Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Madras College Mathematics Higher March Prelim Examination 2010/2011 NATIONAL QUALIFICATIONS Assessing Unit 3 + revision from Units 1 & 2 Time allowed - 1 hour 10 minutes Read carefully 1. Calculators may be used in this paper. 2. Full credit will be given only where the solution contains appropriate working. 3. Answers obtained from readings from scale drawings will not receive any credit. Pegasys 2010 FORMULAE LIST Circle: The equation x 2 y 2 2 gx 2 fy c 0 represents a circle centre ( g , f ) and radius The equation ( x a) 2 ( y b) 2 r 2 represents a circle centre ( a , b ) and radius r. Trigonometric formulae: sin A B sin Acos B cos Asin B cos A B cos Acos B sin Asin B sin 2 A 2sin Acos A cos2 A cos 2 A sin 2 A 2 cos 2 A 1 1 2 sin 2 A Scalar Product: . a b a b cosθ, where θ is the angle between a and b. or . a b a1 b1 a 2 b2 a 3 b3 Table of standard derivatives: Table of standard integrals: f ( x) f ( x) sin ax a cos ax cos ax a sin ax f ( x) dx f ( x) sin ax cos ax Pegasys 2010 a1 b1 where a a 2 and b b 2 a b 3 3 1 cos ax C a 1 sin ax C a g2 f 2 c . SECTION A In this section the correct answer to each question is given by one of the alternatives A, B, C or D. Indicate the correct answer by writing A, B, C or D opposite the number of the question on your answer paper. Rough working may be done on the paper provided. 2 marks will be given for each correct answer. 1. 2. 3. 4. If f ( x) (2 x 1) 4 then f (1) equals A 4 B 1 C 2 D 8 The maximum value of the function g ( x) 3 sin x 2 cos x is A 13 B 5 C 0 D 2 The radius of the circle with equation x 2 y 2 4 x 2 y 4 is A 2 B 3 C 1 D 24 When 5 sin x 3 cos x is written in the form k sin x where 0 360 , tan is equal to 5 3 A B 5 3 C 3 5 D Pegasys 2010 3 5 5. 6. 7. The value of log A 2 B 4 2 C 1 4 D 4 2 4 is 1 p Given that the vectors 4 and 2 are perpendicular, the value of p is 0 3 A 0 B 8 C 4 D 6 Part of the graph of y log 10 x is shown in each diagram below as a broken line. Which diagram also shows, as a full line, part of the graph of y log 10 1x ? A y B y o o x C y o Pegasys 2010 D x x y o x 8. 12 a 12 is a unit vector. Which of the following could be the value of g? g A 1 2 B 1 C 1 D 1 2 [ END OF SECTION A ] SECTION B ALL questions should be attempted 9. Triangle ABC has vertices A( 1, 0 , 3 ), B( 5, 4 , 1 ) and C( 4, 16 , 4 ) respectively. A, B and D are collinear such that A( 1, 0 , 3 ) AB 2 . BD 3 B( 5, 4 , 1 ) D C( 4, 16 , 4 ) (a) Show that the coordinates of D are ( 11, 10, 2 ). 2 (b) Hence show clearly that angle ADC is a right angle. 4 (c) Prove that angle ABC is obtuse. 3 Pegasys 2010 10. A function is defined as f ( x) 6 cos 2 12 x 3 sin x . (a) By using the fact that cos 2 x 12 (cos 2 x 1) show clearly that this function can be expressed in the form f ( x) 3 cos x 3 sin x 3 . (b) (c) 11. 12. 13. 3 Express 3 cos x 3 sin x 3 in the form k cos( x ) 3 where 0 360 and k 0 . 3 Hence solve the equation f ( x) 0 for 200 x 360 . 4 A function f is given by f ( x) x 2 3 2 . 1 (a) Find f (x) (b) Find algebraically the values of x for which f ( x) 3 1 . 2 3 Given that ( x 1) and ( x 3) are both factors of 2 x 3 5 x 2 ax b , find a and b. 4 Solve log 3 ( x 2 4) log 3 ( x 2) 3 5 [ END OF SECTION B ] [ END OF QUESTION PAPER ] Pegasys 2010 Higher Mini Prelim - Unit 3 Section A - Answers 2010/2011 (Answers + Marking Scheme) 1 5 D D 2 6 A B 3 7 B B 4 8 C D 2 marks each (16 marks) Section B - Marking Scheme Give 1 mark for each 9(a) (b) ans: valid method ●1 ●2 answer ●2 ans: proof evidence of using stepping out/section formula (11, – 10, 2) (4 marks) finds DA ●1 10 DA = 10 5 ●2 finds DC ●2 7 DC = 6 2 ●3 ●4 finds DA.DC conclusion ●3 ●4 DA.DC = 70 – 60 – 10 =0 since DA.DC = 0; ADC is right angled finds BA and BC ● 4 BA = 4 2 finds BA.BC conclusion ●2 BA.BC = 4 – 48 – 10 = – 54 ●3 scalar product < 0 so obtuse angle ans: ● 1 ●2 ●3 10(a) ans: (b) (2 marks) ●1 ●1 (c) (11, – 10, 2) Illustration(s) for awarding each mark proof proof (3 marks) 1 1 BC = 12 5 (3 marks) ●1 applies given info to new function ●1 cos 2 12 x 12 (cos x 1) ●2 knows to substitute in function ●2 6[ 12 (cos x 1)] 3 sin x ●3 simplifies to required form ●3 3(cos x 1)] 3 sin x;3 cos x 3 3 sin x ans: ●1 ●2 ●3 √12cos(x – 30)o + 3 finds k finds tan α finds α Pegasys 2010 (3 marks) ●1 ●2 ●3 k 9 3 12 tan 33 30 Give 1 mark for each (c) ans: equates to 0 simplifies ●1 ●2 √12cos(x – 30)o + 3 = 0 cos(x – 30)o = 312 ●3 ●4 finds values discards ●3 ●4 x = 180o; 240o 240o ●1 use of chain rule ●1 ●2 differentiates bracket ●2 ●3 combines and simplifies ●3 1 2 x 3 2 2x 2x ●2 ●3 2 3 1 (3 marks) 2 equates derivative to 1 2 ●1 1 2 x2 3 1 2 2 ans: x = 25 x x 2 3 1 1 2 = 2 1 (4 marks) uses synthetic division to find one equation uses synthetic division to find other eq. knows to use system of equations solves for a and b ●1 b – a = 7 ●2 b + 3a = – 9 ●3 evidence ●4 a = – 4; b = 3 (5 marks) ●1 applies difference of logs ●1 ●2 exponentiates ●2 ●3 difference of two squares ●3 ●4 simplifies ●4 ●5 solves to answer ●5 Pegasys 2010 ●2 2 x x 2 3 2 : 4 x 2 x 2 3 ●3 3 x 2 3 : x 2 1 : x 1 rearranges and squares solves to answer Sect. B (3 marks) ans: a = – 4; b = 3 ●1 ●2 ●3 ●4 13(a) x x ans: x = 1 ●1 12 (4 marks) ●1 ●2 11(a) ans: (b) 240o Illustration(s) for awarding each mark (34 marks) log 3 x2 4 3 x2 x2 4 27 x2 x 2 4 x 2x 2 x 2x 2 x 2 x 2 x 2 27 16 + 34 Total: 50 marks Pegasys 2010