Maths Ext 2 2010 HSC (Sample Answers)

2 k 1 k 1 1 2 1 2 2 2 1 2 1 2 2 2 k 1 k 1 1 2 1 2

Info icon This preview shows pages 19–25. Sign up to view the full content.

View Full Document Right Arrow Icon
2 ) = ( ) k ) k 1 1 + 2 ( 1 + 2 2 + 2 ) + ( 1 2 ( 1 2 2 + 2 ) = ( k + k + 1 1 + 2 ) + ( 1 2 ) where we use ( ) 2 1 + 2 = 1 + 2 2 + 2 and ( 2 1 2 ) = 1 2 2 + 2 Question 6 (c) (i) ( 5 cos θ + i sin θ ) ( 2 3 = cos 5 θ + 5cos 4 θ i sin θ ) + 10 cos 3 θ ( i sin θ ) + 10cos 2 θ ( i sin θ ) + 5cos θ ( ) 4 i sin θ + i 5 sin 5 θ = cos 5 θ + 5 i cos 4 θ sin θ 10 cos 3 θ sin 2 θ 10 i cos 2 θ sin 3 θ + 5cos θ sin 4 θ + i sin 5 θ Question 6 (b) – 19 –
Image of page 19

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
2010 HSC Mathematics Extension 2 Sample Answers Question 6 (c) (ii) ) 5 ( cos θ + i sin θ = cos5 θ + i sin5 θ equating imaginary parts, sin5 θ = 5cos 4 θ sin θ 10cos 2 θ sin 3 θ + sin 5 θ ( ) 2 ( = 5sin θ ( 1 2sin 2 θ + sin 4 θ ) 10sin 3 θ + 10sin 5 θ + sin 5 θ = 5sin θ 10sin 3 θ + 5sin 5 θ 10sin 3 θ + 10sin 5 θ + sin 5 θ = 16sin 5 θ 20sin 3 θ + 5sin θ = 5 1 sin 2 θ .sin θ 10 1 sin 2 θ ) sin 3 θ + sin 5 θ Question 6 (c) (iii) 5 π π π π 16sin 20sin 3 + 5sin = sin 5 × 10 10 10 10 π = sin = 1 2 π sin is a solution to 16 x 5 20 x 3 + 5 x 1 = 0 10 Question 6 (c) (iv) 16 x 5 20 x 3 + 5 x 1 = ( x 1 ) ( 16 x 4 + 16 x 3 4 x 2 4 x + 1 ) Question 6 (c) (v) 16 x 4 + 16 x 3 4 x 2 4 x + 1 2 = ( 4 x 2 + ax 1 ) = 16 x 4 + 8 ax 3 + a = 2 hence q x = 2 + 2 x 1 ( ) 4 x – 20 –
Image of page 20
2010 HSC Mathematics Extension 2 Sample Answers Question 6 (c) (vi) 2 16 x 5 20 x 3 + 5 x 1 = ( x 1 ) ( 4 x 2 + 2 x 1 ) = 0 2 ± 20 1 ± 5 Solutions are x = 1 and x = = 8 4 π 1 + 5 hence the value of sin = 10 4 Question 7 (a) (i) In Δ ADB and Δ KDC ADB = ADK + KDB = CDB + KDB = KDC (given)± ABD = KCD (angles in the same segment)± ∴ Δ ADB ||| Δ KDC (equiangular)± Question 7 (a) (ii) In Δ ADK and Δ BDC AK BC = AD BD AK × BD = AD × BC ................ ± In Δ ADB and Δ KDC KC AB = CD BD KC × BD = CD × AB ................ ² adding ± and ² AK × BD + KC × BD = AD × BC + CD × AB BD AK + KC AD × BC + AB × DC ( ) = BD × AC = AD × BC + AB × DC – 21 –
Image of page 21

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
2010 HSC Mathematics Extension 2 Sample Answers Question 7 (a) (iii) ± By part (ii) x 2 = x + 1 x 2 x 1 = 0 1 ± 5 x = 2 1 + 5 since x > 0, x = 2 1 x x x 1 Question 7 (b)± y x (3, 8) (1, 2) 0 for x > 3, y = 2 x is greater than y = 3 x 1 2 x 3 x 1 for x 3 Question 7 (c) (i) – 22 – ( P ( x ) = 1) n n x n 1 n ( n 1) x n 2 ( ) = ( 1) n P x n n x 2 ( x 1)± P ( x ) = 0 when x = 0 or x = Hence there are exactly two turning points.±
Image of page 22
2010 HSC Mathematics Extension 2 Sample Answers Question 7 (c) (ii) P ( ) 1 = ( n 1 ) × 1 n n × 1 n 1 + 1 = 0 P 1 = 0 (shown in (i)) ( ) P x ( ) has a double root when x = 1. ( ) = ( ) 2 P x n n 1 x n 2 n n ( 1 )( n 2 ) x n 3 ′′ Note: P ′′ ( ) 1 0 P x ( ) does not have a triple root at x = 1 Question 7 (c) (iii) ( ) ( ) P 0 = 1 and P 1 = 0 ( ) ( ) P x → ∞ as x → ∞ and P x → −∞ as x → −∞ (since the degree of P x ( ) is odd) y 1 1 0 x The curve cuts the x -axis in only one place (other than x = 1). there is exactly one real zero of P ( x ) other than 1. – 23 –
Image of page 23

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
2010 HSC Mathematics Extension 2 Sample Answers Question 7 (c) (iv) P ( 1) = ( n 1) × ( 1) n n × ( 1) n 1 + 1 = ( n 1) × ( 1) n × 1 + 1(since n is odd) = n + 1 n + 1 = 2 n + 2 n n 1 1 1 1 P = ( n 1) n + 1 2 2 2 1 = { ( n 1) 2 n + 2 n } 2 n = 2 n 3 n + 1 2 1 n { } > 0 since > 0 and 2 n 3 n + 1 > 0 (from(b)) 2 n 1 1 Hence –1< α <– because P ( x ) changes sign between x = and x = 1.
Image of page 24
Image of page 25
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern