MATHEMATICS HSC Course Assessment Task 3 (Trial Examination) June 21, QUESTION Total MARKS
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1 MATHEMATICS 0 HSC Course Assessment Task (Trial Eamination) June, 0 General instructions Working time hours. (plus 5 minutes reading time) Write using blue or black pen. Where diagrams are to be sketched, these ma be done in pencil. Board approved calculators ma be used. Attempt all questions. At the conclusion of the eamination, bundle the booklets + answer sheet used in the correct order within this paper and hand to eamination supervisors. SECTION I Mark our answers on the answer sheet provided (numbered as page 9) SECTION II Commenceeach newquestion onanew page. Write on both sides of the paper. All necessar working should be shown in ever question. Marks ma be deducted for illegible or incomplete working. STUDENT NUMBER:... # BOOKLETS USED:... Class (please ) MA Mr Berr MC Ms Ziaziaris MD Mr Lowe ME Mr Lam Marker s use onl. QUESTION Total MARKS
2 0 Mathematics HSC Course Assessment Task Trial Eamination Section I: Objective response Mark our answers on the multiple choice sheet provided.. What is the value of 8 0? Marks (A) (B) (C) (D). What is the sum of the eterior angles of a polgon? (A) 90 (B) 80 (C) 60 (D) none of these. Which conditions make the quadratic = a +b+c positive definite? (A) a < 0, < 0 (B) a < 0, > 0 (C) a > 0, < 0 (D) a > 0, > 0 4. Which of the following is not a condition for congruent triangles? (A) SSS (B) AAA (C) SAS (D) AAS 5. What is.996 to two significant figures? (A).0 (B).9 (C).09 (D).0 6. Which of the following is the locus of a point that is equidistant from a fied point and a fied line? (A) a parabola (B) a hperbola (C) a circle (D) an ellipse 7. Evaluate lim (A) (B) 0 (C) (D) 8. Which of the following conditions for dp dt and d P describe the slowing growth of dt a variable P? (A) dp dt > 0 and d P dt (B) dp dt < 0 and d P dt > 0. < 0. (C) dp dt > 0 and d P dt (D) dp dt < 0 and d P dt < 0. > If a > b, which of the following is alwas true? (A) a > b (B) a > b (C) a > b (D) a > b 0. What is the eact value of b if the area beneath the curve = between = and = b (b > ) is equal to units? (A) e (B) e (C) e 5 (D) e JUNE, 0 NORTH SYDNEY BOYS HIGH SCHOOL
3 0 Mathematics HSC Course Assessment Task Trial Eamination Section II: Short answer Question (5 Marks) Commence a NEW page. Marks (a) Solve the equation + 9 = 0. D(0, ), E(4,0) & F(,4) are three points on the number plane. F(,4) D(0, ) E(4, 0) i. Calculate the length of the interval DF. ii. Calculate the gradient of DF. iii. Write the equation of the line DF in general form. iv. Calculate the perpendicular distance from E to the line DF. v. Calculate the area of DEF. (c) ABCDE is a regular pentagon. The diagonals AC and BD intersect at F. A B E F C Cop or trace this diagram into our writing booklet. B giving full reasons for our answer, i. Prove that ABC = 08. ii. Find the size of BAC. (d) Find the eact value of tan0 +sin00. D NORTH SYDNEY BOYS HIGH SCHOOL JUNE, 0
4 4 0 Mathematics HSC Course Assessment Task Trial Eamination Question (5 Marks) Commence a NEW page. Marks (a) Find the values of p, p > 0 for which the roots of the equation p+p = 0 are i. opposite in sign. ii. real i. Sketch the parabola with equation ( ) = (+) Show the verte of the parabola on our sketch. ii. Find the coordinates of the focus and the equation of the directri of the parabola. (c) The sum to n terms of a sequence of numbers is given b S n = 0n n. i. Find an epression for T n, the n-th term of the sequence. ii. What tpe of a sequence is this? (d) Differentiate the following epressions: i. ii. cos4 iii. log e () Question (5 Marks) Commence a NEW page. Marks (a) For the curve = (4 ) i. Find the stationar point(s) and determine their nature. ii. Find the point(s) of infleion. iii. Draw a neat sketch of the curve showing the intercepts with the coordinate aes, an stationar points and an point(s) of infleion. Find f() if f () = + and the curve passes through the point (,). (c) Find the primitive of i. ii. sec JUNE, 0 NORTH SYDNEY BOYS HIGH SCHOOL
5 0 Mathematics HSC Course Assessment Task Trial Eamination 5 Question 4 (5 Marks) Commence a NEW page. Marks (a) Use Simpson s Rule with five function values to evaluate following table:.5.5 f() 0 5 f()d given the The diagram below shows the graphs of = ++8 and = +6. A B = +6 = ++8 i. Show that the coordinate of A and B are = and = respectivel. ii. Hence or otherwise, find the shaded area bounded b the curves and the straight line. (c) i. State the period and amplitude of = sin. ii. Draw a neat sketch of = sin, where 0 π. iii. Hence or otherwise, state the number of solutions to the equation sin = (d) within the domain 0 π. Find the angle subtended at the centre of the circle of sector with radius 4cm and area 0cm. Give our answer correct to the nearest degree. NORTH SYDNEY BOYS HIGH SCHOOL JUNE, 0
6 6 0 Mathematics HSC Course Assessment Task Trial Eamination Question 5 (5 Marks) Commence a NEW page. Marks (a) Solve 8 = In this diagram, BCD+ BED = 80. A m m B E 5 m C D i. Prove that ABE is similar to ADC. ii. Given that AE = m, ED = 5 m and BC = m, calculate the length of AB. (c) i. Evaluate d d (log e(sin)). ii. Hence or otherwise, find cotd. (d) A liquor bottle is obtained b rotating about the ais the part of the curve = between = and =. 4 = Find the eact volume of the bottle. Eamination continues overleaf... JUNE, 0 NORTH SYDNEY BOYS HIGH SCHOOL
7 0 Mathematics HSC Course Assessment Task Trial Eamination 7 Question 6 (5 Marks) Commence a NEW page. Marks (a) ABCD is a quadrilateral inscribed in a quarter of a circle centred at A with radius 00 m. The points B and D lie on the and aes and the point C moves on the circle such that CAB = α as shown in the diagram below. D C A α 00 m B i. Solve the equation sin(+5 ) = cos4. ii. Show that the area of the quadrilateral ABCD can be epressed as A = 5000(sinα+cosα) iii. Show that the maimum area of this quadrilateral is 5000 m. 4 i. Sketch the curve = 4e. ii. Consider the series e +8e +e +. α) Show that this series is geometric. β) Find the values of for which this series has a limiting sum. γ) Find the limiting sum of this series in terms of. End of paper. NORTH SYDNEY BOYS HIGH SCHOOL JUNE, 0
8 STANDARD INTEGRALS n d = n+ n+ +C, n ; 0 if n < 0 d = ln+c, > 0 e a d = a ea +C, a 0 cosad = a sina+c, a 0 sinad = a cosa+c, a 0 sec ad = a tana+c, a 0 secatanad = a seca+c, a 0 a + d = a tan a +C, a 0 a d = sin +C, a > 0, a < < a a ( a d = ln + a )+C, > a > 0 ( +a d = ln + +a )+C NOTE: ln = log e, > 0
9 0 Mathematics HSC Course Assessment Task Trial Eamination 9 Answer sheet for Section I Mark answers to Section I b full blackening the correct circle, e.g STUDENT NUMBER:... Class (please ) MA Mr Berr MC Ms Ziaziaris MD Mr Lowe ME Mr Lam A B C D A B C D A B C D 4 A B C D 5 A B C D 6 A B C D 7 A B C D 8 A B C D 9 A B C D 0 A B C D NORTH SYDNEY BOYS HIGH SCHOOL JUNE, 0
10 0 0 Mathematics HSC Course Assessment Task Trial Eamination SOLUTIONS Suggested Solutions Section I (Lowe). (D). (C). (D) 4. (B) 5. (A) 6. (A) 7. (C) 8. (C) 9. (D) 0. (A) Question (Lowe) (a) ( marks) [] for quartic. [] for final solutions. + 9 = = = 0 ( 9 )( ) = 0 = ±,± (c) v. ( marks) [] for using parts (iii) & (iv) [] for final answer. A = DF d = 0 0 = 0 units i. ( marks) Divide pentagon into five equilateral triangles. B F A E i. ( mark) ii. ( mark) iii. ( mark) DF = ( 0) +(4 ( )) = +6 = 40 = 0 m DF = 6 = 4 = 4 = 6 = 0 iv. ( marks) [] for correctl recalling perpendicular dist formula. [] for final answer. C D Apeangleofoneofthetriangles 60 5 = 7. Angle sum of the two base angles is thus 80 7 = 08 ii. ( mark) BAC is isosceles. BAC = = 6. (d) ( marks) tan0 +sin00 ( ( ) = )+ = = 0 d = a +b +c a +b = (4)+( )(0) + = 0 0 = 0 LAST UPDATED JUNE 6, 0 NORTH SYDNEY BOYS HIGH SCHOOL
11 0 Mathematics HSC Course Assessment Task Trial Eamination SOLUTIONS Question (Lowe) (c) i. ( marks) (a) i. ( mark) p+p = 0 α = β α+β = 0 = b a = p p = 0 But as p > 0, therefore there are no real solutions. ii. ( marks) [] for p 0 or p 4. [] justif wh p 4 onl. S n = 0n n T n = S n S n = 0n n (0(n ) (n ) ) ii. ( mark) = 0n n ( 0n 0 (n n+) ) = n +0+ n 4n+ = 04 4n T = 04 4() = 00 T = 04 4() = 96 T = 04 4() = 9 T T = T T 0 b 4ac = p 4p 0 p(p 4) 0 p 0 or p 4 But as p > 0, hence p 4 onl. (d) Arithmetic sequence. i. ( mark) ii. ( marks) d ( ) = 6 4 d i. ( marks) iii. ( marks) d (cos4) = sin4 d 4 d d (log e) = d d (log e+log e ) = ( ) = (+) ii. ( marks) 4a = a = S ( ), Directri is = 5. NORTH SYDNEY BOYS HIGH SCHOOL LAST UPDATED JUNE 6, 0
12 0 Mathematics HSC Course Assessment Task Trial Eamination SOLUTIONS Question (Berr) (a) i. ( marks) = 4 4 = (4 ) d d = 4 Stationar pts occur when d d = 0: = (4 ) d d 4 ( ) = 0 + = 0, Hence (0,0) is a horizontal point of infleion and (,7) is a local maimum. ii. ( marks) Points of infleion occur when d d = 0: d d = 4 = ( ) d d = 0, When =, = (4 ) = = (4 ) = 6 0 ( marks) (c) [] for correct integral. [] for correct value of C. [] for final answer. (+ f() = ) d = +C f() = +C = C = f() = + i. ( marks) [ ] if missing arbitrar constant. d = 4 4 +C ii. ( marks) [ ] if missing arbitrar constant. sec d = 9tan +C Hence points of infleion occur at (0,0) and (,6) as concavit changes at these two pts. iii. ( marks) [ ] for each omission from requirements of the question, provided graph is correct. LAST UPDATED JUNE 6, 0 NORTH SYDNEY BOYS HIGH SCHOOL
13 0 Mathematics HSC Course Assessment Task Trial Eamination SOLUTIONS Question 4 (Berr) (a) ( marks) f()d = h ( 0 +4 odd + even + l ) = (0++4(+)+(5)) π π π π = 6 i. ( marks) { = ++8 = +6 Solve b equating, ii. ( marks) ++8 = +6 + = 0 ( )(+) = 0 =, ( A = + ) d [ = + ] + = ( ( ) ) + ( ( ) ) +( ( )) = + +6 = 9 iii. ( mark) 4 solutions. (d) ( marks) [] for answer in radians. [] for answer in degrees. A = r θ 0 = 4 θ θ = 5 4 = π 7 (c) i. ( marks) T = π = π a = ii. ( marks) [] for shape. [] for correct period. [] for amplitude. NORTH SYDNEY BOYS HIGH SCHOOL LAST UPDATED JUNE 6, 0
14 4 0 Mathematics HSC Course Assessment Task Trial Eamination SOLUTIONS Question 5 (Ziaziaris) are equal, (a) ( marks) [] for resolving into powers of. [] for final answer. 8 = ( ) = ( 4 ) + ( ) = 4+4 = +4 = +4 = 4 i. ( marks) [] for each correct reason. In ABE and ADC A m C B θ θ 80 θ m E 5 m D CAD (common) Let BCD = θ. From the information, BED = 80 θ Hence AEB = θ(supplementar), and ACD = AEB. ABE = ADC (remaining ) Hence ABE ACD(equiangular) ii. ( marks) [] for ratio of lengths. [] for setting up quadratic. [] for final answer. Let AB =. As the ratio of the side lengths of corresponding sides (c) AB AD = AE AC 8 = + (+) = = 0 (+6)( 4) = 0 = 4, 6 As > 0 (length), = 4 onl. i. ( mark) d d (log e(sin)) = cos sin ii. ( marks) cos cotd = sin d (d) (4 marks) = log e (sin)+c = + = = + V = π d = π [ = π log e ( +) = π(log e 4 log e ) = πlog e 4 ] d + LAST UPDATED JUNE 6, 0 NORTH SYDNEY BOYS HIGH SCHOOL
15 0 Mathematics HSC Course Assessment Task Trial Eamination SOLUTIONS 5 Question 6 (Lam) i. ( marks) (a) i. ( mark) sin(+5 ) = cos4 = sin(90 4 ) +5 = 66 = (log e,) ii. ( marks) A CAB = 00 sinα = 5000sinα ii. (α) ( mark) A CAD = ( π ) e +8e +e 00 sin α = 5000cosα T = 8e A ABCD = 5000(sinα+cosα) T e = 4e iii. (4 marks) A ABCD = 5000(sinα+cosα) da dα = 5000(cosα sinα) Stationar pts occur when da dα = 0, i.e. 5000(cosα sinα) = 0 cosα = sinα cosα cosα tanα = α = π 4 T T = e 8e = 4e T T = T T e + 8e + e is a geometric series with a = e and r = 4e. (β) ( marks) [] for 4e <. [] for justification. A geometric series has a limiting sum when < r < ; i.e. < 4e < α da dα A + π 4 0 B inspecting the graph in the previous part, < 4e < when > log e. limiting sum eists when > log e α < π 4, da dα < 0. α > π 4, da dα > 0. Maimum area occurs when A = 5000(sinα+cosα) α= π 4 ( = ) ( ) = 5000 (γ) ( marks) [] for recalling formula [] for final answer S = a r = e 4e = 5000 m NORTH SYDNEY BOYS HIGH SCHOOL LAST UPDATED JUNE 6, 0
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