2016 Undergraduate Admissions Assessment Mark Scheme: Mathematics Section C and D

Transcription

1 206 Undergraduate Admissions Assessment Mark Scheme: Mathematics Section C and D This is the mark scheme for the most recent Undergraduate Admissions Assessment at LSE. Abbreviations for Mark Scheme: Abbreviation FT CAO R SC FIW TSMIP Meaning Follow through Correct answer only Reason mark Special case From incorrect working Too simply to allow follow through

2 Section C: Test 2 Mathematics Question The table shows the number of injuries in the work place grouped by type and gender in 20. Injury Male Female Wound Fracture Sprain Internal Other How many injuries in total were there in 20? What percentage of Female injuries were classified as Internal? c) What percentage of Wound injuries were from Males? d) What percentage of all injuries were classified as Sprain? e) The number of female employees included in the data was.803 million. What percentage of these were injured? f) The overall injury rate for male employees was 28.6 per thousand. How many male employees were included in the data? (give your answer to the nearest whole number) g) The total number of injuries showed a 5% decrease from 200. How many people were injured in 200? (give your answer to the nearest whole number) allow any correct to 2sf providing method clear allow FT A 7/348 x 00 = M for either correct % numerator or denominatoraa c) 705/( ) x 00 MAA as above 3 = 60.3% d) ( )/40536 x MAA as above 3 00 = 32.4% e) 348/ x 00 = M as above or digits % 74 seen A A f) 278 = x n M for x (implied by later) M for dealing with 000 n = 278/ = SC 9488 M2 MA 4 g) = n x 0.95 MA n = / 0.95 = MA 4 h) x = 3366 SC - year SC3 - year too many29798 MMAA Be generous for students who have done year by year 25 marks 4 h) If this 5% decrease each year continued what would be the expected number of injuries in 206? (give your answer to the nearest whole number)

3 % Male Literacy Rate (y) Question 2 The scatter graph below shows a relationship between male and female literacy rates % Female Literacy rate (x) It is believed that the line of best fit as shown has an equation of the form: y = mx + c Use the graph to find m and c to 2 significant figures. Show all your working carefully. Use your answer to to find: i. the expected Male literacy rate (y) for a Female literacy rate (x) of 60% the expected Female literacy rate (x) for a Male literacy rate (y) of 60% c) If the line of best fit was y = x what could you say about male and female literacy rates? d) Using algebra find x where the line in your answer to meets y = x 2. c = 33 MA m = 55/80 = 0.69±0.03 MMA (or algebraic method) (M substitute point (not necessarily correct, M2 substitute 2 correct points) i) y = 0.69 x = 74.4 SC if x = 60 used as point in ii) x = (60-33) / 0.69 = 39. SC if y = 60 used as point in c) Male and female literacy rates the same/similar MA (FT for 0.5<m<, 20<c<40) MMMA (Ft as above) A (must refer to literacy rates explicitly/implicitly(th ey) not just x and y) d) x = 0.69x + 33 M (Ft as above) 0.3 x = 33 MA x = 06 A 4 6 marks 5 2 4

4 Question 3 This question concerns a loan of 7000 to buy a car. From Arby Bank you would pay each month over 5 years. i. How much would you pay altogether? (Give your answer to the nearest 0.0) What percentage interest would you have paid overall? From Bass Bank you would pay nothing for 2 months and then each month. How much would you pay altogether? (Give your answer to the nearest 0.0) c) From Carly Bank you pay a 9.5 % deposit and the remainder owing has 6% interest added before calculating an equal monthly payment rounded to the nearest 0.0. i. What is the monthly payment? 3. i) x 60 = MA (condone.8 in this part only) ii) /7000 x 00 = 9.37% MMA (allow AWRT 9.4%) MA (806 implies M) x 58 = SC for x 60 = c) i) 7000 x = 65 MA = 5385 MA 5385 x.06 = MA 6308./ 60 = MA (A0 if.8 or 272) ii) x = 7923 MMA = 923 MA 923/7000x00 = 5.43% MA must be correct to 3sf What percentage interest would you have paid overall? 22 marks

5 Question 4 At a cinema sitting in Standard seats it costs 0.50 for an adult ticket and 8.25 for a student ticket. If a is the number of adults and s the number of students find i. a formula for c the total cost in in terms of a and s. a formula for the number of adult tickets a when the total cost of tickets is giving your answer in the form s = ma + c where m and s are fractions in their lowest terms. i hence find a and s In Premier seating it costs for 4 adults and 6 students. i. If x is the cost of adult tickets and y the cost of student tickets show that 8x + 2y = n where n is an integer to be determined 2 of the students forgot their student ID so have to pay adult prices making a total cost of 3. Form a second equation in x and y in the form px + qy = n where n, p an d q are integers to be determined Use an algebraic method to find the cost of adult and student tickets in Premier seating. 4. i) c = 0.5a s ii) a = 8.25s s = 99.75/ /8.25a s = 33/ - 4/a iii) s = 33-4 a s = 7 a = 4 SC2 if 4, 7 seen i) 4x + 6y =08.5 8x + 2y =27 SC2 if 27 seen SC5 if separate terms are x 2 ii) 6x + 4y = 3 MAA (MAA0 if units in algebraic expression) MMA MAA AA M2AA MAA MAA 8 x + 2y = 339 MAA 0x = 22 MA x = y = 27 MA 2y = 9.4 MA y =9.95 MAA allow other letters MA if x/y negative here then no further marks MA (37 marks)

6 Section D: Test Mathematics Question A x 2 + 2x 3 = (x )(x + 3) M Simplify: 2x x 7x + x 2 + 2x 3 Or 2x(x + 3) (7x + ) (x )(x + 3) 2x 3 3x 2 + (x )(x 2 +2x 3) oe 2x 2 x (x )(x + 3) (2x + )(x ) (x )(x + 3) (2x+) or 2 5 (x+3) x+3 SC 2x 3 3x 2 + (x )(x 2 +2x 3) oe MA M for a common denominator A Or MA M A 5 marks Question A2 y = x 3 3x 2 9x Find dy dx Hence find the coordinates of the stationary points of the curve y = x 3 3x 2 9x dy dx = 3x2 6x 9 3x 2 6x 9 = 0 3(x + )(x 3) = 0 x =, x = 3 y = 5, y = 27 M A 2 marks M AA AA 5 marks 7 marks

7 Question A3 Factorise x 3 + x 2 + x 3 into one linear and one quadratic factor Hence explain why x 3 + x 2 + x 3 = 0 has exactly one real solution. Attempt at factor theorem x clearly shown as factor (x )(x 2 + 2x + 3) Use of discriminant (or completing the square or differentiation) b 2 4ac = 8 oe Discriminant < 0 so quadratic has no real roots and cubic equation only has real root where x = 0 M A A 3 marks M A R 3 marks 6 marks Question A4 Solve the following equations exactly giving each answer either as an integer or in the form lna where a and b are integers lnb 2 x+ 5 = 3 2 x ln3 x + ln2 x+ = ln4 2 2x+ 5 2 x = 3 M 2 2 2x 5 2 x 3 = 0 M 2X 2 5X 3 = 0 where X = 2 x (2X + )(X 3) = 0 M 2 x = 2, A A 2x = 3 x = ns or x = ln3 ln2 A (can just omit this solution) A 7 marks ln3 x 2 x+ = ln4 M (correct law of logs to move forward) 3 x 2 x+ = 4 M 6 x 2 = 4 M x = ln2 A ln6 4 marks marks

8 Question A5 Sketch the curve y = a cos 2 x a for 0 x π showing the intercepts with the x and y axes and the coordinates of any turning points. A shape (ignoring stationary points on x axis) Either way up A0 if outside domain Showing all your working carefully find an exact value for the area enclosed betweeny = a cos 2 x a, and the y axis and for 0 x π Intercepts (0, 0) (π, 0) A2 π Turning point ( 2 A acos 2 x a dx or a sin 2 x dx π/2 a 2 cos 2x a 2 dx 0 [ a 4 sin 2x ax 2 ] 0 ±aπ 4 π/2 4 marks M relevant integral wrt x (ignore limits) M A (ignore limits) AA (ignore limits) ±aπ (or 2 ) A (allow A if limits are 0, π and area is doubled) R Area = ±aπ ±5aπ or marks marks

9 Question A6 Find the coordinates of the point on the curve y = e x 2, x 0where the tangent to the curve passes through the origin. Find the area enclosed by this tangent and the normal at this point and the x - axis. A6 dy dx = 2xex2 e x2 MA x = 2xex2 MA x 2 = 2 x = 2 y = e /2 A A (AA0 if 2 solutions given) 6 marks Tangent y = 2e /2 x Normal M -ve reciprocal y e 2 = 2e ( x ) oe M st line 2 2 A allow FT unless TSIMP y = 0, x = 2e + oe M A must be linear and exact for 2 method Area = ( 2e )e/2 oe MA must be linear and exact for method 7 marks 3 marks Question A7 Find the equation of the tangent to the curve x 3 + y 3 = 3 at the point where x = 2. Give your answer in the form ax + by = c where a, b, c are integers A7 x = 2, 8 + y 3 = 9, y = 3x 2 + 3y 2 dy = 0 oe dx dy dx = 4 y = 4 (x 2) 4x + y = 9 A M AA (M for 3x 2 + 3y 2 = 0 ) A watch from FIW M A allow even if FIW earlier 7 marks

10 Question B Differentiate the following with respect to x i) y = e x ii) y = e x2 iii) y = xe x2 iv) y = e x2 For the curve y = xe x2, x 0 i. find the coordinates of the point where the gradient is 0 giving your answers exactly. show that the line y = x is a tangent to the curve i iv. y = xe x2 and give the coordinates of intersection between the line and the curve. find the x coordinate of the point where the gradient of the curve is a minimum giving your answer exactly. Write down the coordinates of the curve where the gradient is a maximum. c) Find the following indefinite integrals i) e x dx ii) xe x dx iii) xe x2 dx iv) x 2 e x dx x i) e x ii) 2xe x2 iii) e x2 2x 2 e x2 iv) 2x2 e x2 e x2 x 2 i) e x2 2x 2 e x2 = 0 allow ft from iii) x = 2, y = ii) x = xe x2 2 e /2 x( e x2) = 0 or ( 2x 2 )e x2) = x = 0, e x2 = x = 0 is a repeated root so a tangent at that point (0, 0) SC3 if (0,0) used to find gradient and NMS to find point iii) d2 y = 0 dx 2 2xe x2 4xe x2 + 4x 3 e x2 = 0 4x 3 6x = 0 2x(2x 2 3) = 0 x = 0, x = 3 2 x = 3 2 iv) (0, 0) c)i) - e x + c (need + c here but condone missing later) ii) xe x + e x dx = xe x e x + c iii) 2 e x2 + c iv) x 2 e x + 2xe x dx = x 2 e x 2xe x 2e x + c A MA MA MAA 8 marks M AA (AA0 if 2 solutions) M M valid method to find 2 solutions or check gradient of solution A2 R A M MA A A A 5 marks A A MAA MA M A A 0 marks

11 d) Use your answer to c) to find the area between the curve y = xe x2 and the x axis bounded by x = 0.5 and x =.5 (Give your answer to 3 sf) e) An architect uses the curve y = xe x2 between x = 0.5 and x =.5 as a scale model for the cross section of a building He wants the maximum height of the building to be 0m. What is the width and area of the cross section of the building? d) [.5 2 e x2 +] = e.52 + M A allow FT 2 e 0.52 = or from c)iii) better 2 marks e) scale factor = 0 2 e /2 = M A Width = 3.5 m Area = = 237m 2 A M A 5 marks 40 marks

12 Question B2 i. Write down + x + x 2 + x 3 + x n as a single algebraic fraction in terms of x Differentiate your answer to show that + 2x + 3x 2 + 4x 3 + (n )x n 2 = (n )xn nx n + (x ) 2 Matthew is saving his money. He saves 50 at the start of the first month and then more each month so at the start of the 2nd month he saves 5 and at the start of the 3rd month 52. i. How much does he save at the start of the 5th month? i How much does he save at the start of the nth month? How much has he saved in total by the start of the 5th month (including the savings in that month) i) x n oe x MMA ii) (x )nx n (x n ) (x ) 2 MA FT if i) is in a correct form (n )x n nx n + (x ) 2 A 6 marks i) 54 A ii) 50 + (n ) A iii) M A 54 = 260 iv) n 2 (00 + (n )) M A v) n (00 + (n )) = 5000 M allow FT if iv) is in a correct 2 form n n 0000 = 0 M allow T&I if correct 63 months A condone n 63 9 marks iv. How much has he saved in total by the start of the nth month (including the savings in that month) v. (Use your answer to iv) to find how the smallest number of months he needs to save to have at least 5000?

13 c) Jonathan is also saving his money and puts 50 in a savings account at the start of each month which has interest added at the end of each month at a rate of 2.3% per month. i. How much has he in his savings account by the start of the 5th month (including the savings in that month and the monthly interest added at the end of each month) i How much has he in his savings account by the start of the nth month (including the savings in that month and the monthly interest added at the end of each month) Use your answer to ii) to find how the smallest number of months he needs to save to have at least 5000? d) Jane combines both methods and puts 50 into the savings account with the 0.3% interest in the first month and 5 the second and so on. i. Show that at the start of the 3rd month she will have i Show that she will have n a.023 r n 2 n r ( r at the start of the nth month where a and b are to be determined Use your answer to ii) to write the expression in d)ii) as the sum of two algebraic fractions in terms of n. iv. Evaluate your expression in iii) when n = 50 C) i) M A A MA A ii) 50 (.023n ) iii) 50 (.023n ) M.023 n = 3.3 A n = ln3.3 M ln.023 A 53 allow T&I if correct and FT ii) if ii) is of a correct form 0 marks d) i) = M AA ii) n n M AA.023 n (50 + (n )) = 50 (.023 n n ) M n n n 4 + M + (n ) n = r n 2 ( (n ) n n n = r n 2 r ( r.023 ) a = 50 b = iii) 50 (.023n ) +.023n 2 ((n )0.978 n n0.978 n + )) ( 0.978) 2 iv) = 6460 to 3 sf Allow FT providing iii) is in a correct from M A A MMA A 5 marks 40 marks

14 Question B3 Find the area of a regular hexagon sides a cm giving your answer in an exact form. For a hexagonal prism with regular hexagonal base sides a cm and height h cm find an expression for i. the volume V in terms of a and h i iv. the surface area S in terms of a and h in the case where the volume is 500 cm 3 find an expression for S in terms of a use calculus to find an exact value for a that minimise S v. use calculus to explain why your answers give a minimum value for S Area of triangle = 2 a2 sin60 = 3 4 a2 Area hexagon = a2 A i) V = a2 h M A 3 marks MA (FT if of form ka 2 ii) S = 3 3a 2 + 6ah MAA (FT if of form ka 2 iii) h = a 2 M (FT if of form ka 2 S = 3 3a a 2000 = 6 3a iv) ds da 3a 2 ds da = 0 6 3a a 2 = 0 8a 3 = 2000 a = v) d2 S 4000 da2 = a 3 d 2 S da 2 > 0 minimum MA (FT if of form ka 2 MAA allow FT M A CAO MA allow FT R 6 marks

15 c) A hexagonal prism forms a can with filled volume 500cm 3 and height 0cm. A hole is punched in the hexagonal base so that the contents leak out at a rate proportional to the height y of liquid remaining. i. Initially the can is full and liquid is leaking out at a rate of 2 cm 3 i iv. per second. Explain why dz = ky where k > 0 and z is the volume of dt liquid remaining Determine the value of k Show that dy dt = 250 y Use calculus to find y in terms of t How long does it take for the volume of liquid in the can to become halved? c) i) Volume going down hence minus sign R Proportional to y so by y R initially dz = 2 and y = 0 so k = M A dt 5 ii) z = 50y M dz dy MA = 50 dt dt dy dt = 250 y A iii) dy = y dt MAA 250 lny = 250 t + c MAA initially y = 0 c = ln0 MA lny ln 0 = 250 t M iv) y = 5 ln y 0 = 250 t A y 0 = e t MA 250 e 250 t = 2 t = 250 ln MA = 73 seconds 2 M M 24 marks 40 marks