AH Maths 2004
#1
Posted 21 May 2004 - 03:44 PM
These are unofficial solutions subject to checking. There may be typing errors for example. Let me know if you find errors.
#2
Posted 21 May 2004 - 05:23 PM
e^tanx(1-sin2x)
0
(4 marks)
Q1 b
(2-8x invtan(2x))/(1+4x^2)^2
(3 marks)
#3
Posted 21 May 2004 - 05:41 PM
4a^8 -12a^6 +54a&4 -108a^2 +81
(3 marks)
#4
Posted 21 May 2004 - 05:42 PM
-cot(theta)
y+x=0.5*root 10
(5 marks)
#5
Posted 21 May 2004 - 05:44 PM
-10+20i
Proof
-5, 1+2i, 1-2i
(6 marks)
#6
Posted 21 May 2004 - 05:47 PM
Very odd question for the second part.
0.2/(x-3) - 0.2/(x+2)
-0.162 (did anyone else get this)
(6 marks)
#7
Posted 21 May 2004 - 05:50 PM
M1 = 0 -1
1 0
M2 = 1 0
0 -1
M2M1 = 0 -1
-1 0
Geometrical effect:
#8
Posted 21 May 2004 - 05:51 PM
f(x) = x +x^2 + (1/3) x^3
(5 marks)
#9
Posted 21 May 2004 - 05:56 PM
Proof
x=-5
y=68
(4 marks)
#10
Posted 21 May 2004 - 06:01 PM
(1/root x +1)[1/(root x+1) -1]
(5 marks)
#11
Posted 21 May 2004 - 06:03 PM
Odd function
(3 marks)
#12
Posted 21 May 2004 - 06:04 PM
0.25Pi(1-e^-4) units^3
(5 marks)
#13
Posted 21 May 2004 - 06:05 PM
Proof by induction using derivatives!
(5 marks)
#14
Posted 21 May 2004 - 06:09 PM
x=-2
y=1
f'(x) = 0
5 = 0 therefore no stationary points
No f'(x) does not equal zero and there is no change in concavity
x=1
y=-2
x member of set of real numbers, x does not equal one
OR x>1 and x<1
#15
Posted 21 May 2004 - 06:14 PM
x=-2
y=1
f'(x) = 0
5 = 0 therefore no stationary points
YES THERE IS A POINT OF INFLEXION!
x=1
y=-2
x member of set of real numbers, x does not equal one
#16
Posted 21 May 2004 - 06:15 PM
A)
2X+3Y+Z=5
51.9 degrees
(3,5,8)
(10 marks)
#17
Posted 21 May 2004 - 06:19 PM
a)
y=x^4+cx^3
y=x^4+x^3
y= root of (.4x^2 +3x^2 +0.6)
(11 marks)
#18
Posted 21 May 2004 - 06:20 PM
a)
544
11
c)
a=12/11
n=7
(10 marks)
#19
Posted 22 May 2004 - 09:30 AM
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