Log tips
When solving, you can often either:
a) Get in the form logab = c. Then rearrange as ac = b
b) Get in the form logab = logac. Then b = c.
Actual exam questions:
Solve
log2 (2x + 1) – log2 x = 2
2 log3 x – log3 7x = 1
log 2 (x + 1) – log 2 x = log 2 7
Solve simultaneous equations:
a = 3b,
log3 a + log3 b = 2
log5 (4 – x) – 2 log5 x = 1
log2y = –3
log2 32  log2 16
 log2 x
log2 x
logx 64 = 2
log2 (11 – 6x) = 2 log2 (x – 1) + 3
Log tips
Other types of question:
3x = 10
52x – 12(5x) + 35 = 0
What if you had 52x+1 here
instead?
Circles
The points A and B have coordinates (–2, 11) and (8, 1) respectively.
Given that AB is a diameter of the circle C,
(a)
show that the centre of C has coordinates (3, 6),
(1)
(b)
find an equation for C.
(4)
(c)
Verify that the point (10, 7) lies on C.
(1)
(d)
Find an equation of the tangent to C at the point (10, 7), giving your answer
in the form y = mx + c, where m and c are constants.
(4)
Remember that you need the centre (a,b) and the radius r, which
gives the equation:
(x-a)2 + (y-b)2 = r2
The tangent is perpendicular to the radius at the point of contact.
Circles
How could you tell if a line and a circle intersect:
0 times
once
twice
x2 + y 2 = 1
Equate the expressions then look at the discriminant:
b2 – 4ac
? <0
b2 – 4ac
? >0
b2 – 4ac
? =0
Circles
8. The circle C, with centre at the point A, has equation x2 + y2 – 10x + 9 = 0.
Find
(a)
the coordinates of A,
(2)
(b)
the radius of C,
(2)
(c)
the coordinates of the points at which C crosses the x-axis.
(2)
Given that the line l with gradient is a tangent to C, and that l touches C at the
point T,
(d)
find an equation of the line which passes through A and T.
(3)
Circles
If PR is the
diameter, how
would you prove
that a = 13?
Circles
Circles
Factor Theorem
(a)
Use the factor theorem to show that (x + 4) is a factor of
2x3 + x2 – 25x + 12.
(2)
(b)
Factorise 2x3 + x2 – 25x + 12 completely.
(4)
f(x) = x3 + ax2 + bx + 3, where a and b are constants.
Given that when f (x) is divided by (x + 2) the remainder is 7,
(a) show that 2a − b = 6
(2)
Given also that when f (x) is divided by (x −1) the remainder is 4,
(b) find the value of a and the value of b.
(4)
Trigonometric Solutions
(a) Find all the values of  , to 1 decimal
place, in the interval 0   < 360 for
which
5 sin ( + 30) = 3.
(4)
(b) Find all the values of  , to 1 decimal
place, in the interval 0   < 360 for
which
tan2  = 4.
(a) Given that sin  = 5 cos , find the
value of tan .
(1)
(b) Hence, or otherwise, find the values of
 in the interval 0   < 360 for which
sin  = 5 cos ,
giving your answers to 1 decimal place.
(3)
(5)
Find all the solutions, in the interval 0 ≤ x < 2, of the equation
2 cos2 x + 1 = 5 sin x,
giving each solution in terms of .
Trigonometric Solutions
(a) Sketch, for 0 ≤ x ≤ 2, the graph of y =
sin ( x + ( /6)).
(2)
(b) Write down the exact coordinates of
the points where the graph meets the
coordinate axes.
(3)
(c) Solve, for 0 ≤ x ≤ 2, the equation
sin (x + ( /6)) = 0.65,
giving your answers in radians to 2
decimal places.
(a) Show that the equation
3 sin2  – 2 cos2  = 1
can be written as
5 sin2  = 3.
(b) Hence solve, for 0   < 360, the
equation
(2)
3 sin2  – 2 cos2  = 1,
giving your answer to 1 decimal place.
Solve, for 0  x < 360°,
(a)
(7)
sin(x – 20) = 1/√2,
(4)
(b)
cos 3x = –1/2 .
(6)
Trigonometric Solutions
(a) Show that the equation
4 sin2 x + 9 cos x – 6 = 0
can be written as
4 cos2 x – 9 cos x + 2 = 0.
(b) Hence solve, for 0  x < 720°,
4 sin2 x + 9 cos x – 6 = 0,
Show that the equation
tan 2x = 5 sin 2x
can be written in the form
(1 – 5 cos 2x) sin 2x = 0
(2)
giving your answers to 1 decimal place.
(6)
(2)
(b) Hence solve, for 0 ≤ x ≤180°,
tan 2x = 5 sin 2x
giving your answers to 1 decimal place
where appropriate.
You must show clearly how you obtained
your answers.
(5)
(i) Find the solutions of the equation sin(3x - 15 ) = ½ for which 0 ≤ x ≤ 180
Trigonometric Solutions
Summary of tips:
• To get all your solutions when you do you inverse sin/cos/tan:
• Remember that sin(180-x) = sin(x) and cos(360-x) = cos(x)
• sin and cos repeat every 360 (i.e. you can add 360 to your
solution as many times as you like).
• But tan repeats every 180.
• If you’re working in radians, then sin(pi – x) = sin(x), etc.
• Look at the range the question gives: if it’s in radians, give your
answers in radians.
• Ensure your calculator is correctly set to either radians or degrees
mode.
• If you have sin2, then make sure you get both positive and
negative solution. Likewise for cos2 and tan2.
• Make sure you write out enough solutions before you start
manipulating: if you had sin(3x) = ½ for example and had the range
0 < x < 360, then you’d initially need to write values up to 1080
since you’re going to be dividing by 3.
Areas of sector/Arc lengths/Sine and Cosine Rule
Areas of sector/Arc lengths/Sine and Cosine Rule
Areas
Only 1 in 36 candidates (across the
country) got this question fully correct.
Optimisation
Optimisation
Optimisation
Optimisation
Integration
Integration
8
Figure 1 shows part of a curve C with equation 𝑦 = 2𝑥 + 𝑥 2 – 5, x > 0.
The points P and Q lie on C and have x-coordinates 1 and 4 respectively. The region
R, shaded in Figure 1, is bounded by C and the straight line joining P and Q.
(a)
Find the exact area of R.
(8)
(b)
Use calculus to show that y is increasing for x > 2.
(4)
Integration
Integration
Examiner’s Report:
(a) A pleasing majority of the candidates were able to differentiate these fractional powers
correctly, but a sizeable group left the constant term on the end. They then put the derivative equal
to zero. Solving the equation which resulted caused more problems as the equation contained
various fractional powers. Some tried squaring to clear away the fractional powers, but often did
not deal well with the square roots afterwards. There were many who expressed 6x-1/2 = 1/(6x1/2)
and tended to get in a muddle after that. Those who took out a factor x1/2 usually ended with x = 0
as well as x = 4 and if it was not discounted, they lost an accuracy mark. Those who obtained the
solution x = 4 sometimes neglected to complete their solution by finding the corresponding y value.
Some weaker candidates did not differentiate at all in part (a), with some integrating, and others
substituting various values into y.
?
Geometric Series
Geometric Series
Geometric Series
Binomial Expansion