Core Mathematics C2, 2 June 2008 – worked answers

1. (a) f(-4) = -128 – 48 + 156 + 20 = 0, hence (x+4) is a factor of f(x) (2)

(b) From above, f(x) =(x + 4)(2x2 + Ax + 5)

Comparing x2 terms: -3 = A + 8, so A = -11

Hence, f(x) = (x + 4)(2x2 – 11x + 5) = (x + 4)(2x – 1)(x – 5) (4)

x / 0 / 0.5 / 1 / 1.5 / 2
y / 1.732 / 2.058 / 2.646 / 3.630 / 5.196

2. (a) (Values in table must be 3dp) (2)

(b) I = = 0.25 x 23.597 = 5.899 (4)

3. (a) (1 + ax)10 =

= (4)

(b) 120a3 = 2 x 45a2, so a = 90/120 = 3/4 (2)

4. (a) Taking logs of both sides: x ln 5 = ln 7, so x = =1.21 (must be 3 sig figs) (2)

(b) Let 5x = y, so y2 – 12y + 35 = 0. Hence (y – 5)(y – 7) = 0.

So, y = 5, which gives 5x = 5, and x = 0

Or, y = 7, which gives 5x = 7, and x = 1.21 from above. (4)

5. (a) Using Pythagoras, radius = Ö[(8-3)2 + (3-1)2] = Ö29Hence (x – 3)2 + (y – 1)2 = 29 (4)

(b) Gradient of line from centre to P (ie radius) is (3-1)/(8-3) = 2/5. Radius and tangent meet at 90º and so the product of their gradients is -1. Hence gradient of tangent is -5/2 and its equation is of the form y = -2.5x + c.

At P(8, 3): 3 = -20 + c, so c = 23

Hence the equation is y =-2.5x+23 or 5x + 2y – 46 = 0 (5)

6. (a) 20th term (using nth term is “arn-1”) is =0.072 (2)

(b) =25 (2)

(c)

Hence, 25 (1 – 0.8k) > 24.95 or 1- 0.8k > 0.998, so 0.8k < 0.002

Taking logs, k log 0.8 < log 0.002, so k > (reverse < sign as log 0.8 <0) (4)

(d) k > 27.85, so the smallest possible value of k = 28 (1)

7. (a) Arc length = “rθ” = 7 × 0.8 = 5.6cm (2)

(b) Area of sector ABC = “½r2θ” = ½ × 0.8 × 72

= 19.6cm2 (2)

(c) Using cosine rule:

BD2 = 3.52 + 72 – 2 × 7 × 3.5 cos 0.8 = 27.1113712….

So BD = 5.207 (to 3dp)

Hence, Perimeter of R = BD + DC + BC

= 5.207 + 3.5 + 5.6 = 14.3cm (must be 3sf) (4)

(d) Area of R = area of sector ABC – area of triangle ADB

Area of sector = 19.6cm2 from above.

Area of triangle = “ ½absinC” = ½ × 7 × 3.5 sin 0.8 = 8.788cm2

Hence, area of R = 19.6 - 8.788 = 10.8cm2 (must be 3sf) (4)

8. (a) dy/dx = 8 + 2x – 3x2 = 0 at maximum point.

So 3x2 – 2x – 8 = 0 or (3x + 4)(x – 2) = 0, so x = 2 or x = -4/3

From diagram, clearly x is positive, so x = 2 (3)

(b) Area under curve = ==342/3

At A, x = 2, so using the equation for the curve y = 10 + 8x + x2 – x3

= 10 + 16 + 4 – 8 = 22

Area under diagonal line = area of triangle = ½ × 2 × 22 = 22

Shaded area = 342/3 – 22 = 122/3 or 38/3. (8)

9. (a) sin(x – 20°) = , so x – 20 = sin-1 = 45 or 135, so x = 65 or 155 degrees (4)

(b) cos 3x = –, so 3x = cos-1 - = 120 or 240 or 480 or 600 or 840 or 960

So, x = 40 or 80 or 160 or 200 or 280 or 320 degrees (6)

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