Unit 2 Practice NAB Marking

1. Show that (x + 1) is a factor of f(x) = x 3 + 2x 2 – 5x – 6, and express f(x) in fully factorised form. Outcome (x + 1) is a factor → f(x) = (x + 1)(x 2 + x – 6) = (x + 1)(x – 2)(x + 3)

2. Use the discriminant to determine the nature of the roots of the equation 2x 2 – 3x + 2 = 0. using b 2 – 4aca = 2, b = -3, c = 2 =(-3) 2 – 4(2)(2) =9 – 16 = -7 →no real roots Threshold 4 out of 6

3.Find ∫ 2 dx x2x2 = ∫ 2x -2 dx = 2x -1 + c = -2x -1 + c

∫ x 2 (4 – x) dx 0 4 ∫( 4x 2 – x 3 ) dx 0 4 = 4x 3 – x [ ] 0 4 = 4(4) 3 – ( ) – 4(0) 3 – ( ) = 64 / 3 units 2 Decimal Acceptable for answer

Finding Limits using y = y x 2 – 2x + 2 = x + 2 x 2 – 3x = 0 x(x – 3) = 0 x = 0 or x = 3 ∫ (x + 2) – (x 2 – 2x + 2) dx 0 3 Threshold 8 out of 11

2cos2x=1 for 0≤x<π6. Solve algebraically the equation 2cosA = 1where A = 2x cosA = 0.5 cos -1 (0.5) = 60 0 A = 60 0 or x = 60 0 or x = 30 0 or = π / 6 or 5π / 6 radians

5 13 a)sin x = 4 / 5 cos x = 3 / 5 sin y = 5 / 13 cos y = 12 / 13 b)cos(x + y) = cosx cosy – sinx siny = 3 / 5 X 12 / / 5 X 5 / 13 = 16 / 65

sinx°cos10°+ cosx°sin10°= 2 / 3 for 0 ≤x< (a) Express sinx ° cos 10° + cos x° sin 10° in the form sin(A + B) °. (b) Use the result of (a) to solve the equation a)sin(x + 10) 0 b) sin(x + 10) 0 = 2 / 3 sin A = 2 / 3 where A = x + 10 sin -1 ( 2 / 3 ) = A = or x + 10 = or x = or Threshold 9 out of 12

using (x – a) 2 + (y – b) 2 = r 2 → (x + 3) 2 + (y – 2) 2 = (a) A circle has radius 4 units and centre (–3, 2). Write down the equation of the circle. (b) A circle has equation x 2 + y 2 + 6x – 8y – 11 = 0. Write down its radius and the coordinates of its centre. using x 2 + y 2 + 2gx+ 2fy + c = 0 g = 3,f = -4, c = -11 centre (-3, 4) radius √(3 2 + (-4) ) = √36 = 6

substitute → x 2 + (2x – 3) 2 + 2x – 4 = 0 x 2 + 4x 2 – 12x x – 4 = 0 5x 2 – 10x + 5 = 0 5(x 2 – 2x + 1) = 0 5(x – 1)(x – 1) = 0 x = 1 only 1 solution → Tangent 10. Show that the straight line with equation y = 2x – 3 is a tangent to the circle with equation x 2 + y 2 + 2x – 4 = 0. Either Or using discriminant (-10) 2 – 4(5)(5) = 0 equal roots → tangent

m PQ = – 1 = 2 m tangent = - ½ (m 1 m 2 = -1) Equation y + 5 = -½(x – 1) etc. Threshold 10 out of 14