Answers: 1(a) 3.84 (b) 2x + 6x2 (c) (d) -6 x 4 (e) (3x + 2)(x – 1) (f) $28.90
2. (a) (b) (6 – 1) = (5 – x) x = 0 (3 – 0) = (1 – y) y = -2 D(0, -2)
(c) AB2 = (6 – 1)2 + (3 – 1)2 = 52 + 22 = 29 AB = units
(d) m = =
(e) y = mx + b y = + b (1, 1) satisfies so 1 = + b b =
y = + 5y = 2x + 3
general form: 2x – 5y + 3 = 0
(f) Perpendicular distance of a point from a line:
=
(g) Area of parallelogram = base x height = x = 13 units2
3.a. (i) (ii) e3x 3(x + 1) (iii) 10 sec22x
(b) (i) After 1 application 35% remains
After 2 applications 35% of 35% remains = .35 x .35 = .1225 = %
(iii) % remaining = where n = number of applications
=
Taking logs of both sides: n log 0.35 = log 0.01
n = = = 4.39
i.e. 5 applications are necessary.
(c) xex = xex + ex = ex (x + 1)
dx = = 1e1 – 0e0 = e
(d) dx = = =
4. a. (i) Since EF||BC & EB || FC then BCFE is a parallelogram.
BC = EF (opposite sides of parallelogram)
EC is common
BCE = CEF (alternate)
?EBC ?ECF (SAS)
(ii) ECF = BEC (alternate)
& BEC = ABE (alternate)
ECF = ABE
(iii) DAB = 120o (co-interior with ADC, co-interior sum = 180o)
BCD = 120o (opposite angles of parallelogram)
& FEC = 120o (alternate)

4. b. (i) P = Poe-kt 45 000 = 60 000 e-5k = e-5k e-5k = 0.75
Taking logs to the base e on both sides gives ln e-5k = ln 0.75
-5k ln e = ln 0.75 but ln e = 1 so -5k = ln 0.75 = - 0.287682
k = 0.0575364
(ii) ln 2020 t = 20 P = 60 000 e-20k = 60 000 x e-1.15073
= 60 000 x 0.3164 = 18 984
(iii) It will decay indefinitely. Its asymptote is zero. It is exponential decay, not a G.P.
1 = 60 000 e -0.0575364t = e -0.0575364t
ln = -0.0575364t ln e = 0.0575364t
11.002 = 0.0575364t t = 191.3 years
It will decay to less than 1 person in 192 years. Since the number of people is an integer and can’t be subdivided then the exponential formula is meaningless for very small populations.

5. a
Domain: |x| 4 i.e. x - 4 or x 4 Range: y 0
b. = o.63
This means that 30.63 = 2
210 = (30.63)10 = 36. 3
Hence x = 6.3
c. arc = r? ? = radians 6 = r r = cm
Area of triangle = ½ ab sin C
Area of ?ABO = . . . = .
Area of sector = = .
Shaded area = ( - ) = 0.0118 x 131.3 = 1.55 cm2
d. (i) Simpson’s Rule: A = {T1 + Tn + 2 Todd + 4 Teven }
h = width of each interval = 2m
A = {0 + 0 + 2(1.7 + 3.0 + 1.6) + 4(2.5 + 2.1)}
= {12.6 + 18.4} = m2
(ii) Volume = speed x time x area of cross section
= 0.25 x 60 x 60 x 24 x = 446400 = 4.464 x 105 m3 / day

6. a. (i) 2s
(ii) approx. 4 to 5 seconds. The concavity changes from concave down to concave up. This indicates that the second derivative of x (acceleration) changes sign. a = 0 at the changeover.
(iii) 9s
b. 5x2 – (2k + 1)x + k = 0 discriminant is b2 – 4ac
(2k + 1)2 – 4x5xk = 49
(2k + 1)2 – 20k = 49
4k2 + 4k + 1 – 20k = 49
4k2 – 16k – 48 = 0
k2 – 4k – 12 = 0
(k – 6)(k + 2) = 0
k = 6 or –2
When k = 6 equation becomes 5x2 – 13x + 6 = 0
(5x – 3)(x – 2) = 0 x = 2 or
When k = -2 equation becomes 5x2 + 3x – 2 = 0
(5x – 2)(x + 1) = 0 x = or -1
k = 6 or –2 roots are -1, 2, - or
c. (i) C = 2?r = 500 r = m ˜ 79.5775m
(ii) = = 0.21043
? = 1.3588 = 1.36 (2 d.p.)
(iii) arc = r? = x 1.36 = 108.128
Perimeter = arc + 2r = x 1.36 + + = x 3.36 = 267.4 (1 d.p.)
If ? is taken to the maximum number of decimal places allowed by the calculator it becomes 1.358779854 radians and the perimeter becomes 267.3 (1 d.p.)

7. a. (i) y’ = 2x when x = 2 y’ = 4
(ii) y = mx + b y = 4x + b (-2, 0) satisfies 0 = -8 + b b = 8
Tangent: y = 4x + 8
(iii) sub. x = 2 into y = 4x + 8 y = 16 (2, 16)
(iv) y = x2 + c (2, 16) satisfies 16 = 4 + c c = 12
parabola: y = x2 + 12
(v) When x = 0, y = c = 12 vertex: (0, 12)
(vi) A = = = units2
(vi) V = = ? dy = ?
(vii) = ??[128 -192] – [72 – 144]}
= ?{-64 - - 72} = 8? units3

a. (i) Let the point be (x, y)
Distance from (2, 1) is
Distance from (5, 4) is
= 2
Squaring both sides gives (x – 2)2 + (y – 1)2 = 4{(x – 5)2 + (y – 4)2}
x2 – 4x + 4 + y2 – 2y + 1 = 4{ x2 – 10x + 25 + y2 – 8y + 16}
x2 – 4x + 4 + y2 – 2y + 1 = 4x2 – 40x + 100 + 4y2 – 32y + 64
3x2 – 36x + 3y2 – 30y + 159 = 0
Dividing both sides by 3 gives x2 – 12x + y2 – 10y + 53 = 0
Completing the squares gives x2 – 12x + 36 – 36+ y2 – 10y + 25 – 25 + 53 = 0
(x – 6)2 – 36 + (y – 5)2 – 25 + 53 = 0
(x – 6)2 + (y – 5)2 = 8
This is the equation of a circle
(ii) radius =
(iii) centre = (6, 5)

8. a. (i)
(ii) 3 cos x = 2 cos x =
cos-1 = 0.841 rad. or – 0.841 rad
(iii) Area = area under the curve y = 3 cos x between – 0.841 rad. and + 0.841 rad. less the area under the line y = 2 between – 0.841 rad. and + 0.841 rad.
= = 2.236 – – 2.236 = 4.472
= = 1.682 – – 1.682 = 3.364
4.472 – 3.364 = 1.11 units2 (2 d.p.)

b. (i) V = ?r2h = 4
A = ?r2 + = ?r2 +
(ii) A = ?r2 + = ?r2 + 8r-1
= 2?r – 8r - 2
Put = 0 2?r – = 0 2?r3 = 8 r3 = r = 1.084m
Area of base = ?r2 = ?1.0842 = 3.69
h = = = 1.0836
Area of wall = 2?rh = 2?x1.084x1.084 = 7.38
Total area = 3.69 + 7.38 = 11.07 m2

c. (i) Sn = {2a + (n-1)d}
S7 = {2a + (7-1)d} = 7a + 21d = 133
S10 = {2a + (10 – 1)d} = 10a + 45d = 250
S7 x 10 70a + 210d = 1330
S10 x 7 70a + 315d = 1750
(S10 x 7) – (S7 x 10) 105d = 420 d = 4
7a + 21d = 133
7a + 84 = 133 7a = 133 – 84 = 49 a = 7
(ii) Tn = a + (n – 1)d 382 = 7 + (n – 1)4 382 = 7 + 4n – 4 4n = 379
n = 94.75
Since n is not a whole number then 382 is not a member of the series.
9. a. (i) or 0.6 (the second digit can be one of 3 odd digits or 2 even digits)
(ii) = = 0.5
There are 5 x 4 = 20 possible numbers. 34 & 35 are greater than 32 as are the four numbers beginning with 4 and the four numbers beginning with 5. i.e. there are 10 numbers greater than 32.
(iii) There are 20 possible numbers but only one of them is 41.
(iv) zero No two digit numbers less than 11 can be made from the digits 1 to 5
(v) The highest possible number is 54. The only perfect squares with 2 digits less than 54 are 16, 25, 36 and 49. 25 is the only one with the digits 1-5.
(vi) = The possible numbers are 15, 51, 24 & 42
b. (i) 4x + 2x – 2 = 0 22x + 2x – 2 = 0 let 2x = a
a2 + a – 2 = 0 (a + 2)(a – 1) = 0 a = 1 or -2
2x = -2 has no solution If 2x = 1 then x = 0
x = 0
(ii) 2 sin 2x = sin 2x = sin-1 =
2x = x =
The total number of x values that satisfy is best seen by sketching the graph y = 2 sin 2x.
It can be seen that there are 8 solutions to 2 sin 2x = (-2? x ???
x = , , , , , ,

10. a. (i) 6% p.a. = 0.5% per month
${(500 000 x 1.005) – M}
(ii) Let A be the amount in the account.
After the 1st month A = (500 000 x 1.005) – M
After the 2nd month A = (500 000 x 1.005 – M)1.005 – M
= 500 000 x 1.0052 – 1.005M – M
After n months A = 500 000x1.005n – 1.005n-1M – 1.005n-2M ……..- 1.005M – M
A = 500 000 x 1.005n – M( 1 + 1.005 + 1.0052 + ….+ 1.005n-1)
After 30 years (i.e. 360 months)
A = 500 000 x 1.005360 – M( 1 + 1.005 + 1.0052 + ….+ 1.005359) = 0
(iii) = = = 1004.515
A = 500 000 x 1.005360 – 1004.515M = 0
3011287.6 = 1004.515M
M = $2997.75 (to nearest cent)

b. (i) Gradient = y’ = = When x = 2, y’ = = ½
(ii) Gradient of normal = -2. (negative reciprocal of ½ )
Equation of normal is y = -2x + c
To find the value of c find the y coordinate when x = 2.
y = 2 loge (22 + 4) = 2 loge 8 = 4.16
So (2, 4.16) satisfies the normal equation y = - 2x + c
4.16 = -4 + c
c = 8.16
Normal: y = - 2x + 8.16