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IGCSE Further Pure Mathematics Revision Guide

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REVISION GUIDE
Pearson
Edexcel
IGCSE
Further Pure
Mathematics
Surds
1. √π‘Ž π‘₯ √𝑏 = √π‘Žπ‘
2. √π‘Ž ÷ √𝑏 = √
π‘Ž
𝑏
3. √a / √b and √a - √b are conjugate surds. The product of conjugate surds is a rational
number.
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Indices
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1. Am x an = a m + n
2. am / an = am – n
6. π‘Ž = n√a
n
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π‘Žπ‘›
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1
𝑛
π‘š
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3. (am)n = amn
4. a0 = 1
1
5. a –n =
m
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7. π‘Ž 𝑛 = √a = ( √a)
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1. log a + log b = logab
π‘Ž
2. log a – log b = log
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Logarithms
𝑏
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3. a log x y = log x y a
4. log a a = 1
5.
logaπ‘₯
log 𝑏π‘₯
=
log π‘π‘Ž
6. loga1 = 0
1
7. log a b = log 𝑏 a
Binomial theorem
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1. (x + y)n = xn + (nc1 * xn-1 * y)+ (nc2 * xn-2 * y2n)+ ( nc3 xn-3 y3 ) + ( nc4 xn-4 y4 ) ………………………
2. (r+1)th term = ( ncr xn-r yr )
Quadratic Equation
Nature of roots
ax2 + bx + c = 0
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1. If b2 – 4ac > 0, roots are real & different / real and distinct and the curve y = ax2 + bx + c
will cut the x axis at two real and distinct points
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2. If b2 – 4ac < 0, roots are not real/ imaginary / complex and the curve y = ax2 + bx + c will
lie entirely above the x axis if a > 0 and entirely below the x axis if a < 0.
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3. If b2 – 4ac = 0, roots are real and equal / repeated / coincident and the curve y = ax2 + bx
+ c touches the x-axis.
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4. If b2 – 4ac ≥ 0, roots are real.
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Solving Quadratic Inequality
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If π‘Žπ‘₯2 + 𝑏π‘₯ + 𝑐 > 0, range of values of π‘₯: π‘₯ < 𝛼, π‘₯ > 𝛽
If π‘Žπ‘₯2 + 𝑏π‘₯ + 𝑐 ≥ 0, range of values of π‘₯: π‘₯ ≤ 𝛼, π‘₯ ≥ 𝛽
If π‘Žπ‘₯2 + 𝑏π‘₯ + 𝑐 < 0, range of values of π‘₯: 𝛼 < π‘₯ < 𝛽
If π‘Žπ‘₯2 + 𝑏π‘₯ + 𝑐 ≤ 0, range of values of π‘₯: 𝛼 ≤ π‘₯ ≤ 𝛽
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1.
2.
3.
4.
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When α and β (α<β) are two roots of π‘Žπ‘₯2 + 𝑏π‘₯ + 𝑐 = 0 (a>0) and
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1. If α and β are two roots of π‘Žπ‘₯2 + 𝑏π‘₯ + 𝑐 = 0,
𝑏
a. 𝛼 + 𝛽 = −
b. 𝛼𝛽 =
𝑐
π‘Ž
π‘Ž
2. If two roots of an unknown equation is given and you want to find the equation, follow the
following steps:
a. Find the sum of the roots.
b. Find the product of the root.
c. Use the following formula, π‘₯2 − (π‘ π‘’π‘š π‘œπ‘“ π‘Ÿπ‘œπ‘œπ‘‘π‘ ) × π‘₯ + π‘π‘Ÿπ‘œπ‘‘π‘’π‘π‘‘ π‘œπ‘“ π‘Ÿπ‘œπ‘œπ‘‘π‘  = 0
d. Simplify the equation if needed.
3. ∝2+ 𝛽2 = (𝛼 + 𝛽)2 − 2𝛼𝛽
4. (𝛼 − 𝛽)2 = (𝛼 + 𝛽)2 − 4𝛼𝛽
5. 𝛼3 + 𝛽3 = (𝛼 + 𝛽)3 − 3𝛼𝛽(𝛼 + 𝛽)
6. 𝛼3 − 𝛽3 = (𝛼 − 𝛽)3 + 3𝛼𝛽(𝛼 − 𝛽)
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7. (𝛼 + 𝛽)3 = 𝛼3 + 𝛽3 + 3𝛼𝛽(𝛼 + 𝛽)
Set, Relation and Function
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1. A set is any well-defined collection, list or class of objects. It may be given by either listing its
members or defining its properties clearly.
2. A relation is a set of ordered pairs.
3. A function is a relation in which every elements in the domain has a unique image in the range.
4. The range of a function f is the set of values f(x) for the given domain.
5. Two functions f and g can be combined to produce composite functions fg or gf such that
fg(x) = f(g(x)) and gf(x) = g(f(x)).
In general, fg and gf are different functions.
6. The absolute value of x, written as |x|, is defined as
x if x > 0
|x| = 0 if x = 0
-x if x < 0
Circular Measure
1. ∏ radian = 1800
2. For a sector of a circle enclosed by two radii that subtend an angle of θ radians at the
centre, the arc length s is given by
s=rθ
and the area of the sector A is given by
1
A = π‘Ÿ2 θ
2
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where r is the radius of the circle.
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Trigonometry
π‘œπ‘π‘
1. Sin θ = β„Žπ‘¦π‘
π‘Žπ‘‘j
2. Cos θ = β„Žπ‘¦π‘
π‘œπ‘π‘
3. Tan θ = π‘Žπ‘‘j
1
4. Sec x = cos π‘₯
1
5. Cosec x = sin π‘₯
1
6. Cot x = tan π‘₯
cos π‘₯
7. Cot x = sin π‘₯
8. Sin2 + cos2 x = 1
9. sec2 x - tan2 x =1
10. cosec2 - cot2 x = 1
11. Sec2 x = 1 + tan2 x
12. Sin(A + B) = sin A cos B + cos A sin B
13. Cos(A+B) = cos A cos B – sin A sin B
tan 𝐴 + tan 𝐡
14. Tan (A+B) = 1 − tan 𝐴 tan 𝐡
15. Sin 2A = 2 sin A cos A
16. Cos 2A = cos2 A – sin2 A = 1 -2sin2 A = 2 cos2 A -1
2 tan 𝐴
17. Tan 2A = 1−tan2 𝐴
sin 𝐴
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18. Tan A = cos 𝐴
Arithmetic Progression (A.P)
1. Nth term = a + (n-1)d
𝑛
2. Sn = {2π‘Ž + (𝑛 − 1)𝑑
2
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,r<1
# -1 < r < 1 or |r| < 1.
The series is convergent. It has sum to infinity.
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π‘Ž
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1−π‘Ÿ
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3. Sn =
π‘Ÿ−1
π‘Ž (1−π‘Ÿπ‘›)
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2. Sn =
o.
1. Nth term = arn-1
π‘Ž (π‘Ÿπ‘›−1)
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Geometric Progression (G.P)
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1. Sα = 1−π‘Ÿ
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Otherwise the series is divergent. It has does not have sum to infinity.
Co – ordinate Geometry
1. The distance between two points A(x1 , y1) and B(x2 , y2)𝑦is
√{(π‘₯2 − π‘₯1) + (𝑦2 − 𝑦1)}
2−𝑦1
2. The gradient of the line joining A(x y ) and B(x y ) is
1, 1
2, 2
π‘₯2−π‘₯1
3. The coordinates of the mid-point of the line joining A(x1 , y1) and B(x2 , y2) are
𝑦 +𝑦
π‘₯1 + π‘₯2
) , ( 1 2) .
2
2
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𝑇+π‘ˆ
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𝑏=
(π‘ˆ × π‘₯1) + (𝑇 × π‘₯2)
𝑇+π‘ˆ
(π‘ˆ × π‘¦1) + (𝑇 × π‘¦2)
rd
π‘Ž=
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4. Finding coordinates when a point divides a line internally.
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5. The equation of the straight line having a gradient m and passing through the point (x1,
y2) is given by : y – y1 = m (x – x1).
6. Two lines are parallel if their gradients are equal.
7. Two lines are perpendicular to each other if the product of their gradients is -1.
Differentiation
1. For a curve 𝑦 = 𝑓(π‘₯) represents the gradient of the tangent to the curve at any point π‘₯.
𝑑𝑦
2. If 𝑦 = π‘Žπ‘₯𝑛, then = π‘Žπ‘›π‘₯𝑛−1, where π‘Ž and n are constants.
3.
𝑑
(𝑒 + 𝑣) =
𝑑π‘₯
𝑑𝑒
𝑑π‘₯
𝑑𝑣
+
𝑑π‘₯
𝑑π‘₯
4. If 𝑦 is a function of 𝑒, and 𝑒 us a function of π‘₯, then
5. If y, 𝑒 and v are functions of π‘₯ and 𝑦 = 𝑒𝑣, then
𝑑𝑦
𝑑𝑦
=
𝑑π‘₯
=𝑒
𝑑π‘₯
𝑑𝑦
×
𝑑𝑒
𝑑𝑒
𝑑𝑣
𝑑π‘₯
𝑑𝑒
𝑑π‘₯
𝑑π‘₯
+𝑣
𝑑𝑣
𝑣𝑑𝑒−
𝑒
𝑒
𝑑π‘₯
(quotient rule).
𝑑π‘₯
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(cos π‘₯) = − sin π‘₯
(tan π‘₯) = sec2 π‘₯
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9.
𝑑π‘₯
𝑑
(sin π‘₯) = cos π‘₯
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8.
𝑑
𝑑π‘₯
𝑑
15.
16.
17.
18.
𝑑π‘₯
𝑑
𝑑π‘₯
𝑑
𝑑π‘₯
𝑑
𝑑π‘₯
𝑑
𝑑π‘₯
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(π‘Ž sin 𝑏π‘₯) = π‘Žπ‘ cos 𝑏π‘₯
by
𝑑π‘₯
𝑑
(tan𝑛 π‘₯) = 𝑛 tan𝑛−1 π‘₯ (sec2 π‘₯)
(π‘Ž cos 𝑏π‘₯) = −π‘Žπ‘ sin 𝑏π‘₯
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14.
𝑑π‘₯
𝑑
(cos𝑛 π‘₯) = 𝑛 cos𝑛−1 π‘₯ (− sin π‘₯)
(π‘Ž tan 𝑏π‘₯) = π‘Žπ‘ sec2 𝑏π‘₯
ct
13.
𝑑π‘₯
𝑑
(sin𝑛 π‘₯) = 𝑛 sin𝑛−1 π‘₯ (π‘π‘œπ‘ π‘₯)
[sin(π‘Žπ‘₯ + 𝑏)] = π‘Ž cos(π‘Žπ‘₯ + 𝑏)
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12.
𝑑π‘₯
𝑑
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11.
𝑑
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Other formulae
10.
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The following are true only when π‘₯ is in radians:
7.
(product rule).
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𝑑𝑦
6. If 𝑦, 𝑒 and 𝑣 are functions of π‘₯ and 𝑦 = , then
= 𝑑π‘₯
𝑣2
𝑑π‘₯
𝑣
(chain rule).
[cos(π‘Žπ‘₯ + 𝑏)] = −π‘Ž sin(π‘Žπ‘₯ + 𝑏)
[tan(π‘Žπ‘₯ + 𝑏)] = π‘Ž sec2(π‘Žπ‘₯ + 𝑏)
Application of Differentiation
19. Stationary points or turning points of a function 𝑦 = 𝑓(π‘₯) occur when
20. The second derivative (
(a) If
𝑑2𝑦
𝑑π‘₯2
𝑑2𝑦
𝑑 2𝑦
𝑑π‘₯ 2
𝑑𝑦
𝑑π‘₯
= 0.
) determines the nature of the stationery points:
is negative, the stationery point is a maximum point.
(b) If 𝑑π‘₯2 is positive, the stationary point is a minimum point.
𝑑2𝑦
(c) If 𝑑π‘₯2 is zero, the nature of the stationery point depends on how the value of
𝑑𝑦
𝑑π‘₯
changes
near the stationary point:
𝑑𝑦
(i) if the sign 𝑑π‘₯ does not change before and after the turning point, the point is a point of
inflexion.
𝑑𝑦
(ii) if the value of 𝑑π‘₯ changes from positive to negative after the turning point, the point is a
maximum point.
(iii) if the value of
𝑑𝑦
changes from negative to positive after the turning point, the point is a
𝑑π‘₯
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minimum point.
21. To sketch a curve, note
(i) the points where π‘₯ = 0 or 𝑦 = 0
(ii) the nature and position of the stationary points
(iii) the direction of the curve as π‘₯ and 𝑦 approach infinity.
(iv) the interval on which the gradient is positive or negative.
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Further Applications of Differentiation
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22. If 𝑦 = 𝑓(π‘₯) and 𝛿π‘₯ and 𝛿𝑦 are small increments in π‘₯ and 𝑦 respectively, then 𝛿𝑦 =
𝑑𝑦
× π›Ώπ‘₯
𝑑π‘₯
𝑑2𝑠
by 𝑑𝑑 and acceleration π‘Ž is given by 𝑑𝑑 or 𝑑𝑑2.
rd
𝑑𝑣
rlo
𝑑𝑠
s.
23. If the displacement of a body is given by 𝑠 = 𝑓(𝑑), where 𝑑 is the time, then velocity v is given
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24. When a body is instantaneously at rest, its velocity is zero.
𝑑𝑦
25. If a curve is given by 𝑦 = 𝑓(π‘₯), then the gradient of the tangent to the curve is given and the
1
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gradient of the normal is − 𝑑𝑦 .
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𝑑π‘₯
𝑑π‘₯
Integration
1. J π‘Žπ‘₯𝑛 𝑑π‘₯ =
π‘Žπ‘₯𝑛+1
𝑛+1
2. J (π‘Žπ‘₯ + 𝑏)𝑛𝑑π‘₯ =
3.
4.
5.
6.
+𝑐
𝑛 ≠ −1
(π‘Žπ‘₯+𝑏)^(𝑛+1)
(𝑛+1)π‘Ž
+𝑐
J cos π‘₯ 𝑑π‘₯ = sin π‘₯ + 𝑐
J sin π‘₯ 𝑑π‘₯ = −π‘π‘œπ‘ π‘₯ + 𝑐
J sec2 π‘₯ 𝑑π‘₯ = tan π‘₯ + 𝑐
1
J cos 𝑏π‘₯ 𝑑π‘₯ = sin 𝑏π‘₯ + 𝑐
𝑏
1
7. J sin 𝑏π‘₯ 𝑑π‘₯ = − cos 𝑏π‘₯ + 𝑐
1
9. J cos(π‘Žπ‘₯ + 𝑏) 𝑑π‘₯ =
π‘Ž
m
𝑏
𝑏
tan 𝑏π‘₯ + 𝑐
co
1
sin(π‘Žπ‘₯ + 𝑏) + 𝑐
o.
8. J sec2 𝑏π‘₯ 𝑑π‘₯ =
1
π‘Ž
ez
π‘Ž
1
tan(π‘Žπ‘₯ + 𝑏) + 𝑐
s.
11. J sec2(π‘Žπ‘₯ + 𝑏) 𝑑π‘₯ =
yr
10. J sin(π‘Žπ‘₯ + 𝑏) 𝑑π‘₯ = − cos(π‘Žπ‘₯ + 𝑏) + 𝑐
rd
12. The area bounded by the curve 𝑦 = 𝑓(π‘₯), the π‘₯-axis and the lines π‘₯ = π‘Ž and π‘₯ = 𝑏 is given by
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∫π‘Ž 𝑦 𝑑π‘₯.
𝑏
∫π‘Ž π‘₯ 𝑑𝑦.
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13. The area bounded by the curve π‘₯ = 𝑓(𝑦), the 𝑦-axis and the lines 𝑦 = π‘Ž and 𝑦 = 𝑏 is given by
𝑏
14. Area between 𝑔(π‘₯) and 𝑓(π‘₯) = ∫π‘Ž |𝑔(π‘₯) − 𝑓(π‘₯)| 𝑑π‘₯
15. When the area bounded by 𝑦 = 𝑓(π‘₯), the π‘₯-axis and the lines π‘₯ = π‘Ž and π‘₯ = 𝑏 is rotated
𝑏
through 360o about the π‘₯-axis, the volume of solid of revolution is given by πœ‹ ∫π‘Ž 𝑦2 𝑑π‘₯.
16. When the area bounded by 𝑦 = 𝑓(π‘₯), the 𝑦-axis and the lines 𝑦 = π‘Ž and 𝑦 = 𝑏 is rotated
𝑏
through 360o about the 𝑦-axis, the volume of solid of revolution is given by πœ‹ ∫π‘Ž π‘₯2 𝑑𝑦.
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17. If the velocity v of a particle is given as a function of time t, i.e. 𝑣 = 𝑓(𝑑), then ∫ 𝑓(𝑑) 𝑑𝑑 will
give the expression for the distance covered.
18. If the acceleration α of a particle is given as function of time t, i.e. α=f(t), then ∫ 𝑓(𝑑) 𝑑𝑑 will give
the expression for the velocity.
Vector
1. A vector quantity possesses both magnitude and direction. The vector −→ has a
0𝐴
magnitude or modulus |−→| and its direction is from O to A.
0𝐴
2. If a fixed point O is taken as a origin, the vector −→ is known as the position vector of A
0𝐴
m
𝐴𝐡
𝑃Q
s.
6.
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5.
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3.
4.
with references to O.
Two vectors are equal if they have the same magnitude and direction.
Vector addition: If two vectors acting at a point are represented both in magnitude and
direction by the adjacent sides of a parallelogram, then their sum is represented both in
magnitude and direction by the diagonal of the parallelogram through the point.
a = b βž” |a| = |b| and a is parallel to b.
ha = kb βž” |ha| =|hb| and a is parallel to b.
or, h = k = 0 if a is not parallel to b.
If −→ = ha + kb and −→ = ma + nb, where h, k, m and n are constants, and a is not parallel
𝑃Q
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𝐴𝐡
rd
to b, then −→ = −→ → β„Ž = m and k = n.
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7. If the position vectors of A and B relative to an origin O are a and b respectively, the
(π‘Ž+𝑏)
position vector of the mid point (M) of AB is −→= π‘š =
.
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8. The scalar product of two vectors a and b is given by
a-b = |a||b| cos θ
where θ is the angle between the two vectors.
9. Commutative law: a x b = b x a
10. Distributive law: a(b+c) = a x b + a x c
11. If a = x1i + y1j and b = x2i + y2j, then
a x b = x1x2 + y1y2
2
Binomial Expansion
1. 𝑛! = 𝑛(𝑛 − 1)(𝑛 − 2)(𝑛 − 3) …
• eg: 4! = 4 * 3 * 2 * 1 = 24
𝑛!
2. (𝑛−2)! = 𝑛(𝑛−1)(𝑛−2) = 𝑛(𝑛 − 1)
(𝑛−2)
3. 𝑛𝑐1 = 𝑛
4. 𝑛𝑐2 =
5. 𝑛𝑐3 =
2!
𝑛(𝑛−1)(𝑛−2)(𝑛−3)
3!
(π‘Ž + π‘₯)𝑛 = π‘Žπ‘› +
𝑛𝑐 1 * π‘Žπ‘›−1 * π‘₯ + 𝑛𝑐 2 * π‘Žπ‘›−2 * π‘₯2 + 𝑛𝑐 3 * π‘Žπ‘›−3 * π‘₯3 + …
m
6.
𝑛(𝑛−1)
2!
yr
1
2
ez
{4(1 + )}
s.
π‘₯2
1
1
➒
= 42 * (1 +
π‘₯2 2
4
1
π‘₯2 2
➒
= 2 (1 + )
lle
ct
ed
by
w
w
w
4
co
rd
4
)
rlo
(4 + π‘₯2 )2 =
pe
1
8.
3!
[In above formula the value of ‘a’ must be ‘1’ ]
.p
a
•
o.
co
• [Note: Above formula cannot be used when ‘n’ is negative or in fraction]
𝑛(𝑛−1) 2
𝑛(𝑛−1)(𝑛−2) 3
7. (π‘Ž + π‘₯)𝑛 = 1 + 𝑛π‘₯ +
π‘₯ +
π‘₯ + …
Graph Sketching
𝑦 = −π‘₯
𝑦=π‘₯+1
𝑦 = −π‘₯ + 1
co
𝑦 = π‘₯2 + 1
lle
ct
ed
by
w
w
w
.p
a
pe
rlo
rd
s.
ez
yr
o.
co
m
𝑦=π‘₯
𝑦 = π‘₯2 - 1
𝑦 = (π‘₯ + 1)2
m
𝑦 = π‘₯2
by
w
w
w
.p
a
pe
rlo
rd
s.
ez
yr
o.
co
𝑦 = (π‘₯ − 1)2
co
lle
ct
ed
π‘Œ = 𝑒 π‘₯ ; 𝑦 = 𝑒−π‘₯; 𝑦 = − 𝑒−π‘₯; 𝑦 = − 𝑒π‘₯
π‘Œ = 𝑒π‘₯ + 1
𝑦 = − π‘₯3
by
w
w
w
.p
a
pe
rlo
rd
s.
ez
yr
o.
𝑦 = π‘₯3
co
m
𝑦 = 𝑒π‘₯ - 1
co
lle
ct
ed
𝑦 = π‘₯3 + 1
𝑦 = π‘₯3 - 1
𝑦 = −π‘₯3 − 1
π‘Œ = (π‘₯ + 1)3
π‘Œ = (π‘₯ - 1)3
co
lle
ct
𝑦 = (π‘₯ + 1)(π‘₯ + 4)(π‘₯ − 2)
ed
by
w
w
w
.p
a
pe
rlo
rd
s.
ez
yr
o.
co
m
𝑦 = −π‘₯3 + 1
𝑦 = (π‘₯ − 1)(π‘₯ − 3)(−π‘₯ + 2)
ed
ct
lle
co
by
yr
ez
s.
rd
rlo
pe
.p
a
w
w
w
m
co
o.
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