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4PM1
REVISION GUIDE
Pearson
Edexcel
IGCSE
Further Pure
Mathematics
Surds
1. √π π₯ √π = √ππ
2. √π ÷ √π = √
π
π
3. √a / √b and √a - √b are conjugate surds. The product of conjugate surds is a rational
number.
m
Indices
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1. Am x an = a m + n
2. am / an = am – n
6. π = n√a
n
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ππ
m
n
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1
π
π
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3. (am)n = amn
4. a0 = 1
1
5. a –n =
m
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7. π π = √a = ( √a)
ct
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1. log a + log b = logab
π
2. log a – log b = log
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Logarithms
π
co
lle
3. a log x y = log x y a
4. log a a = 1
5.
logaπ₯
log ππ₯
=
log ππ
6. loga1 = 0
1
7. log a b = log π 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
by
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If ππ₯2 + ππ₯ + π > 0, range of values of π₯: π₯ < πΌ, π₯ > π½
If ππ₯2 + ππ₯ + π ≥ 0, range of values of π₯: π₯ ≤ πΌ, π₯ ≥ π½
If ππ₯2 + ππ₯ + π < 0, range of values of π₯: πΌ < π₯ < π½
If ππ₯2 + ππ₯ + π ≤ 0, range of values of π₯: πΌ ≤ π₯ ≤ π½
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1.
2.
3.
4.
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When α and β (α<β) are two roots of ππ₯2 + ππ₯ + π = 0 (a>0) and
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α, β
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1. If α and β are two roots of ππ₯2 + ππ₯ + π = 0,
π
a. πΌ + π½ = −
b. πΌπ½ =
π
π
π
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 ππ ππππ‘π ) × π₯ + πππππ’ππ‘ ππ ππππ‘π = 0
d. Simplify the equation if needed.
3. ∝2+ π½2 = (πΌ + π½)2 − 2πΌπ½
4. (πΌ − π½)2 = (πΌ + π½)2 − 4πΌπ½
5. πΌ3 + π½3 = (πΌ + π½)3 − 3πΌπ½(πΌ + π½)
6. πΌ3 − π½3 = (πΌ − π½)3 + 3πΌπ½(πΌ − π½)
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7. (πΌ + π½)3 = πΌ3 + π½3 + 3πΌπ½(πΌ + π½)
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 = π2 θ
2
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where r is the radius of the circle.
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Trigonometry
πππ
1. Sin θ = βπ¦π
ππj
2. Cos θ = βπ¦π
πππ
3. Tan θ = ππj
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 π΄ + tan π΅
14. Tan (A+B) = 1 − tan π΄ tan π΅
15. Sin 2A = 2 sin A cos A
16. Cos 2A = cos2 A – sin2 A = 1 -2sin2 A = 2 cos2 A -1
2 tan π΄
17. Tan 2A = 1−tan2 π΄
sin π΄
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18. Tan A = cos π΄
Arithmetic Progression (A.P)
1. Nth term = a + (n-1)d
π
2. Sn = {2π + (π − 1)π
2
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s.
,r<1
# -1 < r < 1 or |r| < 1.
The series is convergent. It has sum to infinity.
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π
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1−π
,r>1
rlo
3. Sn =
π−1
π (1−ππ)
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2. Sn =
o.
1. Nth term = arn-1
π (ππ−1)
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Geometric Progression (G.P)
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1. Sα = 1−π
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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)π¦is
√{(π₯2 − π₯1) + (π¦2 − π¦1)}
2−π¦1
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
π¦ +π¦
π₯1 + π₯2
) , ( 1 2) .
2
2
(
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π+π
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π=
(π × π₯1) + (π × π₯2)
π+π
(π × π¦1) + (π × π¦2)
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π=
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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 π¦ = π(π₯) represents the gradient of the tangent to the curve at any point π₯.
ππ¦
2. If π¦ = ππ₯π, then = πππ₯π−1, where π and n are constants.
3.
π
(π’ + π£) =
ππ₯
ππ’
ππ₯
ππ£
+
ππ₯
ππ₯
4. If π¦ is a function of π’, and π’ us a function of π₯, then
5. If y, π’ and v are functions of π₯ and π¦ = π’π£, then
ππ¦
ππ¦
=
ππ₯
=π’
ππ₯
ππ¦
×
ππ’
ππ’
ππ£
ππ₯
ππ’
ππ₯
ππ₯
+π£
ππ£
π£ππ’−
π’
π’
ππ₯
(quotient rule).
ππ₯
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(cos π₯) = − sin π₯
(tan π₯) = sec2 π₯
s.
9.
ππ₯
π
(sin π₯) = cos π₯
rd
8.
π
ππ₯
π
15.
16.
17.
18.
ππ₯
π
ππ₯
π
ππ₯
π
ππ₯
π
ππ₯
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(π sin ππ₯) = ππ cos ππ₯
by
ππ₯
π
(tanπ π₯) = π tanπ−1 π₯ (sec2 π₯)
(π cos ππ₯) = −ππ sin ππ₯
ed
14.
ππ₯
π
(cosπ π₯) = π cosπ−1 π₯ (− sin π₯)
(π tan ππ₯) = ππ sec2 ππ₯
ct
13.
ππ₯
π
(sinπ π₯) = π sinπ−1 π₯ (πππ π₯)
[sin(ππ₯ + π)] = π cos(ππ₯ + π)
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12.
ππ₯
π
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11.
π
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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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ππ¦
6. If π¦, π’ and π£ are functions of π₯ and π¦ = , then
= ππ₯
π£2
ππ₯
π£
(chain rule).
[cos(ππ₯ + π)] = −π sin(ππ₯ + π)
[tan(ππ₯ + π)] = π sec2(ππ₯ + π)
Application of Differentiation
19. Stationary points or turning points of a function π¦ = π(π₯) occur when
20. The second derivative (
(a) If
π2π¦
ππ₯2
π2π¦
π 2π¦
ππ₯ 2
ππ¦
ππ₯
= 0.
) determines the nature of the stationery points:
is negative, the stationery point is a maximum point.
(b) If ππ₯2 is positive, the stationary point is a minimum point.
π2π¦
(c) If ππ₯2 is zero, the nature of the stationery point depends on how the value of
ππ¦
ππ₯
changes
near the stationary point:
ππ¦
(i) if the sign ππ₯ does not change before and after the turning point, the point is a point of
inflexion.
ππ¦
(ii) if the value of ππ₯ changes from positive to negative after the turning point, the point is a
maximum point.
(iii) if the value of
ππ¦
changes from negative to positive after the turning point, the point is a
ππ₯
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minimum point.
21. To sketch a curve, note
(i) the points where π₯ = 0 or π¦ = 0
(ii) the nature and position of the stationary points
(iii) the direction of the curve as π₯ and π¦ 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 π¦ = π(π₯) and πΏπ₯ and πΏπ¦ are small increments in π₯ and π¦ respectively, then πΏπ¦ =
ππ¦
× πΏπ₯
ππ₯
π2π
by ππ‘ and acceleration π is given by ππ‘ or ππ‘2.
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ππ£
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ππ
s.
23. If the displacement of a body is given by π = π(π‘), where π‘ is the time, then velocity v is given
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24. When a body is instantaneously at rest, its velocity is zero.
ππ¦
25. If a curve is given by π¦ = π(π₯), then the gradient of the tangent to the curve is given and the
1
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gradient of the normal is − ππ¦ .
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ππ₯
ππ₯
Integration
1. J ππ₯π ππ₯ =
ππ₯π+1
π+1
2. J (ππ₯ + π)πππ₯ =
3.
4.
5.
6.
+π
π ≠ −1
(ππ₯+π)^(π+1)
(π+1)π
+π
J cos π₯ ππ₯ = sin π₯ + π
J sin π₯ ππ₯ = −πππ π₯ + π
J sec2 π₯ ππ₯ = tan π₯ + π
1
J cos ππ₯ ππ₯ = sin ππ₯ + π
π
1
7. J sin ππ₯ ππ₯ = − cos ππ₯ + π
1
9. J cos(ππ₯ + π) ππ₯ =
π
m
π
π
tan ππ₯ + π
co
1
sin(ππ₯ + π) + π
o.
8. J sec2 ππ₯ ππ₯ =
1
π
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π
1
tan(ππ₯ + π) + π
s.
11. J sec2(ππ₯ + π) ππ₯ =
yr
10. J sin(ππ₯ + π) ππ₯ = − cos(ππ₯ + π) + π
rd
12. The area bounded by the curve π¦ = π(π₯), the π₯-axis and the lines π₯ = π and π₯ = π is given by
π
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∫π π¦ ππ₯.
π
∫π π₯ ππ¦.
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13. The area bounded by the curve π₯ = π(π¦), the π¦-axis and the lines π¦ = π and π¦ = π is given by
π
14. Area between π(π₯) and π(π₯) = ∫π |π(π₯) − π(π₯)| ππ₯
15. When the area bounded by π¦ = π(π₯), the π₯-axis and the lines π₯ = π and π₯ = π is rotated
π
through 360o about the π₯-axis, the volume of solid of revolution is given by π ∫π π¦2 ππ₯.
16. When the area bounded by π¦ = π(π₯), the π¦-axis and the lines π¦ = π and π¦ = π is rotated
π
through 360o about the π¦-axis, the volume of solid of revolution is given by π ∫π π₯2 ππ¦.
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17. If the velocity v of a particle is given as a function of time t, i.e. π£ = π(π‘), then ∫ π(π‘) ππ‘ 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 ∫ π(π‘) ππ‘ will give
the expression for the velocity.
Vector
1. A vector quantity possesses both magnitude and direction. The vector −→ has a
0π΄
magnitude or modulus |−→| and its direction is from O to A.
0π΄
2. If a fixed point O is taken as a origin, the vector −→ is known as the position vector of A
0π΄
m
π΄π΅
πQ
s.
6.
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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 β |a| = |b| and a is parallel to b.
ha = kb β |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
πQ
rlo
π΄π΅
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to b, then −→ = −→ → β = 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+π)
position vector of the mid point (M) of AB is −→= π =
.
0π
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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. π! = π(π − 1)(π − 2)(π − 3) …
• eg: 4! = 4 * 3 * 2 * 1 = 24
π!
2. (π−2)! = π(π−1)(π−2) = π(π − 1)
(π−2)
3. ππ1 = π
4. ππ2 =
5. ππ3 =
2!
π(π−1)(π−2)(π−3)
3!
(π + π₯)π = ππ +
ππ 1 * ππ−1 * π₯ + ππ 2 * ππ−2 * π₯2 + ππ 3 * ππ−3 * π₯3 + …
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6.
π(π−1)
2!
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1
2
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{4(1 + )}
s.
π₯2
1
1
β’
= 42 * (1 +
π₯2 2
4
1
π₯2 2
β’
= 2 (1 + )
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4
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4
)
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(4 + π₯2 )2 =
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1
8.
3!
[In above formula the value of ‘a’ must be ‘1’ ]
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• [Note: Above formula cannot be used when ‘n’ is negative or in fraction]
π(π−1) 2
π(π−1)(π−2) 3
7. (π + π₯)π = 1 + ππ₯ +
π₯ +
π₯ + …
Graph Sketching
π¦ = −π₯
π¦=π₯+1
π¦ = −π₯ + 1
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π¦ = π₯2 + 1
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π¦=π₯
π¦ = π₯2 - 1
π¦ = (π₯ + 1)2
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π¦ = π₯2
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π¦ = (π₯ − 1)2
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π = π π₯ ; π¦ = π−π₯; π¦ = − π−π₯; π¦ = − ππ₯
π = ππ₯ + 1
π¦ = − π₯3
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π¦ = π₯3
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π¦ = ππ₯ - 1
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π¦ = π₯3 + 1
π¦ = π₯3 - 1
π¦ = −π₯3 − 1
π = (π₯ + 1)3
π = (π₯ - 1)3
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π¦ = (π₯ + 1)(π₯ + 4)(π₯ − 2)
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π¦ = −π₯3 + 1
π¦ = (π₯ − 1)(π₯ − 3)(−π₯ + 2)
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