THOMAS WHITHAM SIXTH FORM

Additional Mathematics

Revision Guide

S J Cooper

The book contains a number of worked examples covering the topics needed in the Additional Mathematics Specifications. This includes Calculus, Trigonometry, Geometry and the more advanced Algebra. Plus the topics required for Mechanics and statistics. Algebra

 Indices for all rational exponents

n a m  a n  a mn , a m  a n  a mn , a m   a mn , a 0 1 ,

n n 1 1 1 1  a   b  a 2  a , a n  n a , a n  ,  a n ,      a n a n  b   a 

m m a n  n a m  n a 

1  1 3 3 2 1 1 1 Example 8  8  2 Example 9  1   9 2 9 3

3 2 2 2 3 3 Example x 3  4  x 3   4 2  x   4  8 Example Simplify 2x1 3x2  6x3

2 1 18x 2 3 2x 1  3x2  6x3  9x 2    x 6x3 6x 4 x 2

2  Quadratic equations ax  bx  c  0 Examples (i) 4x2  9 (ii) 4x2  9x (iii) 4x2  9x  2

(i) (ii) (iii)

9 2 2 x 2  4x  9x  0 4x  9x  2  0 4 x(4x  9)  0 (4x 1)(x  2)  0 3  x   9 1 2  x  0 or x   x  2 or x  4 4 Example Solve (i) 2x2  x  5  0 using the method of CTS (ii) 5x2  7x 1  0 using the formula. (i) x 5 x 2   2 2 2 2 x  1  5  1  x 2         2  4  2  4  1 2 5 1 x  4   2  16 1 2 41 x  4   16 1 41 x  4   16

1 41 x   4  16 x  1.35, 1.85

 Simultaneous equations Example Solve simultaneously 2x  3y  8, y  x2  x  2 Here we substitute for y from the second equation into the first 2x  3y  8  2x  3x2  x  2  8  3x2  x  2  0

2  3x  2x 1  0  x   3 , 1

2 4 2 28 Solutions, 2 28 when x   3 , y  9  3  2  9 x   3 , y  9 x 1, y  2 when x  1 , y 11 2  2 } The geometrical interpretation here is that the straight line and the

2 2 28 parabola y  x  x  2 intersect at points  3 , 9  1, 2 y

 2 , 28 3 9 (1,2)

0 x

 Intersection points of graphs to ‘solve’ equations. There are many equations which can not be solved analytically. Approximate roots to equations can be found graphically if necessary. Example What straight line drawn on the same axes as the graph of y  x3 will give

the real root of the equation x3  x  3  0?

y

x3  x 3  0  x3  3 x  draw y  3  x 0  x As can be seen from the sketch there is only one real root .

 Expansions and factorisation –extensions Example 2x 1x 2  x  3  2x3  2x 2  6x  x 2  x  3 Expanding  2x3  3x 2  7x  3 Example x3  9x  xx2  9  xx  3x  3 Factorising

 The remainder Theorem

b If the polynomial p(x) be divided by ax  b the remainder will be p a 

Example When p(x)  2x3  3x  5 is divided by 2x 1 the remainder is p 1    1  3  5   15 2 4 2 4 Example Find the remainder when 4푥3 − 3푥2 + 11푥 − 2 is divided by 푥 − 1.

f 1 413  312 111 2 10 remainder is 10

 The factor theorem Following on from the last item

b p a   0  ax  b is a factor of Example Show that x  2 is a factor of 6x3 13x2  x  2 and hence solve the

equation 6x3 13x2  x  2  0 Let p(x)  6x3 13x2  x  2

p(2)  623 1322  2 2  48  52  2  2  0  x  2 is a factor of p(x) p(x)  x  26x2  x 1 …by inspection  x  23x 12x 1

1 1  Solutions to the equation are x  2,  3 , 2

n  Expansion of a  b where n is a positive integer.

This has the same coefficients as the expansion of 1 xn . Each term will have degree

n, powers of a descending from a n

Example 1 x8 1 8x  28x2 ......

a  b8  a8  8a7b  28a6b2 ......

n nr r The general term will be Cr  a b Example

Use the binomial theorem to expand 3  2x4 , simplifying each term of your expansion. 4 4 3 2 2 3 4 a  b  a  4a b  6a b  4ab  b 4 4 3 2 2 3 4  3  2x  3  43 2x 63 2x  432x  2x

2 3 4  81 216x  216x  96x 16x

Geometry

 Gradient/ intercept form of a straight line Equation y  mx  c y

Gradient = m (= tan ) c  0 x

 Distance between two points

Given A x1, y1  B x2 , y2  then

2 2 2 AB  x2  x1   y2  y1 

 Gradient of a line through two points… Ax1, y1  and Bx2 , y2  say y  y m  2 1 x2  x1

 Equation of a line through (x’, y’) of gradient m y  y  mx  x

 Equation of a line through two points

Find the gradient using and use the formula as above.

 Parallel and perpendicular lines

Let two lines have gradients m1 and m2

Lines parallel  m1  m2 1 Lines perpendicular  m1m2  1 or m1   m2

 Mid-point of line joining … and coordinates are 1 x  x , 1 y  y  2 1 2 2 1 2 

 General form of a straight line ax  by  c  0. To find the gradient, rewrite in gradient/intercept form. Example Given points A 2, 3 and B1, 1 find (a) distance AB (b) the coordinates of the mid-point M of AB (c) the gradient of AB (d) the equation of the line through C5, 2 parallel to AB

(a) AB 2  1 22  1 32  9 16  25  AB = 5

1 (b) M  2 , 1

1 3 4 (c) Gradient AB =   1 2 3

4 (d) Point 5, 2 Gradient =  3

4 Equation y  2   3 x  5 3y  6  4x  20

3y  4x  26

Example Find the gradient of the line 2x  3y 12 and the equation of a perpendicular line through the point 0,  4

2 2  3y  2x 12  y   3 x  4  grad =  3

1 Gradient of perpendicular =   3 2 2  3

3 Equation y  2 x  4 y  mx  c

 The Circle

Angles in semicircle is 90

Perpendicular to a chord from centre of circle bisects the chord.

 Centre, radius form of equation

x  a2  y  b2  r 2 Centre (a, b) radius = r

Example Centre (2, -1) radius 3 equation x  22  y 12  9

Example Centre (1, 2) touching 0x Equation x 12  y  22  4

(1, 2)

0

. General form of equation

x  a2  y  b2  r 2 Circle centre 푎, 푏 with radius 푟 To find centre and radius, use the method of CTS to change into centre/radius form.

Example x2  y 2  2x  3y  3  0

x 2  y 2  2x  3y  3  0 x 2  2x y 2  3y  3 2 2 3 2 3 2 x  2x 1 y  3y  2   3 1 2  2 3 2 25 x 1  y  2   4

3 5  Centre 1, 2  radius = 2

 Tangents

Angle between tangent and radius drawn to point of contact is 90

Tangents drawn from extended point

Line of symmetry

Example Find the equation of the tangent to the circle x2  y 2  2x  4y  5  0 at the point P(2, 1) x 2  y 2  2x  4y  5  0 x 2  2x y 2  4y  5

x 2  2x 1 y 2  4y  4  5 1 4 x 12  y  22  10

 Centre at 1, 2 , radius 10

2 1 1 Gradient CP =   (-1, 2) 1 2 3  gradient of tangent at P = 3 Equation P(2, 1) 0 y 1  3x  2

y  3x  5

Calculus

 Differentiation by rule Examples

d d 1  1 1 x  x 2  1 x 2  dy     2 y dx dx 2 x dx d  4  d 4 x n nx n1    4x 1   4x 2   dx  x  dx x 2 ax n anx n1 d  x  d 1 1 ax a    2 x  dx  2  dx 2 a 0 d du dv dw 10  0 u  v  w   dx dx dx dx d 2 3x  x  5  6x 1 dx

 Vocabulary and more notation dy is the derivative of y (with respect to x) dx dy is the differential coefficient of y (with respect to x). dx Example 푦 = 푥3 − 4푥2 + 3푥 − 1 푑푦 = 3푥2 − 8푥 + 3 푑푥

2 2 x  2 x 2 3 1 Example f (x)     x 2  2x  2 x x x

1  3 1 3 2 1 2 3 2 1 3 1  f (x)  2 x   2 2x  2 x  3  x  x 2 2 x x dy  The gradient of a curve at any point is given by the value of at that point. dx Example Find the gradient at the point P(1, 5) on the graph of y  x2  2x  2 . Hence find the equation of the tangent at P.

2 y y  x  2x  2 dy  2x  2 dx P(1, 5)  At P(1, 5) gradient = 4 Tangent at P 0 x y  5  4x 1

 y  4x 1

 Stationary points on the graph of a function are points where the gradient is zero.

STATIONARY POINTS

TURNING POINTS POINTS OF INFLEXION _ + + _ _ + + _

MAXIMUM POINT MINIMUM POINT TANGENT PASSING THROUGH THE CURVE

 To obtain coordinates of a SP. on the graph of y  f (x) (i) Put f (x)  0 and solve for x. (ii) If x  a is a solution of (i) the SP will be a, f (a). (iii) If f (a)  0 there will be a minimum point at x  a

If f (a)  0 there will be a maximum point at

If f (a)  0 there could be max or min or inflexion so the second derivative rule fails. Investigate the gradient to the immediate left and right of the stationary point. (see the + and - signs on the diagrams in the previous section).

Example Find the stationary points on the graphs of

(i) y  x2  2x  2

(ii) y  x3  3x  2 and sketch the graphs.

(i) Here we have a quadratic function, which will have a true max or min. y  x 2  2x  2 dy  2x  2 y dx

 SP at 2x  2  0 2 Check point (0, 2) i.e. at x  1 (-1, 1) i.e. at 1, 1 0 x

d 2 y  2  0 dx 2

 SP is a minimum. y

(ii) (-1, 4) (2, 4) dy  3x 2  3 2 dx

2 For SP 3x  3x  0 (-2, 0) 0 (1, 0) x

 x2 1 x  1

 SPs at (1, 0) (-1, 4) d 2 y  6x dx2

d 2 y At (1, 0)  6  0  Min dx 2 d 2 y At (-1, 4)  6  0  Max dx 2

Check points (0, 2) (2, 4) (-2, 0)

Note that the turning points are Local Max and Local Min

d 2 y  6x dx2

d 2 y At (1, 0)  6  0  Min dx 2

At (-1, 4)  Max

Check points (0, 2) (2, 4) (-2, 0)

Note that the turning points are Local Max and Local Min

 Integration

 Indefinite integrals

y ydx Indefinite integration is  x n1 the reverse of x n  c n  1 differentiation. Every n 1 n1 n ax  indefinite integral must ax  c n 1  Special cases worth have an arbitrary constant  x 2 remembering x  c  added. 2  a ax  c u  v  w udx   vdx   wdx  f (x)dx reads “the (indefinite) integral of f (x) with respect to x” f (x) is called the integrand. dx is the differential of the integration and must never be omitted. x 1 dx Example Find (i) 3x 12 dx (ii) dx (iii)   x  x 2

9x3 6x 2 3x 12 dx  9x 2  6x 1dx    x  c (i)   3 2  3x3  3x 2  x  c 3 1 2 2 x 1  x 1  1  1 x x dx    dx  x 2  x 2 dx    c       3 1 (ii) x  x x  2 2 2  3 x x  2 x  c

dx 1 x 1 (iii) Not a misprint!  .dx  x 2dx   c  x 2  x 2  1 1    c x

 Definite integrals

b If I  f (x)dx  F(x)  c , then the definite integral f (x)dx is the difference in the   a value of I when x  b and x  a .

b i.e. f (x)dx  F(b)  F(a) no constant! a The limits of the definite integral are a (lower limit) and b (upper limit).

b Note the use of square brackets. F(x)a  F(b)  F(a)

3  1  Example Evaluate  x  3 dx  1  x 

3 3 3 2 2  1  3  x x   x  3 dx  x  x dx      1  1  x   2  21 3  x 2 1   9 1   1 1  40    2           2 2x  2 18   2 2  9  1

 Area on a graph as a definite integral (i)

b y A  ydx  a

A 0 a b x

y

a b b 0 ydx  A x  a A i.e. the value of the definite integral will be negative if y is negative for a  x  b (iii) y

A b b 0 a ydx  A  B x  a B

(iv)

b y xdy  A  a b

A a

0 x

y (v) y2

b

y1  y2 dx  A  a A NB y  y for a  x  b a 1 2 0 b x

y1

NB (i) most certainly will be tested, and (iv) could be. (ii) and (iii) most unlikely to be included. (iv) most likely area between a line and a curve.

1 Example Find the area enclosed between the graph of y  1 the x-axis and the x

9 ordinates at x = 1 and x  4

9 9 4  1  4  1 y A  1 dx  1 x 2 dx sketch  1  x   1 9 1 4  x 2  9  x    x  2 x 4 1  1  2  A  1 0 1 9 x 4  9 3  9    2    1 2   4 2  4

Example The diagram shows the sketch of graph of y  x2  2x  3 and y  x 1. Find the x coordinates of the points of intersection P and Q of the graphs. Calculate the shaded area.

For P, Q y

2 y  x  2x  3 2 Q x  2x  3  x 1 y  x 1  x 2  3x  4  0  P x 1x  4  0 0 x  x  1, 4

Shaded area = y  y dx  1 2

4 4  x 1 x 2  2x  3dx  3x  4  x 2 dx 1 1 4 3x 2 x3   64   3 1     4x    24 16      4    2 3  1  3   2 3 1  21 6

Trigonometry

 Trig ratios for 30, 60, 45

30 2 2 2 1 3

60 45

1 1 1 sin30  cos60  1 sin 45  cos 45  1 2 2

3 sin 60  cos30  2 tan45 1

tan60  3 tan30  1 3

 Trig ratios for all angles NB the CAST DIAGRAM For the sign of a trig ratio S A All positive in first quadrant Sine (only) in second quadrant C T Etc…

Example Without using a calculator find (i) cos150 (ii) tan210 (iii) sin 240  (i) (ii) (iii) S A S A S A 150 210 30 60 30 -240 T C T C T C cos150  cos30 tan 210  tan30 sin 240   sin 60

3  1 3   2 3  2

 Trig of Scalene triangles Sine rule B a b  c      A sin A sin B  sinC  a b

C

Given AAS use it to find a second side Given SSA use it to find a second angle (but take care to choose the angle size appropriately –it could be acute or obtuse).

Cosine rule c B a 2  b 2  c 2  2bc cos A b 2  c 2  a 2 A cos A  a 2bc

b Both formulae with two more sets.

C

Given SAS use it to find the third side Given SSS use it to find an angle (no possible ambiguity here).

Example Triangle PQR has PR = 3cm, QR = 7cm and QPˆR  36

Find (i) QR using the cosine rule and then (ii) PQˆR using the sine rule.

Q (i) QR2  9  49  42cos36  24.021...

QR  4.901..  4.90 3 R 7 4.901.. (ii)  36 7 sin PQR sin 36 7sin 36 P sin PQR   0.8394... 4.901..  PQR  57.086.. or PQR 122.914.. It can’t be 57.08.. since R would be 86.92.. and would be the largest angle in the triangle, but R faces the smallest side so is the smallest angle. Hence PQR 122.91

1  Area “ 2 absinC ” rule given SAS Area of triangle =

 Graphs of trig functions (all periodic) 1. Graph of y  sin x

y 1 Period 2

sin(2  x)  sin x

0 sin x  1 x

-1

2. Graph of y  cos x y

1 Period 2

cos(2  x)  cos x 0 x cos x  1

-1

3. Graph of y  tan x

Period  y tan(  x)  tan x

Vertical asymptotes at

 3  0   3 x   2 , x   2 , etc  2 x 2 2 2

Vertical asymptotes

 Boundary values of trig ratios S=1 T- C=0 T Verify these from graphs

S=T=0 S=T=0 C= -1 C=1

T S= -1 T- C=0

 Two important trig identities sin  tan sin 2   cos 2  1 cos

8 Example Given  is obtuse and sin  17 find the values of cos and tan . sin 2   cos 2  1  cos 2  1 sin 2  S 64 A  1 289  225  289

15  cos   17 T C sin 8 tan   tan  17   8 15 15 cos  17 NB Learn how to rearrange the identities sin sin  cos tan cos  tan sin 2  1 cos 2 

1 1 1  Trig equations Remember that from your calculator sin , cos and tan give the principal value (p.v.) Example Solve the equations (i) tan  1.5 for 0   360 (ii) sin 2  0.5 for 180  180

(iii) 2cos 2  1 sin for

(iv) 2sin 2   sin cos for

3 (v) sin  80  for 180  180 2 (i) S A

tan  1.5 PV = -56.30   T   124 , 304 C

(ii) sin 2  0.5 …..first solve for 2 for  360   360 S A 2  30, 150;  210,  330 PV = 30    15 , 75 ; 105 , 165 T C

(iii) (In this example, use cos 2  1 sin 2  )

2cos 2   1 sin 21 sin 2    1 sin 2  2sin 2   1 sin 2sin 2   sin 1  0 S A sin 12sin 1  0 PV = -30 sin  1 sin   1 2 T  or C   90   210 , 330

   90 , 210 , 330

(iv) Don’t cancel out sin . Bring to LHS and factorise

2sin 2   sin cos 2sin 2   sin cos  0 sin 2sin  cos   0

 sin  0 or 2sin  cos S A PV = 26.56 sin 1   0 , 180  cos 2 T C 1 tan  2   27 , 207

   0 , 180 , 27 , 207

3 (v) sin  80  solve first for  260  100 2 S A   80  60 ,  240 PV = 60   140 , 160 T C

In the next example, angles are in radians. The radian sign c is sometimes omitted, but is implied when the interval contains .

Example Solve the following equations

(i) cos x  0.3 for 0  x  2 , answers correct to 2d.p.

x (ii) tan  3 for  2  x  2 , answers in exact form 2

(i) ……put calculator into RAD mode. S A x  1.266..., 2 1.266... PV = 1.2661.. x  1.27, 5.02 T C

(ii) In exact terms means in terms of . The implication is that the angles will be exact form in degrees. So, work in degrees first and then convert to radians.

… solve first for   x  

x  60 ,120 2 S A  x  120 ,240 PV = 60 2 4 T  x  ,  C 3 3

Mechanics

1. Rectilinear motion with constant acceleration a

t = 0 t

u v

s

Remember that s is a displacement, is directed, and should be shown with one arrow head. v  u  at 1 s  ut  at 2 2

v 2  u 2  2as 1 s  u  vt 2 Example A cyclist moves along a straight line passing through points O, A and B with constant acceleration. 2 seconds after passing O he is at A where OA = 9m and after a further 4 seconds he is at B where AB = 36m. Find his constant acceleration, his speed at Oa and his speed at B.

t t = 2 t = 6 B 9m A u v

45m 1 s  ut  at 2 2

O – A 9  2u  2a O – B 45  6u 18a  18  2a  a 1.5m/ s 2 , u  3m/ s v  u  at

O – B v  31.56 12m/ s

Properties of the velocity/time graph Gradient = acceleration Area under graph = distance travelled Example A train starting from rest is uniformly accelerated during the first ½ km of its run, maintains its acquired speed for the next 1½ km, and is then brought to rest with uniform retardation in the last ¼ km. The time for the whole journey is 5 minutes. Find the acceleration in the first part of the run and the retardation in the final stage. velocity v

500 1500

250

time t1 t2 t 3

1 Area t v  500  t v 1000 2 1 1

t2v 1500

1 t v  250  t v  500 2 3 3

t1v  t2v  t3v  3000

 vt1  t2  t3   3000 v  300  3000 v  300  3000 v  10m / s

 t1  100 and t3  50

10 1 Gradient 1st stage acceleration =  m / s 2 100 10

10 1 3rd stage deceleration =  m / s 2 50 5

2. Vertical motion under gravity The assumption here will be that the motion is unrestricted, so that the constant acceleration formulae can be used with g usually given as having magnitude 9.8 m/s2

Example A ball is thrown vertically upwards with speed 14.7 m/s from a platform 19.6m above level ground. Find the time for the ball to reach the ground (g = 9.8 m/s2) State any assumption that you make.

Notice here how the displacement is 14.7m/s shown from the platform downwards 9.8m/s2 and NOT from ground upwards. This is very important. 19.6m

t

1 1 s  ut  at 2 19.6  14.7t  9.8t 2 2 2

 t 2  3t  4  0  t 1t  3  0  t  3s

Assume no air resistance during motion.

Probability Distribution –Binomial

The binomial random variable X is where the probability of success remains constant from trial to trial –say ‘p’. Therefore the probability of failure is ‘q’ where 푞 = 1 − 푝. When the trial is repeated for n trials then a probability can be found using the following formula.

n r nr PX  r Cr  p  q where 푟 = 1,2,3,4, … . 푛

Example When a drawing pin is thrown onto a table, the probability that it will fall ‘point upwards’ is 0.2. Ten drawing pins in a packet are thrown onto a table, work out the probability that (i) All land point down (ii) Exactly three land point up (iii) Less than two land point up.

푝 = 0.2, 푞 = 0.8 (i) P(all land point down) = 0.8010  0.107

10 3 7 (ii) P(exactly three land point up) = PX  3 C3  0.2  0.8

120  0.23  0.87  0.201 (iii) P(less than two land point up) = PX  2 PX  0 PX 1

 0.107 10  0.2  0.89

 0.375