a-level maths question and answers 2015 · length x 200.1 201.8 201.8 202 204.3 205.1 ... = 12 ln...

A-Level Maths Question and Answers 2015

2

Table of Contents Advanced Algebra (Questions) ........................................................................................................... 3

Advanced Algebra (Answers) .............................................................................................................. 4

Basic Algebra (Questions) ................................................................................................................... 7

Basic Algebra (Answers) ...................................................................................................................... 8

Bivariate Data (Questions) .................................................................................................................. 9

Bivariate Data (Questions) ................................................................................................................ 11

Coordinate Geometry ......................................................................................................................... 13

Differentiation (Questions) ................................................................................................................ 15

Differentiation (Questions) ................................................................................................................ 17

Functions (Questions) ........................................................................................................................ 20

Functions (Answers) ........................................................................................................................... 21

Integration (Questions) ..................................................................................................................... 23

Integration (Answers) ........................................................................................................................ 24

Numerical Methods (Questions) ....................................................................................................... 25

Numerical Methods (Answers) .......................................................................................................... 27

Probability (Questions) ...................................................................................................................... 30

Probability (Answers) ......................................................................................................................... 32

Probability Distributions (Questions) ............................................................................................... 34

Probability Distributions (Answers) .................................................................................................. 36

Representation of Data (Questions) ................................................................................................ 38

Representation of Data (Answers) ................................................................................................... 39

Sequences and Series (Questions) .................................................................................................. 40

Sequences and Series (Answers) ..................................................................................................... 41

The Normal Distribution (Questions) ............................................................................................... 42

The Normal Distribution (Answers) .................................................................................................. 43

Trigonometry (Questions) ................................................................................................................. 45

Trigonometry (Answers) .................................................................................................................... 47

Vectors, Lines and Planes (Questions) ............................................................................................ 50

Vectors, Lines and Planes (Answers) ............................................................................................... 51

3

Advanced Algebra (Questions)

1. The function f is given by:

Given that:

where A, B and C are constants,

(i) find the values of B and C,

(ii) show that A = 0.

(Marks available: 4 marks)

2. The function f is defined for the domain by:

a) Deduce that f is a decreasing function for all.

(Marks available: 2)

3. A function is given by:

find the value of the constants A and B.


4

Advanced Algebra (Answers)

Answer outline and marking scheme for question: 1

Give yourself marks for mentioning any of the points below:

(max 3 marks)

(max 1 mark)



a) The function f(x) is decreasing if f '(x) is negative. Therefore we need to show that f '(x) is less

than zero, shown below:

rearranging this and changing back to square roots gives:

rearranging further gives:

which is always true for all.

5

(2 marks)




Adding the two fractions on the right hand side gives:

As the denominators are now the same the numerators must match as well.

Therefore:

Selecting x = -1, will eliminate the constant B.

This gives:

-2 -1 = A (-3 + 2 )

-3 = -A

A = 3

Selecting x = 2/3 to this will eliminate the constant A.

This gives:

-2/3 x 2 -1 = B (-2/3 +1)

-4/3 -1 = B (1/3)

-7/3 = 1/3 B

-7 = B

B = -7

Therefore:

(Marks available: 3

7

Basic Algebra (Questions)

1. It is given that:

f(x) = 8x3 + 18x2 + 3x -2.

a) Use the factor theorem to show that (x + 2) and (4x -1) are factors of f(x).

b) Given that the other factor of f(x) is (a + 1 ), find the roots of the equations.

(i) f(x) = 0,

(ii) f(a) = 0.


8

Basic Algebra (Answers)



a) Use the factor theorem

f (-2) = -64 + 72 - 6 - 2 = 0, thus (x+2) is a factor.

f (1/4) = 1/8 + 9/8 + 3/4 - 2 = 0, thus (4x-1) is a factor.

(2 marks)

b) (i) replacing the given equation with its factors:

f (x) = (x+2) (4x-1) (2x+1)

This gives roots of x = -2, x = 1/4, x = -1/2.

(1 mark)

(ii) change the above equation to a function of 2x

f (2x) = (2x+2) (8x-1) (4x+1)

This gives roots of x = -1, x = 1/8, x = -1/4.

(2 marks)


9

Bivariate Data (Questions)

1. Sketch scatter diagrams to show possible forms (or describe in words) for the following values of the product

moment correlation coefficient, r.

a) r = 0.78

b) r = -1

c) r = -0.25

(3 marks)

d) calculate the regression line x on y for the following summarised data

n = 12 Σx = 1800 Σx2 = 336296 Σy = 36.0 Σy2 = 126.34 Σxy = 6348.6

(5 marks)


2. A wine expert grades 10 bottles of wine on a scale from 0 to 50. He records the results next to their ages

Wine A B C D E F G H I J

Age 15 21 24 28 30 34 36 40 42 44

Score 18 11 14 25 27 17 33 22 37 41

a) calculate the product moment correlation coefficient for the age of the wine against the grade given

b) calculate Spearman's rank correlation coefficient for the same data and comment on the results


3. A physicist wants to find out what happens to a length of metal rod (cm) when put under various

temperatures (°C).

She carries out an experiment and her data is as follows

10

Temp t 105 110 115 120 125 130

Length x 200.1 201.8 201.8 202 204.3 205.1

The physicist would like to calculate a line of regression for this data.

a) advise the physicist on which line to use 'x on t' or 't on x'

b) calculate this line of regression and use it to estimate the length of a bar at 145°C


11

Bivariate Data (Questions)


a) a good positive correlation

b) a perfect negative correlation - all points lie on line

c) a poor negative correlation - a fair degree of scatter

(3 marks)

d) need Sxy = 6348.6 - 1800 x 36 = 948.6

12

Syy = 126.34 - 362 = 18.34

12

so b' = 948.6 = 51.7

18.34

a' = 1800 - 51.7 x 36 = -5.17

12 12

so regression line x on y is x = 51.7y - 5.17

(5 marks)



a) data obtained from calculator

n = 10 Σx = 314 Σx2 = 10678 Σy = 245 Σy2 = 6907 Σxy = 8347 and from calculator r = 0.760 (3sf)

if you prefer to use the formulae then you should obtain the following

Sxy = 654 Sxx = 818.4 Syy = 904.5 and r = 654 = 0.760

12

b) ranking


a) as the temperature appears to be controlled and independent we should calculate the line 'x on t'

b) summarised data from calc

n = 6 Σ t = 705 Σt2 = 83275 Σx = 1214.1 Σx2 = 245692.39 Σtx = 142746

Stx = 89.25 Stt = 437.5 hence b = 89.25 = 0.204

437.5

so a = 1214.1 - 0.204 x 705 = 178.38

6 6

equation of 'x on t' is x = 178.38 + 0.204t

at 1450C x = 178.38 + 0.204 x -145 = 207.96cm


13

Coordinate Geometry

1.

The diagram shows a circular path with centre O and radius r, together with two other paths along the radii AO

and OB. The size of the angle AOB is θ radians, where θ < π.

The widths of the paths may be neglected in the calculations. Peter runs along the radii AO and OB, then along

the minor arc BA. Mary runs along the major arc AB.

a) Given that Peter and Mary run the same distance, show that:

b) Given that they each run 410 metres, find the radius of the circular path correct to the nearest metre.


14

Coordinate Geometry



a) Peter's distance = r + r + r Mary's distance = 2 r - r rearranging this gives = - 1

(2 marks)

b) Rearranging the above equation to show r as a function of gives: This gives a radius of 99m

(3 marks)


15

Differentiation (Questions)

1. A curve C has the equation

a) (i) Show that:

(ii) Hence find the coordinates of the stationary point on the curve C

(iii) Show that this stationary point is a point of inflection.

b) (i) Show that:

where a and b are constants to be determined.

(ii) Deduce that the curve has another point of inflection.

c) Sketch the curve C, indicating the two points of inflection.


2. A piece of wire, of length 20cm, is to be cut into two parts. One of the parts, of length x cm, is to be formed

into a circle and the other part into a square.

a) Show that the sum, A cm2, of the areas of the circle and the square is given by

b) Show that A has a stationary value when


16

3. The variables x and y are related by

y = 4x.

a) Find the value of x when y = 12, giving your answer to two decimal places

b) Show that y = ekx, where k = In 4.

c) Hence find dy/dx.

d) Given that x is a function of a third variable t and that

dy/dx = x, deduce that dy/dx = 12 ln 12, when y = 12.


17

Differentiation (Questions)



a) (i) Using the product rule gives:

dy/dx = 2xe-x - (x2 +1)e-x

= -e-x (x -1)2

(ii) dy/dx = 0 at stationary points, thus:

0 = 2xe-x - (x2 +1)e-x

This gives stationary point at (1, 2e-1)

(iii) dy/dx (i.e. the gradient) remains the same sign either side of the stationary point. Therefore the stationary

point must be a point of inflection.

Mathematically shown below:

dy/dx is less than zero before the stationary point (i.e. at x less than one)

dy/dx equals zero at the stationary point

dy/dx is less than zero after the stationary point (i.e. at x more than one)

(5 marks)

b) (i) Using the product rule on dy/dx gives:

d2y/dx2 = 2e-x - 2xe-x - 2xe-x + (x2 +1)e-x

= e-x (x -1)(x -3)

Therefore a = 1 and b = 3.

(4 marks)

The gradient is always less than zero, so the curve slopes downwards.

From the above there are two points of inflections at x = 1 and x = 3.

The equation never becomes negative; therefore the curve does not cross the x axis.

Therefore the curve looks like:

18

(2 marks)




a) Area of circle

Side of square

Entering these to find a total area, gives:

(3 marks)

b) Differentiating the area equation above, gives:

Solving this equation when dA/dx = 0, gives:

(5 marks)


19



a) 4x = 12, using logs on either side, gives:

x log 4 = log 12

Solving this for x gives, x = 1.79 to 2d.p.

(2 marks)

b) We know that 4 = eln4 therefore, multiplying both side by the power of x, gives:

4x = (eln4)x

From this you can deduce:

y = ekx where k = ln 4.

(1 mark)

c) dy/dx = kekx = (ln 4)exln4 = (ln 4).4x = yln4.

(1 mark)

d) dy/dt = dy/dx . dx/dt

= (y ln 4).x

= (y ln 4).(ln 12/ ln 4)

When y = 12, from (a)

= 12 ln 12

(3 marks)


20

Functions (Questions)

1. A graph has equation

y = cos2x,

where x is a real number.

a) Draw a sketch of that part of the graph for which

b) On your sketch show two of the lines of symmetry which the complete graph possesses.


2. The function f is defined on the domain x > -1 by

a) Write down the equations of the asymptotes to the curve y = f(x).

b) Give the range of the function f.

c) Give the domain and range of the inverse function f -1.

d) Find an expression for f -1(x).


3. The functions f and g are defined for all real numbers by:

a) (i) State whether f is an odd function, an even function or neither.

(ii) State whether g is an odd function, an even function or neither.

b) Given that f and g are periodic functions, write down the periods of f and of g.

c) Solve, for -π < x ≤ π the equations

21


Functions (Answers)



a) The graph would look like:

Note 1: it must be a sine-wave shape - not W shape.

(the curve is only shown in the domain)

Note 2: Stationary Points at (0.1). (π/2.-1) .etc (degrees not allowed here)

(2 marks)

b) Lines of symmetry are:

x = 0 or π/2 or π etc

(you will get a mark for each correct line of symmetry, up to 2 marks).

(2 marks)




a) Asymptotes are defined by the lines

x = -1, y = 2

(2 marks)

b) The range of f is y > 2

22

(1 mark)

c) The domain of f -1 is x > 2

The range of f -1 is x > -1

(1 mark)

d) Changing subject of y = f(x)

(3 marks)




a) f is odd

g is even

(2 marks)

b) Period of f is π

Period of g is 1/2 π

(2 marks)

c) (i) Solving f (x) =1/2, gives:

(ii) Solving g (x) =1/2, gives the same results as above, but with ±.

(6 marks)


23

Integration (Questions)

1. a) By using the substitution u = 4- sin x, or otherwise, find

b) Hence, or otherwise, find


24

Integration (Answers)



a) u = 4 - sin x, therefore du = (-) cos x dx.

Substitute u into the given equation gives:

Integrating this gives:

Therefore:

(4 marks)

b) Use u = 2x, therefore 1/2 du = dx

Substitute u into the given equation and integrating gives:

(2 marks)


25

Numerical Methods (Questions)

1.

Figure 1 shows the graphs of y = tan-l x and xy = 1 intersecting at the point A with x-coordinate.

a) (i) Show that 1.1 < a < 1.2.

(ii) Use linear interpolation to find an approximation for A, giving your answer to two decimal places.

Figure 2 shows the tangent to the curve xy = 1 at A. This tangent meets the x-axis at B. The region between the

arc OA, the line AB and the x-axis is shaded.

b) Show that the x-coordinate of B is 2.


26

2.The variables x and y satisfy the differential equation

and y = 2 when x = 1.

a) Use a local linear approximation to show that, when x = 1.02,

b) By using the iterative equation

find approximations for the values of y when x takes the values 1.02, 1.04 and 1.06, giving each value to three

places of decimals.


3.

Figure 1 above shows sketches of the graphs of

y = e-x and y = x

and their intersection at x = α, where α is approximately 0.57.

(i) Starting with x1 = 0.57, carry out the iteration

xn+1 = e-xn

up to and including x5, recording each value of xn to four decimal places as you proceed.

27

(ii) Write down an estimate of the value of α to three decimal places.


Numerical Methods (Answers)



a) (i) The line intersect at a tan-l a =1.

Let f(x) = a tan-l a -1

Then f(1.1) = -0.0837206 < 0

and f (1.2) = +0.0512696 > 0

Therefore the change of sign (i.e. the x value of a is between 1.1 and 1.2)

(ii) Using linear interpolation to get a more accurate answer (to 2 decimal places).

(4 marks)

b) Rearranging the equation of the straight line, gives:

so

at point A, x = a, so:

The equation to a tangent to the straight line is:

or

At B, y = 0 therefore x = 2a.

28

(3 marks)


29



a) Using the local linear approximation, gives:

(2 marks)

b) Substituting values of x into the iterative equation given gives:

First value is 2.157

Second value is 2.316

Third value is 2.477

(4 marks)




Performing the iteration on the equation given, gives:

x2 = 0.5655

x3 = 0.5681

x4 = 0.5666

x5 = 0.5675

Taking the results above, the closest approximation to α at 3d.p is 0.567.


30

Probability (Questions)

1.Ten pupils are placed at random in a straight line. Find:

a) how many different ways there are of them lining up?

(1 marks)

b) the number of ways of arranging the 10 pupils if the 2 youngest are to stand next to each other

(3 marks)

c) how many different ways the first 4 places can be filled if they were to have a race.

(2 marks)

d) the number of combinations of selecting 2 representatives from the 10

(2 marks)


2. On my way to work the probability that I have to stop at the first set of traffic lights is 0.3. The probability of

stopping at the second is 0.75.

Find the probability that on one morning:

a) I stop at both sets of lights

(2 marks)

b) I stop only once at the second set of lights

(2 marks)

c) I stop at least once

(2 marks)


31

3. In a penalty shoot-out competition 2 players take a penalty to score a goal. Each player has a probability of

0.7 to score.

Calculate the probability of:

a) only one penalty being scored

on further inspection it appears that if the 1st penalty taker misses the probability of the 2nd penalty taker

scoring is increased to 0.85

(2 marks)

b) draw a tree diagram to show all the events of the 2 penalties using this new evidence

(2 marks)

c) find the probability that the second penalty is missed

(2 marks)

d) find the probability that the 1st penalty is missed given that the second is missed.

(3 marks)


32

Probability (Answers)


a) 10! = 3,628,800

(1 mark)

b) the trick is to treat the 2 youngest as 1 item (M1). This leaves us with arranging 9 pupils, i.e. 9! (A1) Ways.

BUT the 2 youngest can be arranged in 2! ways

so total ways = 2! x9! = 725760

(3 marks)

= 10!

6! = 5040

=

10! 8!

2!

= 45

(2 marks)



Although it isn't essential in this case it is advisable to draw a tree diagram to help with the calculations.

a) P (stop and stop) = 0.3 x 0.75 = 0.225

(2 marks)

b) P (go and stop) = 0.7 x 0.75 = 0.525

(2 marks)

33

c) P (stop at least once) = 1 - P (no stop)

= 1 - 0.175 = 0.825

(2 marks)



a) P (only 1 scored) = P (score and miss) or P (miss and score)

= 0.7 x 0.3 + 0.3 x 0.7 = 0.42

(2 marks)

b) see ans sheet 2 for tree diagram

(2 marks)

c) P (2nd miss) = P (miss and miss) or P (score and miss)

= 0.3 x 0.85 + 0.7 x 0.3 = 0.255

(2 marks)

d) P (1st miss|2nd miss) = P (1st and 2nd miss) = 0.045 = 0.176 (3sf)

P (2nd miss) 0.255

(3 marks)


34

Probability Distributions (Questions)

1. Two fair dice are thrown. If the scores are unequal, the larger of the two scores is recorded. If the scores are

equal then that score is recorded. Let X denote the number recorded.

a) show that P(X = 2) = 1/12 and draw up a table showing the probability distribution of X

(4 marks)

b) find the mean and variance of

(4 marks)


2. It is known that 65% of sixth formers think their maths teacher is cool. In a certain sample, 10 pupils are

picked at random and it is required to calculate the probability that less than 3 of the 10 think that their teacher

is cool. Name a probability distribution that can be used for modelling this situation stating one necessary

assumption for this model to be valid.

a) use the model to calculate the required probability

(4 marks)

b) calculate the mean and variance of this distribution

(3 marks)


3. When I ask a sixth form class a question the probability that I get an answer is 0.36. if I don't get an answer

first time I keep trying until I am blessed with an answer. Let X denote the number of attempts I have to make

in order to get an answer. Stating any assumption, identify the probability distribution of X.

(2 marks)

hence calculate:

a) P(X = 5)

(2 marks)

b) P(X = 4)

35

(3 marks)

c) E(X) and Var(X)

(3 marks)


36

Probability Distributions (Answers)


a) P(X = 2) = P (1 and 2) or P (2 and 1) or P (2 and 2)

= (1/6 x 1/6) + (1/6 x 1/6) + (1/6 x 1/6) = 3/36 A1

= 1/12 as required A1

x 1 2 3 4 5 6

P(X=x) 1/36 1/12 5/36 7/36 9/36 11/36

(4 marks)

b) E(X) = (1 x 1/36) + (2 x 1/12) + (3 x 5/36) + (4 x 7/36) + (5 x 9/36) + (6 x 11/36)

= 161/36

= 4.47 (2dp)

for Var(X) we first need

E(X2) = (12 x 1/36) + (22 x 1/12) + (32 x 5/36) + (42 x 7/36) + (52 x 9/36) + (62 x 11/36)

= 791/36 = 21.97 (2dp)

Var(X) = E(X2) - E(X)2

= 21.97 - 4.472

= 1.97

(4 marks)



Binomial distribution with n = 10 and probability of success, p = 0.65

assume each pupil picked is independent of each other

a) using X ~ B(10, 0.65)

37

need P(X < 3) = P(X = 0) + P(X = 1) + P(X =2)

=0.3510 + 10(0.35)9(0.65)10C2 + (0.35)8(0.65)2

= 0.0048 (2sf)

(4 marks)

b) E(X) = n x p = 10 x 0.65 = 6.5 A1

Var(X) = n x p x q = 10 x 0.65 x 0.35 = 2.275

(3 marks)



3. Geometric distribution with probability of success, p = 0.36

assume independence of sixth formers asked

(2 marks)

a) P(X = 5) = (0.64)4 x 0.36 = 0.0604 (3sf)

(2 marks)

b) X ~ Geo (0.36)

P(X = 4) = P(X > 5)

= q5 = 0.645

= 0.107 (3sf)

(3 marks)

c) E(X) = 1/p = 1/0.36 = 2.78 (3sf)

Var(X) = q/p2

= 0.64/0.362 = 4.94

(3 marks)


38

Representation of Data (Questions)

1. The following raw data was obtained from the ages of 31 people asked in a cinema one Saturday afternoon

21 17 24 23 43 42 14 51 22 18 17

15 16 23 33 21 12 13 34 22 15

12 17 22 28 29 32 38 12 11 8

a) By drawing a stem-and-leaf diagram find the median age of cinema goers, and the inter-quartile range.

b) Using the answers obtained in part a) draw a boxplot of the ages of cinema goers that afternoon and

comment on the shape of the distribution, i.e. do you think there is any skew to the distribution?

c) Are there any 'outliers' in this distribution? Use the rule that an 'outlier' is a value more than 1.5 times of the

inter-quartile range from either quartile.

2. 30 students were asked to attempt a maths problem. The time it took them to complete

the problem (to the nearest second) is given in the table.

Time (secs) Frequency

0 = t < 10 3

10 = t < 15 7

15 = t < 20 10

20 = t < 30 6

30 = t < 50 4

a) explain why representing this data on a histogram is appropriate.

b) represent the data on a histogram.


39

Representation of Data (Answers)


1. a) see ans sheet 1

the median is 16th value = 21

upper quartile = 29

lower quartile = 15

inter-quartile range = 29 - 15 = 14

b) see ans sheet 1

there appears to be a slight positive skew to the distribution.

c) inter-quartile range = 14 so 14 x 1.5 = 21

do we have any values that lie 21 away from either quartile?

Upper end 29 + 21 = 50

Lower end 8 - 21 = -13

Hence 1 outlier the value of 51


a) the data is continuous and we have been given unequal class intervals

b)

Time (secs) Frequency Class Width Frequency density

0 = t < 10 3 10 0.3

10 = t < 15 7 5 1.4

15 = t < 20 10 5 2

40

20 = t < 30 6 10 0.6

30 = t < 50 4 20 0.2

Sequences and Series (Questions)

1. A sequence ul, u2, u3,... is defined by

ul = 10,

un+l = 0.9un.

a) Find the value of u4

b) Find an expression for un in terms of n.

c) Find


2. Every year the Queen presents special coins (Maundy Money) to a number of selected people. The number of

people receiving the coins in a year is equal to twice the Queen's age in years.

Given that in 1952, the first year of the Queen’s reign, her age was 26,

a) find an expression for the number of people receiving the coins in the nth year of her reign,

b) calculate the total number of people receiving the coins from 1952 to 1998 inclusive.


41

Sequences and Series (Answers)



a) Simply placing calculating the values of u1, u2, u3 and u4, gives:

u4 = 7.29

(1 mark)

b) Using the series equation nth term = arn-1, gives:

un = 10(0.9)n-l

(2 marks)

c) Using the 'Sum to infinity' equation:

sum = a/(1-r)

Substituting in a and r, gives:

Sum to infinity = 100.

(2 marks)




a) Use a + (n -1) d with a =52 and d=2 The nth term = 52 + 2(n - 1)

Any alternative methods leading to mn+c with, m = 2 and c = 50 is also acceptable.

(3 marks)

b) Establish that n = 47 (No of terms)

Use the equation Marks available: 1/2n x (2a + (n -1) d)

Therefore the No of people = 4606

(3 marks)

42


The Normal Distribution (Questions)

1. The time taken for a paperboy to deliver his papers is normally distributed with mean 52 minutes and

standard deviation 6.5 minutes.

Find the probability on any given day the paperboy takes:

a) longer than 1 hour to deliver

(3 marks)

b) less than 45 minutes

(3 marks)

c) between 48 and 56 minutes

(3 marks)


2. An Olympic high jumper jumps distances which are normally distributed with mean, μ = 2.45m and variance

0.49m.

a) find the probability the jumper manages a height over 2.35m

(2 marks)

b) What distance is exceeded by 5% of his jumps?

(4 marks)


43

The Normal Distribution (Answers)


1. X ~ N(52, 6.52)

a) P(X > 60) = P(Z > 60 - 52) add bell on ans sheet 3 A1

6.5

= P(Z > 1.23) = 1 - F(1.23) A1

= 1 - 0.8907 = 0.1093 A1

(3 marks)

b) P(X < 45) = P(Z < 45 - 52) add bell on ans sheet 3 A1

6.5

= P(Z < -1.08) A1

= 1 - F(1.08)

= 1 - 0.8599 = 0.1401 A1

(3 marks)

c) P(48 < X < 56) = P(48 - 52 < Z < 56 - 52) A1

6.5 6.5

= P(-0.62 < Z < 0.62) bell on ans sheet 3

= F(0.62) - [1-F(0.62)] A1

= 2F(0.62) - 1 = 2 ' 0.7324 - 1

= 0.4648 A1

(3 marks)



X ~ N(2.45,0.49) we are given the variance so s = v0.49= 0.7

44

a) P(X > 2.35) = P(Z > 2.35 - 2.45) add bell on sheet 3

0.7

= P(Z > -0.14) = F(0.14) = 0.5557

(2 marks)

b) we need the distance d where

P(X > d) = 0.05

P(Z > d - 2.45) = 0.05 sheet 3 for bell

0.7

1 - F(d - 2.45) = 0.05 hence F(d - 2.45) = 0.95

0.7 0.7

so d - 2.45 = 1.645 hence d = 3.6m ( a very high jump!)

0.7

(4 marks)


45

Trigonometry (Questions)

1. a) Express 2cos x - sin x in the form Rcos (x + a), where R is a positive constant and α is an angle

between 0° and 360°.

b) Given that 0 ≤ x < 360°

(i) solve 2cos x - sin x = 1

(ii) deduce the solution set of the inequality 2cos x- sin x ≥ 1.


Answer

2.

The diagram shows the triangle ABC in which AB = 7 cm, BC = 9cm and CA = 8cm.

a) Use the cosine rule to find cos C, giving your answer as a fraction in its lowest

terms.

b) Hence show that sin C =

c) Find sinA in the form where p and q are positive integers to be determined.


http://www.s-cool.co.uk/a-level/maths/trigonometry/test-it/exam-style-questions

46

3. a) Express 2 sin θ cos 6θ as a difference of two sines.

b) Hence prove the identity

c) Deduce that


47

Trigonometry (Answers)



a) Using the Rcos formula give:

and

Therefore

(2 marks)

b) (i) input values into the Rcos formula

solving the above equation gives x = 36.9o and x = 270o.

(2 marks)

(ii) solving the given in-equality gives:

(2 marks)




a) Applying the cosine rule gives:

48

(2 marks)

b) Rearranging gives:

(2 marks)

c) Appling the sine rule gives:

(3 marks)

Total 7 marks



a) Using the sine rule:

2sin θ cost 6θ = sin 7θ - sin 5θ

(1 mark)

b) Applying the sine rule again:

2sin θ cost 4θ = sin 5θ - sin 3θ

2sin θ cos 2θ = sin 3θ - sin θ

Adding the three expressions above, gives:

(3 marks)

49

c) Substitute θ = 2π/7.

Putting this into the equation in (b) gives:

.

(3 marks)


50

Vectors, Lines and Planes (Questions)

1. The line L passes through the points A (3, 0, -1) and B (5, -1, 4).

a) Find the vector equation of the line L.

b) Determine whether or not the line L intersects the line with the equations


2. A body of mass 0.5 kg moves so that its velocity at time t seconds is

Find the magnitude of the momentum when t = 0 and t = 2.


3. Two lines A and B, have the following formulas:

and

a) determine whether these two lines intersect

b) find the angle between them.


51

Vectors, Lines and Planes (Answers)



a) The equation for a line should be expressed as:

r = a + λb

Where a is a point on the line, b is a vector parallel to the line and λ is any number.

a = the first point A.

b = point B minus point A.

Putting these values into the equation of the line above, gives:

(2 marks)

b) Consider the point where the x values are the same for both lines, therefore:

3 + 2λ = 5 - 4μ

Consider the point where the y values are the same for both lines, therefore:

0 - 1λ = 1 + 1μ

Solving these equation simultaneously, gives:

λ= -3, μ = 2.

Putting these values into equation for line L, gives:

z = -1 + 5λ = -16

52

Putting these values into equation for line r, gives:

z = 11 + 3μ= 17.

As the value of z is not the same, both the line cannot be at the same point in space (i.e. they do not intersect).

(4 marks)




At t = 0, the vectors equals

The magnitude of the velocity equals

Therefore the momentum at t = 0 equals

0.5 x 12.17 = 6.08 kgms-1.

Performing the same calculation at t = 2, gives the momentum equal to

0.5 x 4.47 = 2.23 kgms-1.

(Marks available: 3 marks)



a) Matching the x-values gives: 4 - 4λ = 6 +2μ

Matching the y-values gives: 0 + 8λ = -10 - 6μ

Matching the z-values gives: -2 -2λ = -10 -2μ

Solving the first two simultaneous equations gives: λ = 1, μ = -3.

53

These values work in the third equation therefore the lines meet.

Substituting λ = 1, μ = -3 into the equation of lines gives the point of intersection as being:

x = 0, y = 4, z = -2.

Therefore the lines meet at (0, 4, -2)

(3 marks)

b) The angle between the lines is the angle between the direction vectors, so using the scalar product we get,

=

= -0.855

This gives θ = 148.8o (or 180o - 148.8o = 31.2o).

(3 marks)


a-level maths question and answers 2015 · length x 200.1 201.8 201.8 202 204.3 205.1 ... = 12 ln...

Documents