abc General Certificate of Education Mathematics Pure Core SPECIMEN UNITS AND MARK SCHEMES ADVANCED SUBSIDIARY MATHEMATICS (56) ADVANCED SUBSIDIARY PURE MATHEMATICS (566) ADVANCED SUBSIDIARY FURTHER MATHEMATICS (57) ADVANCED MATHEMATICS (66) ADVANCED PURE MATHEMATICS (666) ADVANCED FURTHER MATHEMATICS (67)
General Certificate of Education Specimen Unit Advanced Subsidiary Eamination MATHEMATICS Unit Pure Core MPC abc In addition to this paper you will require: an 8-page answer book; the AQA booklet of formulae and statistical tables. You must not use a calculator. Time allowed: hour 0 minutes Instructions Use blue or black ink or ball-point pen. Pencil should only be used for drawing. Write the information required on the front of your answer book. The Eamining Body for this paper is AQA. The Paper Reference is MPC. Answer all questions. All necessary working should be shown; otherwise marks for method may be lost. Calculators (scientific and graphical) are not permitted in this paper. Information The maimum mark for this paper is 75. Mark allocations are shown in brackets. Advice Unless stated otherwise, formulae may be quoted, without proof, from the booklet. []
Answer all questions. The line L has equation and the curve C has equation + y = 9 y = + (a) Sketch on one pair of aes the line L and the curve C. Indicate the coordinates of their points of intersection with the aes. ( marks) (b) Show that the -coordinates of the points of intersection of L and C satisfy the equation + 6 = 0 (c) Hence calculate the coordinates of the points of intersection of L and C. ( marks) ( marks) The line AB has equation 5 y = 7. The point A has coordinates (, ) and the point B has coordinates (, k). (a) (i) Find the value of k. ( mark) (ii) Find the gradient of AB. ( marks) (b) The point C has coordinates ( 6, ). Show that AC has length p, where p is an integer. ( marks) (a) Given that ( ) is a factor of f ( ), where f ( ) = k + 0 show that k = 7. ( marks) (b) Divide f ( ) by ( ) to find a quadratic factor of ( ) f. ( marks) (c) Write f ( ) as a product of three linear factors. ( marks) (d) Calculate the remainder when f ( ) is divided by ( ). ( marks) []
The number satisifies the equation where m is a constant. + m + 6 = 0 Find the values of m for which this equation has: (a) equal roots; (b) two distinct real roots; (c) no real roots. ( marks) ( marks) ( marks) 5 The diagram shows a part of the graph of y = dy (a) (i) Find. d ( marks) (ii) Show that the -coordinate of the stationary point P is. ( marks) (iii) Find the y-coordinate of P. (b) Find the area of the shaded region. ( mark) (5 marks) Turn over []
6 A circle C has equation + y 0 = 0 (a) By completing the square, epress this equation in the form ( a) + y = r (b) Write down the radius and the coordinates of the centre of the circle C. ( marks) ( marks) (c) Describe a geometrical transformation by which C can be obtained from the circle with equation + y = r ( marks) (d) The point P, which has coordinates (9, ), lies on the circle C. (i) Show that the line which passes through P and the centre of C has gradient ( marks) (ii) Find the equation of the tangent to the circle C at the point P. Give your answer in the form y = m + c. ( marks) 7 (a) Epress + in the form a + b, where a and b are integers. ( marks) (b) Solve the inequality ( ) < + ( marks) [5]
5 8 A curve has equation y = 8 + 6 + 8 (a) Find dy d y and. d d (5 marks) (b) Find the three values of for which d y = 0. ( marks) d (c) Determine the coordinates of the point at which y has a maimum value. (5 marks) END OF QUESTIONS [6]
MPC Specimen Question Solution Marks Total Comments (a) Sketches of L and C Coordinates (0, 9), (9, 0) indicated Coordinates (0, ) indicated A A General shape Accept labels on sketch ditto (b) Equating epressions for y + 6 = 0 (c) Solving quadratic = or = Points are (, ) and (, 7) A A AA Total 9 (a)(i) k = B ( ) (ii) Gradient = = 5 (b) Distance formula AC = 50 = 5 (a) Use of factor theorem k + 0 = 0, so k = 7 A A A Total 6 A oe convincingly shown (ag) Two solutions needed A if not clearly paired or use of equation of AB ft wrong value of k stated or used or complete division convincingly shown (ag) (b) Quotient is 0 B B if or 0 correct (c) f() = ( )( + )( 5) B B if signs wrong (d) f() = B so remainder is B ft wrong value for f() (a) m 6 = 0 m = ±8 (b) m 6 > 0 m < 8 or m > 8 (c) m 6 < 0 8 < m < 8 Total 8 A A A Total 6 abc [7]
MPC (cont) Question Solution Marks Total Comments 5(a)(i) y = 8 A if at least one term correct (ii) SP y = 0 PI (iii) = 8 A = convincingly shown A ag; / for verification y = B P 8 (b) y d = (+ c) Substitution of Area = = 80 9 8 80 5 = 6(a) Use of ( 5) = 0 + 5 a = 5, r = 5 (b) Radius 5 Centre (5, 0) (c) Translation 5 units in positive direction (d)(i) Use of formula for gradient Grad convincingly shown (ii) Grad of tangent is Tangent is y = ( 9) 5 A m A A 5 if at least one term correct First A awarded if at least one term correct Total AA Condone RHS = 5 B B ft wrong value for a ditto Condone transformation if clarified A ft wrong value for a A ag A ft wrong gradient i.e. y = +5 A ft wrong gradient Total 7(a) Rationalising denominator Numerator becomes + ma Denom =, so ans is + A ft one small error in numerator (b) LHS = B Allow < + for isolating terms Reasonable attempt at division m + A ft error in epanding LHS < Total 8 B [8]
MPC (cont) Question Solution Marks Total Comments 8(a) y = + at least one term correct A A with one error y = 8 + m at least one term correct A 5 ft numerical error in y (b) y = 0 if = 0 or if 6 + 8 = 0 B A A if only one small error ie = or = (c) y =, 6, A Relationship between sign of y" and maimum/minimum Ma at = y = A A 5 Total 5 TOTAL 75 if at least one correct ft numerical error in y [9]
General Certificate of Education Specimen Unit Advanced Subsidiary Eamination abc MATHEMATICS Unit Pure Core MPC In addition to this paper you will require: an 8-page answer book. the AQA booklet of formulae and statistical tables. You may use a graphics calculator. Time allowed: hour 0 minutes Instructions Use blue or black ink or ball-point pen. Pencil should only be used for drawing. Write the information required on the front of your answer book. The Eamining Body for this paper is AQA. The Paper Reference is MPC. Answer all questions. All necessary working should be shown; otherwise marks for method may be lost. The final answer to questions requiring the use of tables or calculators should normally be given to three significant figures. Information The maimum mark for this paper is 75. Mark allocations are shown in brackets. Advice Unless stated otherwise, formulae may be quoted, without proof, from the booklet. [0]
Answer all questions. The diagram shows triangle ABC. The lengths of AB and AC are 7cm and 8cm respectively. The size of angle BAC is 60. Calculate the length of BC, giving your answer to significant figures. ( marks) The diagrams show a square of side 6 cm and a sector of a circle of radius 6 cm and angle θ radians. The area of the square is three times the area of the sector. (a) Show that θ =. ( marks) (b) Show that the perimeter of the square is times the perimeter of the sector. ( marks) The nth term of an arithmetic sequence is u n, where u n = 0 + 0.5n (a) Find the value of u and the value of u. (b) Write down the common difference of the arithmetic sequence. (c) Find the value of n for which u n = 5. (d) 0 Evaluate u n. n= ( marks) ( mark) ( marks) ( marks) []
(a) Given that log a = log a 5 + log a where a is a positive constant, show that = 5. ( marks) (b) (i) Write down the value of log. ( mark) (ii) Given that log y = log find the value of y. ( marks) 5 The curve C is defined by the equation y = + for > 0 (a) Write in the form k, where k is a fraction. ( mark) (b) Find d y. d ( marks) (c) Find an equation of the tangent to the curve C at the point on the curve where =. ( marks) d y (d) (i) Find. ( marks) d (ii) Hence deduce that the curve C has no maimum points. ( marks) 6 The amount of money which Pauline pays into an insurance scheme is recorded each year. The amount which Pauline pays in during the nth year is A n. The first three values of A n are given by: A = 800 A = 650 A = 50 The recorded amounts may be modelled by a law of the form where p and q are constants. An + = pan + q (a) Find the value of p and the value of q. (5 marks) (b) Given that the amounts converge to a limiting value, V, find an equation for V and hence find the value of V. ( marks) Turn over []
7 (a) Epress 5 + in the form p + q, where p and q are integers. ( marks) 5 + (b) Hence find the eact value of d. (5 marks) 8 The angle θ radians, where 0 θ π, satisfies the equation tan θ = cos θ (a) Show that sin θ = cos θ ( mark) (b) Hence use an appropriate identity to show that sin θ + sin θ = 0 ( marks) (c) (i) Solve the quadratic equation in part (b). Hence eplain why the only possible value of sin θ which will satisfy it is. ( marks) (ii) Find the values of θ for which sinθ = and 0 θ π. ( marks) (d) Hence write down the solutions of the equation tan = cos that lie in the interval 0 80. ( marks) []
5 9 The diagram shows a sketch of the curve with equation y =. (a) (i) Use the trapezium rule with five ordinates (four strips) to find an approimation for d. 0 ( marks) (ii) By considering the graph of y =, eplain with the aid of a diagram whether your approimation will be an overestimate or an underestimate of the true value for d. 0 ( marks) (b) Describe the single transformation by which the curve with equation y = 5 can be obtained from the curve with equation y =. ( marks) (c) Sketch the curve with equation y =. ( mark) (d) The two curves y = 5 and y = intersect at the point P. (i) Show that the -coordinate of the point P is a root of the equation = 0. ( marks) (ii) Solve this equation to find the -coordinate of the point P. Give your answer to 5 significant figures. ( marks) END OF QUESTIONS []
MPC Specimen Question Solution Marks Total Comments o BC = 7 + 8 7 8 cos60 = 9 + 6 56 m = 57 BC = 57 = 7. 55 to sf A Total (a) {Area of square} = 6 θ Use of r θ PI 6 = 6 θ θ = A ag (b) Arc length = 6θ B abc Perimeter of sector = ( + 6θ ) = + 6 = A ag = perimeter of square Total 5 (a) u = 0.5; u = BB sc B for 0, 0.5 (b) Common difference is 0.5 B (c) 0 + 0.5n = 5 0.5n = 5 0 n = 0 A (d) 0 n= u = sum of AP with n = 0 n 0 = ( 0.5 + 5) m = 5.5 A Total 8 (a) log a = log a 5 + log a log a = log a [5 ] m = 5 A (b)(i) log = B (ii) {log y =} log = 0.5 B y = = B Total 6 oe PI ag convincingly found [5]
MPC (cont) Question Solution Marks Total Comments 5(a) 5 = B Accept k =.5 (b) 5 y = + dy d = 5 5 AA One correct inde ft A for each correct term (c) When =, y = + = B When =, y = 5 = Eqn. of tangent: y = ( ) m A Accept any valid form (d)(i) d y 5 0 = + 6 B, ft each term provided equivalent demands d ie indices one fractional one negative (ii) Since >0, y () is >0 so any turning point must be a minimum E for attempt to find the sign of y () ie C has no maimum points E,,0 6(a) 650 = 800 p + q 50 = 650 p + q 67 = 9 Total A m For either equation Need both Full valid method to solve simultaneous equations p = ; q = 0 5 AA 5 (b) V = pv + q q V = p m V = 50 A ft on numerical slip in (a) Total 8 7(a) + One power correct A (b) Inde raised by (either term) A One term correct ft p, q. A All correct m A 5 Total 7 Use of limits Must be an eact value [6]
MPC (cont) Question Solution Marks Total Comments 8(a) sin θ = cosθ sinθ = cos θ cosθ B ag convincingly found (b) sin θ + cos θ = oe seen sinθ = ( sin θ ) sin θ + sin θ = 0 (c)(i) Attempt to solve for sin θ sinθ sinθ + = ( )( ) 0 A A A ag convincingly found M0 for verification oe eg use of the formula. Since sin θ, the only possible value for sin θ is A ag convincingly found and eplained (ii) θ = 0.55 (d) θ =.679 tan = cos sin = = 0 or 50 = 5 or = 75 B B A A Total 9(a)(i) h = 0.5 B Integral = h/ { } {.}= f (0) + [f + f () + f ] + f () {.}=+[++8]+6 A Integral =.5 A (ii) Relevant trapezia drawn on a copy of given graph. Overestimate A` (b) Stretch in y-direction scale factor 5 A (c) Sketch showing the reflection of the graph (d)(i) of the given curve in the y-ais B 5 = 5 = In (ii) accept sf and in terms of π; ft on their 0.55 Links with previous parts 5 = = 0. A ag convincingly found (ii) ln = ln 0. oe using base 0 ln0. = ln A = 0.5808(0 ) A Need 5sf or better Total TOTAL 75 [7]
abc General Certificate of Education Specimen Unit Advanced Level Eamination MATHEMATICS Unit Pure Core MPC In addition to this paper you will require: an 8-page answer book; the AQA booklet of formulae and statistical tables; an insert for use in Question 5 (enclosed). You may use a graphics calculator. Time allowed: hour 0 minutes Instructions Use blue or black ink or ball-point pen. Pencil should only be used for drawing. Write the information required on the front of your answer book. The Eamining Body for this paper is AQA. The Paper Reference is MPC. Answer all questions. All necessary working should be shown; otherwise marks for method may be lost. The final answer to questions requiring the use of tables or calculators should normally be given to three significant figures. Information The maimum mark for this paper is 75. Mark allocations are shown in brackets. Advice Unless stated otherwise, formulae may be quoted, without proof, from the booklet. [8]
Answer all questions. Find d y d when: (a) (b) y = tan ; sin y =. ( marks) ( marks) A curve has equation y = ( + ) from the origin to the point (,0). The region R is bounded by the curve, the -ais and the line =. (a) Eplain why R lies entirely above the -ais. ( mark) (b) (c) Use Simpson s Rule with five ordinates (four strips) to find an approimation for the area of R, giving your answer to significant figures. ( marks) Find the eact value of the volume of the solid formed when R is rotated through π radians about the -ais. (5 marks) A curve has equation y = e. (a) Show that the -coordinate of the stationary point on the curve is ln. Find the corresponding y-coordinate in the form a + b ln, where a and b are integers to be determined. (6 marks) d y (b) Find an epression for and hence determine the nature of the stationary d point. ( marks) (c) Show that the area of the region enclosed by the curve, the -ais and the lines = 0 and = e 5. (5 marks) is ( ) [9]
(a) Describe a sequence of geometrical transformations that maps the graph of y = sin onto the graph of y = + sin. ( marks) (b) Find the gradient of the curve with equation y = + sin at the point where π =. 6 ( marks) (c) (i) Find sin d. ( marks) π π (ii) Hence show that ( ) ( π + ) + sin d =. 0 8 ( marks) 5 [An insert is provided for use in answering this question.] The curve with equation y = intersects the -ais at the point A where = α. (a) Show that α lies between and 5. ( marks) (b) Show that the equation = 0 can be rearranged in the form = +. ( marks) (c) (i) Use the iterative formula n+ = + with = 5 to find, giving your answer n to three significant figures. ( marks) (ii) The sketch shows the graphs of y = + and y = and the position of. On the insert provided, draw a cobweb or staircase diagram to show how convergence takes place, indicating the positions of and. ( marks) [0] Turn over
6 Solve the equation cot + cosec + = 0 giving all values of to the nearest degree in the interval 0 o 60. (7 marks) 7 The functions f and g are defined with their respective domains by f > + g ( ) =, 0 ( ) = 9, (a) Find fg ( ), giving your answer in its simplest form. ( marks) (b) (i) Solve the equation g ( ) =. ( marks) (ii) Eplain why the function g does not have an inverse. ( mark) (c) Solve the equation g ( ) =. ( marks) (d) The inverse of f is f. (i) Find ( ) f. ( marks) (ii) Solve the equation f ( ) = f ( ). ( marks) END OF QUESTIONS []
Surname Centre Number Candidate Signature Other Names Candidate Number abc General Certificate of Education Specimen Unit Advanced Level Eamination MATHEMATICS Unit Pure Core MPC Insert for use in answering Question 5. Fill in the boes at the top of this page. Fasten this insert securely to your answer book. []
MPC Specimen Question Solution Marks Total Comments (a) sec + tan Product Rule sec A Correct (b) cos sin Quotient Rule B d(sin ) d = cos A Correct Total 6 (a) y 0 when 0 so R is above -ais E abc b) Outside multiplier 0. 5 0.5{ y (0) + y() + y ( 0.5) + y(0.75) + y(0.5) [ ] } y(0) = 0; y (0.5) = 0.7609; y(0.5) = 0.00; y (0.75) = 0.89; y() = 0.5775 = 0.69. A Correct to at least sf = 0.5 to sf A (c) V = π d + 0 k ln( +) B ln( ) A Integration correct π (ln ln) m A 5 Total 0 Correct use of limits []
MPC (cont) Question Solution Marks Total Comments (a) y e = ke d A d y 0 e = = d = ln = ln A ag y = ln a = B b = B 6 (b) d y = e d B d y = ln = 8 d Minimum Point A ft if negative for maimum (c) ke A e e = e 0 = ( e 5) A e m Use of limits 0 and A 5 ag Total []
MPC (cont) Question Solution Marks Total Comments (a) One way stretch in direction (b) (c)(i) scaling factor A translation (in y-direction) 0 of A d y cos d A π dy When =, = = 6 d A cos A + cos d = cos + sin (ii) + A cos Correct A (ignore +c) previous result and attempt at limits Ignore [] 0 π π π π + = + 8 8 π(π + ) = 8 Integration by parts attempt cos A ag all integration must be correct for A Total [5]
MPC (cont) Question Solution Marks Total Comments 5(a) f ( ) = ; f ( 5) = ( ) f = change of sign root between and 5 A (b) = 0 Divide by = + A ag (c)(i) = + 5 =.6 A =. (to sf) A (ii) 6 ( cosec ) + cosec + = 0 A A Total 0 Cobweb to curve first Thin line marked Net iteration marked Attempt at cot = cosec cosec + cosec = 0 A or may use cos = sin ± 9 cosec = 8 = 0.05.05 A cosec = sin B sin = 0.09 =98 A A 7 Total 7 after cos + + = 0 etc sin sin [6]
MPC (cont) Question Solution Marks Total Comments 7(a) fg( ) = + 9 A and no further wrong simplification = 6 (b)(i) 9 = = = scores only = ± A (ii) Two values of map onto one not one-one E or many-one (c) Two values from g() = = ± B 9 = = 5 =± 5 A (d)(i) y = y + = + y multiplying out and attempt to make the subject = y A f ( ) = f = f and multiplying up (ii) their ( ) ( ) A + = 0 =, A > 0 only solution is = A Total 5 TOTAL 75 or f ( ) = = ( + ) A ( + )( ) = 0 [7]
General Certificate of Education Specimen Unit Advanced Level Eamination abc MATHEMATICS Unit Pure Core MPC In addition to this paper you will require: an 8-page answer book; the AQA booklet of formulae and statistical tables. You may use a graphics calculator. Time allowed: hour 0 minutes Instructions Use blue or black ink or ball-point pen. Pencil should only be used for drawing. Write the information required on the front of your answer book. The Eamining Body for this paper is AQA. The Paper Reference is MPC. Answer all questions. All necessary working should be shown; otherwise marks for method may be lost. The final answer to questions requiring the use of tables or calculators should normally be given to three significant figures. Information The maimum mark for this paper is 75. Mark allocations are shown in brackets. Advice Unless stated otherwise, formulae may be quoted, without proof, from the booklet. [8]
Answer all questions. The polynomials f() and g() are defined by (a) By considering f () and (), have a common linear factor. f g () = + () = g or otherwise, show that f ( ) and g ( ) ( marks) (b) Hence write f g ( ) ( ) as a simplified algebraic fraction. ( marks) (a) Obtain the binomial epansion of ( ) + as far as the term in. ( marks) (b) (i) Hence, or otherwise, find the series epansion of ( + ) as far as the term in. (ii) Find the range of values of for which this epansion is valid. ( marks) ( mark) (a) Epress sin θ cos θ in the form R sin( θ α), where R is a positive constant and o o 0 < α < 90. Give the value of α to the nearest 0. o. ( marks) (b) Hence find the solutions in the interval o o 0 < θ < 60 of the equation sinθ cosθ = Give each solution to the nearest degree. ( marks) [9]
(a) Epress sin in the form a + bcos, where a and b are constants. ( marks) π (b) Find the value of sin d. 0 ( marks) (c) Hence find the volume generated when the part of the graph of y = sin between = 0 and = π is rotated through one revolution about the -ais. ( marks) 5 The population P of a particular species is modelled by the formula P = Ae -kt where t is the time in years measured from a date when P = 5000. (a) Write down the value of A. ( mark) (b) Given that P = 500 when t = 0, show that k 0.0567. ( marks) (c) Find the value of the population 0 years after the initial date. ( marks) 6 (a) Solve the differential equation d dt 0 = 5 given that = when t = 0. ( marks) (b) Find the value of t for which =, giving your answer to three decimal places. ( marks) Turn over [0]
7 (a) Epress 5 + ( )( + ) in the form A B C + + + + ( ) ( marks) (b) Hence find the value of 5 + + ( )( ) d giving your answer in the form p + qn. (6 marks) 8 A curve is defined by the parametric equations = t +, y = t, t 0 t t (a) (i) Epress + y and y in terms of t. ( marks) (ii) Hence verify that the cartesian equation of the curve is ( + y)( y) = ( marks) (b) (i) By finding d dy and, dt dt calculate the value of dy at the point for which t =. d (5 marks) (ii) Hence find the equation of the tangent to the curve at this point. Give your answer in the form a + by = c, where a, b and c are integers. ( marks) []
[] 5 9 The line l has equation + = t z y. The line l has equation. 0 9 + = s z y (a) Show that the lines l and l intersect and find the coordinates of their point of intersection. (5 marks) (b) The point P on the line l is where t = p, and the point Q has coordinates (5, 9, ). (i) Show that ( marks) (ii) Hence find the coordinates of the foot of the perpendicular from the point Q to the line l. ( marks) END OF QUESTIONS 8. = p QP
MPC Specimen Question Solution Marks Total Comments (a) f( ) = 0, g( ) = 0 A or other complete method A So is a common factor or (b) f ( ) = ( )( + ) B g( ) = ( )( + ) B f ( ) + So = g( ) + B ft numerical error Total 6 (a) ( + ) = + +... 8 A if two terms correct (b)(i) ( + ) = ( + ) B Reasonable attempt... = + +... 6 A (ii) Valid if < < B Total 6 (a) R = 5 B cos α or sin α = or 5 5 α 6.9 A (b) sin( θ α) = 5 PI One solution is α + sin 5 m Solutions 60 and 9 AA Accept awrt 60 or 6, and awrt 9 Total 7 (a) Use of cosa sin A PI A sin cos (b)... d = sin (+ c) π Use of sin = 6 A m if at least one term correct (c) π 0...d = π 6 ( sin ) = sin So volume is π ( π ) 6 A B B Total 8 5(a) A = 5000 B (b) 0k 500 = 5000e B ln 0.7 = 0k A k = ln0.7 0 Accept 0.06 Accept 0.07 ft one error oe; ft wrong value for A oe ft one numerical error... 0.0567 A convincingly shown (ag) 0. P = 5000(0.7) or 5000e A if only one small error... = 50 A Accept awrt 50 Total 8 (c) 7 abc []
MPC (cont) Question Solution Marks Total Comments 6(a) Attempt to separate variables d m d = ±k ln(0 ) (+ c) 0 t = 5 ln(0 ) + c A c = 5 ln 9 A (b) = t = 5 ln 9 5 ln 8... 0.589 A Total 6 7(a) 5 + A( + ) + B( )( + ) + C( ) B A = 6, B =, C = 8 A A if only one error (b)... = ln( )... A ft wrong values for 8 B coeffs throughout this question... ln( + ) (+ c) B + m 8... = ln ln + ln ( )... = + ln A 6 Total 0 8(a)(i) + y = t, y = / t B (ii) ( + y)( y) = (t )(6 / t ) A... = Convincingly shown (ag) (b)(i) d dy A = t, = t + dt t dt t m Use of chain rule d y 9 When t =, = d 7 A, 5 ft numerical or sign error (ii) The point is (5, ) B The tangent is y = 9 ( 5) 7 ie 9 7y = A ft one numerical error Total []
MPC (cont) Question Solution Marks Total Comments 9(a) + t = 9 s oe + t= + s + t = s A Solve two to obtain s =, t = m Check in rd equation A Point of intersection is (7,, ) AF 5 (b)(i) p QP = p A p 0. m QP = ( p ) + ( ) + ( 0) A Convincingly shown (ag)... = p 8 (ii) QP l p 8 = 0... p = Foot of perpendicular is (, 6, 7) A A Total TOTAL 75 [5]