PRE-LEAVING CERTIFICATE EXAMINATION, 04 MARKING SCHEME PROJECT MATHS (Phase ) HIGHER LEVEL Page of 9
Table of Contents SOLUTIONS TO PAPER... PAPER I: Question...4 PAPER I: Question...6 PAPER I: Question...7 PAPER I: Question 4...9 PAPER I: Question 5... PAPER I: Question 6... PAPER I: Question 7...6 PAPER I: Question 8...8 PAPER I: Question 8...0 SOLUTIONS TO PAPER... PAPER II: Question... PAPER II: Question...4 PAPER II: Question...6 PAPER II: Question 4...7 PAPER II: Question 5...9 PAPER II: Question 6...0 PAPER II: Question 7... PAPER II: Question 8...6 PAPER II: Question 9...8 Page of 9
OVERVIEW OF MARKING SCHEME (QUESTIONS TO 6) Scale label A B C D E No of categories 4 5 6 5 mark scale 0, 5 0,, 5 0,, 4, 5 0 mark scale 0, 0 0, 6, 0 0, 5, 8, 0 0,, 5, 8, 0 5 mark scale 0, 5 0, 8, 5 0, 8,, 5 0, 7, 0,, 5 0 mark scale 0, 0 0, 0, 0 0, 0, 8, 0 0, 5, 0, 5, 0 5 mark scale 0,, 5 0, 8, 0, 5 0, 6,, 9, 5 0, 5, 0, 5, 0, 5 A general descriptor of each point on each scale is given below. More specific directions in relation to interpreting the scales in the context of each question are given in the body of the scheme, where necessary. Marking scales level descriptors A-scales (two categories) incorrect response (no credit) correct response (full credit) B-scales (three categories) response of no substantial merit (no credit) partially correct response (partial credit) correct response (full credit) C-scales (four categories) response of no substantial merit (no credit) response with some merit (low partial credit) almost correct response (high partial credit) correct response (full credit) D-scales (five categories) response of no substantial merit (no credit) response with some merit (low partial credit) response about half-right (middle partial credit) almost correct response (high partial credit) correct response (full credit) E-scales (six categories) response of no substantial merit (no credit) response with some merit (low partial credit) response about half-right (lower middle partial credit) response more than half-right (upper middle partial credit) almost correct response (high partial credit) correct response (full credit) Marking categories for all questions are shown throughout the solutions. In certain cases, typically involving rounding or omission of units, a mark that is one mark below the fullcredit mark may also be awarded. Such cases are flagged with an asterisk. Thus, for example, Scale 0C* indicates that 9 marks may be awarded. Page of 9
Part (a) Part (b) Part (c) SOLUTIONS TO PAPER QUESTION Scale 5C Scale 0C Scale 0C (a) If z = a+ bi, where a and b, show that zz. simplifies to a constant. ( a + bi)( a bi) = a abi + abi b i a + b Full Credit (5 Marks) a constant High Partial Credit ( Marks) One error in multiplication Low Partial Credit ( Marks) (b) Examine the complex plane below. Identify on the diagram each complex number. z = z z = iz z 4 = image of z under S y.z.z 4.z.z High Partial Credit (7 Marks) Two correctly identified Low Partial Credit ( Marks) One correctly identified Page 4 of 9
(c) Use De Moivre s theorem to prove that sin sin 4sin θ = θ θ. ( cosθ isinθ ) cosθ + i sin θ = + cos sin cos cos sin cos sin sin isin θ = icos θsinθ isin θ sin θ = sin θ sinθ sin θ + i θ = θ + i θ θ θ θ i θ ( ) θ sin θ = sinθ 4sin θ High Partial Credit (7 Marks) Error in expansion but continues to end Low Partial Credit ( Marks) Page 5 of 9
Part (a) Part (b) QUESTION Scale 0C Scale 5C (a) Show that + + for all values of ab,. a 6 b ab a + 6 + b ab a ab+ b + 6 0 ( a b) ( ) + 4 0 true as all squared numbers are 0 High Partial Credit (8 Marks) One error in multiplication Low Partial Credit (4 Marks) (b) Solve the simultaneous equations: x+ y = 5 x + y = 4 where x, y x+ y = 5 x = y 5 ( ) y y 5 + = 4 9y + 60y+ 9 = 0 y = or x = 4 y = 9 Full Credit (5 Marks) High Partial Credit ( Marks) Error solving the quadratic but continues to end with no further errors but uses the y = and continues to end 9 Low Partial Credit (8 Marks) Page 6 of 9
Part (a) Part (b) Part (c) QUESTION Scale 5C Scale 0C Scale 0C (a) Given that x + x 6 is a factor of x px qx 6 +, find the value of p and q. By inspection: ( x + x 6)( x+ ) x + x x 6 p = and q = *Other acceptable methods warrant full marks Full Credit (5 Marks) High Partial Credit ( Marks) One error only and solves to end Low Partial Credit ( Marks) (b) Prove that the function ( ) f x : x + 7x + 7x+ 5 has only one real root. ( ): + 7 + 7 + 5 ( : ) ( ) 4( ) 7( ) 5 0 ( x ) f x x x x f + + + = + is a factor x + 7x + 7x+ 5 = x + 4x + 5 x + x + 4x+ 5= x+ i x+ + i Complex roots ( ) ( )( ) f x : x + 7x + 7x+ 5 has only one real root *Note: Students may also carry out by calculus. *Note: Candidates may also use b 4ac< 0 from resulting quadratic (algebra) By calculus candidate must have a fully correct solution using the quadratic formula. High Partial Credit (7 Marks) One error in solving but continues to end Omits conclusion with a fully correct solution Differentiates and attempts to solve the resulting quadratic Low Partial Credit ( Marks) Page 7 of 9
(c) Solve for x. x ( ( ) ) x+ + 6 9= 0 x ( ) ( x x ) ( ) x + + 6 9= 0 + 6 9= 0 y + 6y 9= 0 ( y )( y+ 9) y = or y = 9 x = or x 9 x = High Partial Credit (7 Marks) One error in solving Fails to omit the invalid solution Low Partial Credit ( Marks) Page 8 of 9
Part (a) Part (b) QUESTION 4 Scale 5D Scale 0C n n (a) Prove by induction that is a factor of 5 for all n and n. Step : Prove for n = 5 = is divisible by Step : Assume true for n= k k k 5 = A k k 5 = A + Step : Prove true for n= k+ k+ k+ 5 k k 5.5. 5 k k A +. ( ) 5A +. k 5 ( A + k ) is a factor for n= k+ Step 4: This is true for n =. If it is true for n= k it will also be true for n= k+ true for all n by induction. Full Credit (5 Marks) solution High Partial Credit ( Marks) Three correct steps Mid Partial Credit (7 Marks) Two correct steps. Low Partial Credit (4 Marks) Page 9 of 9
(b) Séan wants to save 50,000 over a ten-year period. If the expected interest rate is.5% how much does Séan need to invest at the beginning of each year to save 50,000? ( ) ( ) ( ) ( ) 9 50, 000 = P+ P.05 + P.05 + P.05 + + P.05 a = P r =.05 n = 0 P(.05 0 ) S0 =.05 ( 0 P.05 50, 000 = ).05 50, 000 =.0877P P 4, 46.94 High Partial Credit (7 Marks) Sets up geometric series but error in solving Low Partial Credit ( Marks) Page 0 of 9
QUESTION 5 Part (a) Scale 0C Part (b) Scale 5C* (a) Evaluate 6 x 0 x e dx. 6 x 0 x e dx u = x du x dx = du = 6x dx u u u edu= e = e 0 e High Partial Credit (7 Marks) One error and continues to end without over simplification of the problem Low Partial Credit ( Marks) Page of 9
(b) Find the area bounded by the curve f ( x ), the line g( x ) and the x-axis. ( ) ( )( ) ( ) = x+ f x = x+ x = x + x+ 6 g x f ( x) = g( x) x + x+ = x+ 6 x x = 0 x = or x = ( x) Area = f + Area from to Area = + + 6 + 4 4 ( x x ) ( )( ) x x Area = + + 6x + 8 6 Area = square units 6 *If candidate takes points from graph High Partial at most can be achieved. Full Credit (5 Marks) solution High Partial Credit ( Marks) Reads points from graph and finishes correctly Finds points algebraically and finishes to end with no more than two errors Mid Partial Credit (7 Marks) Reads points form graph and finishes with no more than two errors Low Partial Credit (4 Marks) Page of 9
Part (a) Part (b) Part (c) (i) (ii) QUESTION 6 Scale 0C Scale 5C Scale 5C Scale 5C (a) Differentiate x + x 5 from first principles. ( ) = + 5 ( ) ( ) ( ) ( + ) ( ) = + ( + ) ( ) f x x x f x+ h = x+ h + x+ h 5= x 4xh h + h f x h f x 4hx h h f x h f x 4xh h h = + h h h h f ( x+ h) f ( x) = 4x h + h f ( x+ h) f ( x) lim = 4x 0 + h h f ( x+ h) f ( x) lim = 4x + h h High Partial Credit (7 Marks) Error expanding brackets but continues to end correctly Error handling limits Low Partial Credit ( Marks) Page of 9
(b) Find the equation of the tangent to the curve = + 5 at the point (, ). y x x dy x dx = + + = 5 () m = 5 (, 6) ( ) ( x ) y y = m x x y = 5 5x y = Full Credit (5 Marks) High Partial Credit ( Marks) Error finding slope but continues to end correctly Low Partial Credit ( Marks) Page 4 of 9
(c) The function g( x) = is defined for all x x \. (i) Prove that the function has no turning points. 6 g' = 0 ( x) ( x) g( x) has no turning points Full Credit (5 Marks) High Partial Credit ( Marks) Error differentiation but continues to end correctly Correct with no conclusion Low Partial Credit ( Marks) (ii) Jane states that the graph of g( x ) would be increasing for all values of x. Would you agree with Jane? Explain your answer fully. 6 g' ( x) = ( x) g'( x) is always positive g( x) always increasing for all values of x. Yes. Full Credit (5 Marks) High Partial Credit ( Marks) Error differentiation but continues to end correctly Correct with no conclusion Low Partial Credit ( Marks) Page 5 of 9
QUESTION 7 Part (a) Scale 5A* Part (b) Scale 5C Part (c) Scale 5C Part (d) Scale 0D* (a) Using the information above, write an equation to represent the volume of water in the cone at any time t. V = 0.0t litres/second Full Credit (5 Marks) (b) Write the radius of the cone r, in terms of the height of the water h. tan 0 r = h r = htan 0 or r = h Full Credit (5 Marks) High Partial Credit ( Marks) Incorrect trigonometric ratio and continues Low Partial Credit (8 Marks) (c) Show that the height of the water h at any time t can be expressed as h = 0.8t. π V = π r h= 0.0 h π h 0.0t = h = 0.8t π Full Credit (5 Marks) High Partial Credit ( Marks) No more than two errors transposing the formula Low Partial Credit (8 Marks) Page 6 of 9
(d) Find the rate of change in the height of the water after seconds. h = 0.8t π 0.8 h= t π dh 0.8 = t dt π At t = dh 0.8 ( ) = dt π dh 0.06 dt = litres/second *Accept candidate s h from previous section Full Credit (0 Marks) solution High Partial Credit (5 Marks) Correct differential and stops Incorrect differential and continues to end correctly Mid Partial Credit (0 Marks) Incorrect differential and stops Low Partial Credit (5 Marks) Page 7 of 9
QUESTION 8 Part (a) Scale 0C Part (b) Scale 0C* Part (c) Scale 5C In 95, while working at the California Institute of Technology, Charles Francis Richter invented the Richter scale. The Richter scale is used to measure earthquake intensity on a machine called a seismograph. I The formula for the Richter is given by M = log0 I0 where M is the magnitude on the Richter scale in millimetres, I 0 is the threshold quake intensity, i.e. a movement that can barely be detected, and I is the intensity of the earthquake in millimetres. I 0 = 0 mm. (a) Calculate the magnitude of an earthquake with an intensity of 500 mm. 500 M = log0 0 M = 5.69897 5.7 High Partial Credit (8 Marks) Error in the calculation Low Partial Credit (4 Marks) (b) In 005 an earthquake in Kashmir in Pakistan measured 7.5 on the Richter scale. Calculate the intensity of the earthquake. I 7.5 = log 0 0 7.5 0 0 = I ( )( ) I = 6.7766 mm High Partial Credit (8 Marks) One error in the calculation Low Partial Credit (4 Marks) Page 8 of 9
(c) A heavy goods lorry rumbling on a road can cause a micro quake with a rating of up to. on the Richter scale. An at-home seismograph records an intensity of 40,000 I 0 while the owner is away on holidays. What event was likely to have occurred while the owner was away, an earthquake or a heavy goods lorry passing his house? M log 40,000I 0 = 0 I0 M = 4.6 Likely to have been a small earthquake and not a heavy goods lorry as magnitude is too big for a heavy goods lorry. Full Credit (5 Marks) with explanation High Partial Credit ( Marks) Correct with no explanation Incorrect but with a correct explanation Low Partial Credit (8 Marks) Page 9 of 9
Part (a) Part (b) Part (c) Part (d) Part (e) Paul is creating designs using green blocks. QUESTION 9 Scale 0C Scale 5C Scale 0B Scale 0D Scale 5B (a) Draw the next two patterns in the sequence. Both correct High Partial Credit (7 Marks) One correct Low Partial Credit ( Marks) One partially correct Page 0 of 9
(b) John states the pattern is linear. Mary disagrees. Would you agree with John or Mary? Explain your answer fully. Pattern 5 0 7 6 st Difference 5 7 9 nd Difference Pattern is quadratic as the second differences are the same. Full Credit (5 Marks) with explanation High Partial Credit ( Marks) Correct with no explanation Incorrect but with a correct explanation Low Partial Credit (8 Marks) (c) The general term for the sequence is given by and c. Tn an bn c = + +. Find the values of a, b T: a+ b+ c= 5, T :4a+ b+ c= 0, T:9a+ b+ c Gives a=, b=, c= T = n + n+ n Full Credit (0 Marks) High Partial Credit ( Marks) Forms three correct equations but has no more than two errors in solving Forms three incorrect equations and has no errors in solving Low Partial Credit (7 Marks) Page of 9
(d) What pattern in the sequence will contain green blocks? Verify your solution. Tn n n = + + = n + n 0= 0 n = 0 Tenth pattern Full Credit (5 Marks) and verified High Partial Credit ( Marks) Error setting up the quadratic but continues to end and tries to verify Correct but fails to verify Low Partial Credit (8 Marks) (e) How many blocks would Paul need to create patterns? 85 blocks Full Credit (5 Marks) Partial Credit ( Marks) Any correct attempt to find total sum Page of 9
Part (a) Part (b) Part (c) Part (d) SOLUTIONS TO PAPER QUESTION Scale 5B Scale 5B Scale 5B Scale 0B (a) Explain the term Stratified Sampling. Where the population is split based on certain factors, e.g. Age, Sex Full Credit (5 Marks) Explains with a valid example Partial Credit ( Marks) Partially Correct (b) Explain the term Cluster Sampling. Sample or population splits naturally into clusters, e.g. geographical areas. Full Credit (5 Marks) Explains with a valid example Partial Credit ( Marks) Partially correct (c) Are Stratified and Cluster Sampling random or non-random methods of sampling? Explain your answer fully. Both are random as everyone in the each population has an equal change of being included in the sample. Full Credit (5 Marks) Explains with a valid example Partial Credit ( Marks) Partially correct (d) An estate agent wishes to find out the prices of the houses in a newly developing area of a city. Which type of sampling above would be most suitable to use? Explain your answer fully. Cluster sampling. As he wants to find the prices in a certain area, i.e. a geographical cluster. Explains with a valid example Partial Credit (6 Marks) Partially correct Page of 9
Part (a) (i) (ii) Part (b) (i) (ii) (a) (i) What is a Bernoulli Trial? QUESTION Scale 5B Scala 5B Scala 5C Scale 0C A Bernoulli trial is a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted. Full Credit (5 Marks) Explains with a valid example Partial Credit ( Marks) Partially correct (ii) Give two characteristics of a binomial distribution. Finite number of trials Two outcomes. Success or failure Trials are independent of each other The probability of a success is the same for each trial Full Credit (5 Marks) Two correct Partial Credit ( Marks) One correct One Partially correct Page 4 of 9
(b) At a certain Garda checkpoint one in every 0 cars has a faulty headlamp. (i) Find the probability that the first three cars will have a faulty headlamp. 5 9 8 = 0 0 0,000 Full Credit (5 Marks) with work shown High Partial Credit ( Marks) Correct with no work shown Incorrect with workings shown Low Partial Credit ( Marks) (ii) Find the probability that at least two of the first four cars stopped will have a faulty headlamp. ( nonehavea faultylamporhasa faultylamp) 0 5 4 5 9 5 9 + 5 0 0 0 0 4,07 50, 000 with work shown High Partial Credit (8 Marks) Correct with no work shown Incorrect with workings shown Low Partial Credit (4 Marks) Page 5 of 9
Part (a) Part (b) QUESTION Scale 0C Scale 5C (a) Calculate the shortest distance between the line 6x 7y 0 0 ( ) ( ) 6+ 76 0 0 85 = 7 ( 6) + ( 7) with work shown + = and the point ( ) High Partial Credit (8 Marks) Correct with no work shown Incorrect with workings shown Uses distance between two points correctly to end with a second point on the line, 6. Low Partial Credit (4 Marks) Uses distance between two points correctly to end with a second point not on the line (b) Find the equations of the lines through the point ( 4, ) which make an angle of 45 with the line x+ y = 7. tan 45 = m + m + m+ m = 4 m+ m 4 m + 8m = 0 m = or m = x y = 0 and x+ y+ 4 = 0 Full Credit (5 Marks) solution High Partial Credit ( Marks) Finds the slopes correctly but omits the equations of one or both lines Finds the slopes with no more than two errors and finds the equations of both lines Finds the slopes correctly and has errors in one or both equations of the lines Mid Partial Credit (7 Marks) Finds the slopes with more than two errors and finds the equations of both lines Low Partial Credit (4 Marks) Page 6 of 9
Part (a) Part (b) QUESTION 4 Scale 5D Scale 0C (a) Find the equation of the circle c, which passes through the points (, ) and ( 8, ) whose centre lies on the line 4x+ 5y =. and x y gx fy c + + + + = 0 Equation : ( ) ( ) ( ) ( ) + + g + f + c= 0 g 6f + c= 0 Equation : ( ) ( ) ( ) ( ) 8 + + g 8 + f + c= 0 6g 4 f + c= 68 Equation : 4 g + 5 f = ( ) ( ) 4g 5f = Solving: g = f = c= 8 x + y x y = 6 4 8 0 Full Credit (5 Marks) with work shown High Partial Credit ( Marks) Finds three equations and attempts and stops No more than two errors solving correct equations Mid Partial Credit (7 Marks) More than two errors solving correct equations Finds two correct equations Low Partial Credit (4 Marks) Page 7 of 9
(b) The line 5x 4y+ k = 0 is a tangent to the circle c. Find two possible values for k. ( ) ( ) ( ) + 8 = 4 ( ) ( ) ( 5) + ( 4) 5 4+ k ( k ) 7+ = 4 k + 4k 6 = 0 = 4 ( 4) ( 4) 4( )( 6) () ± k = 5 or k = 49 with work shown High Partial Credit (7 Marks) No more than two errors solving for k Low Partial Credit ( Marks) Page 8 of 9
Part (a) Part (b) QUESTION 5 Scale 0C Scale 5C (a) Find two values for, where sinθ = and 0 θ 60. θ = sin = 60 θ = 40 and θ = 00 with work shown High Partial Credit (8 Marks) Only finds one value for Finds two incorrect values for with work shown Low Partial Credit (4 Marks) (b) If the point D is the midpoint of [ AC ], find [ BCD], correct to the nearest degree. ( 7.6) ( 8) ( 7.6)( 8) cos( 50) AD = + AD = 6.6 ( ) ( ) ( ) ( )( ) 8 = 6.6 + 6.6 cos BCD BCD = 9 Full Credit (5 Marks) with work shown High Partial Credit ( Marks) Solves with nn more than two errors Low Partial Credit (8 Marks) Page 9 of 9
Part (a) Part (b) QUESTION 6A Scale 0C Scale 5B A fire station is being built to service the three towns shown so that it is equidistant from the roads adjoining each town. (a) Show, using a suitable construction, where the fire station should be built. Full Credit (0 Marks) with all constructions shown High Partial Credit (5 Marks) with no/some constructions shown Constructs an incorrect centre fully correct with constructions shown Mid Partial Credit (0 Marks) Constructs an incorrect centre not fully correct with some constructions shown Low Partial Credit (5 Marks) (b) What is the name given to the point where the fire station will be built? Incentre Full Credit (5 Marks) Partial Credit ( Marks) Mentions any other centre Page 0 of 9
Part (a) Part (b) QUESTION 6B Scale 5C Scale 0C (a) [ BC ] and [ DE ] are two parallel chords of a circle with centre A and radius r. [ AF] [ BC] and [ BAC] [ DAE] =. If the area ABC = area ADE, show that π a = rads. 6 ( r)( r) sin 4a= ( r)( r) sin a sinacosa = sina cos a = π a = 6 Full Credit (5 Marks) with work shown High Partial Credit ( Marks) Solves with no more than two errors Low Partial Credit (8 Marks) (check diagram for valid work). Page of 9
(b) Find the value of r, where r, if BC + DE = 6. BC DE sin a = sin a = r r BC = rsina DE = r sina ( r a) ( r a) sin + sin = 6 ( ) sin + sin = 9 r a a () r = 9 r = 9 r = with work shown High Partial Credit (8 Marks) Finds BC and DE correctly and continues to end with errors Finds BC and DE incorrectly and continues to end with no further errors Low Partial Credit (4 Marks) Page of 9
Part (a) (i) (ii) (iii) (iv) (v) (vi) Part (b) (i) (ii) (iii) QUESTION 7 Scale 0C Scale 5A Scale 0 C Scale 5B Scale 5B Scale 5B Scale 5A Scale 0B Scale 0B (a) Each year the Central Statistics Office, or CSO, compile vital statistics about life in Ireland. The following table from the CSO shows the number of people employed and unemployed in the Irish workforce, in thousands, from 987 to 0. (i) Examining the figures in the table discuss the trends in the unemployment figures between 987 and 0. Three factors to correct:. High unemployment in 987. Unemployment decreased to 00 (might mention small rise from in early 90 s). On the rise since 00 Three facts from table High Partial Credit (7 Marks) Two facts from table Low Partial Credit ( Marks) One fact from table (ii) Ireland has seen a dramatic change in the employment figures since the turn of the century. Why, in your opinion, have the figures changed? Recession. Building trade burst. Bank crisis. Full Credit (5 Marks) Any correct reason Page of 9
The following table shows the changes in Ireland s population from 987 to 0. All the figures are in thousands. (iii) Discuss the trends in the Immigration/Emigration figures for the years shown. Three facts correct Three facts from table High Partial Credit (7 Marks) Two facts from table Low Partial Credit ( Marks) One fact from table Paul states Recent emigration figures are affected by the lack of jobs in Ireland. (iv) Would you agree with Paul? Explain your answer fully. Yes with a valid reason. Full Credit (5 Marks) Partial Credit ( Marks) Partially correct reason (v) Paul decides to represent the data on a suitable graph. He wants to investigate if there is a relationship between unemployment figures and emigration in Ireland. Display the data on a suitable graph. Any valid suitable display. Full Credit (5 Marks) display High Partial Credit (7 Marks) Partially correct display (vi) Explain how the graph supports/does not support Paul s claim. Make reference to statistical analysis of the given data. Valid explanation to support display. Full Credit (5 Marks) Partial Credit ( Marks) Partially correct reason Page 4 of 9
(b) (i) What is the relationship between the mean, mode and median of a normal distribution? All equal. Full Credit (5 Marks) (ii) Which of the following would you expect to be normally distributed? Explain your answer fully. The lifespan of a television. OR The number of accidents on a stretch of M4 motorway in a year. Answer: The lifespan of a television. Reason: Quality control procedures ensure they all last roughly the same amount of time. Small few have problems early and small number last a very long time. Partial Credit (6 Marks) Partially correct reason (iii) 400 samples of size 0 are taken from a production line producing jars of gravy. The mean weight of a jar is 00g, with a standard deviation of.5g. Calculate the limits between which you would expect 95% of the sample means to lie..5 00 ± 0 99.05g to 00.95g or.5 00 ± 0 99.5g to 00.47g Partial Credit (6 Marks) Partially correct Page 5 of 9
Part (a) Part (b) Part (c) Part (d) QUESTION 8 Scale 0C Scale 0C Scale 0C Scale 0C A rear windshield wiper has a radius of r. It cleans half the width of the windshield. The highest point of the arc of the semicircle is 60% of the height of the windshield. (a) Show that the distance x can be written as r. 5 h= r 5 h= 5 x = r r x = r r with work shown High Partial Credit (7 Marks) Correct answer with no work shown Finds x and stops Low Partial Credit ( Marks) (check diagram) (b) Show that the total area of the windshield is 0 r. 5 0 = ( 4 r ) r r with work shown High Partial Credit (7 Marks) Correct answer with no work Low Partial Credit ( Marks) (check diagram) Page 6 of 9
(c) Show that the ratio of the area cleaned to the area unclean is π square units. 40 π r 0 40 r = π with work shown High Partial Credit (7 Marks) Correct answer with no work shown Low Partial Credit ( Marks) (check diagram) The design of the windshield is altered so that the width of the bottom of the windshield is reduced by 5% and the angle at the centre of the arc is reduced by 0%. (d) Calculate the area of the sector cleaned in terms of r in radians. 4 ( r ) π 5 5 π r with work shown High Partial Credit (7 Marks) Correct answer with no work shown Low Partial Credit ( Marks) (check diagram) Page 7 of 9
Part (a) Part (b) Part (c) QUESTION 9 Scale 0C Scale 5C Scale 0B A sign is attached to a vertical wall at the point E. [ AE] wall. The sign is then suspended by two wires which are attached to the wall at the points B and C as shown. (a) Calculate the length of the wires. BE = 0 + 50 BE = 0cm BA = BE + EA BA = 0 + 0 BA = 76.98cm BA = 77cm with work shown High Partial Credit (7 Marks) Correct answer with no work shown Works out any correct side that may lead to correct answer Low Partial Credit ( Marks) (check diagram) (b) Calculate the angle that each wire makes with the wall. ( )( ) 77 = 00 + 77 00 77 cos ABC ABC = 7.59 Full Credit (5 Marks) with work shown High Partial Credit (0 Marks) Correct answer with no work shown Works out any correct side or angle correctly that may lead to correct answer (not from earlier parts) Low Partial Credit (5 Marks) (check diagram) Page 8 of 9
(c) Calculate the AEC, correct to two decimal places. sin x sin 90 = 0 77 x = 4.68 with work shown High Partial Credit (7 Marks) Correct answer with no work shown Works out any correct side or angle correctly that may lead to correct answer (not from earlier parts) Low Partial Credit ( Marks) (check diagram) Page 9 of 9
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