Additional Mathematics - O Levels Cheat Sheet
March 26, 2017 | Author: Hassan Nadeem | Category: N/A
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New Additional Mathematics: Cheat Sheet
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New Additional Mathematics: Cheat Sheet For O Levels
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1. Sets A null or empty set is donated by { } or π. P = Q if they have the same elements. P β Q, Q is subset of P. P β Q, P is subset of R. P β Q, Q is proper subset of P. P β Q, P is proper subset of Q. P β Q, Intersection of P and Q. P β Q, union of P and Q. Pβ compliment of P i.e. β-P
2. Simultaneous Equations βπ Β± βπ 2 β 4ππ π₯= 2π
3. Logarithms and Indices Indices 1. π0 = 1 1
2. π βπ = 1 π
ππ
π
3. π = βπ π π
π
4. π = ( βπ )
π
5. ππ Γ ππ = ππ+π 6.
ππ ππ
= ππβπ
7. (ππ )π = πππ 8. ππ Γ π π = (ππ)π 9.
ππ ππ
π π
=( )
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π
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Logarithms 1. π π₯ = π¦ β« π₯ = ππππ π¦ 2. ππππ 1 = 0 3. ππππ π = 1 4. ππππ π₯π¦ = ππππ π₯ + ππππ π¦ 5. ππππ
π₯
π¦
= ππππ π₯ β ππππ π¦
6. ππππ π = 7. ππππ π =
ππππ π ππππ π 1 ππππ π
π¦
8. ππππ π₯ = π¦ππππ π₯ 9. πππππ π₯ = ππππ π₯
1 π
10. log π π₯ = log π πlog π π₯ =
logπ π₯ logπ π
4. Quadratic Expressions and Equations 1. Sketching Graph y-intercept Put x=0
x-intercept Put y=0
Turning point Method 1 x-coordinate: π₯ = y-coordinate: π¦ =
βπ 2π 4ππβπ2 4π
Method 2 Express π¦ = ππ₯ 2 + ππ₯ + π as π¦ = π(π₯ β β)2 + π by completing www.o-alevel.com
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the square. The turning point is(β, π ).
2. Types of roots of πππ + ππ + π = π π 2 β 4ππ β₯ 0 : real roots π 2 β 4ππ < 0 : no real roots π 2 β 4ππ > 0 : distinct real roots π 2 β 4ππ = 0 : equal, coincident or repeated real roots
5. Remainder Factor Theorems Polynomials 1. ax 2 + bx + c is a polynomial of degree 2. 2. ax 3 + bx + c is a polynomial of degree 3.
Identities π(π₯) β‘ π(π₯) βΊ π(π₯) = π(π₯) For all values of x To find unknowns either substitute values of x, or equate coefficients of like powers of x.
Remainder theorem If a polynomial f(x) is defined by (x-a), the remainder is R =f(a)
Factor Theorem (x-a) is a factor of f(x) then f(a) = 0
Solution of cubic Equation I.
Obtain one factor (x-a) by trail and error method.
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II. III.
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Divide the cubic equation with a, by synthetic division to find the quadratic equation. Solve the quadratic equation to find remaining two factors of cubic equation.
For example: I. II.
III. IV. V.
The equation π₯ 3 + 2π₯ 2 β 5π₯ β 6 = 0 has (x-2) as one factor, found by trail and error method. Synthetic division will be done as follows:
The quadratics equation obtained is π₯ 2 + 4π₯ + 3 = 0. Equation is solved by quadratic formula, X=-1 and X=-3. Answer would be (x-2)(x+1)(x+3).
6. Matrices 1. Order of a matrix Order if matrix is stated as its number of rows x number of columns. For example, the matrix (5
6
2) has order 1 x 3.
2. Equality Two matrices are equal if they are of the same order and if their corresponding elements are equal.
3. Addition To add two matrices, we add their corresponding elements. For example, (
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6 3
β2 β4 )+( 5 4
2 2 )=( 1 7
0 ). 6
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4. Subtraction To subtract two matrices, we subtract their corresponding elements. For example: (
6 9
3 14
5 2 )β( β5 β4
7 20
4 5 )=( 12 1
β4 β6
0 ). β6
5. Scalar multiplication To multiply a matrix by k, we multiply each element by k. For example, π (
2 3
4 2π )=( β1 3π
2 6 4π ) or 3 ( ) = ( ). 4 12 βπ
6. Matrix multiplication To multiply two matrices, column of the first matrix must be equal to the row of the second matrix. The product will have order row of first matrix X column of second matrix. π π π π 2 4 3 2 1 4 For example: (1 3 ) ( ) = (π π π β ) 1 5 2 7 π π π π 2 β1 To get the first row of product do following: a = (2 x 3) + (4 X 1) = 10 (1st row of first, 1st column of second) b = (2 x 2) + (4 x 5) = 24 (1st row of first, 2st column of second) c = (2 x 1) + (4 x 2) = 10 (1st row of first, 3st column of second) d = (2 x 4) + (4 x 7) = 36 (1st row of first, 4st column of second) e = (1 x 3) + (3 x 1) = 6 (2st row of first, 1st column of second) f = (1 x 2) + (3 x 5) = 17 (2st row of first, 2st column of second) g = (1 x 1) + (3 x 2) = 7 (2st row of first, 3st column of second) h = (1 x 4) + (3 x 7) = 25 (2st row of first, 4st column of second) i = (2 x 3) + (-1 x 1) = 5 (3st row of first, 1st column of second) j = (2 x 2) + (-1 x 5) = -1 (3st row of first, 2st column of second) k = (2 x 1) + (-1 x 2) = 0 (3st row of first, 3st column of second) l = (2 x 4) + (-1 x 7) = 1 (3st row of first, 4st column of second)
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7. 2 x2 Matrices 1 0 a. The matrix ( ) is called identity matrix. When it is multiplied with 0 1 any matrix X the answer will be X. π π π π b. Determinant of matrix ( | = ππ β ππ ) will be = | π π π π π π π βπ c. Adjoint of matrix ( ) will be = ( ) π π βπ π π π d. Inverse of non-singular matrix (determinant is β 0) ( ) will be : π π πππππππ‘ 1 π βπ = ( ) πππ‘ππππππππ‘ ππ β ππ βπ π
8. Solving simultaneous linear equations by a matrix method ππ₯ + ππ¦ = β π π π₯ β β«β« ( ) (π¦ ) = ( ) ππ₯ + ππ¦ = π π π π β1 π₯ π π β (π¦) = ( ) Γ( ) π π π
7. Coordinate Geometry Formulas π·ππ π‘ππππ π΄π΅ = β(π₯2 β π₯1 )2 + (π¦2 β π¦1 )2 ππππππππ‘ ππ π΄π΅ = (
π₯1 + π₯2 π¦1 + π¦2 , ) 2 2
Parallelogram If ABCD is a parallelogram then diagonals AC and BD have a common midpoint. Equation of Straight line To find the equation of a line of best fit, you need the gradient(m) of the line, and the y-intercept(c) of the line. The gradient can be found by taking any two points on the line and using the following formula: ππππππππ‘ = π =
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π¦2 β π¦1 π₯2 β π₯1
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The y-intercept is the y-coordinate of the point at which the line crosses the y-axis (it may need to be extended). This will give the following equation: π¦ = ππ₯ + π Where y and x are the variables, m is the gradient and c is the y-intercept. Equation of parallel lines Parallel line have equal gradient. If lines π¦ = π1 π1 and π¦ = π2 π2 are parallel then π1 = π2 Equations of perpendicular line If lines π¦ = π1 π1 and π¦ = π2 π2 are perpendicular then π1 = β β
1 π1
1 π2
and π2 =
.
Perpendicular bisector The line that passes through the midpoint of A and B, and perpendicular bisector of AB. For any point P on the line, PA = PB
Points of Intersection The coordinates of point of intersection of a line and a non-parallel line or a curve can be obtained by solving their equations simultaneously.
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8. Linear Law To apply the linear law for a non-linear equation in variables x and y, express the equation in the form π = ππ + π Where X and Y are expressions in x and/or y.
9. Functions Page 196 of Book
10. Trigonometric Functions πππ + π£π 90
Sin 2
All 1
180
0,360
Tan 3
Cos 4 270 πππ β π£π
π is always acute.
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Basics sin π = cos π = tan π = tan π =
ππππππππππ’πππ βπ¦πππ‘πππ’π π πππ π βπ¦πππ‘πππ’π π ππππππππππ’πππ πππ π sin π cos π 1
cosec π = sec π = cot π =
sin π 1
cos π 1 tan π
Rule 1 sin(90 β π) = cos π cos(90 β π) = sin π tan(90 β π) =
1 tan π
= cot ΞΈ
Rule 2 sin(180 β π) = + sin π cos(180 β π) = βcos π tan(180 β π) = βtan π
Rule 3 sin(180 + π) = βsin π cos(180 + π) = βcos π tan(180 + π) = +tan π
Rule 4 sin(360 β π) = β sin π cos(360 β π) = +cos π tan(360 β π) = βtan π
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Rule 5 sin(β π) = βsin π cos(βπ) = +cos π tan(βπ) = βtan π
Trigonometric Ratios of Some Special Angles cos 45 = sin 45 =
1 β2 1
β2 tan 45 = 1
cos 60 = sin 60 =
1 2
β3 2
tan 60 = β3
β3 2 1 sin 30 = 2 1 tan 30 β3
cos 30 =
11. Simple Trigonometric Identities Trigonometric Identities sin2 π + cos 2 π = 1 1 + tan2 π = sec 2 π 1 + cot 2 π = cosec 2 π
12. Circular Measure Relation between Radian and Degree π 2
πππππππ = 90Β°
3π 2
πππππππ = 270Β°
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π πππππππ = 180Β° 2π πππππππ = 360Β°
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π = ππ³ where s is arc length, r is radius and Ο΄ is angle of sector is radians 1
1
2
2
π΄ = ππ = π 2 π³
where A is Area of sector ππππ ππ π πππ‘ππ πππππ ππ π πππ‘ππ = ππππ ππ ππππππ πππππ ππ ππππππ
13. Permutation and Combination π! = π(π β 1)(π β 2) Γ β¦ Γ 3 Γ 2 Γ 1 0! = 1 π! = π(π β 1)!
πππ = ππΆπ =
π! (π β π)!
π! (π β π)! π!
14. Binomial Theorem (π + π)π = ππ + πΆ1π ππβ1 π + πΆ2π ππβ2 π 2 + πΆ3π ππβ3 π3 + β― + π π ππ+1 = ππΆπ ππβπ π π
15. Differentiation π π (π₯ ) = ππ₯ πβ1 ππ₯ π (ππ₯ π + ππ₯ π ) = πππ₯ πβ1 + πππ₯ πβ1 ππ₯ π π ππ’ (π’ ) = ππ’πβ1 ππ₯ ππ₯
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π ππ£ ππ’ (π’π£) = π’ +π£ ππ₯ ππ ππ₯ ππ’ ππ£ π£ βπ’ π π’ ππ₯ ππ₯ ( )= 2 ππ₯ π£ π£ Where βvβ and βuβ are two functions Gradient of a curve at any point P(x,y) is
ππ¦ ππ₯
at x
16. Rate of Change The rate of change of a variable x with respect to time is
ππ₯ ππ‘
ππ¦ ππ¦ ππ₯ = Γ ππ‘ ππ₯ ππ‘ πΏπ¦ ππ¦ β πΏπ₯ ππ₯ πππππππ‘πππ πβππππ ππ π₯ =
πΏπ₯ Γ 100% π₯
π(π₯ + πΏπ₯) = π¦ + πΏπ¦ β π¦ +
ππ¦ πΏπ₯ ππ₯
17. Higher Derivative ππ¦ ππ₯ ππ¦ ππ₯
= 0 when x =a then point (a, f(a)) is a stationary point. = 0 and
π2 π¦ ππ₯ 2
β 0 when x =a then point (a, f(a)) is a turning point.
For a turning point T
I.
If
II.
If
π2 π¦ ππ₯ 2 π2 π¦ ππ₯ 2
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18. Derivative of Trigonometric Functions π (sin π₯) = cos π₯ ππ₯ π (cos π₯) = β sin π₯ ππ₯ π (tan π₯) = sec 2 π₯ ππ₯
π (sinn π₯) = π sinnβ1 π₯ cos π₯ ππ₯ π (cosn π₯) = βπ cos nβ1 π₯ sin π₯ ππ₯ π (tann π₯) = π tannβ1 π₯ sec 2 π₯ ππ₯
19. Exponential and Logarithmic Functions π π’ ππ’ (π ) = π π’ ππ₯ ππ₯ π ππ₯+π (π ) = ππ ππ₯+π ππ₯ A curve defined by y=ln(ax+b) has a domain ax+b>0 and the curve cuts the x-axis at the point where ax+b=1 π 1 (ππ π₯) = ππ₯ π₯ π 1 ππ’ (ln π’) = ππ₯ π’ ππ₯ www.o-alevel.com
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π π [ππ(ππ₯ + π)] = ππ₯ ππ₯ + π
20. Integration ππ¦ = π₯ βΊ π¦ = β« π₯ ππ₯ ππ₯ π 1 2 1 ( π₯ + π) = π₯ βΊ β« π₯ ππ₯ = π₯ 2 + π ππ₯ 2 2 ππ₯ π+1 β« ππ₯ ππ₯ = +π π+1 π
π
β«(ππ₯ + ππ
π )ππ₯
ππ₯ π+1 ππ₯ π+1 = + +π π+1 π+1
(ππ₯ + π)π+1 β«(ππ₯ + π) ππ₯ = +π π(π + 1) π
π π [πΉ(π₯)] = π(π₯) βΊ β« π(π₯) ππ₯ = πΉ(π) β πΉ(π) ππ₯ π π
π
π
β« π(π₯) ππ₯ + β« π(π₯) ππ₯ = β« π(π₯) ππ₯ π
π π
π π
β« π(π₯) ππ₯ = β β« π(π₯) ππ₯ π
π π
β« π(π₯) ππ₯ = 0 π
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π (sin π₯) = cos π₯ βΊ β« cos π₯ ππ₯ = sin π₯ + π ππ₯ π (βcos π₯) = sin π₯ βΊ β« sin π₯ ππ₯ = β cos π₯ + π ππ₯ π (tan π₯) = sec 2 π₯ βΊ β« π ππ 2 π₯ ππ₯ = π‘ππ π₯ + π ππ₯
π 1 1 [ sin(ππ₯ + π)] = cos(ππ₯ + π) βΊ β« cos(ππ₯ + π) ππ₯ = sin(ππ₯ + π) + π ππ₯ π π π 1 1 [β cos(ππ₯ + π)] = sin(ππ₯ + π) βΊ β« sin(ππ₯ + π) ππ₯ = β cos(ππ₯ + π) + π ππ₯ π π π 1 1 [ tan(ππ₯ + π)] = sec 2 (ππ₯ + π) βΊ β« π ππ 2 (ππ₯ + π) ππ₯ = π‘ππ (ππ₯ + π) + π ππ₯ π π
π π₯ (π ) = π π₯ βΊ β« π π₯ ππ₯ = π π₯ + π ππ₯ π (βπ βπ₯ ) = π βπ₯ βΊ β« π βπ₯ ππ₯ = βπ βπ₯ + π ππ₯
21. Applications of Integration For a region R above the x-axis, enclosed by the curve y=f(x), the x-axis and the lines x=a and x=b, the area R is: π
π΄ = β« π(π₯) ππ₯ π
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For a region R below the x-axis, enclosed by the curve y=f(x), the x-axis and the lines x=a and x=b, the area R is: π
π΄ = β« βπ(π₯) ππ₯ π
For a region R enclosed by the curves y=f(x) and y=g(x) and the lines x=a and x=b, the area R is: π
π΄ = β« [π(π₯) β π(π₯) ]ππ₯ π
22. Kinematics π£=
ππ ππ‘
π=
ππ£ ππ‘
π = β« π£ ππ‘ π£ = β« π ππ‘ π΄π£ππππ π ππππ =
π‘ππ‘ππ πππ π‘ππππ π‘πππ£πππππ π‘ππ‘ππ π‘πππ π‘ππππ
π£ = π’ + ππ‘
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1 π = π’π‘ + ππ‘ 2 2 1 π = (π’ + π£)π‘ 2 π£ 2 = π’2 + 2ππ
23. Vectors π₯ βββββ = ( ) then |ππ βββββ | = βπ₯ 2 + π¦ 2 If ππ π¦ π = ππ and k > 0 a and b are in the same direction π = ππ and k < 0 a and b are opposite in direction Vectors expressed in terms of two parallel vectors a and b: ππ + ππ = ππ + π π βΊ p = r and q = s If A, B and C are collinear points βΊ AB=kBC If P has coordinates (x, y) in a Cartesian plane, then the position vector of P is βββββ = π₯π + π¦π ππ where i and j are unit vectors in the positive direction along the x-axis and the y-axis respectively. βββββ is Unit vector is the direction of ππ 1 1 π₯ (π₯π + π¦π) ππ (π¦) βπ₯ 2 + π¦ 2 βπ₯ 2 + π¦ 2
24. Relative velocity
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