Unit 2 Practice Assessment - Solutions

1. 1 2 -3 -2 3

2 -1 -3

2 -1 -3 0 since the remainder is zero (x – 1) is a factor.

(x- 1)(2x² – x –3)

(x – 1)(x + 1)(2x – 3)

2. 2x² + 5x + 9 b² – 4ac = 25 – 36 = -9

Since b² – 4ac < 0 this quadratic has no real roots.

3. =

4. = =

5. To find the limits of integration equate the two curves.

x² – 4 = 2x – x²

2x² – 2x - 4 = 0 let f(x) = 2x – x² negative quadratic so upper curve

2(x² – x – 2) = 0 let g(x) = x² – 4 positive quadratic so lower curve

2(x - 2)(x + 1) = 0

x = -1, x = 2

area between curves is given by:

6. cos(3x) = (3x) = 30^o, 330^o, 390^o, 690^o, 750^o, 1050^o

x = 10⁰, 110^o 130^o 230^o 250^o 350^o

since 0 < x< 180^o solution is x = 10⁰, 110^o 130^o

7. First draw two triangles, and find the length of the remaining side.

sin(A) = 4/5 cos (A) = 3/5 sin(B) = 5/13 cos(B) = 12/13

cos(A+B) = cos(A)cos(B) - sin(A)sin(B)

=

8. sin(x)cos() + cos (x)sin() = sin(x + ) = 1

sin(x + ) = 1

(x + ) =

x =

9a) centre (-4,5) radius 5 Equation: x² + y² + 8x – 10y + 16 = 0

b) x² + y² – 6x + 6y – 18 = 0 centre (3, -3) radius 6

10. x² + y² - 6x – 4y – 4 = 0 y = 4x + 7

substituting for y gives

x² + (4x + 7)² – 6x – 4(4x + 7) = 0

17x² + 34x + 17 = 0

17(x + 1)² = 0 Equal roots imply tangent (also could have used b²-4ac = 0 to prove tangent)

11. x² + y² + 20y + 20 = 0 at the point (8, -6)

centre (0, -10) point (8,-6)

m_radius = 1/2 m_tangent = -2

Equation y + 6 = -2(x – 8)

of tangent: y = - 2x + 10