STEP I (Mathematics)
1:
2: Solution by kabbers
3: Solution by Dystopia
4: Solution by Dystopia
5: Solution by Dystopia
6: Solution by Swayam
7: Solution by squeezebox
8: Solution by squeezebox
9: Solution by nota bene
10:
11: Solution by Glutamic Acid
12:
13:
14:
15:
16:
STEP II (F.Maths A)
1: Solution by squeezebox
2: Solution by kabbers
3: Solution by squeezebox
4:
5: Solution by Dystopia
6: Solution by *bobo*
7:
8:
9:
10:
11:Solution by *bobo*
12:
13:
14:Solution by squeezebox
15:
16:
STEP III (F.Maths B)
1: Solution by squeezebox
2: Solution by *bobo*
3:
4:
5:
6:
7:
8: Solution by Dystopia
9: Solution by squeezebox
10: Solution by squeezebox
11:
12:
13:
14:
15:
16:
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STEP Maths I, II, III 1989 solutions
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 1
 02022008 03:24
Last edited by squeezebox; 28032009 at 19:23. 
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 2
 02022008 03:29
I am no expert at STEP, so don't hesitate to correct me, it is more than likely made that I have made a mistake.
STEP I  Question 8
Using good old de Moivre's theorem;
(using
and similary;
Now, consider:
Where c is .
Now let ,
so:
Which gives;
().
These values of give distinct values of
or
()
Hence the roots of the equation are:
and . 
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 3
 02022008 03:37
STEP III  Question 10
Lets assume that the result is true for n = k;
(*)
This is the same as (*) except k+1 replaces k. Hence if the result is true for n=k, its true for n= k+1.
When n=1,
LHS of (*) = 1x2x3x4x5 = 120
RHS of (*) = (1/6)x1x2x3x4x5x6 = 120.
So (*) is true for n=1.
Hence, by induction;
.
Since;
.
Using (*);
In this case,
From the previous parts, we know that:
and clearly, also:
as ,
and
So,
Using a similar arguement, we can show that:
We have shown that the limits exist and are equal to .
Hence;
Last edited by squeezebox; 02022008 at 14:28. 
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 4
 02022008 13:23
STEP I, Q2
The For , find
Using integration by parts ():
By approximating the area corresponding to by n rectangles of equal width and with their top right hand vertices on the curve , show that, as ,
Firstly, note that
(by the above indefinite integral)
The sum of the areas of our rectangles is going to be
Claim:
So for k + 1, we must prove that is the equation we arrive at.
Basis case (let k=1):
Inductive step:
So,
As , our sum tends to the integral . So,
is this ok? have i missed anything out or not explained each step in enough detail?Last edited by kabbers; 02022008 at 19:49. 
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 5
 02022008 14:01
Question 4, STEP I
With six points, each point is attached to five other points, so at least three lines radiating from A must be of the same colour (gold, say). The points which are joined to A must then be joined to each other by silver lines, or an entirely gold triangle would be made. However, then a triangle with entirely silver edges can be made from these three points.
Diagram showing five points is attached. 
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 6
 02022008 14:25
Question 5, STEP I
a)
Let x = 1.
Let x = 1
As n is even. So
As their sum is
b) Suppose that
Then
Note that and that
So, upon adding these two expressions, we get
Also,
So true by induction.
Last edited by Dystopia; 02022008 at 16:27. 
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 7
 02022008 14:36
STEP II Q1
Substituting x= ya into the equation:
Notice that the reduced form shown in the question has no term, so we need to find the value of a which will get rid of the term.
Expanding out and collectiing terms we get:
So the value of a to get rid of the is 1.
So we now have:
(*)
Substituting y=z/b into (*);
dividing both sides by 2 and multiplying by ;
.
And so to make this equal to;
we make b = .
Lets take b = and let z =
so;
and (these give distinct values of , and hence, three distinct solutions.)
we know that; x = y  a = (z/b)  a.
Hence the solutions of the equation are:
andLast edited by squeezebox; 26082008 at 22:10.Post rating:1 
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 8
 02022008 14:51
Question 11, STEP I.
Vertical component =
Using s = ut + 1/2at^2.
Quadratic in t. Ignoring the negative root.
R = horizontal component x time.
Horizontal component =
Multiply both numerator and denominator by 2.
Using
We can simplify it.
Splitting up Mr. Big Fraction.
Ignore the first fraction for the moment, we'll work on the second one.
Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.\dfrac{2v^2 \cos \theta \sin \theta}{2g} \times \sqrt{1 + \dfrac{2hg}{v^2 \sin^2 \theta}
We can use our old friend Mr. Double Angle Formula to simplify it:
The whole fraction looks like:
Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.R = (\dfrac{v^2\sin 2\theta}{2g}) + (\dfrac{v^2 \sin 2\theta}{2g}) \sqrt{1 + \dfrac{2hg}{v^2 \sin^2 \theta}
And lo and behold, there's a common factor to both fractions. Let's take it out:
And that's what we're looking for.
*I'll finish off the last small bit in a few minutes*Last edited by Glutamic Acid; 02022008 at 15:23.Post rating:1 
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 9
 02022008 15:09
STEP I Question 6
The normal to the curve will therefore have gradient 1/f'(x). The equation of the normal would be
(where x' is the general x coordinate of the normal).
At the point Q, x' = 0 as it cuts the y axis.
The distance PQ can be worked out using Pythagoras' theorem.
It's given that
So
(use a substitution of u = x^2 if you can't see why)
Last edited by Swayum; 02022008 at 16:38. 
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 10
 02022008 16:23

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 11
 02022008 19:50
Hang on a sec, mine was mislabelled, should be Step I not step II.. sorry about that!
edit: thanks.. now typing up II/2Last edited by kabbers; 02022008 at 20:48. 
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 12
 02022008 22:10
STEP II/2
Therefore, by (*),
we can also use (*) to express 2cot2x as
Therefore,
we can compare the coefficients of the respective summations and say
for the second part, note that
we have to find an identity for 2csc2x
(using partial fractions)
Also,
, so:
Comparing coefficients of the polynomials (in x) of the expansion,
Rearranging and substituting in the earlier definition of ,
any corrections welcome !Last edited by kabbers; 02022008 at 22:20. 
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 13
 03022008 00:40
STEP II, (11) attached
attachments 2&3 are in the wrong orderLast edited by *bobo*; 03022008 at 00:43. 
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 14
 03022008 01:53
Step III (2)

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 15
 03022008 14:37
STEP II (6) attached

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 16
 03022008 15:07
STEP II Question 3
Equating real and imaginary parts:
(*)
(**)
if x=a, then:
(1)
(2)
Now, subbing (1) and (2) into:
We get:
As required.
Substituting y=b into (*) and (**):
(3)
(4).
This time, using , we end up with:
.
Since and , . Hence, () corresponds to a single hyeprbola branch, which is above the line v=0.
One point of intersection is: and .
Differentiating implicitly we get:
.
Which is equal to: at and .
Differentiating implicitly we get:
.
Which is, when and .
X
Hence the curves intersect and right angles at this point.
Regarding the sketch, I think the elipse becomes a vertical straight line from v = 1 to v = 1, and the hyperbola becomes the line v = 1.Last edited by squeezebox; 15022008 at 01:53. 
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 17
 10022008 17:10
STEP 1, Q3
(By the ratio theorem.)
These vectors are perpendicular, so the scalar product is zero.
(Note that the scalar product is commutative and distributive.)
Let
Minimum and maximum values occur at the endpoint of a range or at a turning point. There is a turning point when x = 1/2.
So
Note that
Where
The maximum value is occurs when
Suppose that
Then
So
By the cosine rule,
Let F be the midpoint of AB. By Pythagoras,
Last edited by Dystopia; 10022008 at 17:15. 
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 18
 15022008 01:45
*bump* (just to remind people that this thread is still here )  I have solutions that I can type up if no one else wants to, but I really don't want to take over the thread..
Last edited by squeezebox; 15022008 at 02:12. 
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 19
 16022008 01:41
STEP I  Question 7
*Graph Attached*
If we work out the area of a quarter of the rectangle, which is in the top right hand bit of the axes, and maximise this, it make calculation soooo much eaiser. (Since we can forget about modulus signs and y being negative) :
Let the area of the rectangle be .
Base of smaller rectangle = 1x
height =
so
which we want to maximise.
Which is zero when
Finding the second derivative and subsituting x = in shows is a maximum at this point.
So .
________
We can do the second bit in the same way, since its a closed curve that is symmetrical in both axes.
Let the area of the second rectangle be .
Working in the top right quadrant again;
Base = 1x
height = .
So .
Which is zero when x = 0 or x = .
Clearly the area is a minimum when x= 0, and so is a maximum when x =.
. 
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 20
 26022008 19:16
I've got a solution to I question 9 , but don't have a scanner available so will add graphs when I get the opportunity.
Question 9 STEP I
and Setting y'=0
f(2)=2, f(2)=2, also for graphing purposes it may be worth to note that f(0)=0 and f(4)=2 and f(4)=2. y=0 has solutions x=0 or f''(2)= i.e. local max f''(2)=+ i.e. local min
Graph, see attachment
(a) i.e. therefore the equation of y in this XY plane is
setting it equal to 0 gives f(1)=2 ans f(1)=2
For graphing it can also be good to see that f(2)=2, f(2)=2 and
Graph, see attachment
(b) i.e. therefore the equation of y in this XY plane is
Setting this equal to 0 gives . f(2)=1 and f(2)=1
For graphing, also see that , f(4)=1 and f(4)=1
Graph, see attachment.
(c) i.e. therefore the equation in the XY plane is
Setting this equal to 0 gives f(0)=2, f(2)=2.
For graphing, also see that and f(1)=2, f(3)=2
(d) i.e. therefore the equation in the XY plane is
setting this equal to 0 gives f(2)=0 f(2)=2
For graphing I'll check f(4) and f(4) too, which give 2 and 0 respectively.
For the last part, we're looking for a graph in the XY plane with local min (0,0) and local max (1,1). To obtain this on the form X=ax+b and Y=cy+d, first observe that for a graph of a cubic, changing the sign means a reflection in the xaxis i.e. local max and min swap places, this is what we need to do. That means either a or c is negative (but not both!). a squeezes the graph along the Xaxis and c squeezes it along the Yaxis, c moves the graph to the left/right and d moves the graph up/down. Knowing these we can see that and will produce the desired graph.
Okay, so I hope I've interpreted everything correct in the question. Seems pretty nice as a question.Last edited by nota bene; 26022008 at 19:58.
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Updated: November 28, 2015
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