# STEP Maths I, II, III 1989 solutions

Watch
Announcements
Thread starter 13 years ago
#1
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:

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 - 2007
0
reply
Thread starter 13 years ago
#2
0
reply
Thread starter 13 years ago
#3
0
reply
13 years ago
#5
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.
0
reply
13 years ago
#6
0
reply
Thread starter 13 years ago
#7
2
reply
13 years ago
#8
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*
1
reply
13 years ago
#9
1
reply
13 years ago
#10
0
reply
13 years ago
#11
Hang on a sec, mine was mislabelled, should be Step I not step II.. sorry about that!

edit: thanks.. now typing up II/2 0
reply
13 years ago
#13
STEP II, (11) attached

attachments 2&3 are in the wrong order
0
reply
13 years ago
#14
Step III (2)
0
reply
13 years ago
#15
STEP II (6) attached
0
reply
Thread starter 13 years ago
#16
0
reply
13 years ago
#17
0
reply
Thread starter 13 years ago
#18
*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..
0
reply
Thread starter 13 years ago
#19
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 = 1-x
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 = 1-x
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 = . .
1
reply
13 years ago
#20
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 X-Y 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 X-Y 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 X-Y 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 X-Y 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 X-Y 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 x-axis 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 X-axis and c squeezes it along the Y-axis, 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.
0
reply
X

### Quick Reply

Write a reply...
Reply
new posts Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### Do you think receiving Teacher Assessed Grades will impact your future?

I'm worried it will negatively impact me getting into university/college (139)
42.77%
I'm worried that I’m not academically prepared for the next stage in my educational journey (38)
11.69%
I'm worried it will impact my future career (27)
8.31%
I'm worried that my grades will be seen as ‘lesser’ because I didn’t take exams (68)
20.92%
I don’t think that receiving these grades will impact my future (34)
10.46%
I think that receiving these grades will affect me in another way (let us know in the discussion!) (19)
5.85%

View All
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.