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(^1) (a) x^5 (y^2 z^3 )^2 x−^2 y^4 z^5
= x
(^5) y (^4) z 6 x−^2 y^4 z^5
= x^7 z [2 marks]
(b) b
b^2 − 6 b + 9
= (b^ −^ 3)(b^ + 3) (b − 3)^2
= b^ + 3 b − 3
. [2 marks]
(^2) x (^2) + 3^3 x + (^) x + 3^1 = (^) x(3 +x + 3)^ x =^1 x [4 marks]
3 (a) x^2 − x − 20 = (x − 5)(x + 4) so solutions are x = 5, x = −4. [2 marks]
(b) Using the quadratic formula x =
√ 169 − 4 × 10 × (−3) 6 =
6 = 5 or^ − 2 3 [2 marks]
4 (a) y = 3x − 6 represents a straight line with slope 3 meeting the y-axis at y = −6. [2 marks] (b) y = x^2 + 4x − 5 is a quadratic curve, which is U-shaped, crossing the y-axis at y = −5 and the x-axis at x = 1, x = −5. The curve is symmetric about the line x = − 4 /2 = −2. The bottom point is at x = − 2 , y = −9. [3 marks] (c) y = |x^2 + 4x − 5 | is given from (b) by reflecting the part below the x-axis in the x-axis. [3 marks]
5 Put y = 1 + 3 2 x − x 5 , and solve for x in terms of y = f (x). Then (2x − 5)y = 1 + 3x so
2 xy − 5 y = 1 + 3x, giving x(2y − 3) = 1 + 5y.
Thus x = f −^1 (y) = 1 + 5 2 y − y 3 and so f −^1 (x) = 1 + 5 2 x − x 3. [3 marks]
(^6) (a) Either use the formula a^1 −^ rn 1 − r
with r = 3, a = 3, n = 6, giving 3× 3
or simply add up the 6 terms. [3 marks]
(b) The formula is (^1) −a r. [1 marks]
Here a = 58 , r = 58 , giving the sum as
5 8 1 − 58 =
5 (^83) 8
= 53. [2 marks]
(^7) (a) lim n→∞
2 − 3 n + 5n^2 5 − 2 n^2
= lim n→∞
2 n^2 −^
3 5 n^ + 5 n^2 −^2
= −^5
as n → ∞. [2 marks]
(b) Putting x = −2 in the bottom of the fraction gives 0, as also in the top. Factorise to write x^2 − x − 6 x^2 − 4 =
(x − 3)(x + 2) (x − 2)(x + 2) =^
x − 3 x − 2
Now put x = −2 to get the limit 54. [2 marks]
8 (a) Put u = 3x − 5. Then y = (3x − 5)^5 = u^5 so dy dx
= dy du
du dx
= 5u^4 × 3 = 15(3x − 5)^4.
[2 marks] (b) Put u = x^3 + 2. Then y = (x^3 + 2)^34 = u 43 and dy dx =^
4 3 u^
(^13) × du dx =^
4 3 u^
(^13) × 3 x (^2) = 4x (^2) (x (^3) + 2) (^13).
[3 marks] (c) y = x^6 sin x = uv with u = x^6 , v = sin x. dy dx =^ u
dv dx +^ v
du dx =^ x
(^6) cos x + 6x (^5) sin x.
[3 marks]
9 The slope of the tangent is the value of dy dx at x = 1. Now (^) dxdy = 6x^2 , so the slope is
- When x = 1 we have y = −3, so the tangent line has equation y + 3 = 6(x − 1), giving y = 6x − 9. [3 marks]
(d) Put u = sin^3 x and w = sin x. Then u = w^3 and du dx = (^) dwdu × dw dx = 3w^2 ×(cos x) =
3 sin^2 x cos x. Setting also v = 2x^2 − 3, we have
d dx
sin^3 x 2 x^2 − 3
= d dx
u v
= v^
du dx −^ u^
dv dx v^2
= (2x
(^2) − 3) × 3 sin^2 x cos x − sin^3 x × 4 x (2x^2 − 3)^2 [4 marks]
14 (i) The stationary points occur where f ′(x) = 0 and the inflection points where f ′′(x) = 0. Now f ′(x) = 3x^2 + 2x − 2 and f ′′(x) = 6x + 2. [3 marks] The inflection point occurs where x = −^13 = 0.33 and f (x) = −^2027 = − 0 .74. [1 marks]
The stationary points are given by the quadratic formula as x = −^2 ±
3 = 0.55 or^ −^1 .22.^ These are respectively a local minimum, where^ f^
′′(x) > 0,
and a local maximum, where f ′′(x) < 0. The corresponding values of f (x) are − 0 .63 and 2.11. [2 marks] (ii) The curve y = x^3 + x^2 − 2 x crosses the x-axis when x^3 + x^2 − 2 x = 0. This happens when x = 0 or when x^2 + x − 2 = 0, giving x = −2 or x = 1 as well. [2 marks] (iii) Using the information from (a) and (b), sketch the curve y = x^3 + x^2 − 2 x. [3 marks] (iv) The total area bounded by the curve and the x-axis is made up of two pieces, one between∣ x = −2 and x = 0, and the other between x = 0 and x = 1. These are found as ∣∣ ∣
∫ (^0) − 2 f^ (x)^ dx
∣∣ ∣∣ and
∣∣ ∣∣
∫ (^1) 0 f^ (x)^ dx
∣∣ ∣∣. Now ∫^ f (x) dx = ∫^ (x^3 + x^2 − 2 x) dx = x
4 4 +
3 x
(^3) − x (^2) , giving
the first area as | 4 − 8 / 3 − 4 | = 8/3 and the second as | 1 /4 + 1/ 3 − 1 | = 5/12 making a total of 37/12 = 3.08. [4 marks]
15 (a) Substitute u = x^4 + 3. Then du = 4x^3 dx, so
I =
∫ x^3 sin(x^4 + 3) dx =
∫ 1 4 sin^ u du^ =^ −
1 4 cos^ u^ +^ C^ =^ −
1 4 cos(x
4 + 3) + C.
[4 marks] (b) Substitute t = tan x. Then dt = sec^2 x dx, so
I =
∫ tan^5 x sec^2 x dx =
∫ t^5 dt = t 66 + C = 16 tan^6 x + C.
[4 marks] (c) Substitute x = 2 sin t. Then dx = 2 cos t dt. Therefore
I =
∫ (^1) 0
√^ dx 4 − x^2
∫ (^) x= x=
√^ 2 cos^ t 4 − 4 sin^2 t
dt =
∫ (^) x= x=
√^ 2 cos^ t 4 cos^2 t
dt =
∫ (^) x= x=
dt = [t]x x=1=
Now when x = 0 we have t = 0 and when x = 1 we have 1 = 2 sin t so that sin t = 12 and
t = π 6. Consequently I = [t]t t==0π/ 6 = (
π 6 −^ 0) = 0.^52.^ [7 marks]
