PaperPractice [0 marks] Consider the expansion of (x + ) 10. 1a. Write down the number of terms in this expansion. [1 mark] 11 terms N1 [1 mark] 1b. Find the term containing x. evidence of binomial expansion ( n ) a n r b r, attempt to expand r evidence of choosing correct term () 8 th 10 term, r = 7, ( ), (x) () 7 7 correct working () 10 ( ) (x) () 7 10, ( )(x) () 7, 7 6440 x (accept 6000 x ) N The following table shows the average weights ( y kg) for given heights (x cm) in a population of men. Heights (x cm) 165 170 175 180 185 Weights (y kg) 67.8 70.0 7.7 75.5 77. a. The relationship between the variables is modelled by the rression equation y = ax + b. Write down the value of a and of b. a = 0.486 (exact) N1 b = 1.41 (exact), 1.4 N1 b. The relationship between the variables is modelled by the rression equation y = ax + b. Hence, estimate the weight of a man whose height is 17 cm.
correct substitution () 0.486(17) 1.41 71.18 71. (kg) N c. Write down the correlation coefficient. [1 mark] r = 0.99776 r = 0.997 [1 mark] N1 d. State which two of the following describe the correlation between the variables. strong zero positive native no correlation weak strong, positive (must have both correct) A N The following diagram shows two perpendicular vectors u and v. a. Let w = u v. Represent w on the diagram above.
METHOD 1 N Note: Award for sment connecting endpoints and for direction (must see arrow). METHOD N Notes: Award for sment connecting endpoints and for direction (must see arrow). Additional lines not required. b. Given that u = and, where, find \(n\). v = 5 1 n n Z evidence of setting scalar product equal to zero (seen anywhere) u v = 0, 15 + n + = 0 correct expression for scalar product () 5 + n + 1, n + 18 = 0 attempt to solve equation n = 18 n = 9 N R1 4. Let f(x) = g(x), where g() = 18, h() = 6, g () = 5, and h () =. Find the equation of the normal to the graph of f at x =. h(x) [7 marks]
recognizing need to find f() or f () (R1) f() = 18 (seen anywhere) () 6 correct substitution into the quotient rule gradient of normal is 6 () () attempt to use the point and gradient to find equation of straight line correct equation in any form N4 [7 marks] 6(5) 18() 6 f 6 () = 6 1 y f() = (x ) f () y = 6(x ), y = 6x 9 Let f(x) = 5 x. Part of the graph of fis shown in the following diagram. The graph crosses the x-axis at the points A and B. 5a. Find the x-coordinate of A and of B. recognizing f(x) = 0 f = 0, x = 5 x = ±.606 x = ± 5 (exact), x = ±.4 N 5b. The rion enclosed by the graph of f and the x-axis is revolved 60 about the x-axis. Find the volume of the solid formed. attempt to substitute either limits or the function into formula involving f π (5 x ) dx, π.4 ( x 4 10 x + 5), π.4 187.8 volume = 187 A N 0 f
The following table shows the amount of fuel ( y litres) used by a car to travel certain distances ( x km). Distance (x km) 40 75 10 150 195 Amount of fuel (y litres).6 6.5 9.9 1.1 16. y = ax + b This data can be modelled by the rression line with equation. 6a. Write down the value of a and of b. a = 0.08604, b = 0.06186 a = 0.084, b = 0.06 N 6b. Explain what the gradient a represents. [1 mark] correct explanation with reference to number of litres required for 1 km N1 a represents the (average) amount of fuel (litres) required to drive 1 km, (average) litres per kilometre, (average) rate of change in fuel used for each km travelled [1 marks] 6c. Use the model to estimate the amount of fuel the car would use if it is driven 110 km. valid approach y = 0.084(110) + 0.06, sketch 9.658 9.7 (litres) N Let A and B be independent events, where P(A) = 0. and P(B) = 0.6. 7a. Find P(A B). correct substitution 0. 0.6 P(A B) = 0.18 () N 7b. Find P(A B).
correct substitution () P(A B) = 0. + 0.6 0.18 P(A B) = 0.7 N 7c. On the following Venn diagram, shade the rion that represents A B. [1 mark] N1 7d. Find P(A B ). appropriate approach 0. 0.18, P(A) P( B ) P(A B ) = 0.1 (may be seen in Venn diagram) N In triangle ABC, AB = 6 cm and AC = 8 cm. The area of the triangle is 16 cm. 8a. Find the two possible values for A. correct substitution into area formula 1 (6)(8)sin A = 16, sin A = () correct working () A = arcsin( ) A = 0.7977656,.41186499 ; ( 41.810149, 18.1896851 ) A = 0.70;.41 N (accept drees ie 41.8 ; 18 ) 16 4
8b. Given that A is obtuse, find BC. evidence of choosing cosine rule correct substitution into RHS (angle must be obtuse) () B C = 6 + 8 (6)(8) cos.41, 6 + 8 (6)(8) cos18, BC = 171.55 B C = A B + AC (AB)(AC) cosa, a + b ab cosc BC = 1.09786 BC = 1.1 cm N Let f(x) = p cos(q(x + r)) + 10, for 0 x 0. The following diagram shows the graph of f. The graph has a maximum at (4, 18) and a minimum at (16, ). 9a. Write down the value of r. r = 4 A N Note: Award for r = 4. 9b. Find p. evidence of valid approach p = 8 max y value -- y value N, distance from y = 10 9c. Find q.
valid approach period is,, substitute a point into their f(x) π 4 π 1 q = (, exact), 0.6 (do not accept drees) N 4 60 4 9d. Solve f(x) = 7. valid approach line on graph at x = 11.4688 x = 11.5 (accept (11.5,7)) N π y = 7, 8 cos( (x 4)) + 10 = 7 4 Note: Do not award the final if additional values are given. If an incorrect value of q leads to multiple solutions, award the final only if all solutions within the domain are given. The velocity of a particle in ms 1 is given by v = e sin t 1, for 0 t 5. On the grid below, sketch the graph of v. 10a.
N Note: Award for approximately correct shape crossing x-axis with < x <.5. Only if this is awarded, award the following: for maximum in circle, for endpoints in circle. Find the total distance travelled by the particle in the first five seconds. 10b. [1 mark] t = π (exact),.14 N1 [1 mark] Write down the positive t-intercept. 10c. recognizing distance is area under velocity curve s = v, shading on diagram, attempt to intrate valid approach to find the total area area A + area B, vdt vdt, vdt+ vdt, v.14 correct working with intration and limits (accept dx or missing dt ) () vdt+ vdt,.067 + 0.878, 1 5 0 e sin t.14.14 0 distance =.95 (m) N.14 5 0 5 11. Consider the expansion of x ( x 8 k + ). The constant term is 16 18. x Find k. [7 marks]
valid approach ( 8 ), r ( x) 8 r r ( ) ( x ) 8 + ( 8 ) ( ) + ( ) +, Pascal s triangle to line 1 ( x) 7 k 8 x ( x) 6 ( k ) 9 th x attempt to find value of r which gives term in x 0 exponent in binomial must give x, x ( x ) 8 r ( ) = correct working () (8 r) r =, 18 r = 0, r + ( 8 + r) = evidence of correct term () 8 8 ( ), ( ), r = 6, r = 6 ( x) ( k ) x equating their term and 1618 to solve for k x 8 ( ) = 1618, = 6 ( x) ( k 6 ) x k = ± N k x k 6 1618 8(9) M1 k x r x 0 Note: If no working shown, award N0 for k =. Total [7 marks] Let f(x) = x 4 1. 1a. Find f (x). expressing f as x 4 f 4 x1 4 (x) = (= x) N 1b. Find f(x)dx. attempt to intrate 4 x +1 4 +1 7 x7 x 4 x f(x)dx = + c N4 Two events A and B are such that P(A) = 0. and P(A B) = 0.5. 1a. Given that A and B are mutually exclusive, find P(B). correct approach () 0.5 = 0. + P(B), P(A B) = 0 P(B) = 0. N
1b. Given that A and B are independent, find P(B). Correct expression for P(A B) (seen anywhere) attempt to substitute into correct formula for P(A B) correct working P(A B) = 0.P(B), 0.x () P(B) = (= 0.75, exact) 8 N P(A B) = 0. + P(B) P(A B), P(A B) = 0. + x 0.x 0.5 = 0. + P(B) 0.P(B), 0.8x = 0. A particle moves along a straight line such that its velocity, v ms 1, is given by v(t) = 10te 1.7t, for t 0. 14a. On the grid below, sketch the graph of v, for 0 t 4. A N Notes: Award for approximately correct domain 0 t 4. The shape must be approximately correct, with maximum skewed left. Only if the shape is approximately correct, award A for all the following approximately correct features, in circle of tolerance where drawn (accept seeing correct coordinates for the maximum, even if point outside circle): Maximum point, passes through origin, asymptotic to t-axis (but must not touch the axis). If only two of these features are correct, award. 14b.
valid approach (including 0 and ) 10te dt, f(x), area from 0 to (may be shaded in diagram) 0 0 distance =. (m) N 14c. Find the velocity of the particle when its acceleration is zero. recognizing acceleration is derivative of velocity (R1) a = dv dv, attempt to find, reference to maximum on the graph of v dt dt valid approach to find v when a = 0 (may be seen on graph) dv = 0, 10e 1.7t 17t e 1.7t = 0, t = 0.588 dt velocity =.16 (m s 1 ) N Note: Award R1M1A0 for (0.588, 16) if velocity is not identified as final answer The time taken for a student to complete a task is normally distributed with a mean of 0 minutes and a standard deviation of 1.5 minutes. 15a. A student is selected at random. Find the probability that the student completes the task in less than 1.8 minutes. Note: There may be slight differences in answers, depending on whether candidates use tables or GDCs, or their sf answers in subsequent parts. Do not penalise answers that are consistent with their working and check carefully for FT. attempt to standardize 1.8 0 z =, 1.44 1.5 P(T < 1.8) = 0.95 N 15b. The probability that a student takes between k and 1.8 minutes is 0.. Find the value of k. [5 marks] Note: There may be slight differences in answers, depending on whether candidates use tables or GDCs, or their sf answers in subsequent parts. Do not penalise answers that are consistent with their working and check carefully for FT. attempt to subtract probabilities P(T < 1.8) P(T < k) = 0., 0.95 0. P(T < k) = 0.65 EITHER finding the z-value for 0.65 () z = 0.186 (from tables), z = 0.188 attempt to set up equation using their z-value k 0 0.186 =, 0.54 1.5 = k 0 1.5 k = 0.4 N OR k = 0.4 A N [5 marks]
Consider the graph of the semicircle given by f(x) = 6x x, for 0 x 6. A rectangle PQRS is drawn with upper vertices R and S on the graph of f, and PQ on the x-axis, as shown in the following diagram. Let OP = x. 16a. (i) Find PQ, giving your answer in terms of x. (ii) Hence, write down an expression for the area of the rectangle, giving your answer in terms of x. (i) valid approach (may be seen on diagram) Q to 6 is x PQ = 6 x N (ii) A = (6 x) 6x x N1 16b. Find the rate of change of area when x =. recognising at x = needed (must be the derivative of area) da dx da dx 7 =, 4.95 N 16c. The area is decreasing for a < x < b. Find the value of a and of b. a = 0.879 b = N The first three terms of an arithmetic sequence are 6, 40, 44,. (i) Write down the value of d. 17a. (ii) Find. u 8 (i) d = 4 N1 (ii) evidence of valid approach e.g. = 6 + 7(4), repeated addition of d from 6 u 8 u 8 = 64 N
(i) Show that S. 17b. n = n + 4n (ii) Hence, write down the value of S14. (i) correct substitution into sum formula e.g. = { (6) + (n 1)(4)}, evidence of simplifying e.g. S n (ii) 868 N1 n n {4n + 68} S n = n + 4n AG N0 n {7 + 4n 4} Let f and g be functions such that g(x) = f(x + 1) + 5. (a) The graph of f is mapped to the graph of g under the following transformations: 18a. [6 marks] vertical stretch by a factor of k, followed by a translation ( p ). q Write down the value of (i) k ; (ii) p ; (iii) q. (b) Let h(x) = g(x). The point A( 6, 5) on the graph of g is mapped to the point A on the graph of h. Find A. (a) (i) k = N1 (ii) p = 1 N1 (iii) q = 5 N1 (b) recognizing one transformation horizontal stretch by A is (, 5) N 1, reflection in x-axis Total [6 marks] The graph of f is mapped to the graph of g under the following transformations: 18b. vertical stretch by a factor of k, followed by a translation ( p ). q Write down the value of (i) k ; (ii) p ; (iii) q.
(i) k = N1 (ii) p = 1 N1 (iii) q = 5 N1 Let h(x) = g(x). The point A( 6, 5) on the graph of g is mapped to the point A on the graph of h. Find A. 18c. recognizing one transformation horizontal stretch by A is (, 5) N 1, reflection in x-axis Total [6 marks] A random variable X is normally distributed with μ = 150 and σ = 10. 19. Find the interquartile range of X. [7 marks] recognizing one quartile probability (may be seen in a sketch) P(X < Q ) = 0.75, 0.5 finding standardized value for either quartile z = 0.67448, z = 0.67448 () attempt to set up equation (must be with z-values) 0.67 = Q 150, one correct quartile = 156.74, Q correct working () other correct quartile, valid approach for IQR (seen anywhere) Q Q 1, ( μ) IQR = 1.5 N4 [7 marks] 10 Q 0.67448 = x 150 10 Q 1 = 14.5 Q μ = 6.744 () The random variable X is normally distributed with mean 0 and standard deviation 5. Find P(X.9). 0a.
evidence of appropriate approach z =.9 0 5 z = 0.58 () P(X.9) = 0.719 N Given that P(X < k) = 0.55, find the value of k. 0b. z-score for 0.55 is 0.1566 () valid approach (must be with z-values) using inverse normal, k = 0.6 N 0.157 = k 0 5 The following diagram shows a triangle ABC. The area of triangle ABC is 80 cm, AB = 18 cm, AC = x cm and BAC = 50. Find x. 1a. correct substitution into area formula setting their area expression equal to 80 1 (18x) sin 50 9xsin 50 = 80 x = 11.6 N () Find BC. 1b.
evidence of choosing cosine rule c = a + b + ab sin C correct substitution into right hand side (may be in terms of x) 11.6 + 18 (11.6)(18) cos50 BC = 1.8 N () The sum of the first three terms of a geometric sequence is 6.755, and the sum of the infinite sequence is 440. Find the. common ratio. [6 marks] correct substitution into sum of a geometric sequence 6.755 = ( ), u 1 + u 1 r + u 1 r = 6.755 correct substitution into sum to infinity u 1 1 r = 440 attempt to eliminate one variable substituting 1 r u 1 1 r u 1 = 440(1 r) correct equation in one variable 1 r 1 r 6.755 = 440(1 r)( ), () 440(1 r)(1 + r + r ) = 6.755 evidence of attempting to solve the equation in a single variable sketch, setting equation equal to zero, r = 0.95 = 19 0 [6 marks] N4 6.755 = 440(1 r ) The following diagram shows a circle with centre O and radius r cm. Points A and B are on the circumference of the circle and AOB = 1.4 radians. The point C is on [OA] such that BCO = radians. π Show that OC = rcos1.4. a. [1 mark]
use right triangle trigonometry cos1.4 = OC r OC = rcos1.4 [1 mark] AG N0 The area of the shaded rion is 5 cm. Find the value of r. b. [7 marks] correct value for BC BC = rsin 1.4, r (rcos1.4) () area of ΔOBC = area of sector attempt to subtract in any order sector triangle, correct equation 1rsin 1.4 rcos1.4 1 OAB = 1.4 r 1 sin 1.4 cos1.4 0.7 r r 0.7r 1 rsin 1.4 rcos1.4 = 5 (= 1 sin 1.4 cos1.4) r attempt to solve their equation sketch, writing as quadratic, r = 6.7 [7 marks] N4 5 0.616 Note: Exception to FT rule. Award FT for a correct FT answer from a quadratic equation involving two trigonometric functions. Consider the points A( 5,, 1), B( 6, 5, ), and C( 7, 6, a + 1), a R. Find 4a. (i) AB ; (ii) AC. (i) appropriate approach AO+ OB, B A AB = 1 N (ii) AC = 4 N1 a
Let q be the angle between AB and AC. Find the value of a for which q = π. 4b. valid reasoning (seen anywhere) scalar product is zero, R1 correct scalar product of their AB and AC (may be seen in part (c)) () 1() + (4) + (a) correct working for their AB and AC () a + 14, a = 14 a = 7 N cos π = u v u v 4c. i. Show that cosq = a+14. 14 a +80 ii. Hence, find the value of a for which q = 1.. [8 marks] correct magnitudes (may be seen in (b)) ()() 1 + + (= 14 ), + 4 + a (= 0 + a ) substitution into formula cosθ, simplification leading to required answer 1 + 4+ a 1 + + + 4 + a cosθ = 14+a cosθ = a+14 14 0+a 14 a +80 AG N0 14+a 14 4+16+a correct setup () cos1. = a+14 valid attempt to solve sketch, a =.5 14 a +80 a+14 14 a +80 A cos1. = 0, attempt to square N 4d. Hence, find the value of a for which q = 1..
correct setup () cos1. = a+14 valid attempt to solve sketch, a =.5 14 a +80 a+14 14 a +80 A cos1. = 0, attempt to square N The heights of a group of seven-year-old children are normally distributed with mean 117 cm and standard deviation 5 cm. A 5a. child is chosen at random from the group. Find the probability that this child is taller than 1.5 cm. evidence of appropriate method e.g. z = 1.5 117, sketch of normal curve showing mean and 1.5, 1.1 P(Z < 1.1) = 0.864 0.15666 P(H > 1.5) = 0.16 5 () N The heights of a group of seven-year-old children are normally distributed with mean 117 cm and standard deviation 5 cm. A 5b. child is chosen at random from the group. The probability that this child is shorter than k cm is 0.65. Find the value of k. z = 0.85 set up equation e.g. X 117 5 k = 118.9660 k = 199 () = 0.85, sketch N A particle moves in a straight line with velocity v = 1t t 1, for t 0, where v is in centimetres per second and t is in seconds. Find the acceleration of the particle after.7 seconds. 6a.
recognizing that acceleration is the derivative of velocity (seen anywhere) (R1) e.g. correctly substituting.7 into their expression for a (not into v) e.g. d a = s, v,1 6t dt s (.7) acceleration = 1.74 (exact), 1.7 N () Find the displacement of the particle after 1. seconds. 6b. recognizing that displacement is the intral of velocity e.g. s = v correctly substituting 1. e.g. 1. 0 () displacement = 7.41195 (exact), 7.41 (cm) N vdt R1 The following diagram shows the cuboid (rectangular solid) OABCDEFG, where O is the origin, and OA = 4i, OC = j, OD = k. 7a. (i) Find OB. (ii) Find OF. (iii) Show that AG = 4i + j + k. [5 marks]
(i) valid approach e.g. OA + OB OB = 4i + j (ii) valid approach N e.g. OA+ AB + BF ; OB+ BF ; OC+ CG + GF OF = 4i + j + k (iii) correct approach N e.g. AO+ OC + CG ; AB+ BF + FG ; AB+ BC + CG AG = 4i + j + k [5 marks] AG N0 Write down a vector equation for 7b. (i) the line OF; (ii) the line AG. (i) any correct equation for (OF) in the form r = a + tb A N where a is 0 or 4i + j + k, and b is a scalar multiple of 4i + j + k e.g. r = t(4,, ), r = 4t, t r = 4i + j + k + t(4i + j + k) t (ii) any correct equation for (AG) in the form r = a + sb A N where a is 4i or j + k and b is a scalar multiple of 4i + j + k e.g. r = (4, 0, 0) + s( 4,, ), r = 4 4s, s r = j + k + s( 4i + j + k) s Find the obtuse angle between the lines OF and AG. 7c. [7 marks] choosing correct direction vectors, OF and AG ()() scalar product = 16 + 9 + 4 (= ) () magnitudes 4 + +, ( 4) + +, ( 9, 9 ) ()() substitution into formula M1 16+9+4 e.g. cosθ = = ( ) ( 4 + + ) ( 4) + + 95.9777, 1.6744 radians θ = 95.9 or 1.67 N4 [7 marks] 9
The probability of obtaining tails when a biased coin is tossed is 0.57. The coin is tossed ten times. Find the probability of 8a. obtaining at least four tails. evidence of recognizing binomial distribution e.g. X B(10,0.57), p = 0.57, q = 0.4 EITHER evidence of using complement P(X ) =.16 10 4 + 0.0086 + 0.01709 + 0.06041 (= 0.08057) e.g. 1 any probability, P(X 4) = 1 P(X ) 0.9194 P(X 4) = 0.919 OR N summing the probabilities from X = 4 to X = 10 correct expression or values 10 10 () () e.g. ( )(0.57) r (0.4) 10 r, 0.1401 + 0.9 + + 0.071 + 0.006 r r=4 0.91944 P(X 4) = 0.919 N 8b. The probability of obtaining tails when a biased coin is tossed is 0.57. The coin is tossed ten times. Find the probability of obtaining the fourth tail on the tenth toss. evidence of valid approach 9 e.g. three tails in nine tosses, ( ) (0.57) (0.4) correct calculation 9 e.g. ( ) (0.57) (0.4) 0.57, 0.0984 0.57 () 0.05605178 P(4th tail on 10th toss) = 0.0561 N b x 8 Consider the expansion of ( x + ) = 56 x 4 + 07 x 0 + + k x 0 +. Find b. 9a.
valid attempt to find term in x 0 8 e.g. ( ) ( 7 )(b), ( x ) 7 b ( ) = 07 x 1 correct equation 8 e.g. ( ) ( 7 )(b) = 07 1 b = N 9b. Find k. evidence of choosing correct term e.g. 7th term, r = 6 correct expression 8 e.g. ( ) (x ) ( ) x 6 k = 81648 (accept 81600 ) N 6 Jose takes medication. After t minutes, the concentration of medication left in his bloodstream is given by A(t) = 10(0.5) 0.014t, where A is in milligrams per litre. 0a. Write down A(0). [1 mark] A(0) = 10 N1 [1 mark] Find the concentration of medication left in his bloodstream after 50 minutes. 0b. substitution into formula () e.g. 10(0.5) 0.014(50), A(50) A(50) = 6.16 N At 1:00, when there is no medication in Jose s bloodstream, he takes his first dose of medication. He can take his medication [5 marks] 0c. again when the concentration of medication reaches 0.95 milligrams per litre. What time will Jose be able to take his medication again?
set up equation e.g. A(t) = 0.95 attempting to solve e.g. graph, use of logs correct working () e.g. sketch of intersection, 0.014t log0.5 = log0.095 t =.0005 correct time 18: or 18:4 (accept 6: or 6:4 but nothing else) N [5 marks] Let f(t) = t + 7, where t > 0. The function v is obtained when the graph of f is transformed by a stretch by a scale factor of parallel to the y-axis, followed by a translation by the vector ( ). 4 1 1a. Find v(t), giving your answer in the form a(t b ) + c. applies vertical stretch parallel to the y-axis factor of 1 e.g. multiply by, 1 f(t), applies horizontal shift units to the right e.g. f(t ), t applies a vertical shift 4 units down e.g. subtracting 4, f(t) 4, 4 v(t) = (t ) 5 1 7 N4 1 1b. A particle moves along a straight line so that its velocity in ms 1, at time t seconds, is given by v. Find the distance the particle travels between t = 5.0 and t = 6.8.
recognizing that distance travelled is area under the curve e.g. v, (t 9 ) 5 t, sketch M1 distance = 15.576 (accept 15.6) A N A random variable X is distributed normally with a mean of 0 and variance 9. Find P(X 4.5). a. σ = () evidence of attempt to find P(X 4.5) e.g. z = 1.5, 4.5 0 P(X 4.5) = 0.9 N Let P(X k) = 0.85. b. (i) Represent this information on the following diagram. [5 marks] (ii) Find the value of k.
N Note: Award with shading that clearly extends to right of the mean, for any correct label, either k, area or their value of k. (ii) z = 1.0(648) () attempt to set up an equation k 0 e.g. = 1.064, k =.1 [5 marks] N k 0 = 0.85 A box holds 40 gs. The probability that an g is brown is 0.05. a. Find the expected number of brown gs in the box. correct substitution into formula for E(X) e.g. 0.05 40 E(X) = 1 N () Find the probability that there are 15 brown gs in the box. b. evidence of recognizing binomial probability (may be seen in part (a)) 40 e.g. ( ) (0.05) (0.95), X B(40,0.05) 15 P(X = 15) = 0.07 N c. Find the probability that there are at least 10 brown gs in the box. P(X 9) = 0.6 () evidence of valid approach e.g. using complement, summing probabilities P(X 10) = 0.764 N
Let f(x) = x, g(x) = x 5 and h(x) = (f g)(x). Find h(x). 4a. attempt to form composite e.g. f(x 5) h(x) = 6x 15 N 4b. Find h 1 (x). interchanging x and y evidence of correct manipulation e.g. y + 15 = 6x, h 1 x+15 (x) = 6 x 6 = y N 5 () Let the random variable X be normally distributed with mean 5, as shown in the following diagram. The shaded rion between 5 and 7 represents 0% of the distribution. Find P(X > 7). 5a. symmetry of normal curve e.g. P(X < 5) = 0.5 P(X > 7) = 0. N 5b. Find the standard deviation of X. [5 marks]
METHOD 1 finding standardized value e.g. 7 5 σ evidence of complement e.g. 1 p, P(X < 7), 0.8 finding z-score e.g. z = 0.84 () () attempt to set up equation involving the standardized value e.g. 0.84 = 7 5, σ =.8 METHOD N set up using normal CDF function and probability e.g. P(5 < X < 7) = 0., P(X < 7) = 0.8 correct equation A e.g. P(5 < X < 7) = 0., P(X > 7) = 0. attempt to solve the equation using GDC e.g. solver, graph, trial and error (more than two trials must be shown) σ =.8 [5 marks] σ 0.84 = X μ σ N M1 A standard die is rolled 6 times. The results are shown in the following table. 6a. Write down the standard deviation. σ = 1.61 A N Write down the median score. 6b. [1 mark] median = 4.5 N1 [1 mark] 6c. Find the interquartile range.
Q 1 =, = 5 (may be seen in a box plot) ()() Q IQR = (accept any notation that suggests the interval to 5) N th The n term of an arithmetic sequence is given by = 5 + n. u n 7a. Write down the common difference. [1 mark] d = N1 [1 mark] 7b. (i) Given that the n th term of this sequence is 115, find the value of n. (ii) For this value of n, find the sum of the sequence. [5 marks] (i) 5 + n = 115 () n = 55 N (ii) = 7 (may be seen in above) () u 1 correct substitution into formula for sum of arithmetic series () e.g. = (7 + 115), = ((7) + 54()), (5 + k) 55 S 55 55 S 55 S 55 = 55 (accept 60) N [5 marks] 55 k=1 Let f (x) = 4 x + 9 x + x + 1. 8a. There are two points of inflexion on the graph of f. Write down the x-coordinates of these points. valid approach R1 e.g. f (x) = 0, the max and min of f gives the points of inflexion on f 0.114, 0.64 (accept ( 0.114, 0.811) and ( 0.64,.1) ) N1N1 Let. Explain why the graph of g has no points of inflexion. 8b. g(x) = f (x)
METHOD 1 graph of g is a quadratic function R1 N1 a quadratic function does not have any points of inflexion R1 N1 METHOD graph of g is concave down over entire domain R1 N1 therefore no change in concavity R1 N1 METHOD (x) = 144 R1 N1 therefore no points of inflexion as g (x) 0 R1 N1 The following frequency distribution of marks has mean 4.5. 9a. Find the value of x. fx = 1() + (4) + + 7(4), fx = 146 + 5x (seen anywhere) evidence of substituting into mean correct equation fx f 146+5x 4+x e.g. = 4.5, 146 + 5x = 4.5(4 + x) x = 14 N 9b. Write down the standard deviation. σ = 1.54 A N
Let f(x) = A e kx +. Part of the graph of f is shown below. The y-intercept is at (0, 1). Show that A = 10. 40a. substituting (0, 1) into function e.g. 1 = A e 0 + 1 = A + A = 10 AG N0 M1 40b. Given that f(15) =.49 (correct to significant figures), find the value of k. substituting into f(15) =.49 e.g..49 = 10 e 15k +, 0.049 = e 15k evidence of solving equation e.g. sketch, using ln k = 0.01 (accept ) N ln 0.049 15 (i) 40c. Using your value of k, find f (x). (ii) Hence, explain why f is a decreasing function. (iii) Write down the equation of the horizontal asymptote of the graph f. [5 marks]
(i) f(x) = 10 e 0.01x + f(x) = 10 e 0.01x 0.01 (=.01 e 0.01x ) N Note: Award for 10e 0.01x, for 0.01, for the derivative of is zero. (ii) valid reason with reference to derivative R1 N1 e.g. f (x) < 0, derivative always native (iii) y = N1 [5 marks] 40d. Let g(x) = x + 1x 4. Find the area enclosed by the graphs of f and g. [6 marks] finding limits.895, 8.6940 (seen anywhere) evidence of intrating and subtracting functions correct expression e.g. g(x) f(x)dx, [( x + 1x 4) (10 + )]dx.90.90 e 0.01x 8.69 area = 19.5 A N4 [6 marks] 8.69 Consider f(x) = x, for x and g(x) = sin e x, for x. The graph of f is given below. 41a. On the diagram above, sketch the graph of g.
N 41b. Solve f(x) = g(x). x = 1., x = 1.68 (accept x = 1.41, x = 1.9 if working in drees) N 41c. Write down the set of values of x such that f(x) > g(x). 1. < x < 1.68 (accept 1.41 < x < 1.9 if working in drees) A N International Baccalaureate Organization 016 International Baccalaureate - Baccalauréat International - Bachillerato Internacional Printed for Atherton High School