Robbie10538
13: Solution by Unbounded
14: Solution by brianeverit
STEP II:
1: Solution by sonofdot
2: Solution by dadeyemi
3: Solution by sonofdot
4: Solution by dadeyemi
5: Solution by dadeyemi
6: Solution by welshenglish
7: Solution by Glutamic Acid
8: Solution by Glutamic Acid
9: Solution by tommm
10: Solution by tommm
11: Solution by tommm
12: Solution by brianeverit
13: Solution by brianeverit
14: Solution by brianeverit
STEP III:
1: Solution by SimonM
2: Solution by Dadeyemi
3: Solution by Daniel Freedman
4: Solution by Dadeyemi
5: Solution by Dadeyemi
6: Solution by Elongar
7: Solution by Dadeyemi
8: Solution by Dadeyemi
9: Solution by brianeverit
10: Solution by brianeverit
11: Solution by brianeverit
12: Solution by brianeverit
13: Solution by brianeverit
14: Solution by brianeverit
Solutions written by TSR members:
1987 - 1988 - 1989 - 1990 - 1991 - 1992 - 1993 - 1994 - 1995 - 1996 - 1997 - 1998 - 1999 - 2000 - 2001 - 2002 - 2003 - 2004 - 2005 - 2006 - 2007Last edited by SimonM; 26-06-2011 at 20:45. -
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Re: STEP I, II, III 2002 SolutionsSTEP III, Question 1
a)
Spoiler:Show
![\displaystyle \int^{a}_{1} \frac{ln(x)}{x} dx = \left(\frac{[ln(x)]^2}{2}\left)^{a}_{1} = \frac{[ln(a)]^2}{2}](http://www.thestudentroom.co.uk/latexrender/pictures/73/73aa039824a69525039cf5b42cdf9695.png "\displaystyle \int^{a}_{1} \frac{ln(x)}{x} dx = \left(\frac{[ln(x)]^2}{2}\left)^{a}_{1} = \frac{[ln(a)]^2}{2}")
ln(a) tends to infinty as a tends to infinty;
b)
Spoiler:Show
Let
and it follows that the integral is transformed to;
And IBP twice yields;
As a tends to infinty the volume tends to
Last edited by Oh I Really Don't Care; 05-04-2009 at 01:16. -
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Re: STEP I, II, III 2002 SolutionsSTEP II 2002 Question 3
Spoiler:Show
If k=1, then
and 
, therefore it is true for k=1.
If k=2, then
and 
, therefore it is true for k=2.
If k=3, then
and 
, therefore it is true for k=3.
Assume, for some k
Therefore by induction, for all positive integers k.
Now consider
and 
with r<s. Assume that and have a common factor p
Since all Fermat numbers are odd, and 
cannot share a common factor greater than one, hence by contradiction, no two Fermat numbers have a common factor greater than 1.
All Fermat numbers are either prime or made up of prime factors. Since no Fermat numbers share a common factor greater than one, no Fermat numbers can have the same prime factor. Since there are infinitely many Fermat numbers, there must therefore be infinitely many prime numbers.Last edited by sonofdot; 03-04-2009 at 19:50. -
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Re: STEP I, II, III 2002 SolutionsQuestion 1, STEP I, 2002
Spoiler:ShowWe need to find the points of intersection between the two ellipses:
- ellipse 1
- ellipse 2
making y^2 the subject in the equation for ellipse 1:
and substitution into the equation for ellipse 2:
Subbing back into the equation for y^2:
Therefore the coordinates of intersection are (2,1) and (2,-1)
From a sketch, we can see that this means that the centre of a circle passing through those two points must lie on the x-axis.
Letting the coordinates of the centre of the circle be (a,0):
where r is the radius of the circle.
Substitute the point (2,1) in:
So the equation of the circle now becomes:
Last edited by Unbounded; 12-04-2009 at 09:09. -
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Re: STEP I, II, III 2002 SolutionsQuestion 6, STEP I, 2002
Let the height of the equilateral triangle base be x.
From pythag:
Therefore the area of this equilateral triangle base is given by:
Drawing a line down from the apex of the pyramid to the triangle base, letting h be the height, from pythagoras, we can see that:
The volume of the cone, V is given by,
The area of the new base is to be found:
The height of the isoceles base, y, is given by:
And so the area of this isoceles base is
Now
Let the new height of the pyramid be k
Considering the right angled triangle formed now with the height of this pyramid and the height of the equilateral triangle. Let
denote the angle between the equilateral triangle and the horizontal.
Last edited by Unbounded; 13-04-2009 at 04:21. -
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Re: STEP I, II, III 2002 SolutionsSTEP I 2002 Question 7
and 
with 
(Note that in the interval
, 
)
![\begin{array}{rl}
I+J & \displaystyle = \int_0^a \frac{\sin x + \cos x}{\sin x + \cos x} \, dx \\ \br \\
& \displaystyle = \int_0^a 1 \, dx \\ \br \\
& \displaystyle = \left[ x \right]_0^a = a \end{array}](http://www.thestudentroom.co.uk/latexrender/pictures/6d/6d5769791396694f7beccfa3bd36ace3.png "\begin{array}{rl}
I+J & \displaystyle = \int_0^a \frac{\sin x + \cos x}{\sin x + \cos x} \, dx \\ \br \\
& \displaystyle = \int_0^a 1 \, dx \\ \br \\
& \displaystyle = \left[ x \right]_0^a = a \end{array}")
![\begin{array}{rl}
I-J & \displaystyle = \int_0^a \frac{\cos x - \sin x}{\sin x + \cos x} \, dx \\ \br \\
& \displaystyle = \left[ \ln (\sin x + \cos x) \right]_0^a \\ \br \\
& \displaystyle = \ln (\sin a + \cos a)\end{array}](http://www.thestudentroom.co.uk/latexrender/pictures/79/79d14131b4413750d94662214312463e.png "\begin{array}{rl}
I-J & \displaystyle = \int_0^a \frac{\cos x - \sin x}{\sin x + \cos x} \, dx \\ \br \\
& \displaystyle = \left[ \ln (\sin x + \cos x) \right]_0^a \\ \br \\
& \displaystyle = \ln (\sin a + \cos a)\end{array}")
Let
(Note again that for
sin(x) and cos(x) are never negative, so psin(x) + qcos(x) is never equal to 0)
Let
![\begin{array}{rl}p^2 I_1 - pq J_1
& \displaystyle = p \int_0^{\frac{\pi}{2}} \frac{p\cos x - q\sin x}{p \sin x + q \cos x} \, dx \\ \br \\
& \displaystyle = p \left[ \ln (p \sin x + q\cos x) \right]_0^{\frac{\pi}{2}} \\ \br \\
& \displaystyle = p\ln \frac{p}{q}\end{array}](http://www.thestudentroom.co.uk/latexrender/pictures/5b/5b95b965fc68bb1c113f5973f9675d09.png "\begin{array}{rl}p^2 I_1 - pq J_1
& \displaystyle = p \int_0^{\frac{\pi}{2}} \frac{p\cos x - q\sin x}{p \sin x + q \cos x} \, dx \\ \br \\
& \displaystyle = p \left[ \ln (p \sin x + q\cos x) \right]_0^{\frac{\pi}{2}} \\ \br \\
& \displaystyle = p\ln \frac{p}{q}\end{array}")
Let
and let 
![\begin{array}{rl}3(3I_2 - 4J_2)
& \displaystyle = 3\int_0^{\frac{\pi}{2}} \frac{3\cos x+ 12 - 4\sin x - 12}{3\sin x + 4 \cos x +25} \, dx \\ \br \\
& \displaystyle = 3\int_0^{\frac{\pi}{2}} \frac{3\cos x - 4\sin x}{3\sin x + 4 \cos x +25} \, dx \\ \br \\
& \displaystyle = 3\left[ \ln ( 3\sin x + 4 \cos x + 25) \right]_0^{\frac{\pi}{2}} \\ \br \\
& \displaystyle = 3\ln \frac{28}{29}\end{array}](http://www.thestudentroom.co.uk/latexrender/pictures/e8/e8a36df41389fab88f894091bc8c274f.png "\begin{array}{rl}3(3I_2 - 4J_2)
& \displaystyle = 3\int_0^{\frac{\pi}{2}} \frac{3\cos x+ 12 - 4\sin x - 12}{3\sin x + 4 \cos x +25} \, dx \\ \br \\
& \displaystyle = 3\int_0^{\frac{\pi}{2}} \frac{3\cos x - 4\sin x}{3\sin x + 4 \cos x +25} \, dx \\ \br \\
& \displaystyle = 3\left[ \ln ( 3\sin x + 4 \cos x + 25) \right]_0^{\frac{\pi}{2}} \\ \br \\
& \displaystyle = 3\ln \frac{28}{29}\end{array}")
Last edited by sonofdot; 03-04-2009 at 19:54. -
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Re: STEP I, II, III 2002 SolutionsII/8:
For x < 0, dy/dx = -y ==> 1/y dy/dx = -1 ==> ln y = -x +c ==> y = Ae^(-x); as y = a when x = -1, a = Ae so A = a/e; therefore
for x < 0.
For > 0, dy/dx = y ==> 1/y dy/dx = 1 ==> ln y = x + c ==> y = Be^x; as y = b when x = 1, b = Be so B = b/e therefore
for x > 0.
If a = b, there is no jump.
Note that e^x - 1 is positive if x > 0, and negative if x < 0.
for x > 0.
Given that y = e^e when x = 1, e^e = Ae^(e - 1) so A = e;
for x < 0.
At x = 0, y =
using the solution for x > 0. So for continuity, this must the same result when substituted into x < 0, so
(i)
(ii)
.
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Re: STEP I, II, III 2002 SolutionsII/7: (Scary scary vectors.)
Let the lines have direction vector
which I'll denote by r, and without loss of generality let |r| = 1 so a^2 + b^2 + c^2 = 1;
.
b = 1 - a; c = 1 - a so
The solutions are a =1, b = 0, c = 0 and a = 1/3, b = 2/3, c = 2/3.
m3 has direction
; m4 direction 
.
Considering
gives 
.
(i) A has position vector
; B 
.
Let P =, and Q =
AQ . BP = 0 so
. Showing that the latter is > 0; 1 - sqrt(6)/3 > 0 therefore 1 > sqrt(6)/3 therefore 1 > 6/9, which is true.
(ii)
So
The second equation gives
. Substituting into the third gives 
. Substituting these into the first:
; multiplying out gives 
, so no non-zero solutions. -
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Re: STEP I, II, III 2002 SolutionsQuestion 2, STEP I, 2002
At stationary points, f'(x) = 0
which is the point that satisfies the conditions.
Now the first and third brackets are always positive. The middle bracket alternates signs. When n is even, the middle bracket is negative, which implies f''(x) < 0 which implies it is a maximum. When n is odd, the middle bracket is positive, which implies it is a minimum. Q.E.D.Last edited by Unbounded; 05-04-2009 at 01:18. -
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Re: STEP I, II, III 2002 Solutions2002 II Question 11
Total weight (to check our calculus)![\displaystyle = \int_0^l \alpha Wl^{-1}(x/l)^{\alpha-1}\,\mathrm{d}x = aWl^{-\alpha}\left[\frac{1}{\alpha}x^\alpha \right]_0^l = Wl^{-\alpha}l^\alpha = W](http://www.thestudentroom.co.uk/latexrender/pictures/e8/e80711111f6c7935c3741c11ced7294e.png "\displaystyle = \int_0^l \alpha Wl^{-1}(x/l)^{\alpha-1}\,\mathrm{d}x = aWl^{-\alpha}\left[\frac{1}{\alpha}x^\alpha \right]_0^l = Wl^{-\alpha}l^\alpha = W")
Total moment
.
We know that total moment
, so 
, as required.
Draw out a diagram: A touching the ground, with R_A going up, F_A going right. B touching the wall, with F_B going up, R_B going left. Somewhere ebtween A and B, we have the centre of mass. We denote
to be the distance of c.o.m. from A. Angle 
going from the horizontal to the line AB.
Equating the horizontal forces:
. Noting that 
, we get 
. Noting that 
, we get 
Equating the vertical forces:
.
Equating the moments:
as required.
Not sure if I'm right, but I'll give it a go:
, and 
Last edited by Aurel-Aqua; 10-04-2009 at 11:22. -
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Re: STEP I, II, III 2002 Solutions2002 II Question 12
probability such that more than l coins of the L coins will produce less than m heads
Estimating q:
, where z is the probability of a coin producing less than m heads.
We can say that
, which gives:
for small z.
We then substitute
to get 
as required.
z is the probability that less than m heads are made after M throws.
.
as required.
K = 7, k = 2, L = 500, l = 4, M = 100, m = 48, p = 0.6.
, which is a large z, so we are not allowed to approximate as needed.
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Re: STEP I, II, III 2002 SolutionsQuestion 10, STEP I, 2002
By Newton's law of restitution, for the first collision:
Assume for some n=k it is true, ie:
Looking at the (k+1)th collision:
And so by induction, we have shown it to be true:
[With a diagram] we can see that:
[Also with a diagram] we can see that:
substituting for
into 
, we get:
u = v,
Similarly,
By
we have:
Last edited by Unbounded; 11-04-2009 at 00:14. -
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Re: STEP I, II, III 2002 SolutionsQuestion 11, STEP I, 2002
Let the initial speed of
be u and let and 
leave the collision at speeds 
and 
respectively.
Looking at momentum in the direction of initially:
Looking at momentum perpendicular to the initial direction of
:
And considering the energy of the system:
Squaring we get:
And substituting in
Making the subject in 
we get:
Substituting this result into
we get:
![\iff \sin \theta = k \left [ \cos 2 \phi \sin \theta + \sin 2\phi \cos \theta \right ]](http://www.thestudentroom.co.uk/latexrender/pictures/f9/f9ca21f6d1a1a875acd5ab1329c8ea89.png "\iff \sin \theta = k \left [ \cos 2 \phi \sin \theta + \sin 2\phi \cos \theta \right ]")
Applying product-sum formulae:
By double angle formulae:
I fear I've taken some unnecessarily long route
If the angle between the particles after the collision is a right angle:
And also
and 
Applying this to the equation
we get
Last edited by Unbounded; 04-05-2009 at 11:16.
![\displaystyle \int_1^a \frac{\ln x}{x} \, dx = \left [ \frac{(\ln x)^2}{2} \right ]_1^a = \frac{(\ln a)^2}{2}](http://www.thestudentroom.co.uk/latexrender/pictures/76/7673d89701f436f38777242898b43a4e.png "\displaystyle \int_1^a \frac{\ln x}{x} \, dx = \left [ \frac{(\ln x)^2}{2} \right ]_1^a = \frac{(\ln a)^2}{2}")
![\displaystyle \pi \int_1^a \left ( \frac{\ln x}{x} \right )^2 \, dx = \pi \left ( \left [ - \frac{(\ln x)^2}{x} \right ]_1^a + \int_1^a \frac{2 \ln x}{x^2} \, dx \right) =](http://www.thestudentroom.co.uk/latexrender/pictures/99/99f28ffef32a5bec0f7e47f1fedd40ee.png "\displaystyle \pi \int_1^a \left ( \frac{\ln x}{x} \right )^2 \, dx = \pi \left ( \left [ - \frac{(\ln x)^2}{x} \right ]_1^a + \int_1^a \frac{2 \ln x}{x^2} \, dx \right) =")
![\displaystyle \pi \left [ - \frac{(\ln x)^2}{x} \right ]_1^a + \pi \left [ - \frac{2 \ln x}{x} \right ]_1^a +\int_1^a \frac{2}{x^2} \, dx =](http://www.thestudentroom.co.uk/latexrender/pictures/0c/0c3eeb2c8ed1b3a74b54739fb4d2b6b4.png "\displaystyle \pi \left [ - \frac{(\ln x)^2}{x} \right ]_1^a + \pi \left [ - \frac{2 \ln x}{x} \right ]_1^a +\int_1^a \frac{2}{x^2} \, dx =")
![\displaystyle \pi \left [ - \frac{(\ln x)^2}{x} - \frac{\ln x}{x} - \frac{2}{x} \right ]_1^a =](http://www.thestudentroom.co.uk/latexrender/pictures/01/0124d89e87a7bdec4eab54cf1411ac13.png "\displaystyle \pi \left [ - \frac{(\ln x)^2}{x} - \frac{\ln x}{x} - \frac{2}{x} \right ]_1^a =")
![\begin{array}{rl}
\displaystyle\int_{\frac{\pi}{6} }^{\frac{\pi}{4}} \frac{1}{1-\cos 2\theta} \, d\theta
& \displaystyle = \int_{\frac{\pi}{6}}^{\frac{\pi} {4}} \frac{1}{1-(1-2\sin^2 \theta)} \, d\theta \\ \br \\
& \displaystyle = \int_{\frac{\pi}{6}}^{\frac{\pi} {4}} \frac{1}{2\sin^2 \theta} \, d\theta \\ \br \\
& \displaystyle = \frac12 \int_{\frac{\pi}{6}}^{\frac{\pi} {4}} \mathrm{cosec}^2 \theta \, d\theta \\ \br \\
& \displaystyle = \frac12 \left[ -\cot \theta \right]_{\frac{\pi}{6}}^{\frac{\pi}{4}} \\ \br \\
& \displaystyle = \frac12 \left( \frac{1}{\tan \frac{\pi}{6}} - \frac{1}{\tan \frac{\pi}{4}} \right) \\ \br \\
& \displaystyle = \frac12 \left( \sqrt3 - 1 \right) \\ \br \\
& \displaystyle = \boxed{\frac{\sqrt3}{2} - \frac{1}{2}}
\end{array}](http://www.thestudentroom.co.uk/latexrender/pictures/62/6283bdc74c9091e6d004106834dd7869.png "\begin{array}{rl}
\displaystyle\int_{\frac{\pi}{6} }^{\frac{\pi}{4}} \frac{1}{1-\cos 2\theta} \, d\theta
& \displaystyle = \int_{\frac{\pi}{6}}^{\frac{\pi} {4}} \frac{1}{1-(1-2\sin^2 \theta)} \, d\theta \\ \br \\
& \displaystyle = \int_{\frac{\pi}{6}}^{\frac{\pi} {4}} \frac{1}{2\sin^2 \theta} \, d\theta \\ \br \\
& \displaystyle = \frac12 \int_{\frac{\pi}{6}}^{\frac{\pi} {4}} \mathrm{cosec}^2 \theta \, d\theta \\ \br \\
& \displaystyle = \frac12 \left[ -\cot \theta \right]_{\frac{\pi}{6}}^{\frac{\pi}{4}} \\ \br \\
& \displaystyle = \frac12 \left( \frac{1}{\tan \frac{\pi}{6}} - \frac{1}{\tan \frac{\pi}{4}} \right) \\ \br \\
& \displaystyle = \frac12 \left( \sqrt3 - 1 \right) \\ \br \\
& \displaystyle = \boxed{\frac{\sqrt3}{2} - \frac{1}{2}}
\end{array}")
![\displaystyle =\frac{1}{2}\left[-\cot^2\theta\right]_{\pi/6}^{\pi/4} =](http://www.thestudentroom.co.uk/latexrender/pictures/d5/d5af8c8817a627099ea66058f3b62eae.png "\displaystyle =\frac{1}{2}\left[-\cot^2\theta\right]_{\pi/6}^{\pi/4} =")
![\displaystyle E[Y] = \int_a^b xG'(x)\,\mathrm{d}x = \int_a^b \frac{2xF'(x)}{(2-F(x))^2}\, \mathrm{d}x = \int_a^b xF'(x) \times \frac{2}{(2-F(x))^2}\, \mathrm{d}x](http://www.thestudentroom.co.uk/latexrender/pictures/88/8863fdfe24de45f14abdea073a7b1bd7.png "\displaystyle E[Y] = \int_a^b xG'(x)\,\mathrm{d}x = \int_a^b \frac{2xF'(x)}{(2-F(x))^2}\, \mathrm{d}x = \int_a^b xF'(x) \times \frac{2}{(2-F(x))^2}\, \mathrm{d}x")
![\displaystyle \implies \frac{1}{2}E[X] \leq E[Y] \leq 2E[Y]](http://www.thestudentroom.co.uk/latexrender/pictures/e8/e8ecde5ba00b7d033b03a9e0c682374b.png "\displaystyle \implies \frac{1}{2}E[X] \leq E[Y] \leq 2E[Y]")
![\displaystyle Var[Y] = E[Y^2] - E[Y]^2](http://www.thestudentroom.co.uk/latexrender/pictures/24/247dff57aa639ee275f5bd899c3541df.png "\displaystyle Var[Y] = E[Y^2] - E[Y]^2")
![\displaystyle E[Y^2]\leq 2E[X^2]](http://www.thestudentroom.co.uk/latexrender/pictures/84/84966b767557e43ac9839c5fb2489065.png "\displaystyle E[Y^2]\leq 2E[X^2]")
![E[Y^2] \leq 2E[X^2] = 2E[X]^2+2Var[X]](http://www.thestudentroom.co.uk/latexrender/pictures/8b/8b92b78b8794069e7b5ab957066ed119.png "E[Y^2] \leq 2E[X^2] = 2E[X]^2+2Var[X]")
![\displaystyle (E[Y])^2 \geq \frac{1}{4}E[X]^2 \implies -E[Y]^2 \leq -\frac{1}{4}E[X]^2](http://www.thestudentroom.co.uk/latexrender/pictures/f3/f3c8098090c3564dafdd30e96031603d.png "\displaystyle (E[Y])^2 \geq \frac{1}{4}E[X]^2 \implies -E[Y]^2 \leq -\frac{1}{4}E[X]^2")
![\displaystyle Var[Y] \leq 2Var[X] + 2E[X]^2 - E[Y]^2 \leq 2Var[X] + 2E[X]^2 - \frac{1}{4}E[X]^2 =](http://www.thestudentroom.co.uk/latexrender/pictures/9a/9a4a9cc1dd5a8be120f7e444c08dea86.png "\displaystyle Var[Y] \leq 2Var[X] + 2E[X]^2 - E[Y]^2 \leq 2Var[X] + 2E[X]^2 - \frac{1}{4}E[X]^2 =")
![\displaystyle = 2Var[X] + \frac{7}{4}E[X]^2](http://www.thestudentroom.co.uk/latexrender/pictures/34/34d327511a5246b76c37dcfa91573ce6.png "\displaystyle = 2Var[X] + \frac{7}{4}E[X]^2")
