Download Exam Solutions for Math011: September 2006 and more Exams Mathematics in PDF only on Docsity!
MATH011 Exam Solutions Sept 2006
- (a) a^9 b−^3 c^8 a^7 (bc^2 )^4
a^9 b−^3 c^8 a^7 b^4 c^8
a^2 b^7
[2]
(b) 16 x^2 − 1 16 x^2 + 8x + 1
(4x − 1)(4x + 1) (4x + 1)^2
4 x − 1 4 x + 1
. [2]
y − 7
y^2 − 7 y
y − 7 y(y − 7)
y
[4]
- (a) x^2 − x − 72 = (x − 9)(x + 8) so solutions are x = 9, x = −8. [2]
(b) Using the quadratic formula x =
√ 4 − 4 × 8 × (−21) 16
or −
[2]
- (a) y = 3x − 6 represents a straight line with slope 3 meeting the y-axis at y = −6. [2] (b) y = x^2 + 6x + 8 is a quadratic curve, which is U-shaped, crossing the y-axis at y = 8 and the x-axis at x = − 2 , x = −4. The curve is symmetric about the line x = − 6 /2 = −3. The vertex is at x = − 3 , y = −1. [3] (c) y = |x^2 + 6x + 8| is given from (b) by reflecting the part below the x-axis in the x-axis. [2]
- Put y = 2 x + 7 1 − 3 x
, and solve for x in terms of y = f (x). Then y(1 − 3 x) = 2x + 7 so
y − 3 xy = 2x + 7, giving x(3y + 2) = y − 7.
Thus x = f −^1 (y) = y − 7 3 y + 2
and so f −^1 (x) = x − 7 3 x + 2
. [3]
- (a) Either use the formula a 1 − rn 1 − r with r = − 5 , a = − 5 , n = 5, giving − 5 ×
(−5)^5 − 1
−2605 or simply add up the 5 terms. [3]
(b) The formula is a 1 − r
. [1]
Here a = 107 , r = 107 , giving the sum as
7 10 1 − 107
7 10 3 10
. [2]
- (a) (^) nlim→∞ 2 n^2 − 7 n 5 n^2 + 6n − 1 = limn→∞ 2 − (^7) n 5 + (^6) n − (^) n^12
as n → ∞. [2]
(b) Putting x = 4 in the bottom of the fraction gives 0, as also in the top. Factorise to write
x^2 − x − 12 x^2 − 16
(x − 4)(x + 3) (x − 4)(x + 4)
x + 3 x + 4
Now put x = 4 to get the limit 78. [2]
- (a) Put u = 3x − 8. Then y = (3x − 8)^7 = u^7 so
dy dx
dy du
du dx = 7u^6 × 3 = 21(3x − 8)^6. [2]
(b) Put u = x^4 + 3. Then y = (x^4 + 3) (^54) = u (^54) and dy dx
u
(^14) × du dx
u
(^14) × 4 x^3 = 5x^3 (x^4 + 3)
(^14)
. [3]
(c) y = x^8 cos x = uv with u = x^8 , v = cos x.
dy dx = u dv dx
- v du dx = x^8 (− sin x) + 8x^7 cos x = −x^8 sin x + 8x^7 cos x. [3]
- The slope of the tangent is the value of dy dx at x = −1. Now dy dx = 12x^3 , so the slope is
−12. When x = −1 we have y = 5, so the tangent line has equation y − 5 = −12(x + 1), giving y = − 12 x − 7. [3]
- (a)
∫ (6 − 3 x^5 − sin x) dx = 6
∫ 1 dx − 3
∫ x^5 dx −
∫ sin x dx = 6x −
x^6 + cos x + C [4]
(b)
∫ e−^5 x^ dx =
e−^5 x^ + C = −
e−^5 x^ + C. [2]
- (a)
∫ (^) π/ 14 0
cos 7x dx =
[ 1
sin 7x
]π/ 14 0
sin (π/2) −
sin 0 =
. [3]
(b) Substitute u = 5x − 14 so that du = 5dx. Then
I =
∫ (^6) 3
5 x − 14 dx =
∫ (^) x= x=
u du =
∫ (^) u= u=
u du = [ln u]^161 = ln 16 − ln 1 = ln 16 (^) [3]
- (i) Differentiate the LHS to get
3 × 3 x^2 + (x × 2 y dy dx
[4]
(iv) The total area bounded by the curve and the x-axis is made up of two pieces, one between∣ x = −4 and x = 0, and the other between x = 0 and x = 3. These are found as ∣∣ ∣
∫ (^0) − 4
f (x) dx
∣∣ ∣∣ and
∣∣ ∣∣
∫ (^3) 0
f (x) dx
∣∣ ∣∣. Now ∫^ f (x) dx = ∫^ (x^3 + x^2 − 12 x) dx = x
4 4
x^3 − 6 x^2 , giving
the first area as | − 64 + 64/3 + 96| = 160/3 and the second as | 81 /4 + 9 − 54 | = 99/4 making a total of 937/12 = 78.08. [4]
- (a) Substitute u = x^6 + 4. Then du = 6x^5 dx, so
I =
∫ x^5 cos(x^6 + 4) dx =
cos u du =
sin u + C =
sin(x^6 + 4) + C. [4]
(b) Substitute t = tan x. Then dt = sec^2 x dx, so
I =
∫ tan^8 x sec^2 x dx =
∫ t^8 dt =
t^9 9
+ C =
tan^9 x + C. [4]
(c) Substitute x = 25 sin t. Then dx = 25 cos t dt. Therefore
I =
∫ √ 2 / 5 0
√^ dx 4 − 25 x^2
∫ (^) x=√ 2 / 5 x=
2 √^5 cos^ t 4 − 4 sin^2 t
dt =
∫ (^) x=√ 2 / 5 x=
2 √^5 cos^ t 4 cos^2 t
dt =
∫ (^) x=√ 2 / 5 x=
dt = [t]x=
√ 2 / 5 x=
Now when x = 0 we have t = 0 and when x =
2 /5 we have
√ 2 5 =^
2 5 sin^ t^ so that sin^ t^ =^
√ 2 2 and t = π 4. Consequently I =
[t]t t==0π/ 4 =
π 4
π 20
= 0. 16. [7]
