PRE-LEAVING CERTIFICATE EXAMINATION, 2015 MARKING SCHEME MATHEMATICS HIGHER LEVEL

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PRE-LEAVING CERTIFICATE EXAMINATION, 05 MARKING SCHEME MATHEMATICS HIGHER LEVEL Page of 40

OVERVIEW OF MARKING SCHEME Scale label A B C D E No of categories 4 5 6 5 mark scale 0, 5 0,, 5 0,,, 5 0 mark scale 0, 0 0, 5, 0 0,, 7, 0 0,, 5, 8, 0 5 mark scale 0, 5 0, 7, 5 0, 5, 0, 5 0, 4, 7,, 5 0 mark scale 0, 0 0, 0, 0 0, 7,, 0 0, 5, 0, 5, 0 5 mark scale 0, 5 0,, 5 0, 8, 7, 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 full-credit 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 40

Part (a) Part (b) Part (c) Part (d) PAPER QUESTION Scale 0C (a) What is a proof by contradiction? A proof by contradiction is a proof that establishes the validity of a proposition by showing that the proposition being false would imply a contradiction Or similar. Partial Credit ( Marks) Partially correct (b) What is an irrational number? Give an example of an irrational number. Cannot be written in the form b a where a, b Z and b 0. i.e. cannot be written as a fraction Partial Credit ( Marks) Partially correct Page of 40

(c) Shade in the region of the following Venn diagram that represents the set of irrational numbers. R Z N Partial Credit ( Marks) Partially correct (d) Use a proof by contradiction to prove that is an irrational number. Full Credit (0 Marks) High Partial Credit (7 Marks) error in proof Low Partial Credit ( Marks) in proof Page 4 of 40

Part (a) Part (b) Part (c) QUESTION Scale 5C Scale 0C Scale 0C (a) Simplify: x y 4 y y y + y x + x 6 x 4 y y y y + y x + x 6 ( x )( x + ) y y + y y( y + ) ( x + )( x ) ( x + )( y ) x + ( ) ( )( ) High Partial Credit ( Marks) No more than error Low Partial Credit ( Marks) x (b) Solve the inequality, <, x 4 x + 4 and x R. ( x )( x + 4) ( x + 4) < x + 4 x + 0x 8 < x + 6x + x 6x 40 < 0 x = 4 x = 0 4 < x < 0 or on a graph correctly indicated Full Credit (0 Marks) High Partial Credit (7 Marks) error and continues to end correctly Low Partial Credit ( Marks) x + 4 but continues Does not multiply by ( ) Page 5 of 40

(c) Solve the simultaneous equations: x + y 4z = x + y + z = x y + z = x =, y =, z = Full Credit (0 Marks) High Partial Credit (7 Marks) No more than errors and continues to end Low Partial Credit ( Marks) Page 6 of 40

Part (a) Part (b) Part (c) QUESTION Scale 0C Scale 0C (a) Examine the graph of the polynomial shown. Write down the polynomial for the graph in 4 the form ax + bx + cx + dx + e, where a, b, c, d, e R. x =, x =, x =, x = 4 ( x + )( x )( x )( x 4) = 0 4 x 4x x + 4x 8 = 0 Full Credit (0 Marks) High Partial Credit (7 Marks) No more than errors and continues to end Low Partial Credit ( Marks) (b) Investigate if x is a factor of 6x 5x + 4x + 5. x = 6 5 x is not a factor Partial Credit ( Marks) Partially correct No conclusion + 4 + 5 = 6 Page 7 of 40

(c) Given that x + ax is a factor of + px + qx + r = 0 ( x + ax )( x + t) = x + px + qx + r x + tx + ax + atx x t = x + px + qx + r Equating Coefficients t + a = p t = p a at = q t = r q = a r = ar + ( ) ( ) ***Accept other valid methods for full credit Full Credit (0 Marks) x, show that = ( ar +) High Partial Credit (7 Marks) Correct long division but fails to find correct form Sets up correct equation, multiplies but final answer not in correct form Low Partial Credit ( Marks) q. Page 8 of 40

Part (a) Part (b) (i) (ii) QUESTION 4 Scale 5C Scale 0C Scale 0C (a) Label each of the complex numbers given the following information. z = z z 4 = iz z = z High Partial Credit ( Marks) correct Low Partial Credit ( Marks) correct (b) (i) Given ω = + i, write ω in Polar Form. + π π i = 4cos + i sin Full Credit (0 Marks) High Partial Credit (7 Marks) Correct Modulus Correct Angle Low Partial Credit ( Marks) Sketch of ω Page 9 of 40

Page 0 of 40 (ii) Hence solve the equation ω = 4 z. + = sin cos 4 4 π π i z 4 sin cos 4 + + + = π π π π n i n z + + + = sin cos π π π π n i n z If = 0 n + + + = i z If = n + + = i z If = n + + = i z If = n + + = i z Full Credit (0 Marks) High Partial Credit (7 Marks) No more than errors and continues to end Low Partial Credit ( Marks)

Part (a) Part (b) (i) (ii) Part (c) QUESTION 5 Scale 0C (a) Identify the following functions as injective, surjective or bijective. x f ( x) x f ( x) x f ( x) Bijective Injective Surjective Partial Credit ( Marks) One correct (b) The growth of a certain species of bacteria can be modelled with the equation N = Re kt. A science teacher puts 00 bacteria into nutrient agar plates (agar will act as a food source for the bacteria). Five hours later, there are 600 bacteria. (i) Calculate the value of k correct to four decimal places. 600 = 00e ln6 = k 5 k = 0.584 k ( 5) Partial Credit ( Marks) Page of 40

(ii) After how many hours will the number of bacteria in the plate be,000? ( )t e 0. 584,000 = 00 ln0 = k 0.584 k =.6 hours Partial Credit ( Marks) (c) Solve for x correct to two decimal places: ( ) 8 0 x+ x 0 = x+ x 0( ) 8 = 0 x x ( ) 0( ) 8 = 0 y 0y 8 = 0 ( y + )( y 4) y = or y = 4 x x or = 4 x =.6 Full Credit (0 Marks) High Partial Credit (7 Marks) Correct but fails to cancel incorrect solution No more than error in solving and cancels incorrect solution Low Partial Credit ( Marks) Page of 40

Part (a) (i) (ii) Part (b) Part (c) QUESTION 6 Scale 0C (a) Find: (i) ( x 4x ) + dx x + x x + c Partial Credit ( Marks) dx (ii) ( sin x cosx) cosx sin x + c Partial Credit ( Marks) x 8 (b) Evaluate dx x. x 8 = x ( x + x 4) + x + x + 4x ( 9 + 9 + ) + + 4 = Partial Credit ( Marks) ( x )( x + x + 4) ( x ) 4 Page of 40

(c) Find the area enclosed by the x-axis and the curve y = x + x +. x + x + = 0 x = or x = x + x + x + x + ( 9 + 9 + 9) + = 0 Full Credit (0 Marks) High Partial Credit (7 Marks) Error in solving but continues to end with no further error error in integration Low Partial Credit ( Marks) Page 4 of 40

Part (a) Part (b) Part (c) (i) (ii) (iii) QUESTION 7 Scale 0B Scale 5C Scale 0C Scale 0C Scale 5C (a) John wishes to borrow X from his bank over n years. Interest of r% APR will be applied to the loan. He will make an annual repayment of A. Show the amount owing at the end of the first year, P, can be written as: P = X ( + r)a F = P + F = X + P = ( r) t ( r) t X ( + r) A Full Credit (0 Marks) Partial Credit (5 Marks) (b) If P = 0 after the final year of the mortgage, show that the annual repayment A can be n written as P = X P n n = 0 0 = n Xr( + r) ( + r) A. = n n n n ( + r) A( + r) A( + r) A( + r) A n n n X ( + r) A( + r) A( + r) A( + r) A ( ) ( ) ( ) n ( ) n + r + A + r + + A + r + A + r = A( r) n ( + r) n A + A + A A ( ) ( ) n r Xr( + r) + r = n ( ) = X + r n High Partial Credit (0 Marks) Error in geometric progression and continues correctly to end Low Partial Credit (5 Marks) Page 5 of 40

(c) John borrows 50,000 over 0 years at 4% compound interest per annum. (i) Calculate the monthly interest rate correct to five significant figures. (.04) = 0. 007 Full Credit (0 Marks) High Partial Credit (7 Marks) Correct calculation but incorrect significant figures Low Partial Credit ( Marks) (ii) Calculate John s monthly repayment correct to the nearest euro. A = ( 50,000)( 0.007)( + 0.007) 40 ( + 0.007) A = 90 40 = 90.97 Full Credit (0 Marks) High Partial Credit (7 Marks) error in calculation Treats as power of 0 with no other error Low Partial Credit ( Marks) Page 6 of 40

(iii) After years, John decides to pay off the remainder of the loan in one lump sum. Calculate the remaining balance on the loan correct to the nearest euro. 8 years left = 96 months 90 90 90 90 90 + + + + +.007.007.007. 007 a = 90 and r =.007 96 90.007 S 96 =.007 S = 74. 540 96 ( ) ( ) ( ) 95 High Partial Credit (0 Marks) Error in geometric progression and continues correctly to end Low Partial Credit (5 Marks) Page 7 of 40

QUESTION 8 Part (a) Scale 0B* Part (b) Scale 0B* Part (c) Scale 0B* Part (d) Scale 0B* Part (e) Scale 0C* A water trough is in the shape of a prism as shown. (a) Calculate the internal volume of the water trough. Vol = = Full Credit (0 Marks) ( )(.5)( 6) 9 m Partial Credit (5 Marks) (b) Write an equation to represent the volume ( V ) of water in the tank at any time t in terms of h and w. V = V = wh m ( w)( h)( 6) Full Credit (0 Marks) Partial Credit (5 Marks) 4 (c) Show that w = h metres. w h =.5 4 w = h metres Full Credit (0 Marks) Partial Credit (5 Marks) Page 8 of 40

(d) Show that the volume of the water tank can be expressed as: V = 4h. V = wh 4 V = h h V = 4h m Full Credit (0 Marks) Partial Credit (Marks) (e) If water is being pumped in at a constant rate of 5 m /s, at what rate is the height of the water changing when the water has a height of 0 cm? dh dv = dt dt dv = 8 h dh dh = dt dh dt () 5 dh dv 8. ( ) = 0.5 m/s Full Credit (0 Marks) High Partial Credit (7 Marks) error in calculation Low Partial Credit ( Marks) Page 9 of 40

Part (a) Part (b) Part (c) Part (d) Part (e) QUESTION 9 Scale 5C Scale 0B Scale 0C Scale 0B (a) Complete the following table Shape 4 Number of blocks 7 9 7 High Partial Credit ( Marks) incorrect Low Partial Credit ( Marks) Any correct (b) Describe how the number of blocks is changing. Full Credit (0 Marks) Partial Credit (5 Marks) (c) How many blocks would be in the 5 th shape? 6 Partial Credit ( Marks) Page 0 of 40

(d) Write an expression in n for the number of blocks in the n th pattern in the sequence. nd Diff = 6 a = T n = an + bn + c T n = n + bn + c n = T = () + b() + c = b + c = n = T = ( ) + b( ) + c = 7 b + c = 5 b =, c = = n n + Full Credit (0 Marks) T n High Partial Credit (7 Marks) error solving to end Low Partial Credit ( Marks) (e) What shape will require 97 blocks to build it? n n + = 97 n n 96 = 0 n n = 0 ( n )( n +) n = th shape Full Credit (0 Marks) Partial Credit (5 Marks) Page of 40

Part (a) Part (b) Part (c) PAPER QUESTION Scale 0C Scale 0C Two lines make an angle of tan with the line t, where t : x + y =. 4 (a) Write down the slope of the line t. m = Partial Credit ( Marks) Some correct step towards finding the slope (b) Find the values of m, where m is the slope of the lines that intersect the line t. 4 4 m = ± + m = or ( )( m) m + ( )( m) + ( )( m) ( m) = 4( m) or ( m) = 4( m) 4 = m = 8 4m or m = 8 + 4m 9 7 m = or m = 6 Full Credit (0 Marks) High Partial Credit (7 Marks) error and continues to end correctly Low Partial Credit ( Marks) Page of 40

(c) Find the equations of both lines given that they both contain the point (, 5). 9 7 y 5 = ( x ) or y 5 = ( x ) 6 y 0 = 9x + 9 or 6 y 0 = 7x + 7 9 x + y 9 = 0 or 7 x + 6y 7 = 0 Full Credit (0 Marks) High Partial Credit (7 Marks) error and continues to end correctly Low Partial Credit ( Marks) Page of 40

Part (a) Part (b) Part (c) Part (d) QUESTION Scale 0C In a survey of 80 men and 60 women, 60% of each gender said that they smoked cigarettes. (a) Complete the following tree diagram. Full Credit (0 Marks) High Partial Credit (7 Marks) Correct tree diagram with no probabilities shown Correct tree diagram with incorrect consistent probabilities shown Low Partial Credit ( Marks) A person is chosen at random and sent a follow-up questionnaire. (b) Find the probability that the person chosen is a male who does not smoke. 8 5 Partial Credit ( Marks) Some correct step towards finding the slope 75% of the non-smokers having never smoked. 50% of those who have smoked are males. Page 4 of 40

(c) How many females were smokers but have quit? 7 Partial Credit ( Marks) Some correct step towards finding the slope The smoking ban was introduced in Ireland in 004. A government official claims that the number of smokers has fallen steadily since 004 and is significantly lower now than in 004. (d) Discuss the validity of his claim. Regular and heavy smokers numbers on the decline Light and occasional on the increase. Need further figures to determine if numbers have fallen significantly. Partial Credit ( Marks) Partial explanation Page 5 of 40

QUESTION Scale 5D Find the equation of a circle which contains the points (, 5) and (,) line x 5y = 4. x + y + gx + fy + c = 0 ( ) + () 5 + g ( ) + f () 5 + c = 0 6g + 0 f + c = 4 eg ( 7) + ( ) + g ( 7) + g( ) + c = 0 Centre ( g, f ) ( g ) 5( f ) = 4 0 + f = 6 + 5 f = 4 4g + f + c = 70 eg g + 5 f = 4 eg g ( ) g ( 0) High Partial Credit (9 Marks) Error in equations but continues to end Equation of circle not written at end Mid Partial Credit ( Marks) Sets up two correct equations Low Partial Credit (6 Marks) 60 g + 6 f = 408 60g + 00 f = 680 6 f = 7 f = 7, and whose centre lies on the Page 6 of 40

Part (a) Part (b) (i) (ii) (iii) QUESTION 4 Scale 0C The probability of a certain make of car breaking down within two years after it is purchased is 0.8. (a) Write the probability that the car will not break down within the two years. 0.8 = 8% = 4 50 Partial Credit ( Marks) (b) A local car dealer sold 0 of these cars in January 04. Calculate the probability that: (i) None of the cars will break down within the two year period. 0 0 0 0 ( 0.8) ( 0.8) = 0. 0889 Partial Credit ( Marks) (ii) No more than two of the cars will break down within the two year period. 0 0 0 0 0 Full Credit (0 Marks) 0 0 9 8 ( 0.8) ( 0.8) + ( 0.8) ( 0.8) + ( 0.8) ( 0.8) = 0. 747 High Partial Credit (7 Marks) One error Low Partial Credit ( Marks) Page 7 of 40

(iii) At least three of the cars will break down within the two year period. 0.747 = 0.75 Partial Credit ( Marks) Page 8 of 40

Part (a) Part (b) Part (c) QUESTION 5 Scale 5C Scale 5C (a) If sin A = 0. 69, find two values of A, correct to the nearest degree, where 0 A 60. A = 9, High Partial Credit ( Marks) Correct reference angle found with work towards finding two values of A Low Partial Credit ( Marks) (b) The lines q makes an acute angle with the x-axis as shown. (i) Write down the slope of the line q. m = 5 Partial Credit ( Marks) Some correct step towards finding the slope Page 9 of 40

tan α =. (ii) The line p makes an angle β with the x-axis such that ( + β ) 5 + tan β = β tan 5 Find the slope of the line p. + tan β = tan β 5 5 6 5 tan β = 5 5 tan β = m = tan β = p High Partial Credit (0 Marks) Solves with no more than error Low Partial Credit (5 Marks) Page 0 of 40

Part (a) Part (b) QUESTION 6 Scale 0D (a) Prove that the angle at the centre of a circle standing on a given arc is twice the angle at any point of the circle standing on the same arc. Examine five sections of answer Diagram: To Prove: Given: CAD = CBD A circle with centre A and Points B, C and D on the circle as shown Construction: Join B to A and extend to E Proof: In the triangle BCE Full Credit (0 Marks) 5 sections fully correct AB = AC Radii ABC = ACB Theorem (isosceles ) X = ABC + ACB Theorem 6 (exterior angle) X = ABC + ABC X = ABC Similarly EAD = ABD CAD = CBD High Partial Credit (5 Marks) or 4 sections fully correct Mid Partial Credit (0 Marks) sections fully correct Low Partial Credit (5 Marks) Page of 40

(b) Hence find CGD. CGD = 08 Partial Credit ( Marks) Partially correct Page of 40

QUESTION 7 Part (a) Scale 0C* Part (b) Scale 5C Part (c) Scale 5A Part (d) Scale 5C Part (e) Scale 0B Part (f) Scale 5D (a) Calculate the distance from Ballyline to Dun Mór correct to one decimal place. x 6.6 = sin 75 sin 47. x = 8.7 km Full Credit (0 Marks) High Partial Credit (7 Marks) One error Low Partial Credit ( Marks) Due to the installation of water meters, the road from Ballyline to Dun Mór will be closed for a week. Traffic will be diverted via Deepdale. (b) Calculate the percentage increase in the distance motorists will have to travel due to the diversion if travelling from Ballyline to Dun Mór via Deepdale. x = sin 57.8 x = 7.6 % increase = 6.6 sin 47. ( 6.6 + 7.6) High Partial Credit (0 Marks) One error Low Partial Credit (5 Marks) 8.7 00 = 6.% 8.7 Each town will receive its water from a central reservoir after the water meters are installed. The reservoir will be equidistant from each town. Page of 40

(c) What name is given to the point equidistant from the three vertices of a triangle? Circumcentre. (d) Using a suitable construction, show where the reservoir will be situated on the diagram above. High Partial Credit (0 Marks) One bisector found correctly Low Partial Credit (5 Marks) Page 4 of 40

(e) Taking a scale of cm : km, estimate the shortest distance from the reservoir to each town. Accept range of 4. km to 4.8 km Full Credit (0 Marks) Partial Credit (5 Marks) Page 5 of 40

The County Planning Officer decides to double-check the above information by overlapping the information with a coordinate plane. (f) By choosing (0., 4.5) as coordinates for Ballyline, find the coordinates of the reservoir and verify your answer from part (e). **Note: Students answers may vary slightly due to rounding of decimals. Correct in Steps:. Two Midpoints. Slopes. equations of perpendicular bisectors 4. Simultaneous equation leading to a suitable x and y value 5. Distance within acceptable limits Deepdale to Dun Mór: Deepdale to Ballyline: + 8 + + 0. + 4.5 + 7 4.5 + 65 m = = m = m = = m = 8 7 0. 9 7 9 y 0.5 = ( x 4.5) y.5 = ( x 0.55) 65 y.5 = 7x +.5 65y 8.5 = 9x 4. 95 7 x + y = 9 x + 65y = 76. Mid =, = ( 4.5, 0.5) Mid =, = ( 0.55,.5) High Partial Credit ( Marks) or 4 correct steps Low Partial Credit (7 Marks) or correct steps 7 x + y = 9 x + 65y = 76. 6 x + 7y = 97 6 x + 455y = 54. 48 y = 8. y =.6 to.8 x =.5 to x =. 9 ( 8.9) + (. 7 ) 4. km Accept range 4. km to 4.8 km 9 65 Low Partial Credit (4 Marks) Any partially correct step Page 6 of 40

Part (a) (i) (ii) (iii) Part (b) (i) (ii) (iii) QUESTION 8 Scale 5C Scale 0B Scale 5C (a) A light bulb manufacturer claims that the lengths of the life of its bulbs have a mean of 6 months and a standard deviation of months. An independent research company tests a random sample of 60 bulbs to investigate the company s claim. (i) Calculate an approximate value for the standard deviation of the sample means. = 60 0.87 Partial Credit ( Marks) Partially correct (ii) Assuming the company s claim is true, what is the probability that the research company s sample has a mean life of 5 months or less? x μ z = σ n 5 6 z = =.58 0.87 P z.58 = P z <.58 = 0.995 = 0.0049 = 0.49% ( ) ( ) High Partial Credit (0 Marks) Error in z value but continues to end correctly Low Partial Credit (5 Marks) Page 7 of 40

(iii) Based on this information, would you agree or disagree with the manufacturing company s claims? Explain your answer fully. As z >, the company s claim we do not agree with the company s claims. The calculated z score suggests the true mean is lower. Full Credit (0 Marks) Partial Credit (5 Marks) Partially correct (b) The manufacturing company claims that 90% of its customers find their bulbs good value for money. Out of 400 customers surveyed, 46 said that the light bulbs were good value for money. (i) State the null hypothesis. H : P 0 = 0.9 Partial Credit ( Marks) Partially correct (ii) State the alternate hypothesis. H : P 0.9 Partial Credit ( Marks) Partially correct Page 8 of 40

(iii) Use a hypothesis test to investigate the company s claim. 46 Step : p ˆ = = 0. 865 400 Step : E = = 0. 05 400 Step : 0.865 0.05 < p < 0.865 + 0. 05 0.85 < p < 0.95 Step 4: population proportion is within the confidence interval. No evidence to reject the company s claims. ***Deduct one mark from a fully correct solution if a candidate states We accept the company s claim High Partial Credit (0 Marks) or correct steps Low Partial Credit (5 Marks) Page 9 of 40

QUESTION 9 Part (a) Scale 5C* Part (b) Scale 0C* A ranger in a forest can view two campsites from the top of the lookout tower which is 60 m high. The angle of depression to Campsite is 7. while the angle of depression with Campsite is 8.4. (a) Calculate the distance from the bottom of the tower to each campsite correct to one decimal place. 60 60 tan 7. = tan 8.4 = x x x = 78.8 m x = 80. 4 m High Partial Credit (0 Marks) Both calculated with one error Low Partial Credit (5 Marks) (b) It is intended to open a walking trail that extends through each campsite. Calculate the distance from Campsite to Campsite correct to one decimal place. ( 78.8) + ( 80.4) ( 78.8)( 80.4) cos( ) x = x =. m Full Credit (0 Marks) High Partial Credit (7 Marks) Calculated with one error to end Low Partial Credit ( Marks) Page 40 of 40