Core Mathematics 2, May 2007: worked answers.

Time: 1 hour 30 minutes

Question 1

=== 2Ö8 – 2 = -2 + 4Ö2

Question 2

(a) f(2) = 24 – 20 – 32 + 12 = -16

(b) 3x3 – 5x2 – 16x + 12 = (x + 2)(3x2 +Ax + 6) {or you can use long division}

Comparing x terms: - 16 = 2A + 6, so A = - 11.

Hence, f(x) = (x + 2)( 3x2 - 11x + 6) = (x + 2)( 3x - 2)(x - 3)

Question 3

(a) (1+kx)6 = = 1 + 6kx + 15k2x2 + 20k3x3 + …

(b) 6k = 15k2, so 6k - 15k2 = 0, or 3k(2 – 5k) = 0, giving k = 2/5 (as question says k is non-zero)

(c) 20k3 = 20 x 8/125 = 160/125 = 32/25

Question 4

(a) Using the cosine rule: 42 = 62 + 52 – 2 × 6 × 5 × cos A

Hence, 16 = 61 – 60 cos A, or – 45 = - 60cos A

So, cos A = -45/-60 = ¾

(b) sin A = = ==

Question 5

(a) Missing values are 1.414 and 3.137 (answers must be 3dp)

(b) Area = (0.5/2) [ 0 + 6 + 2( 0.530 + 1.414 + 3.137) ] = 4.0405 = 4.04 (answer must be 3sf)

(c) Shaded area = area of triangle below the line – area under curve

= ½ × 2 × 6 – 4.0405 = 1.96 (answer must be 3sf)

Question 6

(a) Taking log of both sides: x log 8 = log 0.8, so x = = -0.107 (answer must be 3sf)

(b) 2 log3 x – log3 7x = 1. Hence, log3 x2 – log3 7x = 1

Hence, = 1 or = 1. Hence, x / 7 = 31 = 3, so x = 21

Question 7

(a) Gradient of AM is 3/2, so gradient of line l is –2/3 (since l is perpendicular to AM)

Hence, y = -2/3 x + c. At (3, 1): 1 = -2 + c, so c = 3.

The equation is y = -2/3 x + 3

(b) Substitute x = 6 into y = -2/3 x + 3 to get y = -4 + 3 = -1

(c) Distance AP2 = (6 – 1)2 + (-1 - -2)2 = 25 + 1 = 26 = square of radius

Hence, equation of circle is (x – 6)2 + (y - -1)2 = 26, or (x – 6)2 + (y + 1)2 = 26

Question 8

(a) £50,000rn-1

(b) 50,000rn-1 > 200,000.

Dividing by 50,000: rn - 1 > 4

Taking logs: (n - 1) log r > log 4

n – 1 > , so n > (as required)

(c) n > log 4 / log 1.09 + 1, so n > 17.08, so n = 18.

(d) = £759,646 = £ 760,000 (to the nearest £10,000 as asked).

Question 9

(a) It’s y = sinx moved left by π/6

(b) The graph meets the y-axis at (0, ½) and it meets the x-axis at (5p/6, 0) and (11p/6, 0).

(c) = 0.65, so = sin-1 0.65 = 0.7076 rads or p - 0.7076 = 2.4340 rads

Hence, x = 0.7076 – (p/6) = 0.18

Or x = 2.4340 - (p/6) = 1.91 (answers must be 2dp)

Question 10

(a) Surface area = 2 (xy + 2xy + 2x2) = 600cm2

Hence, 3xy + 2x2 = 300, so y =

Volume = 2x2y = =

(b) for maximum value of V

Hence, x = Ö(200/4) = Ö50, and V = 943cm3 (must be nearest cm)

(c) , which is negative when x = Ö50, so it is a maximum value.