MATHEMATICS (Extension 1)

Transcription

1 009 Year 11 Assessment Task MATHEMATICS (Extension 1) General instructions Working time 50 minutes. Write on your own A4 paper. Each question is to commence on a new page. Write using blue or black pen. Board approved calculators may be used. All necessary working should be shown in every question (Insufficient/illegible working may cause a deduction of marks). Class teacher (please ) Mr Barrett Mr Lowe Mr Rezcallah Mr Trenwith Mr Ireland Mr Weiss Mr Lam Attempt all questions. NAME:... PAGES USED:... Marker s use only. QUESTION Total % MARKS

2 009 Mathematics (Extension 1) Assessment Task Question 1 (9 Marks) Commence a NEW page. Marks Solve, giving exact values: x + 6 = x. 1 x + 5 x 5 = 1 x 5. (x + 1) = 8. (d) x(x + 1) = 1. (e) x + = 1 x. Question (1 Marks) Commence a NEW page. Marks Write the exact values of i. cos ii. cot iii. tan i. Write 75 in radians as an exact value. 1 ii. Evaluate cos. correct to decimal places. Find the value of the pronumeral in each of the following diagrams. 4 cm i. x 59 6 cm ii. 4 cm θ 9 cm (d) A ship leaves port A on a bearing of 10 T travelling for hours at 0 knots (nautical miles/hr). It changes to a new bearing of 0 T for 1 hour at 0 knots. How far is it from port A?

3 009 Mathematics (Extension 1) Assessment Task Question (18 Marks) Commence a NEW page. Marks Solve the following inequalities: x i. 4 7x. ii. x + x >. iii. x + 1 <. iv. x + x 1. Form an equation and solve: i. There are 450 students at a particular school. If there are 40 more boys than girls, how many boys and girls are there? ii. A box contains 50 coins made up of 10 cent coins and 0 cent coins. How many of each are in the box if the value of the coins is $7? Find the coordinates of the points of intersection of the graphs y = x & x + y 4y 1 = 0 Question 4 (11 Marks) Commence a NEW page. Marks If tan θ = 4 and 90 < θ < 180, find the exact value of i. sin θ. ii. cot θ. 1 Simplify fully sinθ i. 1 + cos θ cos θ. sin θ ii. 1 sin (180 θ). Prove the following identities: i. tan θ ( 1 cot θ ) + cot θ ( 1 tan θ ) = 0. ii. sec θ + tan θ = 1 + sin θ. cos θ

4 4 009 Mathematics (Extension 1) Assessment Task Question 5 (17 Marks) Commence a NEW page. Marks Consider the function y = 4sin x, where x is in degrees. i. Find the period. 1 ii. Find the amplitude. 1 iii. Sketch the function for 0 x 60. Solve for 0 x 60 i. cos x = 1. ii. tan x = 1 iii. cos x + 5sin x = 1. 4 The points A and B are 500m apart on the ground and D is the top of a tower. BAD and DBA are 59 and 54 respectively. The elevation of D from A is 5. D C B A Copy the diagram into your writing booklet and mark on the figure all the angles stated above. i. Show that the height h metres of the tower is given by h = 500sin 5 sin 54 sin 67 ii. Find h to the nearest metre. 1 End of paper.

5 009 Mathematics (Extension 1) Assessment Task SOLUTIONS 5 Suggested Solutions Question 1 (1 mark) x + 6 = x x = x = 1 4. ( 1 4) + = 4 + = 5 4 ( 1 1 ) = (d) (e) x = ( marks) [1] for removing denominators. [1] for x = 6. x + 5 x 5 (x 5)(x+5) (x 5)(x+5) ( marks) ( marks) ( marks) = (x 5) (x + 5) = 1 x 10 x 15 = 1 x 5 = 1 x = 6 (x + 1) = 8 x + 1 = ± 8 x = 1 ± 8 = 1 ± x + x 1 = 0 x = 1 ± = 1 ± 5 1 x 5 (x 5)(x+5) x + = 1 x { x + x + 0 x + = (x + ) x + < 0 x + = 1 x 4x = 1 x = 1 4 Test values: (x + ) = 1 x x = 1 x x = x = x =. ( ) + = 9 + = 5 = 5 ( 1 ) = 5 Question i. (1 mark) ii. (1 mark) cot 40 = iii. (1 mark) cos 180 = 1 1 tan( ) = 1 tan 15 = tan(60 45 ) = 1 i. (1 mark) ii. ( marks) i. ( marks) ii. ( marks) π = 5π 1 (cos.) = x = sin 59 4 x = 4sin 59 =.4 ( d.p.) cos θ = b + c a bc = = 61 7 θ = 5 LAST UPDATED MAY, 009

6 6 009 Mathematics (Extension 1) Assessment Task SOLUTIONS (d) ( marks) [1] for correct diagram. [1] for correct application of cosine rule. [1] for b = 59.9 to 1 d.p. A b 0 60 n.m. 0 C 50 0 n.m. b = a + c accos B B = cos 80 = 58.4 b = 59.9 n.m. (1 d.p.) iii. ( marks) x + 1 < { x + 1 x x + 1 = (x + 1) x + 1 < 0 x + 1 < iv. ( marks) x < x < 1 < x < 1 (x + 1) < x + 1 x (x ) (x ) x < 4 x > (x + )(x ) (x ) (x ) (x + )(x ) 0 (x )((x ) (x + )) 0 y (x )(x 5) 0 Question i. ( marks) x ii. ( marks) x 4 7x 9x 8x x x 4 19 x + x > 0 (x + )(x 1) > 0 y i. ( marks) x < or x 5 [1] for solving equation, resulting in x = 05. [1] for conclusion. Let the # of girls be x, (x + 40) + x = x x = 410 x = 05 x < or x > 1. Hence there are 05 girls & 45 boys. LAST UPDATED MAY, 009

7 009 Mathematics (Extension 1) Assessment Task SOLUTIONS 7 ii. ( marks) [1] for setting up simultaneous equations. [1] for solving simultaneous equations. [1] for correct final conclusion. Let the # of 10c coins be x, and # of 0c coins be y. { x + y = 50 (1) 10x + 0y = 700 () () [(1) 10] 10y = 00 y = 0 x = 0 Hence there are thirty 10c coins and twenty 0c coins. ( marks) [1] for correct substitution [1] for solution of x = 1, x = 4. [1] for correct points of intersection. { y = x (1) x + y 4y 1 = 0 () Substitute (1) to () x + (x ) 4(x ) 1 = 0 x + x 6x + 9 4x = 0 x 10x + 8 = 0 x 5x + 4 = 0 (x 4)(x 1) = 0 x = 1 or 4 Points of intersection are (1, ) and (4,1). ii. (1 mark) i. ( marks) cot θ = 4 sinθ 1 + cos θ cos θ sin θ = sin θ + (1 + cos θ + cos θ) sin θ(1 + cos θ) = + cos θ sin θ(1 + cos θ) = (1 + cos θ) sin θ (1 + cos θ) = sin θ = cosec θ ii. ( marks) 1 sin (180 θ) = 1 sin θ = cos θ i. ( marks) tan θ ( 1 cot θ ) + cot θ ( 1 tan θ ) = tan θ 1 tan θ + 1 tan θ tan θ = 0 ii. ( marks) sec θ + tan θ = 1 cos θ + sin θ cos θ = 1 + sin θ cos θ Question 4 i. ( marks) 5 θ 4 sin θ = 4 5 LAST UPDATED MAY, 009

8 8 009 Mathematics (Extension 1) Assessment Task SOLUTIONS Question 5 Let m = sin x, i. (1 mark) ii. (1 mark) T = 60 a = 4 = 10 m 5m = 0 (m + 1)(m ) = 0 m = 1,m = iii. ( marks) y sin x = 1 solution as sin x = has no 4 x = or θ i. ( marks) 4 D i. ( marks) 67 h cos x = 1 x = 10,40 ii. ( marks) [1] for 0 x 70. [1] for finding the values of x. [1] for finding the values of x. B A C 0 x 70 tan x = 1 x = 0,10,90,570 x = 15,105,195,85 iii. (4 marks) [1] for changing cos x 1 sin x. [1] for rearranging to the quadratic sin x 5 sinx = 0. [1] for solving the quadratic. [1] for final answer. cos x + 5sin x = 1 (1 sin x) + 5sin x = 1 sin x + 5sin x = 1 sin x 5sin x = 0 h AD = sin5 AD = h sin5 Applying the sine rule to DBA, ii. (1 mark) AD sin 54 = 500 sin 67 h sin 5 sin54 = 500 sin 67 h = 500sin 5 sin 54 sin 67 h = 8.m = 8m (nearest m) LAST UPDATED MAY, 009