Name: Inde Number: Class: CATHOLIC HIGH SCHOOL Preliminary Eamination 3 Secondary 4 ADDITIONAL MATHEMATICS 4047/1 READ THESE INSTRUCTIONS FIRST Write your name, register number and class on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. 16 September 014 hour Answer All questions. Attempt Questions 1 to 4 in Answer Booklet A, Questions 5 to 8 in Answer Booklet B and Questions 9 to 13 in Answer Booklet C. Give non-eact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. The use of a scientific calculator is epected, where appropriate. You are reminded of the need for clear presentation in your answers. At the end of the eamination, fasten all your work securely together. 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 80. This document consists of 6 printed pages, including this cover page. 014 CHS Preliminary Eaminations III Additional Mathematics Paper 1 1
Mathematical Formulae 1. ALGEBRA Quadratic Equation For the quadratic equation a + b + c = 0, Binomial Epansion Identities b b 4ac a n n n n n n n nr r n a b a a b a b a b b 1 r, 1 where n is a positive integer and Formulae for ABC 1 1 n n! n n n r r r! n r! r!. TRIGONOMETRY sin A + cos A = 1 sec A = 1 + tan A cosec A = 1 + cot A sin (A ± B) = sin A cos B ± cos A sin B cos (A ± B) = cos A cos B sin A sin B tan ( A ± B ) = tan A tan B 1 tan A tan B sin A = sin A cos A cos A = cos A sin A = cos A 1 = 1 sin A tan A = tan A 1 tan A a b c sin A sin B sin C a = b + c bc cos A = 1 ab sin C 014 CHS Preliminary Eaminations III Additional Mathematics Paper 1
1 Solve the equations Answer Questions 1 to 4 in Answer Booklet 1A y 4y y 6 8 [4] 3 1 3 Given that 3 5 9 5, evaluate 15. [3] ln 3 (a) Given that y, find the range of such that it is an increasing function. [4] (b) Differentiate y e 3 and hence find the indefinite integral 4 Find all angles between 0 and 360 which satisfy the equation 3 cot 4 1 cosec e 3 d. [4]. [5] 5 Given that Answer Questions 5 to 8 in Answer Booklet 1B n 5 1 a 1 1... where n is a positive integer, find the value of a and of n. [6] y 6 P The diagram shows part of the graph of y 5 1. Q (a) Find the coordinates of P and Q. [] (b) Find the number of solutions of 5 1 m if (i) m 3. [1] (ii) m 1 [1] (iii) Solve 5 1. [] 014 CHS Preliminary Eaminations III Additional Mathematics Paper 1 3
7 Solutions to this question by accurate drawing will not be accepted. y B (k, 9) The diagram shows a trapezium ABCD in which AD is parallel to BC and AB is perpendicular to BC. The coordinates of A, B and C are ( 1, 3), (k, 9) and (6, 7) respectively and k 3. (i) Find the value of k. [3] (ii) P is a point such that ACBP is a parallelogram, find P. [] Given that the ratio of the area of the triangle BCD to the area of the triangle ABD is 1:3. Find A ( 1, 3) O C (6, 7) (iii) the point D. [] Given further that the line AB produced meets DC produced at the point E. Find (iv) the point E. [] 8 The variables and y are connected by the equation y ab where a and b are constants. The diagram shows the straight line graph obtained from plotting lg y against lg. Calculate the value of a and b. [4] (-3,5) lg y D (-1,1) 014 CHS Preliminary Eaminations III Additional Mathematics Paper 1 4
Answer Questions 9 to 13 in Answer Booklet 1C 9 The temperature of a piece of meat taken out of the freezer increased continuously so that after a period of t mins, the temperature of the meat is T 18 8e kt C. After a period of 90 min the temperature of the meat was found to be1 C. (i) What was the temperature of the meat when it was first taken out of the freezer? [] (ii) Find k. [] (iii) What would the temperature of the piece of meat be after a long period of time? [1] ln 10 Given that y, find 3 5 dy (a), d (b) the rate of change of when 1, given that y is decreasing at a constant rate of 0. units per second. [3] 11 A curve has the equation y 4 3. (i) (ii) 1 The acceleration, dy 4 3 Show that. [3] d 4 3 Find the -coordinates of the points on the curve where the tangent to the curve are perpendicular to the line 4. [3] a ms leaving a fied point O with a velocity of, of a particle travelling in a straight line, at time t seconds after 1 ms, is given by a 3cos t, for 0 t. (a) Find the value of t when the particle achieves its maimum positive acceleration. [] (b) Find the value of t when the particle first comes to instantaneous rest. [4] (c) Calculate the distance travelled in the first seconds. [3] [] 014 CHS Preliminary Eaminations III Additional Mathematics Paper 1 5
13 P Q 7 cm 1 cm B A L cm (90 ) The diagram shows two rods, OP and PQ, of length 1 cm and 7 cm respectively. The rods are fied at P such that angle OPQ 90 and hinged at O so as to rotate in a vertical plane. The rod OP makes acute angle 90 with the line OA, where 0 45. Given that L cm is the perpendicular distance of Q from line OB, (a) Show that L 1sin 7cos. [] (b) Epress L in the form R cos O, where R 0 and 0 90. [3] (c) State the minimum value of L and the corresponding value of. [] (d) Find the value of when L 10. [3] END OF PAPER 014 CHS Preliminary Eaminations III Additional Mathematics Paper 1 6
Answer Key 014 Prelim 3 Add Math Paper 1 Questio n 1 Answer y 1 4 or or y Questio n 9(i) 9(ii) 9(iii) 7 10(a) 5 10(b) 3(a) 0 e 11(ii) 3(b) 3 3 e e c 3 9 4 5.4, 154.6, 70 1(a) 1(b) 1(c) 5 a=, n=4 13(a) 13(b) 13(c) 6(a) 6(b) 1 P(, 0), Q(, 5) 1 solution solutions 6(c) =, =-4 7 k= P(-5, 5) D(11, -3) E(3.5, 1) 8 1 a, 10 1 100 b Danyal 13(d) Answer -10 1 3 k ln 90 14 kt As t, e 0 hence T18 3 5 3ln (3 5) 0.4 units 6 3 3 1.3 radians 6.34 m L 1sin 7cos L 193 cos( 59.7 ) 0 15.8 Education
Mathematical Formulae Identities Formulae for ABC 1. TRIGONOMETRY sin A + cos A = 1. sec A = 1 + tan A. cosec A = 1 + cot A. sin (A ± B) = sin A cos B ± cos A sin B cos (A ± B) = cos A cos B sin A sin B tan A tan B tan (A ± B) = 1 tan A tan B sin A = sin A cos A. cos A = cos A sin A = cos A 1 = 1 sin A tan A tan A = 1 tan A a b c. sin A sin B sin C a = b + c bc cos A. = 1 bc sin A
3 1 Solve the equations y 4y y 6 8 [4] 6 y (1) ( 6 36 4 y y ( y y y ) 1 1 4 4 3 y 1)( y or or y ( 6 y 4 y 8 ) y 0 y ) y 0 0 1 4 y y 4 y 8 4 y ( ) [] [A1] [A1] 8 [] 3 1 3 Given that 3 5 9 5, evaluate 15. [3]. 5 3 3 73 3 5 [-addition/subtraction law, -5 ] 5 3 7 3 5 5 [] 7 3 5 15 5 [A1]
4 3 (a) Given that ln y, find the range of such that it is an increasing function. [4] (b) Differentiate y e ln 1 0 e 3 and hence find the indefinite integral 1 ln dy ( a) [] d 1- ln [A1] dy 1- ln For decreasing function, 0 [] d Since 0, 1- ln 0 ( b) e 3 y e dy 3e d Hence 3 3 e 3 3 e (3e d 3 3 e e 3 [] 3 d e [A1] ) d e 3 3 c c 3 3 e 3 3 e d e c 3 [] 3 3 3 e e e d c 3 9 [A1] 3 3 e e d d c 3 9 [] e 3 d. [4]
5 4 Find all angles between 0 and 360 which satisfy the equation 3 cot 4 1 cosec 3 cot 4 1 cosec 3 cosec 1 4 4 cosec - Apply 1 cot cosec 3cosec 4cosec 7 0 3cosec 7 cosec 1 0 - Epansion to get Quad. Form 7 cosec 3 or cosec 1 3 sin 7 or sin 1 - Factorise & convert into sin Basic = 5.37 70 A1 5.4, 154.6 A1 [5]
6 1 a 5 1 1... where n is a positive integer, find the value of a and of n. [6] 5 Given that n n 5 1 a 1 1... n 1 a1 n n( n 1) n( n 1) 8 48 n n( n 1) an 5 a 0 8 n n n n 5 a 8 4 n n 5 8 8 n n 0 0 [B1] 3 1 a 1... n n 5 4 0 n 5 rejected or n 4 A1 a A1
7 6 y Q P The diagram shows part of the graph of y 5 1. (a) Find the coordinates of P and Q. [] (b) Find the number of solutions of 5 1 m if (i) m 3. [1] (ii) m 1 [1] (iii) Solve 5 1. [] (a) P(, 0), 1 Q(, 5) [A1][A1] (b) (i) m 3, 1 solution [A1] (i) m 1, solutions [A1] (iii) 5 1 1 5 1 5 or 1 5 or 4 A1
8 7 Solutions to this question by accurate drawing will not be accepted. y B (k, 9) The diagram shows a trapezium ABCD in which AD is parallel to BC and AB is perpendicular to BC. The coordinates of A, B and C are ( 1, 3), (k, 9) and (6, 7) respectively and k 3. (i) Find the value of k. [3] (ii) P is a point such that ACBP is a parallelogram, find P. [] Given that the ratio of the area of the triangle BCD to the area of the triangle ABD is 1:3. Find (iii) the point D. [] Given further that the line AB produced meets DC produced at the point E. Find (iv) the point E. [] 9 7 9 3 (i) 1 [] k 6 k 1 k 5k 6 0 [] (k - )(k - 3 ) 0 k or k 3 (rej) [A1] (ii) P(-1-4, 3 ) P( 5, 5) [B] (iii) D(- 11, 36 ) D(11, 3) [B] (iv) A ( 1, 3) O C (6, 7) E( 1 4.5, 3 9) E(3.5, 1) [B] D
9 8 The variables and y are connected by the equation y ab where a and b are constants. The diagram shows the straight line graph obtained from plotting lg y against lg. Calculate the value of a and b. [4] y ab lg y lg b lg a Y lg y X m lg b c lg a From the 5-1 4 m - 3 - (-1) Y -X c Taking (-1,1), 1 -(-1) c c -1 Hence, lgb, and lg y (-3,5) (-1,1) diagram, lg a 1, 1 b 100 1 a 10 [] [] [A1] [A1] lg
10 9 The temperature of a piece of meat taken out of the freezer increased continuously so that after a period of t mins, the temperature of the meat is T 18 8e kt C. After a period of 90 min the temperature of the meat was found to be1 C. (i) What was the temperature of the meat when it was first taken out of the freezer?[] (ii) Find k. [] (iii) What would the temperature of the piece of meat be after a long period of time? [1] ( i) (ii) (iii) t 0, T 18 8e -10 When t 90, T 1 18-8e 1, -90k -90k 3 e 14 3-90k ln 14 1 3 k ln 90 14 kt As t, e 0 0 [] [A1] [] [A1] hence T 18 [A1]
11 10 Given that ln y, 3 5 find dy (a), d (b) the rate of change of when 1, given that y is decreasing at a constant rate of 0. units per second. [3] (a) ln y 3 5 1 3 5 3ln dy d 3 5 3 5 3 ln = A1 3 5 (b) when 1, dy 1 d A1 dy dy d dt d dt d 0. 0.5 dt B1 d 0.4 units/s dt A1 []
1 11 A curve has the equation y 4 3. (i) (ii) dy 4 3 Show that. [3] d 4 3 Find the -coordinates of the points on the curve where the tangent to the curve are perpendicular to the line 4. [3] (i) y 4 3 1 dy 1 4 3 6 4 3 B1 d 6 4 3 = 4 3 4 3 = (shown) A1 4 3 (ii) dy Tangent to the line 4, 0 B1 d 4 3 4 3 3 0 0 3 6 or 3 3 A1
13 1 The acceleration, a ms, after leaving a fied point O with a velocity of a 3cos t, for 0 t. of a particle travelling in a straight line, at time t seconds 1 ms, is given by (a) Find the value of t when the particle achieves its maimum positive acceleration. [] (b) Find the value of t when the particle first comes to instantaneous rest. [4] (c) Calculate the distance travelled in the first seconds. [3] (b) (a) cost 1 B1 basic angle = 0 radian t 0, t, 3 t radians A1
14 (c) a 3cost v 3cost dt =3sin t c when t 0, v 3sin c c 1 v 3sin t 1 B1 At instantaneous rest, v 0 3sin t 1 0 1 sin t 3 basic angle =0.3398 radians t 0.3398, 0.3398 3 t 0.3398, 0.3398 t 1.309( NA),1.309 t 1.3 radians 1.3 s 3sin t 1dt 3sin t 1dt 0 1.3 1.3 = 3cost t 3cost t B1 = 1.59746+4.73906 = 6.3365 6.34 m 0 1.3 A1 A1
15 13 P B Q 7 cm 1 cm A L cm (90 ) The diagram shows two rods, OP and PQ, of length 1 cm and 7 cm respectively. The rods are fied at P such that angle OPQ 90 and hinged at O so as to rotate in a vertical plane. The rod OP makes acute angle 90 with the line OA, where 0 45. Given that L cm is the perpendicular distance of Q from line OB, (a) Show that L 1sin 7cos. [] (b) Epress L in the form R cos O, where R 0 and 0 90. [3] (c) State the minimum value of L and the corresponding value of. [] (d) Find the value of when L 10. [3]
16 OT ST (a) cos90 sin 90 1 7 OT 1sin ST 7cos L 1sin 7 cos (b) 1sin 7 cos R cos 1sin 7 cos R cos cos R sin sin 1 R sin R 1 7 193 1 7 R cos tan 59.74 7 L 193 cos 59.7 (c) Minimum value of L 7 B1 when 90 90 0 B1 (d) when L 10, 193 cos 59.74 10 0 45 A1 10 cos 59.74 193 59.74 59.74 14.74 Basic angle 43.96 59.74 43.96 15.8
Name: Inde Number: Class: CATHOLIC HIGH SCHOOL Preliminary Eamination 3 Secondary 4 ADDITIONAL MATHEMATICS 4047/ READ THESE INSTRUCTIONS FIRST Write your name, register number and class on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer All questions. 17 September 014 hour 30 min Give non-eact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. The use of a scientific calculator is epected, where appropriate. You are reminded of the need for clear presentation in your answers. At the end of the eamination, fasten all your work securely together. 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 100. This document consists of 6 printed pages, including this cover page. 014 CHS PRELIM 3 ADD MATH 4047/ Page 1
Mathematical Formulae 1. ALGEBRA Quadratic Equation For the quadratic equation a + b + c = 0, Binomial Epansion Identities b b 4ac a 1 n n n n n n n a b a a b a b a n r b r b n 1 r, where n is a positive integer and Formulae for ABC 1 1 n n! n n n r r r! n r! r!. TRIGONOMETRY sin A + cos A = 1 sec A = 1 + tan A cosec A = 1 + cot A sin (A ± B) = sin A cos B ± cos A sin B cos (A ± B) = cos A cos B sin A sin B tan ( A ± B ) = tan A tan B 1 tan Atan B sin A = sin A cos A cos A = cos A sin A = cos A 1 = 1 sin A tan A = tan A 1 tan A a b c sin A sin B sin C a = b + c bc cos A = 1 ab sin C 014 CHS PRELIM 3 ADD MATH 4047/ Page
1 Find the range of values of p for which the epression ( ) 1 p p is always positive for all real values of. [4] Find the values of m for which the line y m is a tangent to the curve 3 The epression 3 a b c is divisible by both and 3 4 of 40 when divided by. 1 y. [4] ( 3) but leaves a remainder (i) Determine the value of a, of b and of c. [5] (ii) Factorise the epression completely. [1] (iii) Find the remainder when the epression is divided by 3 1. [1] The diagram shows a cyclic quadrilateral ABCD and that its diagonals intersect at F. The tangent to the circle at C meets AD etend at E. BD is parallel to CE and AD = DE. (i) Show that BCD is an isosceles triangle. [] (ii) B A F By showing triangle CDE is similar to triangle ACE, prove that C CE AE DE. [4] CE (iii) Show that DE. [] D E 014 CHS PRELIM 3 ADD MATH 4047/ Page 3
75 5 (a) Given that p log 5 and q log 3, epress log in terms of p and of q. [3] 16 (b) Solve the following equations (i) (3 ) e, [3] (ii) log log 104 9. [4] 6 The roots of the equation (i) 4 41 10 0 are and. Show that 9. [3] (ii) The roots of the equation p q 0 are and. Given that both roots are negative, find the value of p and of q. [4] 7 The function f is defined, for all values of by f cos. (i) State the period of f. [1] (ii) Sketch the graph of y f for 0 3. [3] (iii) On the same diagram in part (ii), insert the graph of 4 y for 0 3. [1] (iv) Hence, deduce the number of roots of the equation 4cos for (a) 0 3, [1] (b) 3. [1] 8 A circle passes through the points A(9, ) and B(1, 3) and cuts the y ais at P and Q. A line y 7 passes through the centre of the circle. Find (i) the equation of the perpendicular bisector of AB, [3] (ii) the centre of the circle, [3] (iii) the equation of the circle, [] (iv) the eact value of the y coordinate of P and of Q. [] A second circle, centre C, also passes through P and Q. The coordinate of C is negative and the radius of the second circle is 197 units. (v) Find the coordinates of C. [] 014 CHS PRELIM 3 ADD MATH 4047/ Page 4
9 r cm 40 cm h cm 1 cm The diagram shows a hollow cone of height 40 cm and base radius 1 cm and a solid cylinder of radius r cm and height h cm. Both stand on a horizontal surface with the cylinder inside the cone. The upper circular edge of the cylinder is in contact with the cone. (i) Epress h in terms of r and hence show that the volume, V cm 3, of the cylinder is given by V 10 r 4 r 3. [3] 1 Given that r can vary, (ii) determine the value of r for which the volume of the cylinder is a maimum, [4] (iii) show that at its maimum volume, the cylinder occupies 4 of the volume of the cone. [] 9 1 [The volume, V, of a cone of height H and radius R is given by V R H.] 3 014 CHS PRELIM 3 ADD MATH 4047/ Page 5
10 y = The diagram shows part of the curve y 4sin. The line 4 3 intersect the curve at Q and the normal to the curve at P cuts the ais at R. Given that the normal at P is parallel to the line y 3 5, (i) show that the coordinate of R is 4 3 3, [4] (ii) find the total area of the shaded regions. [6] 11 (i) Given that (ii) (iii) 3 18 14 6 ( 1) ( 1) 1 ( 1) 1 A B C D, where A, B, C and D are constants, find the value of A, of B and of C and show that D 0. [5] Differentiate ln( 1) with respect to. [] Using the results from parts (i) and (ii), find 3 18 14 6 d. [4] ( 1) ( 1) 1 (i) Prove the identity sin tan cos tan. [] (ii) Hence, O P (a) without using a calculator, show that tan 67.5 1. [5] (b) find all the angles between 0 and 6 which satisfy the equation R Q 1 3cot sin tan cos. [4] ~ End of Paper ~ 014 CHS PRELIM 3 ADD MATH 4047/ Page 6
014 Prelim 3 AM P Answer Key 1 p 4 m 4 or m 8 3 (i) a 5, b 6, c 0 (ii) ( )( 3) (iii) 16 7 4 Solutions (i) (ii) (iii) CBD DCE BDC DEC (Tangent Chord Theorem / angles in alternate segment) (alternate angles) triangle BCD is isosceles. ECD EAC (Tangent Chord Theorem) CED AEC ( Common Angles) CDE ACE ( angle sum of triangle) triangle CDE is similar to triangle ACE (AAA Similarity) CE AE DE CE DE AE CE CE AE DE CE DE DE AD DE AE DE CE (Given, ) DE CE DE 5 (a) p q 4 (b)(i) 7.03 (ii) 1 104, 6 (i) 1 Show that 9. (ii) p 3, q 3 7 (i) period = (iv) (a) 3 solutions (b) solution (ii) y 1 0.5 1 1.5.5 3 4 1 8 (i) 3 34 y (ii)3, 5 5 5
(iii) 3 y 5 85 or y 6 10y 51 0 (iv) 5 19 (v) C 11, 5 9 (ii) r 8 10 (ii) 0 units 11 (i) A 1, B 3, C 4 (ii) 1 1 (ii) (b) 3 7 0.983, 4.1,, 4 4 1 3 ln 1 ln( 1) c 1 (iii)
014 Prelim 3 AM P Marking Scheme 1 Find the range of values of p for which the epression ( ) 1 p p is always positive for all real values of. [4] Solutions ( ) 1 p p ( p ) 1 p 0 ( 1) 4( p )( p 1) 0 D 0 and p 0 p - Conditions for Discriminant 144 8( p p ) 0 144 8p 8p 16 0 8p 8p 160 0 8p 8p 160 0 p p 0 0 ( p 4)( p 5) 0 p 5 or p 4 Since p, we have p 4 - Quadratic Inequality A1- Solutions for Quadratic Inequality A1 Find the values of m for which the line y m is a tangent to the curve 1 y. ( 3) [4] Solutions 1 y m 1 & y ( 3) 1 1 m ( 3) ( 3)( m ) 1 ( m 3m 6 ) 1 4 ( m 6) 6m 1 0 - Quadratic equation after simultaneous equation [ ( m 6)] 4(4)(6m 1) 0 4m 48m 18 0 m D 0 1m 3 0 ( m 4)( m 8) 0 m 4 or m 8 A
Alternatively, 1 y 3 1 3 dy 1 d 3 7 5 or y 1 or y 1 7 5 Sub,1 and, 1 into y m m 8 or m 4 A
3 The epression 3 a b c is divisible by both and 3 but leaves a remainder of 40 when divided by. (i) Determine the value of a, of b and of c. [5] (ii) Factorise the epression completely. [1] (iii) Find the remainder when the epression is divided by 3 1. [1] Solutions (a) Let f ( ) 3 a b c f (0) 0 c 0 f (3) 0 f ( ) 40 7 9a 3b 0 8 4a b 40 3 (b) f ( ) 5 6 3a b 9 (1) a b 16 () ( (1) (): 5a 5 Subst a 5 into (), 5 6) ( )( 3) a 5 ( 5) b 16 b 6 a 5, b 6, c 0 A1 1 1 1 16 (c) f ( ) ( ) 3 3 3 3 7 A1 A(,1, 0)
4 A F D E The diagram shows a cyclic quadrilateral ABCD and that its diagonals intersect at F. The tangent to the circle at C meets AD etend at E. BD is parallel to CE and AD = DE. (i) Show that BCD is an isosceles triangle. [] (ii) By showing triangle CDE is similar to triangle ACE, prove that CE AE DE. [4] CE (iii) Show that DE. [] (i) (ii) (iii) Solutions CBD DCE BDC DEC (Tangent Chord Theorem / angles in alternate segment) (alternate angles) triangle BCD is isosceles. ECD EAC (Tangent Chord Theorem) CED AEC ( Common Angles) M CDE ACE ( angle sum of triangle) triangle CDE is similar to triangle ACE (AAA Similarity) CE DE AE CE CE AE DE A1 CE AE DE CE DE DE AD DE AE DE CE CE DE (Given, ) DE B A1 C
Alternatively, Area of CDE CE 1 Area of ACE AE CE 1 AE CE 1 DE CE A1 DE
75 5 (a) Given that p log 5 and q log 3, epress log in terms of p and of q. [3] 16 (b) Solve the following equations Solutions (i) (3 ) e, [3] (ii) log log 104 9. [4] 75 5 3 4 (a) log log 16 (b) (i) (3 ) e 4 log 5 log 3 log ----- p q 4 ----- A1 - p q, A1-4 ln ln 3 ln e ln ln 3 (1 ln 3) ln ln 7.03 1 ln 3 (ii) log log 104 9 a a log 9 log log 104 10 log 9 log Let log a, 10 a 9 a a 9a 10 0 10 1 0 a 10 or a 1 - grouping terms A1 log 10 or log 1 10 1 = 104 A1 - change of base - Product / Quotient Law - Convert to Quadratic Form 1 = A1
6 The roots of the equation 4 41 10 0 are and. (i) Show that 9. [3] (ii) The roots of the equation p q 0 are and. Given that both roots are Solutions negative, find the value of p and of q. [4] (i) 4 41 10 0 41 10 + = 4 4 5 51 ( ) 4 4 5 51 4 4 51 5 4 = 9 (shown) (ii) Since roots are both negative, = 3 5 = 3 4 49 = 4 7 (roots are both negative) 7 3 0 7 1 p / 3, q 3 A B1 A1
7 The function f is defined, for all values of by f cos. (i) State the period of f. [1] (ii) Sketch the graph of y f for 0 3. [3] (iii) On the same diagram in part (ii), insert the graph of 4 y for 0 3. [1] (iv) Hence, deduce the number of roots of the equation 4cos for Solutions (i) (ii) 1 y 1 (a) 0 3, [1] (b) 3. [1] period = B1 0.5 1 1.5.5 3 (iii) (a) 3 solutions A1 (b) solution A1 4 1 mark for curve y = cos 1 mark for correct turning 1 mark for 1.5 cycle, 1 mark for straight line
8 A circle passes through the points A(9, ) and B(1, 3) and cuts the y ais at P and Q. A line y 7 passes through the centre of the circle. Find (i) the equation of the perpendicular bisector of AB, [3] (ii) the centre of the circle, [3] (iii) the equation of the circle. [] (iv) the eact values of the y coordinate of P and of Q. [] A second circle, centre C, also passes through P and Q. The coordinate of C is negative and the radius of the second circle is 197 units. (v) Find the coordinates of C. [] Solutions 9 1 3 1 1 (i) Mid point of AB,, 3 5 3 m AB Perpendicular gradient 9 1 3 5 1 1 3 Subst, into y c 5 3 34 Equation of Perpendicular bisector: y 5 5 (ii) 1 3 34 y 71 y 5 5 Subst 1 into 1 : 1 3 34 7 5 5 35 6 68 (iii) Radius = 9 3 5 85 (v) At 0, y 10y 51 0 A1 3 y 5 85 or y 6 10y 51 0 10 100 4( 51) 304 4 19 y 5 5 5 19 3, y 5 Centre of Circle = 3, 5 A y coordinate of P / Q = 5 19, 5 19 (iv) coordinate 197 19 & y coordinate 5 11 = 11 C 11, 5 A1, B1 A A1
9 r cm 40 cm h cm 1 cm The diagram shows a hollow cone of height 40 cm and base radius 1 cm and a solid cylinder of radius r cm and height h cm. Both stand on a horizontal surface with the cylinder inside the cone. The upper circular edge of the cylinder is in contact with the cone. (i) Epress h in terms of r and hence show that the volume, V cm 3, of the cylinder is given by V 10 r 4 r 3. [3] 1 Given that r can vary, (ii) determine the value of r for which the volume of the cylinder is a maimum, [4] (iii) show that at its maimum volume, the cylinder occupies 4 of the volume of the cone. [] 9 1 [The volume, V, of a cone of height H and radius R is given by V R H.] 3 Solutions
40 h 40 10 3 (i) (ii) V 40 r r r 1 3 40 h 10 dv - similar triangles 80 r 10 r r 3 dr dv 3(40 h) 10r At stationary volume, 0 dr 3h 10 10r 80 r 10 r 0 10(1 r) h A1 10 r 8 r 0 3 10 V r (1 r ) r 8 3 A1 V 10 r 4 3 1 r (iii) At 8, cylinder 10 (8) 4 3 3 d V dr 80 0 r d V r V 8 560 1 5760 Vcone (1) 40 3 3 - Both volumes Volume of cylinder 560 4 = = Volume of cone 5760 9 At r 8, 80 160 51 0 dr maimum volume
10 y = The diagram shows part of the curve y 4sin. The line 4 3 intersect the curve at Q and the normal to the curve at P cuts the ais at R. Given that the normal at P is parallel to the line y 3 5, (i) show that the coordinate of R is 4 3 3, [4] (ii) find the total area of the shaded regions. [6] Solutions O P R Q
3 i Gradient of normal Gradient of tangent = 3 dy 4 cos d 3 3 4 cos 3 3 3 Let the -coordinate of 1 0 3 3 cos 3 4 3 basic angle R 3 3 R R be 4 3 rad R (shown) 3 y 4sin 3 3 P, 3 - for finding coordinates of (ii) When y 0, 4 sin 0 3 0, 3, 6 Area of shaded region 1 4 3 = 3 4sin d 3 3 4 3 4sin d 3 3 3 4 = 1 cos 4 1 cos sin 3 3 - Integration of 3 4 1 cos cos 4 1 cos cos 3 3 M3 0 units A1 3 P R
11 (i) Given that 3 18 14 6 ( 1) ( 1) 1 ( 1) 1 A B C D, where A, B, C and D are constants, find the value of A, of B and of C and show that D 0. [5] (ii) Differentiate ln( 1) with respect to. [] (iii) Using the results from parts (i) and (ii), find Solutions A B C D (i), 3 18 14 6 ( 1) ( 1) 1 ( 1) 1 3 18 14 6 d 18 14 6 A( 1)( 1) B( 1) ( C D)( 1) 3 1 15 5 When, B, B 3 A1 4 4 When 0, A 3 D A D 1 (1) 3 Comparing coeff of, we have 18 4 A C A 9 C () Subst () into (1): 9 C D 1 C D 8 (3) Comparing coeff of, we have 14 4 4 A B C D A 4C 4D 17 (4) Subst () into (3): C 4D 8 Subst (5) into (3): d C 4 D (5) 8 5D 8 D 0 From A 9 C A 1. [4] ( 1) ( 1) C 4 A1 A1 (ii) ln( 1) B d 1 3 18 14 6 1 3 4 ( 1) ( 1) 1 ( 1) ( 1) (iii) d d 1 ( 1) ( 1) - Showing D 0 1 3 d d + d = ln 1 ln( 1) 1 1 3 c A3 ( if c is missing) 1
1 (i) Prove the identity sin tan cos tan. [] (ii) Solutions Hence, (a) without using a calculator, show that tan 67.5 1. [5] (b) find all the angles between 0 and 6 which satisfy the equation (i) sin tan cos tan LHS sin tan cos sin cos sin = sin cos sin cos cos sin = tan cos 1 3cot sin tan cos. [4] sin cos = sin cos cos 1 sin sin cos, tan, cos cos 1 - epansion (ii) tan 67.5 sin 67.5 tan 67.5 cos 67.5 tan 67.5 = sin135 tan 67.5 cos 135 - substituting 67.5 tan 67.5 = sin 180 45 tan 67.5 cos 180 45 tan 67.5 = sin 45 tan 67.5 cos 45 - for rewriting angles in terms of 45 tan 67.5 = tan 67.5 sin 4 - for evaluating 5 / cos 45 tan 67.5 1 tan 67.5 1 - for factorising tan 67.5 tan 67.5 1 1 3cot (iii) sin tan cos 1 3cot tan tan 1 3cot tan tan 3 0 tan 3 tan 1 0 - for rationalising - Apply part (i) identity - Obtaining quadratic form 3 tan or tan 1 Basic Angle 0.987 Basic Angle 4 0.983, 4.1 A1 3 7, 4 4 A1 (Accept.36, 5.50)