Download Sample 1 - Chartwell

“Teach A Level Maths”
Yr1/AS Pure Maths
Sample 1
© Christine Crisp
Explanation of Clip-art images
An important result, example or
summary that students might want to
note.
It would be a good idea for students
to check they can use their
calculators correctly to get the
result shown.
An exercise for students to do
without help.
The slides that follow are samples from 7 of the 46
presentations that make up the work for Year 1/AS Pure
Maths.
Roots, Surds and Discriminant
Simultaneous Equations and Intersections
Linear and Quadratic Inequalities
Stationary Points: applications
Circle Problems
Quadratic Trig Equations
Indices and Laws of Logarithms
Roots, Surds and Discriminant
Students have already met the discriminant in
solving quadratic equations. On the following slide
the calculation is shown and the link is made with
the graph of the quadratic function.
The Discriminant of a Quadratic Function
For the equation x 2  4 x  7  0
. . .
. . . the discriminant b 2  4ac  16  28
  12  0
y = x2 – 4x +7
There are no real roots
as the function is never
equal to zero
If we try to solve x 2  4 x  7  0 , we get
4   12
x
2
The square of any real number is positive so there
are no real solutions to  12
Simultaneous Equations and
Intersections
The following slide shows an example of solving a
linear and a quadratic equation simultaneously. The
discriminant ( met earlier ) is revised and the
solution to the equations is interpreted graphically.
y  x 2  3       (1)
y  4 x  1       (2)
e.g. 2
Eliminate y:
x 2  3  4 x  1
 x2  4x  4  0
The discriminant, b 2  4ac  4 2  4(1)( 4 )  0
The quadratic equation
has equal roots.
y  x2  3
2
Solving x  4 x  4  0
 ( x  2)( x  2)  0
 x  2 (twice)
x  2  y  7
y  4 x  1
The line is a tangent to the curve.
Linear and Quadratic Inequalities
Students are shown how to solve quadratic
inequalities using earlier work on sketching the
quadratic function. The following slide shows one of
the two types of solutions that arise.
The notepad icon indicates that this is an
important example that students may want to copy.
e.g.2 Find the values of x that satisfy x 2  4 x  5  0
Solution:
Find the zeros of f ( x ) where f ( x )  x 2  4 x  5
x 2  4 x  5  0  ( x  5 )( x  1 )  0

x5
or x   1
x 2  4 x  5 is greater than or
equal to 0 above the x-axis
There are 2 sets of values of x
 x   1 or x  5
These represent 2 separate intervals and
CANNOT be combined
y  x2  4x  5
Stationary Points: applications
Mathematical modelling is one of the
overarching themes in the 2017 A level.
The following slides show an example of
the application of stationary points to a
simple problem. Students are asked to
think about the assumptions made in
setting up the model.
e.g.2 I want to make a fruit cage from 2 pieces of
netting. The smaller piece will make the top and I’ll
use the longer piece, which is 10m long, for the
sides. The cage is to be rectangular and to enclose
the largest possible area. One side won’t need to
be netted as there is a wall on that side. How long
should the 3 sides be?
Solution: We draw a diagram and choose letters
for the unknown lengths and area.
The lengths are x
metres and y metres.
The area is A m2.
wall
x
A
y
x
We want to find the
maximum value of A, so we
need an expression for A
that we can differentiate.
wall
x
A
x
y
The area of a rectangle  length  breadth
A  xy       (1)
can’t
with
so
we
Can
youdifferentiate
thinkthat
of one
assumption
I have
in
IWe
have
assumed
there
is 3novariables
wastage
atmade
theneed
wall

to or
substitute
for
one
of them.
setting
up this
model?
effect
will large.
it have
corners.
The
area
I What
find will
be too
Since
length
10m, assumptions)
we know that
onnetting
the answer?
(Youthe
may
haveof
thought
ofisother
Rearranging,
2 x  y  10
y  10  2 x       ( 2)
wall
A  xy
y  10  2 x
x
      (1)
      ( 2)
Substituting for y in (1),
 A  x(10  2 x )  A  10 x  2 x 2
We can now find the stationary points.
dA
 10  4 x
dx
dA
 0  10  4 x  0
dx

10  4 x

x  2 5
Substitute x  2  5 in ( 2)  y  5
Substitute in (1),
A  xy  A  12  5
A
y
x
wall
We have
x
x  2  5, y  5 and A  12  5
A
x
y
We know that we have found the maximum since
A  10 x  2 x 2
and we recognise
the shape of this
quadratic.
A  10 x  2 x 2
Circle Problems
There is also a greater emphasis on problem
solving. Questions involving circles provide
useful practice.
Here students are reminded of the
properties of circles, with worked examples
encouraging them to solve problems and
emphasising the need to draw diagrams.
Diagrams are essential when solving problems involving
circles. They don’t need to be accurate !
e.g.1 Find the equation of the tangent at the point
(5, 7) on a circle with centre (2, 3)
x (5, 7)
(2, 3) x
tangent
Decide with your partner
how you would solve this
problem
My reasoning (in order)
is on the next slide.
e.g.1 Find the equation of the tangent at the point
(5, 7) on a circle with centre (2, 3)
gradient m
gradient m1 x (5, 7)
(2, 3) x
tangent
• A tangent is a straight
line so we need the
equation of a straight
line: y  y1  m ( x  x1 )
• We know a point on the
tangent so x1 = 5 and y1 = 7
• We need the gradient m
• The gradients of tangent
and radius are
1
perpendicular so m  
y2  y1
• We can find m1 using m1 
x2  x1
m1
We need to reverse the order of my logic to do
the calculation.
y2  y1
m1 
gradient m
x2  x1
gradient m1 x (5, 7)
(2, 3) x
tangent
73
 m1 
52
4
 m1 
3
3
1
 m 
m
4
m1
y  y1  m ( x  x1 )
3
 y  7    x  5
4
Multiply by 4: 4y – 28 = – 3(x – 5)  3x + 4y = 43
Quadratic Trig Equations
By the time students meet quadratic trig
equations they have practised solving
linear trig equations and have seen a
proof of a Pythagorean identity.
e.g. 3 Solve the equation 2 cos 2   3 cos   2  0 for
the interval 0    360o.
Solution: Let c  cos  . Then,
2c 2  3c  2  0
Factorising:
( 2 c  1 )(c  2 )  0  c  1 or
2
cos    2
cos   1
or

c  2
2
The graph of y  cos  . . .
shows that cos  always
lies between -1 and +1 so,
cos    2 has no solutions
for  .
y  cos 
Solving cos   1 for 0    360o.
2
Principal Solution:   60o
y  05

60 
300 
y  cos x
The 2nd solution is
x  360  60
 300 
Indices and Laws of Logarithms
The approach to solving the equation a x  b
started with a = 10 and b an integer power of 10.
The word logarithm has been introduced and here
the students are shown how to use their calculators
to solve when x is not an integer.
The calculator icon indicates that students should
do the calculation.
Indices and Laws of Logarithms
A logarithm is just an index.
To solve an equation where the index is unknown, we
can use logarithms.
e.g. Solve the equation 10 x  4 giving the answer
correct to 3 significant figures.
x is the logarithm of 4 with a base of 10
We write 10 x  4 
x  log 10 4
 0 602 ( 3 s.f. )
In general if
10 x  b then x  log 10 b
index  log
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