Question
3 Paper 1 1999
Question 3 Paper
1 1998
Question 3 Paper 1 1997
Question 3 Paper 1 1996
Question 3 Paper 1 1995
Question 5
Paper 1. 1999
Question
3 Paper 1 1998 Ordinary
Level
(A) Express
p in terms of q and t when 
.
Here we are asked to write one letter in terms of another
or to make p a subject of a formula.
Remember this is just an equation and you are looking
for p on it's own.
Solution
All we did here was multiply everything by the common
denominator (5t) and bring p to one side.
(B)(i) If (x - 2) is a factor
of 
find k.
Solution
If (x - 2) is a factor then x = 2 is a root if we sub
in 2 for x we will get zero.
.
Be careful when you sub in numbers
remember 
.
(B)(ii) Write down an equation,
which has the three roots of value - 3,1,and 5.
Solution
We use the rule that if a is a root of an equation, x
- a is a factor therefore the three factors
are (x - - 3), (x - 1) and (x - 5) to find the equation
we multiply the factors
(C)(i) Write 
as
a single fraction.
Solution
This is just like adding any two fractions just get the
common denominator and add! The common denominator
here is (x + 1)(x - 3)
.
(10 marks)
(C) (ii) Hence or otherwise
solve
.
Solution
We can use our result from above (hence) or as we did
in part a multiply everything by the common
denominator (otherwise). We will this time use the result
from part (i) 
This is a quadratic equation, which has no factors so
we must use the formula:
and this gives
(10 marks)
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Question
5 Paper 1. 1999 Ordinary
Level
(A) Given 
Find 
, show 
.
Answer: 
(B)
Of a GP are 2 and 2/3. Find r the common ratio, find
show 
Answer: 
. 
To find 
we
use the formula for
now a = ½, r =1/3, n = 6
(C)
Find 
.
show that the series is Arithmetic .
A series is Arithmetic if 
(a
constant)
4n + 1 -( 4n - 3 ) = 4n+1- 4n +3 = 4 .The series is Arithmetic
.
Find
we use 
,
In this case n = 20, d = 4,
a = 
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Ordinary
Level Question 3 1997 paper 1
(a)
Express p in terms of q and t when 2p - q = 3(p - t).
Same as '98 just keep your eye on p.
Solution
2p
- q = 3p -3t, get all the p's to the same side 3t-q =
3p - 2p = 3t - q = p.
(Pray
that there are lots of problems like this on the exam)
(b)
Solve 
,
this means find the values of x, which solve the equation
(Make the equation zero)
Solution
We
find the first value by trial and error to get an idea
as to what values of x to try first divide -6 (end term)
by the number in front of the x cubed
,
This means that 1, or -1 or 3 or -3 is a root (these are
the factors of -3)
We
use synthetic* division to see if - 1 is a root as can
be seen the remainder is zero and the quotient is 
We
now solve 
,
by factors or the formula
*Synthetic
division is probably the best way to do this type of question
as it avoids the "dreaded" long division in algebra!
Note one of the roots of the cubic must be a whole number.
(
c ) Let f (x) = (2 + x)(3- x), Write down the solutions
(roots) of f (x) = 0.
Solution
Just
let (2 + x)(3 - x) = 0, 2 + x = 0, x = -2, 3 - x = 0,
x = 3.
Let
g (x) = 3x-k. The equation f (x) + g (x)=0has equal roots
find one value for k
This
was quite difficult for ordinary students, most would
have no idea how to find k, ex higher level students would
probably have been able to find k, the fact that the question
was a wash out was reflected in the marking scheme where
they gave 7 out of the 10 marks for writing out the question!
Solution
If
this has equal roots
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Ordinary
Level Question
3 1999
paper 1
(a)
Express p in terms of q and r when 
.
Same as '98 and '97. Just multiply everything by the common
denominator (q) and get p on it's own.
Solution
(10 marks).
(b)
Solve for x and y 
a
linear and a quadratic equation and we are asked to solve
them simultaneously. This type of problem comes up regularly
in the question on the circle. The standard method is
as follows.
(1)
Write the linear equation as x equals or y equals.
(2)
Sub the result into the quadratic and simplify
it.
(3)
Solve the quadratic equation using factors or the
formula.
(4)
Substitute your answer(s) back into the line to
get the values of the second letter.
Solution
(1)
(2)
substitute this result into the quadratic to get
Now
find x when y= 1 x = 6-2(1), x =4 our first result is
(4,1)
When
y = 4.8, x = 6 -2(4.8) x = 6 - 9.6 x -3.6 our second result
is (-3.6,4.8)
İShow
is
a factor of 
Solution
Nothing
for it but to divide it in
Since the remainder is zero then 
is
a factor.
To
find the three roots we set the two factors equal to zero
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Ordinary Level paper 1 Question
3 1996
(a)
Express q in terms of p and t when 
.
Again
this is just an equation we want q on it's own this time.
Solution
;
(10
marks)
(b)
Find the roots of the equation 
.
Solution
Find
the first root by trial and error to get an idea of what
the first root might be divide 2 by 2 = 1 (last number
divided by first number) Try x = 1 using Synthetic division.
The
remainder is zero therefore x = 1 is a root the quotient
is 
to
find the other root set 
The
three roots are -1/2,2,1
(20 marks)
(
c ) Let f (x) = (1-x)(2+x) Write down the solutions
of f (x) = 0.
Solution
.
Find
the range of values of x for which f (x) > 0.
This
is a quadratic inequality (unusual for Ordinary level
Maths)
To
sole this we use the following method
(1)
Solve f (x)=0, (2) Plot the results on the number
line, plot zero also on the number line.
(2)
Sub zero into f (x) > 0 if true then zero is
part of the solution, if false then zero is not part of
the solution.
(3)
-2
0 1
(4)
f(0)=(1)(2) > 0, therefore zero is part
of the solution so our solution is
.
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1995
Ordinary Level paper 1 Question 3
(a)
Solve for x the equation 
.
Solution:
This
is an index equation, so change both sides to powers of
the same number and set the indices equal to each other.
(b)
Solve for x and y. 
Solution:
Simultaneous
equations where one equation is linear, the second is
a quadratic we use the exact same method as in the'99
question.
Now
find x
We
know x = 3 -2y, if y = - 1, x = 3 - (2)(-1) = 5 (5,
-1)
Y = 3.4 x = 3 -2(3.4) = x = -3.8 (-3.8,3.4).
(
c ) 
,
If f (x) = 0 when x = -3 and 1 find
b and c
Solution
.
These
are simultaneous equations
Multiply
the bottom line by - 1 
add
-4b = -8, b = 2
b
+ c = - 1 , 2 + c = -1 c = - 3 , b =2 .
If
f (-1) = k find k.
Solution
Solve
f (x) - k = 0. 
Part
c was really more awkward than difficult but it should
have been possible to get most of the question out if
you read it carefully. The function notation is popular
in question 3 c.
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