GCSE_AQA_Further_Maths_Revision_Answers

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Further Maths Revision ANSWERS
1*
2*
Simplify by rationalising
Rationalise
1
√3
√3
3
2√3
5+√2
3
Factorise x² - 64
4
Factorise 3x² + 8x - 3
10√3 − 2√6
23
(x+8)(x-8)
(3x-1)(x+3)
5
Factorise fully x³ - 4x
X(x²-4) = x(x+2)(x-2)
6
7
Make h the subject
A = πr²+2πrh
Make x the subject
h=
y=
10
11
2−𝑦
3𝑦−1
Prove that
(y+6)(y+3) - y² is a multiple of 9
When y> 0
9
2πœ‹π‘Ÿ
π‘₯+2
1+3π‘₯
X=
8
𝐴−πœ‹π‘Ÿ²
Simplify
3π‘Ž²π‘
2𝑐
x
9y+18 = 9(y+2)
4𝑐³
9π‘Žπ‘
Find a and b such that
x² + 8x + 10 ≡ (π‘₯ + π‘Ž)² + b
f(x) = x² + 5x -1 work out
i) f(-3) ii)f(3x) iii)f(x-2)
2
π‘Žπ‘²
3
(x+4)² - 6 a=4 b=-6
i)
ii)
iii)
-7
9x²+15x-1
x²+x-7
12
13
Work out the Range,
where Domain is −2 ≤ π‘₯ ≤ 2
f(x) = x²
0 ≤ 𝑓(π‘₯) ≤ 4
Calculate
i) the gradient between A(-3,7) B(2,-6)
ii) the equation of this line
iii)the gradient of the normal
−13
i)
ii)
𝑦=
iii)
14
Find the gradient & intercept of
2x + y – 4 = 0
15
Draw the graph of y = f(x) where
f(x) = 4
−4 ≤ π‘₯ ≤ −2
= x²
−2 < π‘₯ < 2
= 8 – 2x
2≤π‘₯ ≤4
16
Solve the following, using
−𝑏 ± √𝑏 2 − 4π‘Žπ‘
π‘₯=
2π‘Ž
5
−13
5
5
x-
13
m = -2 c = 4
a=3 b=5 c=1
3x² + 5x + 1 = 0
x = -0.23 or -1.4
17
Solve
2x + y = 8
5x + 2y = 21
X = 5 y = -2
4
5
18
Expand
(π‘₯ − 1)(π‘₯ + 1)(π‘₯ + 3)
x³ + 3x² - x -3
19
20
21*
22
Given that
f(x) = x³+x²-4x-4
i)
show (x+1) is a factor
ii)
factorise f(x)
iii)
solve f(x) = 0
i)
ii)
iii)
x=-1 f(x)=0
(x+1)(x-2)(x+2)
X=-1, 2, -2
Solve
Y – 4 > 3y – 12
Y<4
3
Solve
π‘₯ 2 = 27
3
(√27)² = 9
Find the nth term of 3 6 13 24 39
2n²-3n+4
23
The nth term of a sequence is
i)
ii)
𝑛+1
2𝑛+1
work out the first 3 terms
find the limiting value as n→ ∞
i)
ii)
2
3
1
2
,
3 4
,
5 7
24
A straight line DEF has points
D(4,3) E(8,5) DE:EF is 2:3
Find co-ordinates of F
F ( 14, 8 )
25
(x-3)² + (y+2)² = 36
find the Centre &
Radius of the circle
Centre (3, -2) r = 6
26
Work out x
Use angle in a semi-circle = 90°
So is 90-x
Opp angle = 112°
∴ 2π‘₯ − 60 + 112 + (90 − π‘₯) = 180°
AB is a diameter
X = 38°
27
Prove that triangle ABD is isosceles
Using alt. Segment theory
ABD = BCD
As AB=BC (isosceles)
BAD = BCD
∴ 𝐡𝐴𝐷 = 𝐴𝐡𝐷 (π‘–π‘ π‘œπ‘ π‘π‘’π‘™π‘’π‘ )
AB is a tangent at B
ADC is a straight line AB = BC
28*
Demonstrate cos 60 = 0.5
sin 60 =
√3
2
2
√3
60°
1
cos 60 =
29
1
2
sin 60 =
√3
2
Find x
X = 4.6
Use
π‘Ž
𝑏
𝑐
=
=
𝑆𝑖𝑛𝐴
𝑆𝑖𝑛𝐡
𝑆𝑖𝑛𝐢
30
Find θ
Angle A = 41.4°
Use
π‘Ž2 = 𝑏 2 + 𝑐 2 − 2π‘π‘πΆπ‘œπ‘ π΄
31
Solve for 0≤ π‘₯ ≤ 360
i)tan x = √3
ii)3Cos π‘₯ = 2
iii)√2 πΆπ‘œπ‘  π‘₯ − 1 = 0
32
X= 60°, 240°
X = 48°, 312°
X = 45°, 315°
Solve for 0≤ πœƒ ≤ 360
2Sinπœƒ + Cosπœƒ = 0
π‘†π‘–π‘›πœƒ
Use [πΆπ‘œπ‘ πœƒ = Tanπœƒ]
33
i)
ii)
iii)
p.v = -26.6° θ = 333.4° , 153.4°
Re-write the equation as a quadratic
with only one trigonometric function
Sin²πœƒ − 2πΆπ‘œπ‘ πœƒ + 1 = 0
Use
[𝑆𝑖𝑛2 πœƒ + πΆπ‘œπ‘  2 πœƒ = 1]
Cos²θ + 2Cosθ – 2 = 0
34
Prove the identity
1 − πΆπ‘œπ‘  2 πœƒ
≡ π‘‡π‘Žπ‘›2 πœƒ
1 − 𝑆𝑖𝑛2 πœƒ
𝑠𝑖𝑛2 πœƒ
π‘π‘œπ‘ 2 πœƒ
Use
35
37
38
π‘ π‘–π‘›πœƒ
π‘π‘œπ‘ πœƒ
)² = (tanθ)² = tan²θ
[𝑆𝑖𝑛2 πœƒ + πΆπ‘œπ‘  2 πœƒ = 1]
Differentiate
i) y=3x⁴
ii)y=4x³ + 2x – 5
i)
ii)
36
=(
Find the gradient of
Y=x³+2x²+4 at a tangent at A(3,2)
𝑑𝑦
𝑑π‘₯
𝑑𝑦
𝑑π‘₯
𝑑𝑦
𝑑π‘₯
= 12π‘₯³
= 12x² + 2
= 3π‘₯² + 4π‘₯ when x =3
Is 39
Y=x²-7 find the co-ordinates when
gradient = 6
(3,2)
The sketch shows y=5x-x²
Find
𝑑𝑦
i)the gradient function 𝑑π‘₯
ii)the gradient of curve at P
iii)the equation of the normal at P
i)
ii)
iii)
5-2x
Gradient = -1
Y = -x + 9
39
2
𝐴= (
1
3
2
) B=( )
1
−1
i)
Find i) 4A ii) AB
iii) Write the Identity matrix
ii)
8 12
(
)
4 4
1
( )
1
I=
(
40
Sketch the graph
y=Sin πœƒ
1
0
0
1)
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