Advanced Algebra (Questions)

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1 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) Coordinate Geometry Differentiation (Questions) Differentiation (Questions) Functions (Questions) Functions (Answers) Integration (Questions) Integration (Answers) Numerical Methods (Questions) Numerical Methods (Answers) Probability (Questions) Probability (Answers) Probability Distributions (Questions) Probability Distributions (Answers) Representation of Data (Questions) Representation of Data (Answers) Sequences and Series (Questions) Sequences and Series (Answers) The Normal Distribution (Questions) The Normal Distribution (Answers) Trigonometry (Questions) Trigonometry (Answers) Vectors, Lines and Planes (Questions) Vectors, Lines and Planes (Answers)

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. (Marks available: 3) 3

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) (Marks available: 4) Answer outline and marking scheme for question: 2 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. 4

5 (Marks available: 2) Answer outline and marking scheme for question: 3 Give yourself marks for mentioning any of the points below: 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 = -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 5

6 6

7 Basic Algebra (Questions) 1. It is given that: f(x) = 8x x 2 + 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. (Marks available: 5) 7

8 Basic Algebra (Answers) Answer outline and marking scheme for question: 1 Give yourself marks for mentioning any of the points below: a) Use the factor theorem f (-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. 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. (Marks available: 5) 8

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 = d) calculate the regression line x on y for the following summarised data n = 12 Σx = 1800 Σx2 = Σy = 36.0 Σy2 = Σxy = (5 marks) (Marks available: 8) 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 Score 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 (Marks available: 9) 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 9

10 Temp t Length x 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 (Marks available: 9) 10

11 Bivariate Data (Questions) Answer outline and marking scheme for question: 1 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 d) need S xy = x 36 = S yy = = so b' = = a' = x 36 = so regression line x on y is x = 51.7y (5 marks) (Marks available: 8) Answer outline and marking scheme for question: 2 a) data obtained from calculator n = 10 Σx = 314 Σx2 = Σy = 245 Σy2 = 6907 Σxy = 8347 and from calculator r = (3sf) if you prefer to use the formulae then you should obtain the following S xy = 654 S xx = S yy = and r = 654 =

12 b) ranking Answer outline and marking scheme for question: 3 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 = Σx = Σx2 = Σtx = S tx = S tt = hence b = = so a = x 705 = equation of 'x on t' is x = t at C x = x -145 = cm (Marks available: 9) 12

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. (Marks available: 5) 13

14 Coordinate Geometry Answer outline and marking scheme for question: 1 Give yourself marks for mentioning any of the points below: a) Peter's distance = r + r + r Mary's distance = 2 r - r rearranging this gives = - 1 b) Rearranging the above equation to show r as a function of gives: This gives a radius of 99m (Marks available: 5) 14

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. (Marks available: 11) 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 cm 2, of the areas of the circle and the square is given by b) Show that A has a stationary value when (Marks available: 8) 15

16 3. The variables x and y are related by y = 4 x. a) Find the value of x when y = 12, giving your answer to two decimal places b) Show that y = e kx, 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. (Marks available: 7) 16

17 Differentiation (Questions) Answer outline and marking scheme for question: 1 Give yourself marks for mentioning any of the points below: a) (i) Using the product rule gives: dy/dx = 2xe -x - (x 2 +1)e -x = -e -x (x -1) 2 (ii) dy/dx = 0 at stationary points, thus: 0 = 2xe -x - (x 2 +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: d 2 y/dx 2 = 2e -x - 2xe -x - 2xe -x + (x 2 +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: 17

18 (Marks available: 11) Answer outline and marking scheme for question: 2 Give yourself marks for mentioning any of the points below: a) Area of circle Side of square Entering these to find a total area, gives: b) Differentiating the area equation above, gives: Solving this equation when da/dx = 0, gives: (5 marks) (Marks available: 8) 18

19 Answer outline and marking scheme for question: 3 Give yourself marks for mentioning any of the points below: 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. b) We know that 4 = e ln4 therefore, multiplying both side by the power of x, gives: 4 x = (e ln4 ) x From this you can deduce: y = e kx where k = ln 4. (1 mark) c) dy/dx = ke kx = (ln 4)e xln4 = (ln 4).4 x = 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 (Marks available: 7) 19

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. (Marks available: 4) 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). (Marks available: 7) 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 20

21 (Marks available: 10) Functions (Answers) Answer outline and marking scheme for question: 1 Give yourself marks for mentioning any of the points below: 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) 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). (Marks available: 4) Answer outline and marking scheme for question: 2 Give yourself marks for mentioning any of the points below: a) Asymptotes are defined by the lines x = -1, y = 2 b) The range of f is y > 2 21

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) (Marks available: 7) Answer outline and marking scheme for question: 3 Give yourself marks for mentioning any of the points below: a) f is odd g is even b) Period of f is π Period of g is 1/2 π c) (i) Solving f (x) =1/2, gives: (ii) Solving g (x) =1/2, gives the same results as above, but with ±. (6 marks) (Marks available: 10) 22

23 Integration (Questions) 1. a) By using the substitution u = 4- sin x, or otherwise, find b) Hence, or otherwise, find (Marks available: 6) 23

24 Integration (Answers) Answer outline and marking scheme for question: 1 Give yourself marks for mentioning any of the points below: 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: (Marks available: 6) 24

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. (Marks available: 7) 25

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. (Marks available: 6) 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 (i) Starting with x 1 = 0.57, carry out the iteration x n+1 = e -xn up to and including x 5, recording each value of x n to four decimal places as you proceed. 26

27 (ii) Write down an estimate of the value of α to three decimal places. (Marks available: 4) Numerical Methods (Answers) Answer outline and marking scheme for question: 1 Give yourself marks for mentioning any of the points below: a) (i) The line intersect at a tan -l a =1. Let f(x) = a tan -l a -1 Then f(1.1) = < 0 and f (1.2) = > 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. 27

28 (Marks available: 7) 28

29 Answer outline and marking scheme for question: 2 Give yourself marks for mentioning any of the points below: a) Using the local linear approximation, gives: b) Substituting values of x into the iterative equation given gives: First value is Second value is Third value is (4 marks) (Marks available: 6) Answer outline and marking scheme for question: 3 Give yourself marks for mentioning any of the points below: Performing the iteration on the equation given, gives: x 2 = x 3 = x 4 = x 5 = Taking the results above, the closest approximation to α at 3d.p is (Marks available: 4) 29

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 c) how many different ways the first 4 places can be filled if they were to have a race. d) the number of combinations of selecting 2 representatives from the 10 (Marks available: 8) 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 Find the probability that on one morning: a) I stop at both sets of lights b) I stop only once at the second set of lights c) I stop at least once (Marks available: 6) 30

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 b) draw a tree diagram to show all the events of the 2 penalties using this new evidence c) find the probability that the second penalty is missed d) find the probability that the 1st penalty is missed given that the second is missed. (Marks available: 9) 31

32 Probability (Answers) Answer outline and marking scheme for question: 1 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! = = 10! 6! = 5040 = 10! 8! = 45 2! (Marks available: 8) Answer outline and marking scheme for question: 2 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 = b) P (go and stop) = 0.7 x 0.75 =

33 c) P (stop at least once) = 1 - P (no stop) = = (Marks available: 6) Answer outline and marking scheme for question: 3 a) P (only 1 scored) = P (score and miss) or P (miss and score) = 0.7 x x 0.7 = 0.42 b) see ans sheet 2 for tree diagram c) P (2nd miss) = P (miss and miss) or P (score and miss) = 0.3 x x 0.3 = d) P (1st miss 2nd miss) = P (1st and 2nd miss) = = (3sf) P (2nd miss) (Marks available: 9) 33

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) (Marks available: 8) 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 (Marks available: 7) 3. When I ask a sixth form class a question the probability that I get an answer is 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. hence calculate: a) P(X = 5) b) P(X = 4) 34

35 c) E(X) and Var(X) (Marks available: 10) 35

36 Probability Distributions (Answers) Answer outline and marking scheme for question: 1 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 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 = (2dp) Var(X) = E(X2) - E(X)2 = = 1.97 (4 marks) (Marks available: 8) Answer outline and marking scheme for question: 2 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) 36

37 need P(X < 3) = P(X = 0) + P(X = 1) + P(X =2) = (0.35)9(0.65) 10 C 2 + (0.35)8(0.65)2 = (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 = (Marks available: 7) Answer outline and marking scheme for question: 3 3. Geometric distribution with probability of success, p = 0.36 assume independence of sixth formers asked a) P(X = 5) = (0.64) 4 x 0.36 = (3sf) b) X ~ Geo (0.36) P(X = 4) = P(X > 5) = q 5 = = (3sf) c) E(X) = 1/p = 1/0.36 = 2.78 (3sf) Var(X) = q/p 2 = 0.64/0.362 = 4.94 (Marks available: 10) 37

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 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 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 < = t < = t < = t < = t < 50 4 a) explain why representing this data on a histogram is appropriate. b) represent the data on a histogram. (Marks available: 5) 38

39 Representation of Data (Answers) Answer outline and marking scheme for question: 1 1. a) see ans sheet 1 the median is 16 th value = 21 upper quartile = 29 lower quartile = 15 inter-quartile range = = 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 = 50 Lower end 8-21 = -13 Hence 1 outlier the value of 51 Answer outline and marking scheme for question: 2 a) the data is continuous and we have been given unequal class intervals b) Time (secs) Frequency Class Width Frequency density 0 = t < = t < = t <

40 20 = t < = t < Sequences and Series (Questions) 1. A sequence u l, u 2, u 3,... is defined by u l = 10, u n+l = 0.9un. a) Find the value of u 4 b) Find an expression for u n in terms of n. c) Find (Marks available: 5) 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. (Marks available: 6) 40

41 Sequences and Series (Answers) Answer outline and marking scheme for question: 1 Give yourself marks for mentioning any of the points below: a) Simply placing calculating the values of u 1, u 2, u 3 and u 4, gives: u 4 = 7.29 (1 mark) b) Using the series equation n th term = ar n-1, gives: u n = 10(0.9) n-l c) Using the 'Sum to infinity' equation: sum = a/(1-r) Substituting in a and r, gives: Sum to infinity = 100. (Marks available: 5) Answer outline and marking scheme for question: 2 Give yourself marks for mentioning any of the points below: a) Use a + (n -1) d with a =52 and d=2 The n th term = (n - 1) Any alternative methods leading to mn+c with, m = 2 and c = 50 is also acceptable. 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 =

42 (Marks available: 6) 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 b) less than 45 minutes c) between 48 and 56 minutes (Marks available: 9) 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 b) What distance is exceeded by 5% of his jumps? (4 marks) (Marks available: 6) 42

43 The Normal Distribution (Answers) Answer outline and marking scheme for question: 1 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 = = A1 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) = = A1 c) P(48 < X < 56) = P(48-52 < Z < 56-52) A = P(-0.62 < Z < 0.62) bell on ans sheet 3 = F(0.62) - [1-F(0.62)] A1 = 2F(0.62) - 1 = 2 ' = A1 (Marks available: 9) Answer outline and marking scheme for question: 2 X ~ N(2.45,0.49) we are given the variance so s = v0.49=

44 a) P(X > 2.35) = P(Z > ) add bell on sheet = P(Z > -0.14) = F(0.14) = b) we need the distance d where P(X > d) = 0.05 P(Z > d ) = 0.05 sheet 3 for bell F(d ) = 0.05 hence F(d ) = so d = hence d = 3.6m ( a very high jump!) 0.7 (4 marks) (Marks available: 6) 44

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. (Marks available: 6) 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. (Marks available: 7) 45

46 3. a) Express 2 sin θ cos 6θ as a difference of two sines. b) Hence prove the identity c) Deduce that (Marks available: 7) 46

47 Trigonometry (Answers) Answer outline and marking scheme for question: 1 Give yourself marks for mentioning any of the points below: a) Using the Rcos formula give: and Therefore b) (i) input values into the Rcos formula solving the above equation gives x = 36.9 o and x = 270 o. (ii) solving the given in-equality gives: (Marks available: 6) Answer outline and marking scheme for question: 2 Give yourself marks for mentioning any of the points below: a) Applying the cosine rule gives: 47

48 b) Rearranging gives: c) Appling the sine rule gives: Total 7 marks Answer outline and marking scheme for question: 3 Give yourself marks for mentioning any of the points below: 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: 48

49 c) Substitute θ = 2π/7. Putting this into the equation in (b) gives:. (Marks available: 7) 49

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 (Marks available: 6) 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. (Marks available: 3) 3. Two lines A and B, have the following formulas: and a) determine whether these two lines intersect b) find the angle between them. (Marks available: 6) 50

51 Vectors, Lines and Planes (Answers) Answer outline and marking scheme for question: 1 Give yourself marks for mentioning any of the points below: 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: 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 = λ =

52 Putting these values into equation for line r, gives: z = μ= 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) (Marks available: 6) Answer outline and marking scheme for question: 2 Give yourself marks for mentioning any of the points below: At t = 0, the vectors equals The magnitude of the velocity equals Therefore the momentum at t = 0 equals 0.5 x = 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) Answer outline and marking scheme for question: 3 Give yourself marks for mentioning any of the points below: 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 =

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) b) The angle between the lines is the angle between the direction vectors, so using the scalar product we get, = = This gives θ = o (or 180 o o = 31.2 o ). (Marks available: 6) 53

A-Level Maths Revision notes 2014

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