ADVANCED GCE UNIT MATHEMATICS (MEI) Applications of Advanced Mathematics (C) Paper A TUESDAY 3 JANUARY 007 Additional materials: Answer booklet (8 pages) Graph paper MEI Examination Formulae and Tables (MF) 75(A)/0 Afternoon Time: hour 30 minutes INSTRUCTIONS TO CANDIDATES Write our name, centre number and candidate number in the spaces provided on the answer booklet. Answer all the questions. You are permitted to use a graphical calculator in this paper. Final answers should be given to a degree of accurac appropriate to the context. INFORMATION FOR CANDIDATES The number of marks is given in brackets [ ] at the end of each question or part question. The total number of marks for this paper is 7. ADVICE TO CANDIDATES Read each question carefull and make sure ou know what ou have to do before starting our answer. You are advised that an answer ma receive no marks unless ou show sufficient detail of the working to indicate that a correct method is being used. NOTE This paper will be followed b Paper B: Comprehension. This document consists of 6 printed pages and blank pages. HN/7 OCR 007 [T/0/653] OCR is an exempt Charit [Turn over
Section A (36 marks) Solve the equation [] x x. x Fig. shows part of the curve + x 3. 0 x Fig. Û 3 (i) Use the trapezium rule with strips to estimate Ù + x d, x giving our answer correct to ı0 3 significant figures. [3] (ii) Chris and Dave each estimate the value of this integral using the trapezium rule with 8 strips. Chris gets a result of 3.5, and Dave gets 3.30. One of these results is correct. Without performing the calculation, state with a reason which is correct. [] OCR 007 75A/0 Jan 07
3 (i) Use the formula for sin (q f), with and to show that sin 05 3 +. [] (ii) In triangle ABC, angle BAC 5, angle ACB 30 and AB unit (see Fig. 3). C 3 q 5 f 60, 30 A 5 B Fig. 3 Using the sine rule, together with the result in part (i), show that AC 3 +. [3] tan q Show that tan q sec q. tan q Hence, or otherwise, solve the equation for 0 [7] tan q, q 80. 5 Find the first four terms in the binomial expansion of ( + 3x ) 3. State the range of values of x for which the expansion is valid. 6 (i) Express in partial fractions. [3] (x )(x ) (ii) A curve passes through the point (0, ) and satisfies the differential equation d dx (x )(x ). x Show b integration that x. OCR 007 75A/0 Jan 07 [Turn over
Section B (36 marks) 7 Fig. 7 shows the curve with parametric equations x cos q, sin q 8 sin q, 0 q p. The curve crosses the x-axis at points A(, 0) and B(, 0), and the positive -axis at C. D is the maximum point of the curve, and E is the minimum point. The solid of revolution formed when this curve is rotated through 360 about the x-axis is used to model the shape of an egg. D C B(, 0) A(, 0) x E Fig. 7 (i) Show that, at the point A, Write down the value of q at the point B, and find the coordinates of C. [] q 0. d (ii) Find in terms of q. dx Hence show that, at the point D, cos q cos q 0. (iii) Solve this equation, and hence find the -coordinate of D, giving our answer correct to decimal places. The cartesian equation of the curve (for 0 (iv) Show that the volume of the solid of revolution of this curve about the x-axis is given b 3 p Û 6 Ù 6-8x - 5x + 8x - x d. x ı- Evaluate this integral. [6] ) is ( - x) - x. q p ( ) OCR 007 75A/0 Jan 07
5 8 A pipeline is to be drilled under a river (see Fig. 8). With respect to axes Oxz, with the x-axis pointing East, the -axis North and the z-axis vertical, the pipeline is to consist of a straight section AB from the point A(0, 0, 0) to the point B(0, 0, 0) directl under the river, and another straight section BC. All lengths are in metres. z (N) C(a, b, 0) O x(e) A(0, 0, 0) B(0, 0, 0) Fig. 8 (i) Calculate the distance AB. [] The section BC is to be drilled in the direction of the vector 3i j k. (ii) Find the angle ABC between the sections AB and BC. [] The section BC reaches ground level at the point C(a, b, 0). (iii) Write down a vector equation of the line BC. Hence find a and b. (iv) Show that the vector 6i 5j k is perpendicular to the plane ABC. Hence find the cartesian equation of this plane. OCR 007 75A/0 Jan 07
75 Mark Scheme Jan 007 Paper A Section A x + x x + x+ + x x( x+ ) x + x x D [] Clearing fractions solving cao (i) x 0 0.5.5.060..09 3 7 7 A 0.5[(+3)/ +.0607 +. +.097] 3.8 (3 s.f.) B [3] At least one value calculated correctl or 3.3 or 6.566 seen (ii) 3.5 (or Chris) area should decrease with the number of strips used. B B [] ft (i) or area should decrease as concave upwards 3(i) sin 60 3/, cos 60 /, sin 5 /, cos 5 / sin(05 ) sin(60 +5 ) sin60 cos5 + cos60 sin5 3. +. 3+ * (ii) Angle B 05 B the sine rule: AC sin B sin 30 AC sin05 3 +. sin 30 3+ * sin θ + + tan θ cos θ tan θ sin θ cos θ cos θ + sin θ cos θ sin θ cos θ secθ secθ cosθ ½ θ 60, 300 θ 30, 50 E [] E [3] E B B [7] splitting into 60 and 5, and using the compound angle formulae Sine rule with exact values www sinθ tanθ or +tan²θsec²θ used cosθ simplifing to a simple fraction in terms of sinθ and/or cos θ onl cos θ sin θ cos θ oe used or +tan θ ( tan θ) tan θ ±/ 3oe 30 50 and no others in range 8
75 Mark Scheme Jan 007 5 (+ 3 x) 3 5.( ) ( )( ) 3 3 3 3 3 3 + (3 x) + (3 x) + (3 x) +... 3! 3! 5 3 + x x + x +... 3 Valid for < 3x < -/3 < x < /3 B A,,0 B binomial expansion (at least 3 terms) correct binomial coefficients (all) x, x, 5x 3 /3 6(i) A B + (x + )( x+ ) x+ x+ A(x + ) + B(x + ) x : B B x ½ : ½ A A [3] or cover up rule for either value (ii) d dx (x + )( x + ) d dx (x+ )( x+ ) ( ) x + x + dx ln ln(x + ) ln(x + ) + c When x 0, ln ln ln + c c ln ln ln(x + ) ln(x + ) + ln (x + ) ln x + x + x + * Bft E separating variables correctl condone omission of c. ft A,B from (i) calculating c, no incorrect log rules combining lns www 9
75 Mark Scheme Jan 007 Section B 7(i) At A, cos θ θ 0 At B, cos θ θ π At C x 0, cos θ 0 θ π/ π sin sin π 8 B B [] or subst in both x and allow 80 (ii) d d / dθ dx dx / dθ cosθ cos θ sin θ cos θ cosθ sinθ d/dx 0 when cos θ cosθ 0 cos θ cosθ 0 cos θ cosθ 0* E finding d/dθ and dx/dθ correct numerator correct denominator 0 or their num0 (iii) ± 6+ 8 cosθ ± 6 ( + ½ 6 > so no solution) θ.7975 sin θ sin θ.09 8 ft cao cao ± 6 or (.7,-.7) both or ve their quadratic equation.80 or 03 their angle.03 or better (iv) V π dx π (6 8 x x )( x ) dx 6 + 3 π (6 8x x 6x 8 x x ) dx 6 + + 3 π (6 8x 5x + 8 x x ) dx* 6 3 5 π 6x x 5x x x 6 + 5 π (3 0 ) 6 5.35π. E B cao [6] correct integral and limits expanding brackets correctl integrated substituting limits 0
75 Mark Scheme Jan 007 8 (i) (0 0) + (0+ 0) + ( 0 0) 60 m [] 0 (ii) BA 0 0 0 3. 3 cosθ 9 6 3 6 θ 8 [] or AB 3 oe eg -60 9 6 oe eg 60 6 cao (or radians) 0 3 (iii) r 0 + λ 0 At C, z 0 λ 0 a 0 + 3 0 00 b 0 + 0 80 B B 0 0 +... 0 3... + λ 00 80 a 0 or.+λ b 0 6 (iv) 5. + 0+ 0 6 3 5. 8 0+ 0 6 5 is perpendicular to plane. Equation of plane is 6x 5 + z c At B (sa) 6 0 5 0 + 0 c c 00 so 6x 5 + z 00 B B ( alt. method finding vector equation of plane eliminating both parameters D correct equation stating Normal hence perpendicular B)