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STPM MATHEMATICS T SYLLABUS
Chapter 1 Functions
1.1 Functions
• (a) state the domain and range of a function, and find composite functions;
The domain of a unction is all the possible input values, and the ran!e is all possibleoutput values."omain# set of all possible input values (usually x)$an!e# set of all possible output values (usually y)
a) State the domain and range of the following relation%&'( )*+( &,( -+( &1( )1+( &*( +/
The domain is all the x-values, and the range is all the y-values "omain# %'( ,( 1( */ $an!e# %)*( -( 01( /
a) State the domain and range of the following relation%&'( ,+( &,( ,+( &1( ,+( &*( ,+/
"omain# %'( ,( 1( */
$an!e# %,/
hat is a Function2 A unction( &3+ re4ates an input to an output. Each input is
re4ated to e3act45 one output.
This is a unction. There is only one y for each x
This is a unction. There is only one arrow coming from each x; there is only one y foreach x
http://coolmathsolutions.blogspot.com/2013/02/functions-domain-and-range.htmlhttp://coolmathsolutions.blogspot.com/2013/02/what-is-function.htmlhttp://coolmathsolutions.blogspot.com/2013/02/what-is-function.htmlhttp://coolmathsolutions.blogspot.com/2013/02/functions-domain-and-range.html
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This is not a unction. is in the domain, but it has no range element that corresponds toit
This is not a unction. ! is associated with two different range elements
Functions# "omain and $an!e
The unction composition of two functions ta"es the output of one function as the input
of a second one
#n in6erse unction for f, denoted by f -$, is a function in the opposite direction.- See more at% http%&&coolmathsolutions.blogspot.com&'$!&'&what-is-function.htmlsthash.t!*ri+g.dpuf
(b) determine whether a function is one-to-one, and find the inverse of a one-to-onefunction;
# one-to-one function is a function in which every element in the range of the function
corresponds with one and only one element in the domain.
# function, f(x), has an inverse function if f(x) is one-to-one.
The oriontal /ine Test% 0f you can draw a horiontal line so that it hits the graph inmore than one spot, then it is 12T one-to-one.
a) 0s below function one-to-one3
http://coolmathsolutions.blogspot.com/2013/02/functions-domain-and-range.htmlhttp://coolmathsolutions.blogspot.com/2013/02/one-to-one-function.htmlhttp://coolmathsolutions.blogspot.com/2013/02/one-to-one-function.htmlhttp://coolmathsolutions.blogspot.com/2013/02/inverse-one-to-one-function.htmlhttp://coolmathsolutions.blogspot.com/2013/02/inverse-one-to-one-function.htmlhttp://coolmathsolutions.blogspot.com/2013/02/functions-domain-and-range.htmlhttp://coolmathsolutions.blogspot.com/2013/02/one-to-one-function.htmlhttp://coolmathsolutions.blogspot.com/2013/02/inverse-one-to-one-function.htmlhttp://coolmathsolutions.blogspot.com/2013/02/inverse-one-to-one-function.html
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f(x)=x3
f(x)4x! is one-to-one function and has inverse function. (The horiontal line cuts thegraph of function f at $ point, therefore f is a one-to-one function)
b) 0s below function one-to-one3
f(x)=x2
f(x)4x' is 12T one-to-one function and does 12T has an inverse function. (Thehoriontal line cuts the graph of function f at ' points, therefore f is 12T a one-to-onefunction)
0f we restricted x greater than or e5ual to . The horiontal line cuts the graph of functionf once and f(x)4x' is one-to-one function and has an inverse function.
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(c) s"etch the graphs of simple functions, including piecewise-defined functions;
ow to find the inverse of one-to-one function below3
f(x)=3x−46raw the graph of f(x)4!x-7
The horizontal line cuts the graph of function f once, therefore f is a one-to-onefunction and it has has inverse function.
8ind the inverse function
f(y)=x3y−4=x
y=x+43
f−1(x)=x+43
http://coolmathsolutions.blogspot.com/2013/02/piecewise-defined-functions.htmlhttp://coolmathsolutions.blogspot.com/2013/02/one-to-one-function.htmlhttp://coolmathsolutions.blogspot.com/2013/02/one-to-one-function.htmlhttp://coolmathsolutions.blogspot.com/2013/02/piecewise-defined-functions.htmlhttp://coolmathsolutions.blogspot.com/2013/02/one-to-one-function.htmlhttp://coolmathsolutions.blogspot.com/2013/02/one-to-one-function.html
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The graph of an inverse relation is the reflection of the original graph over the identityline, y 4 x.
9iecewise-defined function is a function which is defined by multiple sub-
functions, each sub-function applying to a certain interval of the mainfunction:s domain.
E3amp4e 1
f(x)={2x−1,x
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f(x)=4x+1
f(x)=2x3−4x2+5x+11
f(x)=x7−6x4+3x2
*ational function is division of two polynomial functions.
f(x)=P(x)Q(x) where 9 and are polynomial functions in x.
y=2x+5x−1f(x) is not defined at x4$.
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• (d) use the factor theorem and the remainder theorem;
The Factor Theorem
0f the remainder of a polynomial, f(x), when divide by (x-a) is ero, then (x-a) is a factor.Show that (x>!) is a factor of
x3+5x2+5x−3
f(−3)=(−3)3+5(−3)2+5(−3)−3
f(−3)=0
?y factor theorem (x>!) is factor of x!>x'>x-!*emainder Theorem
0f a polynomial f(x) is divided by (x @ r) and a remainder * is obtained, then f(r) 4 *.f(x)=(x−r)⋅q(x)+R
Axample% &3+73-8'3'0*3*8309 di6ide :5 &301+
#fter dividing by x-$, there is a remainder of -7.
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(e) solve polynomial and rational e5uations and ine5ualities;
So46in! $ationa4 Ine;ua4ities
# rational function is a 5uotient of two polynomials.
x2+3x+2x2−16≥0
x2+3x+2x2−16=(x+2)(x+1)(x−4)(x+4)
as an example for an ine5uality involving a rational function.
This polynomial fraction will be ero wherever numerator is ero
(x+2)(x+1)=0
x=−2,x=−1
The fraction will be undefined wherever the denominator is ero
(x−4)(x+4)=0
x=4,x=−4
Bonse5uently, the set
(−∞,−4),[−2,−1],(4,+∞)
Solution in Cine5ualityC notation
x4
http://coolmathsolutions.blogspot.com/2013/02/rational-equations-and-inequalities-01.htmlhttp://coolmathsolutions.blogspot.com/2013/02/rational-function.htmlhttp://coolmathsolutions.blogspot.com/2013/02/rational-function.htmlhttp://coolmathsolutions.blogspot.com/2013/02/rational-equations-and-inequalities-01.htmlhttp://coolmathsolutions.blogspot.com/2013/02/rational-function.html
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Find the inter6a4
*ational e5uations and ine5ualities '
So46in! $ationa4 Ine;ua4ities
A rationa4 unction is a ;uotient o t
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Find the inter6a4
(f) solve e5uations and ine5ualities involving modulus signs in simple cases;
So46e the ine;ua4it5 3 * 3 8 1
• S5uare both sides of each e5uation to omit the modulus signs.
• *earrange the ine5uality into an e5uivalent form.
So the solution set is
(g) decompose a rational expression into partial fractions in cases where the denominator
has two distinct linear factors, or a linear factor and a prime 5uadratic factor;A rationa4 unction P&3+D&3+ can :e re
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rite the partia4 raction decomposition o
11x+21(2x−3)(x+6)=A2x−3+Bx+6
"istri:ute A and B
11x+21=(x+6)A+(2x−3)B0f x4-D
11(−6)+21=[2(−6)−3]B
B=30f x4,
21=6A−3B
A=5
The partia4 raction
11x+21(2x−3)(x+6)=52x−3+3x+6
This is the E3ponentia4 Function%
f(x)=ax a is any value greater than
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For > G a G 1#
• #s x increases, f(x) heads to
• #s x decreases, f(x) heads to infinity
• 0t is a Strictly 6ecreasing function
• 0t has a oriontal #symptote along the x-axis (y4).
For a = 1#
• #s x increases, f(x) heads to infinity
• #s x decreases, f(x) heads to
• it is a Strictly 0ncreasing function
• 0t has a oriontal #symptote along the x-axis (y4).
The atura4 E3ponentia4 Function is the function
f(x)=exwhere e is the number (approximately '.E$+'+$+'+)
1.' E3ponentia4 and 4o!arithmic unctions
• (h) relate exponential and logarithmic functions, algebraically and graphically;
This is the E3ponentia4 Function%
f(x)=ax a is any value greater than
http://coolmathsolutions.blogspot.com/2013/02/asymptote-in-rational-function.htmlhttp://coolmathsolutions.blogspot.com/2013/02/asymptote-in-rational-function.htmlhttp://coolmathsolutions.blogspot.com/2013/02/exponential-function.htmlhttp://coolmathsolutions.blogspot.com/2013/02/logarithmic-function.htmlhttp://coolmathsolutions.blogspot.com/2013/02/asymptote-in-rational-function.htmlhttp://coolmathsolutions.blogspot.com/2013/02/asymptote-in-rational-function.htmlhttp://coolmathsolutions.blogspot.com/2013/02/exponential-function.htmlhttp://coolmathsolutions.blogspot.com/2013/02/logarithmic-function.html
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For > G a G 1#
• #s x increases, f(x) heads to
• #s x decreases, f(x) heads to infinity
• 0t is a Strictly 6ecreasing function
• 0t has a oriontal #symptote along the x-axis (y4).
For a = 1#
• #s x increases, f(x) heads to infinity
• #s x decreases, f(x) heads to
• it is a Strictly 0ncreasing function
• 0t has a oriontal #symptote along the x-axis (y4).
The atura4 E3ponentia4 Function is the function
f(x)=exwhere e is the number (approximately '.E$+'+$+'+)
The logarithmic function is the function y 4 logax, where a is any number such that a F ,a G $, and x F .
http://coolmathsolutions.blogspot.com/2013/02/asymptote-in-rational-function.htmlhttp://coolmathsolutions.blogspot.com/2013/02/asymptote-in-rational-function.htmlhttp://coolmathsolutions.blogspot.com/2013/02/what-is-function.htmlhttp://coolmathsolutions.blogspot.com/2013/02/what-is-function.htmlhttp://coolmathsolutions.blogspot.com/2013/02/asymptote-in-rational-function.htmlhttp://coolmathsolutions.blogspot.com/2013/02/asymptote-in-rational-function.htmlhttp://coolmathsolutions.blogspot.com/2013/02/what-is-function.html
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y 4 logax is e5uivalent to x 4 ay
The inverse of an exponential function is a logarithmic function.
Since y 4 log'x is a one-to-one function, we "now that its inverse will also be a function.
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(i) use the properties of exponents and logarithms;
$. 0ndices
an,a≠0
a0=1,a≠0a−n=1/an
'. /aw of 0ndices
am×an=am+n
am÷an=am−n
(am)n=am×n
am×bm=(a×b)m
am÷bm=(ab)m
!. /ogarithms
Ifa=bc,then logba=c
loga1=0
logaa=1
alogab=1
7. /aw of /ogarithms
logaxy=logax+logay
loga(xy)=logax−logay
logaxn=nlogax
Bhange of base of /ogarithms
logab=1logba
logab=lognblogna
(H) solve e5uations and ine5ualities involving exponential or logarithmic expressions;
So46in! $ationa4 Ine;ua4ities
http://coolmathsolutions.blogspot.com/2013/01/rules-of-logarithm.htmlhttp://coolmathsolutions.blogspot.com/2013/02/rational-equations-and-inequalities-01.htmlhttp://coolmathsolutions.blogspot.com/2013/02/rational-equations-and-inequalities-01.htmlhttp://coolmathsolutions.blogspot.com/2013/01/rules-of-logarithm.htmlhttp://coolmathsolutions.blogspot.com/2013/02/rational-equations-and-inequalities-01.html
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A rationa4 unction is a ;uotient o t
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1.- Tri!onometric unctions
• (") relate the periodicity and symmetries of the sine, cosine and tangent functions to their
graphs, and identify the inverse sine, inverse cosine and inverse tangent functions andtheir graphs;
• (l) use basic trigonometric identities and the formulae for sin ( A I B), cos ( A I B) and tan( A I B), including sin 2A, cos 2A and tan 2A;
• (m) express a sin θ > b cos θ in the forms r sin ( θ I J) and r cos ( θ I J);
• (n) find the solutions, within specified intervals, of trigonometric e5uations andine5ualities.
Trigonometric Identities
sin(A+B)=sinAcosB+cosAsinB
sin(A−B)=sinAcosB−cosAsinB
cos(A+B)=cosAcosB−sinAsinB
cos(A−B)=cosAcosB+sinAsinB
tan(A+B)=tanA+tanB1−tanAtanB
tan(A−B)=tanA−tanB1+tanAtanB
sin(2A)=2sinAcosA
cos(2A)=cos2(A)−sin2(A)
=2cos2(A)−1
=1−2sin2
(A)
tan2A=2tanA1−tan2A
E3press a sin J : cos in the orm $ sin & J K+
http://coolmathsolutions.blogspot.com/2013/02/trigonometric-identities.htmlhttp://coolmathsolutions.blogspot.com/2013/02/express-sin-b-cos-in-form-r-sin.htmlhttp://coolmathsolutions.blogspot.com/2013/02/express-sin-b-cos-in-form-r-sin.htmlhttp://coolmathsolutions.blogspot.com/2013/02/express-sin-b-cos-in-form-r-sin.htmlhttp://coolmathsolutions.blogspot.com/2013/02/express-sin-b-cos-in-form-r-sin.htmlhttp://coolmathsolutions.blogspot.com/2013/02/express-sin-b-cos-in-form-r-sin.htmlhttp://coolmathsolutions.blogspot.com/2013/02/express-sin-b-cos-in-form-r-sin.htmlhttp://coolmathsolutions.blogspot.com/2013/02/express-sin-b-cos-in-form-r-cos.htmlhttp://coolmathsolutions.blogspot.com/2013/02/express-sin-b-cos-in-form-r-cos.htmlhttp://coolmathsolutions.blogspot.com/2013/02/express-sin-b-cos-in-form-r-cos.htmlhttp://coolmathsolutions.blogspot.com/2013/02/express-sin-b-cos-in-form-r-cos.htmlhttp://coolmathsolutions.blogspot.com/2013/02/express-sin-b-cos-in-form-r-cos.htmlhttp://coolmathsolutions.blogspot.com/2013/02/trigonometric-identities.htmlhttp://coolmathsolutions.blogspot.com/2013/02/express-sin-b-cos-in-form-r-sin.htmlhttp://coolmathsolutions.blogspot.com/2013/02/express-sin-b-cos-in-form-r-cos.html
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E3press a sin J : cos in the orm $ sin& J K+(
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AxampleAxpress - sin 8' cos in the form $ sin& 8 K+
R=42+32−−−−−−√ =25−−√=5
α=tan−1 34=36.87∘So
4sinθ+3cosα=5sin(θ+36.87∘)
Summar5 o the e3pressions and conditions
E3press a sin J : cos in the orm $ cos & J K+
E3press a sin J : cos in the orm $ cos& J K+(
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=tanαSo
α=tan−1ab
R=a2+b2−−−−−−√
Then we have expressed a sin K > b cos K in the form re5uired
asinθ+bcosθ≡Rcos(θ−α)
Summar5 o the e3pressions and conditions
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Chapter *
Se;uences and Series
*.1 Se;uences
• (a) use an explicit formula and a recursive formula for a se5uence;
• (b) find the limit of a convergent se5uence;
Con6er!ent Se;uence
# se5uence is said to be convergent if it approaches some limit. 8ormally, a se5uenceconverges to the limit
limn→∞Sn=S
E3amp4e
• The se5uence '.$, '.$, '.$, '.$, . . . has limit ', so the se5uence converges to '.
• The se5uence $, ', !, 7, , D, . . . has a limit of infinity (L). This is not a real number, so
the se5uence does not converge. 0t is a divergent se5uence.
Con6er!ent se;uences ha6e a inite 4imit.
1,12, 13,14,15,16....,1n
Limit=0
1,12, 23,34,45,56....,nn+1
Limit=1
*.* Series
http://coolmathsolutions.blogspot.com/2013/03/convergent-sequence.htmlhttp://coolmathsolutions.blogspot.com/2013/03/convergent-sequence.html
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(c) use the formulae for the nth term and for the sum of the first n terms of anarithmeticseries and of a geometric series;
An arithmetic progression or arithmetic sequence is a sequence of numberssuch that the dierence between the consecutive terms is constant.
The nth term of the sequence an! is given b"
an=a1+(n−1)d
The sum of the sequence of the #rst n terms is then given b"
Sn=n2[2a+(n−1)d]or
Sn=n2(a1+a2)
eometric Series is a series
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a+ar+ar2+ar3+ar4.....=∑ k=0∞ark=a1−r
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(x+y)2=x2+2xy+y2
(x+y)3=x3+3x3y+3xy2+y3
(x+y)4=x4+4x3y+6x2y2+4xy3+y4
It is possib&e to e'pand an" power of ' ( " into a sum of the form
(x+y)n=(n0)xny0+(n1)xn−1y1+(n2)xn−2y2+...+
(nn−1)x1yn−1+(nn)x0yn It can be written as
(x+y)n=∑ k=0n(nk)xn−kyk%inomia& formu&a ) invo&ves on&" a sing&e variab&e
(1+x)n=(n0)x0+(n1)x1+(n2)x2+...+(nn−1)xn−1+(nn)xn(1+x)n=∑ k=0n(nk)xk
According to the ratio test for series convergence a series converges when)
if limk→∞(ak+1ak)1,series diverges
iflimk→∞(ak+1ak)=1,result is indeterminate The %inomia& Theorem converges when *'*+,
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(g) expand ($>x) n , wheren$Q, and identify the condition M x M N $ for the validity of thisexpansion;
(h) use binomial expansions in approximations.
Chapter '
Matrices
'.1 Matrices
(a) identify null, identity, diagonal, triangular and symmetric matrices;
http://coolmathsolutions.blogspot.com/2013/03/binomial-expansions.htmlhttp://coolmathsolutions.blogspot.com/2013/03/binomial-expansions.htmlhttp://coolmathsolutions.blogspot.com/2013/01/inverse-of-diagonal-matrix.htmlhttp://coolmathsolutions.blogspot.com/2013/02/what-is-triangular-matrix.htmlhttp://coolmathsolutions.blogspot.com/2013/02/what-is-triangular-matrix.htmlhttp://coolmathsolutions.blogspot.com/2013/03/binomial-expansions.htmlhttp://coolmathsolutions.blogspot.com/2013/01/inverse-of-diagonal-matrix.htmlhttp://coolmathsolutions.blogspot.com/2013/02/what-is-triangular-matrix.html
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In6erse o "ia!ona4 Matri3
6iagonal matrix is a matrix in which the entries outside the main diagonal are all ero.
A=- a11000a22 0 00a33/0
A−1=-11111111a11000 1a22 0 00 1a33/02222222
Axample
A=- 5000 3 0 00 1/0
A−1=-111115000 13 0 001/02222
(b) use the conditions for the e5uality of two matrices;
(c) perform scalar multiplication, addition, subtraction and multiplication of matriceswith at most three rows and three columns;
hat is Trian!u4ar Matri3
# matrix is a trian!u4ar matri3 if all the elements either above or below the diagonal areero. # matrix that is both upper and lower triangular is a diagonal matrix.
Properties o Trian!u4ar Matri3
• The determinant of a triangular matrix is the product of the diagonal elements.
• The inverse of a triangular matrix is a triangular matrix.
• The product of two (upper or lower) triangular matrices is a triangular matrix.
Upper Trian!u4ar Matri3
A=33333130250013333=3.2.1=6
A−1=33331/3−1/6−1/601/2−5/250013333
Lo
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B−1=33331/2003/2−10−11613333
(d) use the properties of matrix operations;
Properties o Matrices
AB≠BA
A+B=B+A
A(B+C)=AB+AC
(A+B)C=AC+BC
A(BC)=(AB)C
AI=IA=A
A0=I
Scalar Oultiplication
λ(AB)=(λA)B
(AB)λ=A(Bλ)
Transpose
(AB)T=BTAT
9roperties of 0nverse Oatrix
AA−1=A−1A=I
(A−1)−1=A
(AT)−1=(A−1)T
(An)−1=(A−1)n
(cA)−1=c−1A−1=1cA−1
(e) find the inverse of a non-singular matrix using elementary row operations;
http://coolmathsolutions.blogspot.com/2013/01/properties-of-matrix.htmlhttp://coolmathsolutions.blogspot.com/search/label/Elementary%20Row%20Operation%20Matrixhttp://coolmathsolutions.blogspot.com/search/label/Elementary%20Row%20Operation%20Matrixhttp://coolmathsolutions.blogspot.com/2013/01/properties-of-matrix.htmlhttp://coolmathsolutions.blogspot.com/search/label/Elementary%20Row%20Operation%20Matrix
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8ind the inverse of a non-singular matrix using elementary row operations
4shirts +2 pants +2 pairofshoes=$190
3shirts +4 pants +3 pairofshoes=$295
2shirts +4 pants +2 pairofshoes=$190
/et x 4 price of shirt , y4 price of pant, 4 price for a pair of shoe
-4322 4 4 232/0-xyz/0=-190295250/0/et
A=-4322 4 4 232/0,B=-190295250/0
A-xyz/0=-190295250/0
AA−1=A−1A=I
A−1A-xyz/0=A−1-190295250/0
-xyz/0=-0.50−0.5−0.5 −0.5 1.50.250.75−1.25/0-190295250/0
-xyz/0=-104035/0
A4ternati6e(
a1x+b1y+c1z=d1
a2x+b2y+c2z=d2
a3x+b3y+c3z=d3
Oode F A1 F Pn"nowns4!, input data below
a1=4,b1=2,c1=2,d1=190
a2=3,b2=4,c2=3,d2=295
a3=2,b3=4,c3=2,d3=250
x=10,y=40,z=35
(f) evaluate the determinant of a matrix;
http://coolmathsolutions.blogspot.com/2013/01/inverse-of-matrices-3x3_6.htmlhttp://coolmathsolutions.blogspot.com/2013/01/inverse-of-matrices-3x3_6.html
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In6erse o a Matrices ''
A=-adgbeh cfi/0
$. 6eterminant of !x! Oatrices
4A4=4444adgbehcfi4444=a444 e hfi444@b444dgfi444+c444dgeh444
=a(ei−fh)−b(di−fg)+c(dh−eg)
'. 0nverse of !x! Oatrices
A−1=-adgbe h cfi/0−1=14A4-A D GBEHCFI/0T=14A4-A BC
DEFGHI/0A=444ehfk444,B=−444dgfk444,C=444dgeh444
D=−444bhck444,E=444agck444,F=−444agbh444
G=444becf444,H=−444adcf444,K=444adbe444
-+−+− +−+−+/0
(g) use the properties of determinants;
Properties o "eterminant
$. M#M 4 if it has two e5ual line
33333132 223133333=0
'. M#M 4 if all elements of a line are ero
33333032023033333=0
!. M#M 4 if the elements of a line are a linear combination of the others. (row ! 4 row $ >row ')
33332133252463333=0
7. The determinant of matrix # and its transpose # are e5ual. M#TM4M#M
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A=33332131213423333, |AT|=33332131243123333
|A|=|AT|=−5. # triangular determinant is the product of the diagonal elements.
A=33333130 250013333=3.2.1=6D. The determinant of a product e5uals the product of the determinants.
|A.B|=|A|.|B|
E. 0f a determinant switches two parallel lines its determinant changes sign.
33332131213423333=@5,33331232114323333=5
'.* S5stems o 4inear e;uations
(h) reduce an augmented matrix to row-echelon form, and determine whether a system oflinear e5uations has a uni5ue solution, infinitely many solutions or no solution;
(i) apply the =aussian elimination to solve a system of linear e5uations;
aussian e4imination is a method for solving matrix e5uations of the form
Ax=b
=aussian elimination starting with the system of e5uations
Bompose the Caugmented matrix e5uationC
9erform elementary row operations to put the augmented matrix into the uppertriangular form
# matrix that has undergone =aussian elimination is said to be in echelon form=iven below matrix e5uation
0n au!mented orm, this becomes
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Switching the first and second rows (without switching the elements in the right-hand column vector) gives
8irst row times ! and minus third row gives
8irst row times ' and minus second row gives
8inally, second row times -E&! and minus third row gives (augmented matrix has been reduced to ro
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Oode F A1 F Pn"nowns4!, input data below
a1=20,b1=0,c1=−10,d1=−100
a2=0,b2=20,c2=10,d2=300
a3=−10,b3=10,c3=20,d3=200
x=−5,y=15,z=0
Chapter -
Comp4e3 um:ers
- Comp4e3 um:ers
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• (a) identify the real and imaginary parts of a complex number ;
The rea4 num:ers include all inte!er, rationa4 num:er (number that can be expressed asthe 5uotient or fraction p&5 of two integers, with the denominator 5 not e5ual to ero) and irrationa4 num:ers (numbers cannot be represented as a simple fraction)
0ntergers
...−4,−3,−2−1,0,1,2,3,4....*ational 1umber
12,−12,1537,....0rrational 1umber
2√,3√3,π,e,log23,....
Ima!inar5 num:er is a number than can be written as a real number multiplied by the
imaginary unit i
i=−1−−−√
i2=−1
# complex number is a number that can be put in the form
a+biwhere a and : are real numbers and i is the s5uare root of -$, the ima!inar5 unit
i=−1−−−√
i2=−1
The a:so4ute 6a4ue or modu4us of a complex number 4 a > bi is
|z|=a2+b2−−−−−−√
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(b) use the conditions for the e5uality of two complex numbers;
(c) find the modulus and argument of a complex number in cartesian form and expressthe complex number in polar form;
hat is Ar!ument o Comp4e3 um:er
Bomplex 1umber , 4a>biOodulus is the length of the line segment, that is 29 (modulus of can Qnd using9ythagorasR theorem)
|z|=a2+b2−−−−−−√
|z|=42+32−−−−−−√
|z|=25−−√=5
The angle theta is called the argument of the complex number,
tanΘ=ba
tanΘ=34
Θ=tan−134
arg z=0.644rad
ow to change theta4!D.+E to radian
rad=Θ6π1800
rad=Θ6π1800
36.8706π1800=0.644rad
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(d) represent a complex number geometrically by means of an #rgand diagram;
Ar!and "ia!ram
The #rgand diagram is used to represent complex numbers.#rgand diagram form by y-axis which represents the ima!inar5 part and x-axisrepresent the rea4 part.
Axample%
Z1=3+2i
Z2=−4+i
BonHugate of
Z61=3−2i
Z62=−4−i
Z1+Z2=−1+3i
(e) find the complex roots of a polynomial e5uation with real coefficients;
(f) perform elementary operations on two complex numbers expressed in cartesian form;
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(g) perform multiplication and division of two complex numbers expressed in polar form;
(h) use de OoivreRs theorem to find the powers and roots of a complex number
"e Moi6reNs ormu4a 0 STPM Mathematics
"e Moi6reNs
6e Ooivre:s formula states that for any real number x and any integer n,
(cosx+isinx)n=cos(nx)+isin(nx)
From Eu4erNs ormu4a
eix=cosx+isinx
(eix)n=einx
ei(nx)=cos(nx)+isin(nx)
"e Moi6reNs Theorem in Comp4e3 um:er
0f 4 x > yi 4 reiK, and n is a natural number. Then
zn=(x+yi)n=(reiθ)n
E3amp4e 1# Pse 6e OoivreRs theorem to Qnd ($>i)$.
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=32(cos450∘+isin450∘)
=32(0+i)
=32i
E3amp4e *# Pse 6e OoivreRs theorem to Qnd (! > i ) E .
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Chapter , Ana45tic eometr5
, Ana45tic eometr5
(a) transform a given e5uation of a conic into the standard form; Transform 7onic 8ection into 8tandard 9orm
A conic section is the intersection of a p&ane and a doub&e right circu&arcone. %" changing the ang&e and &ocation of the intersection: we can producedierent t"pes of conics: such as circles, ellipses, hyperbolas and parabolas.
Equation of a Circle in Standard Form
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(x−h)2+(y−k)2=r2Equation of an Ellipse in Standard Form
(x−h)2a2+(y−k)2b2=1Equation of a Hyperbolas in Standard Form
(x−h)2a2−(y−k)2b2=1Equation of a Parabolas in Standard Form
a(x−h)2+k
(b) find the vertex, focus and directrix of a parabola;
Oerte3( Focus and "irectri3 o a Para:o4a
# para:o4a is the set of all points 9 in the plane that are e5uidistant from a fixed point 8&ocus+ and a fixed line d &directri3+.The 6erte3 of the parabola is at e5ual distance between focus and the directrix.
9arabolas are fre5uently encountered as graphs of ;uadratic unctions, such as
y=x2
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(c) find the vertices, centre and foci of an ellipse;
Oertices( Centre and Foci o an E44ipse
E44ipse is a planar curve which in some Bartesian system of coordinates is described bythe e5uation%
(x−h)2a2+(y−k)2b2=1
Oerte3 o an e44ipse. The points at which an ellipse ma"es its sharpest turns. The verticesare on the maHor axis.Centre o an e44ipse # point inside the ellipse which is the midpoint of the line segmentlin"ing the two foci.Foci o an e44ipse 0 The foci, c is always lie on the maHor (longest) axis, spaced e5uallyeach side of the centre.
c2=a2−b2
E3amp4e
16x2+25y2=400
16x2400+25y2400=1
x225+y216=1
x252+y242=1
(x2−0)252+(y2−0)242=1
Oerte3 is &0,(>+ and &,(>+( the maor &4on!est+ a3is.
Co06erte3 is &>(0-+( &>( -+ minor &shortest+ a3is
The centre is at &h(+7&>(>+
Foci o an e44ipse. The foci are three unit to either side of the centre, at &0'(>+ and &'(>+
c2=a2−b2
c2=52−42
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c2=±3
• 8ind ertices, Bentre, and 8oci of Allipse $
• 8ind ertices, Bentre, and 8oci of Allipse '
• 8ind ertices, Bentre, and 8oci of Allipse !
(d) find the vertices, centre, foci and asymptotes of a hyperbola;
(e) find the e5uations of parabolas, ellipses and hyperbolas satisfying prescribed
conditions (excluding eccentricity);
(f) s"etch conics;
(g) find the cartesian e5uation of a conic defined by parametric e5uations;
(h) use the parametric e5uations of conics.
Chapter Oectors
.1 Oectors in t
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A unit vector, or direction vector is a vector which has &ength of , ormagnitude of ,.
b!u =̂u4u4
c!d!
=osition ! is ca&&ed a position vector.
=oint A has the position vector a, if
A=(35) =oint % has the position vector b, if
B=
(62
)
e!
(b) perform scalar multiplication, addition and subtraction of vectors;
8ca&ar ?u&tip&ication of
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c[a1a2]=[ca1ca2]E3amp4e# Oultiply the vector u 4 N ', ! F of by the scalars ', U!, and $&'
2u @ =2[23]=[46]12u @ =12[23]=[13/2]−3u @ =@3[23]=[−6−9]
(c) find the scalar product of two vectors, and determine the angle between two vectors;
The scalar product( or dot product is an a4!e:raic operation that taes t
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• u(v>w)4uv>uw
• u.u4MMuMMVW'X
• c(u.v)4cu.v4u.cv
Sca4ar product can :e o:tained :5 ormu4a
a.b=|a||b|cosθMaM means the magnitude (length) of vector a. and
a.b=axbx+ayby
Balculate the scalar product of vectors a and b, given K 4 Y.
a=[−68],b=[512]
|a|=(−6)2+82−−−−−−−−−√ =10|b|=52+122−−−−−−−√ =13
a.b=|a||b|cosθ
a.b=10612cos59.5∘or
a.b=axbx+ayby
a.b=(−6)(5)+(8)12)=66
8ind the angle between the two vectors.
a=-234/0,b=-1−23/0
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a.b=axbx+ayby
a.b=|a||b|cosθ
a.b=(2)(1)+(3)(−2)+(4)(3)=8
|a|=22+32+42−−−−−−−−−−√ =29−−√
|a|=12+(−2)2+32−−−−−−−−−−−−√ =14−−√8=29−−√14−−√cosθ
cosθ=0.397
θ=66.6∘
(d) find the vector product of two vectors, and determine the area a parallelogram and ofa triangle;
Oector Product o T
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=−3i+6 j−3k
u @ =@3,6,−3B
ector product, c 4 a x b 4 (-!, D, -!)
.* Oector !eometr5
(e) find and use the vector and cartesian e5uations of lines;
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a=EFaxayazGH,b=EFbxbybzGHJ,r=EFxyzGH7artesian Cquation wi&& be
x−axbx−ax=y−ayby−ay=z−azbz−az
C'amp&e) 7ompute the cartesian equation of a straight &ine through point Aand % with position vector)
a=EF1−32GH,b=EF3−15GH,r=EFxyzGH
x−13−1=y−(−3)−1−(−3)=z−25−2
x−12=y+32=z−23
(f) find and use the vector and cartesian e5uations of planes;
Oector E;uations o P4anes
The vector e5uation of the plane
r=a+λu+µv
Axample% Bompute the vector e5uation of a plane through point #, ? and B with position
vector%a=EF2−13GH,b=EF14−1GH,c=EF0−21GH
u=b−a=EF14−1GH−EF2−13GH=EF−15−4GH
v=c−a=EF0−21GH−EF2−13GH=EF−2−1−2GH
r=a+λu+µv
r=EF2−13GH+λEF−15−4GH+µEF−2−1−2GH
Cartesian E;uations o P4anes
8ormula for Bartesian A5uations of 9lanes
n=(b−a)6(c−a)
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r.n=a.n
a=EF2−13GH,b=EF14−1GH,c=EF0−21GH,r=EFxyzGH
n=(b−a)6(c−a)
n=EF−15−4GHEF−2−1−2GH=EF−14611GH
r.n=a.n
EFxyzGH.EF−14611GH=EF2−13GH.EF−14611GH
−14x+6y+11z=(2)(−14)+(−1)(6)+(3)(11)Bartesian A5uations of 9lanes
−14x+6y+11z=−1
(g) calculate the angle between two lines, between a line and a plane, and between two planes;
Ang&e %etween Two Dines
Ang&e %etween Two Dines
θ=β−α
tanθ=tan(β−α)=tanβ−tanα1+tanβtanα
=m1−m21+m1m29or two &ine of gradient m,: mK te acute ang&e between them is a&wa"spositive
tanθ=333m1−m21+m1m2333
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m1m ! "1: this formu&a doesnLt worM for perpendicu&ar &ines.
E#ample 1) 9ind the acute ang&e between the &ines " N O' P , and " N PK' (O.
tanθ=333m1−m21+m1m2333tanθ=3333−(−2)1+(3)(−2)333
tanθ=|−1|=1
θ=tan−1(1)=45∘
E#ample ) 9ind the acute ang&e between the &ines Q' P " ( R N S and PO'P,," (,S N S
earrange the equation
y=6x+8
y=−311x−1011
m,NQ: mKNPOU,,
tanθ=333m1−m21+m1m2333
tanθ=33336−(−311)1+(6)(−311)3333
tanθ=333−697333θ=tan−1697=84.2∘
(h) find the point of intersection of two lines, and of a line and a plane;
9ind the =oint of Intersection %etween Two Dines
At the point of intersecting &ines: the points are equa&.
C'amp&e) 9ind the point of intersection between &ines " N O' P V and " NPK'(O.
y=3x−7−−−−−(1)
y=−2x+3−−−−−(1)8ubstitute ,! into K!
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3x−7=−2x+3
5x=10
x=2
'NK:y=3(2)−7
y=−1• Wence: the intersecting point is K: P,!
• =oogle
• P 8ee more at) http)UUcoo&mathso&utions.b&ogspot.comUKS,OUSOU#ndPpointPofPintersectionPbetweenPtwo.htm&Xsthash.h"vY
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Find the 6ectorDcross product o these norma4 6ectors
n1→6n2→=3333i23 j−54k3−33333n1→6n2→=333@543−3333i−333233−3333 j+33323−54333k
=3i+15 j+23kector product is N!, $, '!F
Find the position 6ector rom the ori!in
8ind some point which lies on both the planes because then it must lie on their line of
intersection. #ny point which lies on both planes will do. Bould be plane- xy, yz, xz.0f x4,
−5y+3z=12
4y−3z=6
y=−18,z=−269oint with position vector (, -$+, -'D) lies on the line of intersection.
The e5uation of the line of intersection is
r=(0,−18,−26)+t(3,15,23)
To chec" that point that we get does really lie on both planes and so on their line ofintersection. 0f t4$
r=(3,−3,−3)Substitute into the planes e5uations
2(3)−5(−3)+3(−3)=12
3(3)+4(−3)−3(−3)=6
I 57>
2x+3z=12
3x−3z=6
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x=3.6,z=1.69oint with position vector (, !.D, $.D) lies on the line of intersection.
The e5uation of the line of intersection is
r=(3.6,0,1.6)+t(3,15,23)
To chec" that point that we get does really lie on both planes and so on their line ofintersection. 0f t4$
r=(6.6,15,24.6)Substitute into the planes e5uations
2(6.6)−5(15)+3(24.6)=12
3(6.6)+4(15)−3(24.6)=6
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Chapter 9 Limits and Continuit5
9.1 Limits
(a) determine the existence and values of the left-hand limit, right-hand limit and limit ofa function;
DeftPhand Dimit and ightPhand Dimit
A &imit is the va&ue that a function or sequence [approaches[ as the input orinde' approaches some va&ue.
The &imit of f(x) as ' approaches a from the right.
limx→a+f(x)
The &imit of f(x) as ' approaches a from the left .
limx→a−f(x)
C'amp&e) 9ind
limx→2−(x3−1)
limx→2−(x3−1)=limx→2−(23−1)=7As ' approaches K from the &eft: 'Opproaches R and 'OP, approaches V.
9ind
limx→2+(x3−1)
limx→2+(x3−1)=limx→2+(23−1)=7
•
(b) use the properties of limits;=roperties of &imits
The &imit of a constant is the constant itse&f.
limx→ak=k
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The &imit of a function mu&tip&ied b" a constant is equa& to the va&ue of thefunction mu&tip&ied b" the constant.
limx→ak.f(x)=k.limx→af(x)
The &imit of a sum or dierence! of the functions is the sum or dierence! ofthe &imits of the individua& functions.
limx→a[f(x)+g(x)]=limx→af(x)+limx→ag(x)
limx→a[f(x)−g(x)]=limx→af(x)−limx→ag(x)
The &imit of a product is the product of the &imits.
limx→af(x).g(x)=limx→af(x).limx→ag(x)
The &imit of a quotient is the quotient of the &imits
limx→af(x)g(x)=limx→af(x)limx→ag(x)
The &imit of a power is the power of the &imit.
limx→axn=an
9.* Continuit5
(c) determine the continuity of a function at a point and on an interval;
Continuit5 o a Function at A Point and Qn An Inter6a
Continuit5 at a Point
# function f (x) is continuous at a if the following three conditions are valid%
i) The function is de fined at a% That is, a is in the domain of definition of f (x)
ii) if
limx→af(x)exists.
iii) if
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limx→af(x)=f(a)
0f any of the three conditions in the definition of continuity fails when x 4 c, the functionis discontinuous at that point.
Continuit5 on An Inter6a4
# function which is continuous at every point of an open interval I is called continuouson I .
(d) use the intermediate value theorem.
/et f (x) be a continuous function on the interval Ra :
0f d is between &a+ and &:+, then a corresponding c between a and :, exists, so that
f(c)=d
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Chapter "ierentiation
.1 "eri6ati6es
(a) identify the derivative of a function as a limit;
\erivative of a 9unction as a Dimit
9or a function f'!: its derivative is de#ned as
f′(x)=lim]x→0f(x+]x)−f(x)]x
C'amp&e ,) 7ompute the derivative of f'! b" using &imit de#nition
f(x)=13x−25
f′(x)=lim]x→0f(x+]x)−f(x)]x
f′(x)=lim]x→013(x+]x)−25−{13x−25}]x=lim]x→013x+13]x−25−13x+25]x
=lim]x→013]x]x
=13
C'amp&e ,) 7ompute the derivative of f'! b" using &imit de#nition
f(x)=5−x+2−−−−√
f′(x)=lim]x→0f(x+]x)−f(x)]x
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=lim]x→0{5−(x+]x)+2−−−−−−−−−−−√ }−{5−x+2−−−−√}]x
=lim]x→0x+2−−−−√−x+]x+2−−−−−−−−−√]x
=lim]x→0x+2−−−−√−x+]x+2−−−−−−−−−√]x.x+2−−−
−√+x+]x+2−−−−−−−−−√x+2−−−−√+x+]x+2−−−−−−
−−−√
=lim]x→0(x+3)−(x+]x+3)]x{x+3−−−−√+x+]x+3−−−−−−
−−−√}
=lim]x→0−]x]x{x+3−−−−√+x+]x+3−−−−−−−−−√}
=lim]x→0−1x+3−−−−√+x+]x+3−−−−−−−−−√
=−1x+3−−−−√+x+3−−−−√
=−12x+3−−−−√
• (b) find the derivatives of xn (n $ Q ), e x , ln x( sin x, cos x, tan x, sin -1 x, cos -1 x, tan-1 x, with constant multiples, sums, differences, products, 5uotients and composites;
Common "eri6ati6es
Ta:4e o "eri6ati6es
ddx(x)=1
ddx(xn)=nxn−1
ddx(ex)=ex
ddx(lnx)=1x,x>0
ddx(sinx)=cosx
ddx(cosx)=−sinx
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ddx(tanx)=sec2x
ddx(sin−1x)=11−x2−−−−−√
ddx(cos−1x)=−11−x2−−−−−√
ddx(secx)=secxtanx
ddx(cscx)=−cscxcotx
ddx(cotx)=−csc2x
(c) perform implicit differentiation;
C'p&icit and Imp&icit \ierentiation
There are two wa"s to de#ne functions: implicitly and e#plicitly. ?ost of the equations we have dea&t with have been e'p&icit equations: such as " NO'PK. This is ca&&ed e#plicit because given an ': "ou can direct&" get f'!.
The technique of implicit di$erentiation a&&ows "ou to #nd the derivative of " with respect to ' without having to so&ve the given equation for ". Ziven
x2+y2=10
=erforming a chain ru&e to get d"Ud': so&ve d"Ud' in terms of ' and ".9ind d"Ud' for
x2+y2=10
2x+2ydydx=0
2ydydx=−2x
dydx=−xy
(d) find the first derivatives of functions defined parametrically;
First "eri6ati6e o Parametric Functions
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Parametric deri6ati6e is a derivative in calculus that is ta"en when both the x and yvariables (independent and dependent, respectively) depend on an independent thirdvariable t, usually thought of as CtimeC.
The irst deri6ati6e o the parametric e;uations is
dydx=dydtdxdt
dydx=dydt.dtdx
x=f(t),y=g(t)
E3amp4e# 8ind the first derivative, given
x=t+cost
y=sint
dydx=dydtdxdt=cost1−sint
E3amp4e# 8ind the first derivative, given
x=t4−4t2
y=t3
dydx=dydtdxdt=2t23t3−8t
.* App4ications o dierentiation
(e) determine where a function is increasing, decreasing, concave upward and concavedownward
here is a unction increasin! or decreasin!2
• Increasin! unction# if f :(x)F, function is increasing.
• "ecreasin! unction# if f :(x)N, function is decreasing
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here is unction conca6e up or conca6e do
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C'tremum =oints
?a'ima and minima are points where a function reaches a highest or
&owest va&ue: respective&"g!
h!i!
Second %erivative
If fLLL'! is positive: then it is a minimum point.If fLLL'! is ne&ative: then it is a ma#imum point.
If fLLL'!N ^ero: then it cou&d be a ma'imum: minimum or point of in_e'ion.=oint of In_e'ionAn in_ection point is a point on a curve at which the sign of thecurvature changes.
d2ydx2=0
'hird %erivative
If fLLL'! ` S There is an in(e#ion point
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(g) s"etch the graphs of functions, including asymptotes parallel to the coordinateaxes;
(h) find the e5uations of tangents and normals to curves, including parametric curves;
Tan!ents and orma4s to a Cur6e
#t point (x$,y$) on the curve y4f(x) the e;uation o tan!ent is
y−y1=m1(x−x1)where
m1=f′(x)=dydxthe gradient to the function of f(x).
Tangent is perpendicular to norma4, thus
m1m2=−1
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E3amp4e# 8ind the e5uations of the tangent line and the normal line for the curve at t4$.
x=t2
y=2t+1
dxdt=2t
dydt=2
dydx=dydt.dtdx
=2t2t4$
dydx=2Since t 4 $, x 4 $, y 4 ! A5uation of tangent
y−y1=m1(x−x1)
y−3=1(x−1)
y=x+2A5uation of normal
m1m2=−1
m2=−1y−3=−1(x−1)
y=−x+4
(i) solve problems concerning rates of change, including related rates;
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ates of 7hange and e&ated ates
e&ated rates prob&ems invo&ve #nding a rate at which a quantit" changesb" re&ating that quantit" to other quantities whose rates of change areMnown. The rate of change is usua&&" with respect to time.
E#ample) Air is being pumped into a spherica& ba&&oon such that itsradius increases at a rate of .RS cmUmin. 9ind the rate of change of itsvo&ume when the radius is cm.
The vo&ume
V=43πr3\ierentiating above equation with respect to t
dVdt=43π.3r2.drdt
dVdt=4πr2.drdt The rate of change of the radius drUdt N S.RS cmUmin because the radius isincreasing with respect to time.
dVdt=4π(5)2(0.80)
dVdt=80π cm3/min
Wence: the vo&ume is increasing at a rate of RS cmOUmin when the radiushas a &ength of cm.
(H) solve optimisation problems.
The differentiation and its applications can be used to solve practical problems. Thisinclude minimiing costs, maximiing areas, minimiing distances and so on.
$. "ia!ram - 6raw a diagram.'. oa4 - Oaximie or minimie which un"nown3!. "ata - 0ntroduce variable names.
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• Pse the Qrst or second derivative test to determine whether the critical points are localmaxima, local minima, or neither.
• Bhec" end points of f, if applicable.
Chapter Inte!ration
.1 Indeinite inte!ra4s
(a) identify integration as the reverse of differentiation;
(b) integrate xn (n $ Q ), e x , sin x, cos x, sec2 x, with constant multiples, sums anddifferences;
Integra& 7ommon 9unction
Constant
∫adx=ax+C*ariable
∫xdx=x22+CPo+er
∫xndx=xn+1n+1+Ceciprocal
∫1xdx=ln|x|+CE#ponential
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∫exdx=ex+C∫axdx=axlna+C∫ln(x)dx=x(ln(x)−1)+C'ri&onometry
∫cos(x)dx=sin(x)+C∫sin(x)dx=−cos(x)+C∫sec2(x)dx=tan(x)+C
(c) integrate rational functions by means of decomposition into partial fractions;
Inte!ration :5 Partia4 Fractions "ecomposition
Ho< to inte!rate the rationa4 unction( ;uotient o t
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∫1x2+5x+6dx8actor (x) into linear and&or 5uadratic (irreducible) factors Z find the partial fractiondecomposition
1(x+2)(x+3)=Ax+2+Bx+31=A(x+3)+B(x+2)
x=−2,A=1
x=−3,B=−1
1(x+2)(x+3)=1x+2−1x+30ntegrate the result
∫1x2+5x+6dx=∫1(x+2)(x+3)dx=∫1x+2−1x+3dx=ln|x+2|−ln|x+3|+C
E3amp4e *# Avaluate the indefinite integral
1(x−1)(x+2)(x+1)=Ax−1+Bx+2+Cx+1
6ecomposition of rational functions into partial fractions
1=A(x+2)(x+1)+B(x−1)(x+1)+C(x−1)(x+2)
x=1,6A=1,A=16
x=−1−2C=1,C=−12
x=−23B=1,B=13
1(x−1)(x+2)(x+1)=16
(1x−1
)+13
(1x+2
)−12
(1x+1
)0ntegrate the result
∫1(x−1)(x+2)(x+1)dx
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=∫16(1x−1)+13(1x+2)−12(1x+1)=16ln|x−1|+13ln|x+2|−12ln|x+1|+C
(d) use trigonometric identities to facilitate the integration of trigonometric functions;
(e) use algebraic and trigonometric substitutions to find integrals;
Inte!ration :5 Tri!onometric Su:stitution V Identities
Su:stitute one o the o44o
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.* "einite inte!ra4s
(g) identify a definite integral as the area under a curve;
(h) use the properties of definite integrals;
Properties o Inte!ra4s
Additi6e Properties
Split a definite integral up into two integrals with the same integrand but different limits
∫baf(x)dx+∫cbf(x)dx=∫caf(x)dx0f the upper and lower bound are the same, the area is .
∫aaf(x)dx=00f an interval is bac"wards, the area is the opposite sign.
∫ba
f(x)dx=−∫ab
f(x)dx
Inte!ra4 o Sum
The integral of a sum can be split up into two integrands
∫ba[f(x)+g(x)]dx=∫baf(x)dx+∫bag(x)dx
Sca4in! :5 a constant
Bonstants can be distributed out of the integrand and multiplied afterwards.
∫bacf(x)dx=c∫baf(x)dx
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Tota4 Area ithin an Inter6a4
∫baf(x)dx=F(b)−F(a)∫ba|f(x)|dx=F(b)+F(a)
Inte!ra4 ine;ua4ities
0f
f(x)≥0and a
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0f y 4 f(x) is continuous and f(x) N on [a, b\, then the area under the curve from a to bis%
Area=−∫baf(x)dx
0f y 4 f(x) is continuous and f(x) N and f(x) F
Area=∫ba
f(x)d(x)+333∫cb
f(x)d(x)333+∫dc
f(x)d(x)
0f x 4 g (y ) is continuous and non-negative on [c, d\, then the area under the curve of gfrom c to d is%
Area=∫dcg(y)dy
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(") calculate volumes of solids of revolution about one of the coordinate axes.
Chapter 1>
"ierentia4 E;uations
• (a) find the general solution of a first order differential e5uation with separable variables;
• (b) find the general solution of a first order linear differential e5uation by means of anintegrating factor;
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• (c) transform, by a given substitution, a first order differential e5uation into one withseparable
• variables or one which is linear;
• (d) use a boundary condition to find a particular solution;
• (e) solve problems, related to science and technology, that can be modelled by differentiale5uations.
Chapter 11
Mac4aurin Series
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(a) find the Oaclaurin series for a function and the interval of convergence;
?ac&aurin 8eries
'aylor series is a representation of a function as an in#nite sum of termsthat are ca&cu&ated from the va&ues of the functionLs derivatives at a sing&epoint.'aylor Series
f(x)=∑ x=0∞fn(a)n!(x−a)n=f(a)+f′(a)(x−a)+f′(a)2!(x−a)2+f′′(a)3!(x−a)3+...+f(n)(a)n!
(x−a)n+...
-acLaurin series is the Ta"&or series of the function about #./0
f(x)=∑ x=0∞fn(0)n!xnf(x)=f(0)+f′(0)x+f′′(0)2!x2+f′′′(0)3!x3+...+f(n)(0)n!xn+...
E#ample ) 9ind the ?ac&aurin 8eries e'pansion of e'
f(x)=e5x
f′(x)=5e5x
f′′(x)=52e5x
f′′′(x)=53e5x
f(4)(x)=54e5x
f(0)=1
f′(0)=5
f′′(0)=52
f′′′(0)=53
f(4)=54
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f(x)=f(0)+f′(0)x+f′′(0)2!x2+f′′′(0)3!x3+...+f(n)(0)n!xn+...
e5x=1+5x1!+52x22!+53x33!+...+5nxnn!
=∑ n=0∞5nxnn!
Inter6a4 o Con6er!ence
The set of points where the series converges is called the inter6a4 o con6er!enceFindin! inter6a4 o con6er!ence$) perform ratio test to test for the convergence of a series.') Bhec" endpoint
$atio test
L=limn→∞333an+1an333
if / N $ then the series converges.if / F $ then the series does not converge;if / 4 $ or the limit fails to exist, then the test is inconclusive
E3amp4e# 6etermine the interval of convergence for the series
∑ n=1∞
(x−2)
nn.5n
#pply ratio test
limn→∞333an+1an333=333(x−2)n+1(n+1).5n+1.n.5n(x−2)n333
=limn→∞333x−25.nn+1333
=15|x−2|limn→∞333nn+1333
=15|x−2|#s n approcaches infinity, n&n>$ aprpoaches $
limn→∞333nn+1333]1 The series converges for
15|x−2|
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Bhec" end point x4E,
∑ n=1∞(5)nn.5n=∑ n=1∞1nThis is the harmonic series, and it diverges.
Bhec" end point x4-!,∑ n=1∞(−5)nn.5n=∑ n=1∞(−1)n1nThe series converges by the #lternating Series Test.
The inter6a4 o con6er!ence is
−3≤x
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x4+x2*ewrite the function as
x4+x2=x(14+x2)=x4EF11+x24GH
=x4EF11−(−x24)GHJThis is the sum of the infinite geometric series with the first term x&7 and ratio x'&7
=x4∑ n=0∞(−x24)=x4∑ n=0∞(−1)nx2n4n=∑ n=0∞(−1)nx2n+14n+1
(c) perform differentiation and integration of a power series;
(d) use series expansions to find the limit of a function.
Chapter 1* umerica4 Method
1*.1 umerica4 so4ution o e;uations
(a) locate a root of an e5uation approximately by means of graphical considerations and by searching for a sign change;
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(b) use an iterative formula of the form xn>$ 4f(xn) to find a root of an e5uation to a prescribed degree of accuracy;
(c) identify an iteration which converges or diverges;
(d) use the 1ewton-*aphson method;
1*.* umerica4 inte!ration
(e) use the trapeium rule;
(f) use s"etch graphs to determine whether the trapeium rule gives an over-estimate oran under-estimate in simple cases.
Chapter 1'
"ata "escription
(a) identify discrete, continuous, ungrouped and grouped data;
"iscrete( Continuous( Un!rouped and rouped "ata
"iscrete "ata
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6iscrete 6ata is counted and can only ta"e certain values (whole numbers).Ag 1umber of students in a class, 1umber of children in a playground, etc.
Continuous "ata
Bontinuous 6ata is data that can ta"e any value (within a range)Ag. eight of children, weights of car, etc.
rouped "ata
6ata that has been organied into groups (into a fre5uency distribution).
Ag.Class
0−5
6−10
11−15
16−20
Frequency
10
20
17
4
Un!rouped "ata
6ata that has not been organied into groups$ $ ' 7 + 7 DE E+ $ $$ '! D Y + $
(b) construct and interpret stem-and-leaf diagrams, box-and-whis"er plots, histogramsand cumulative fre5uency curves;
Stem0and0Lea "ia!rams
# stem0and04ea dia!rams presents 5uantitative data in a graphical format, similar to ahistogram, to assist in visualiing the shape of a distribution, giving the reader a 5uic"overview of distribution.
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-, - - - ' , 9' 9' 9, 9 1 , 1>9
7 M D + Y M
D M ! D + +E M ! ! D+ M $ +$ M E"ey 7 M means 7
Bo30and0hiser P4ots
Bo30and0hiser P4ots displays of the spread of a set of data through five-number
summaries% the minimum, lower 5uartile ($), median ('), upper 5uartile(!), andmaximum.
• The first and third 5uartiles are at the ends of the box,
• The median is indicated with a vertical line in the interior of the box
• Ands of the whis"ers indicated the maximum and minimum.
Histo!rams and Cumu4ati6e Fre;uenc5 Histo!rams
Histo!rams and Cumu4ati6e Fre;uenc5 Histo!rams
# histo!ram is constructed from a fre5uency tableThe cumu4ati6e re;uenc5 is the running total of the fre5uencies.
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(c) state the mode and range of ungrouped data;
Mode o un!rouped data
#n observation occurring most fre5uently in the data is called mode of the data
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E3amp4e# 8ind the median of the following observations
$, $7, $D, ', '7, '+, '+, !, !', 7
0n the given data, the observation '+ occurs maximum. So the mode is *.
$an!e o un!rouped data
$an!e 7 Hi!hest Oa4ue ) Lo
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E3amp4e 1# 8ind the median, lower 5uartile and upper 5uartile of the following numbers.
1*( ,( **( '>( 9( '( 1( -*( 1,( -'( ',
*earrange the data in ascending order%
$ 44 $&7 ($$>$) 4 !th observation, $4$'' 44 $&' ($$>$) 4 Dth observation, '4''! 44 !&7 ($$>$) 4 Yth observation, !4!+
Inter;uarti4e $an!e 4 ! -$ 4 !+ - $' 4 'D$an!e 4 /argest value - smallest value 4 7! - 4 !+
E3amp4e *# 8ind the median, lower 5uartile and upper 5uartile of the following numbers.
1*( ,( **( '>( 9( '( 1( -*( 1,( -'( ',( ,>
*earrange the data in ascending order%
Q1=(12+152)=13.5Q2=(22+302)=26Q3=
(38+422
)=40
Inter;uarti4e $an!e 4 ! -$ 4 7 - $!. 4 'D.$an!e 4 /argest value - smallest value 4 - 4 7
Median and Inter;uarti4e $an!e o rouped "ata
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Median=Lm+(n2−Ffm)Cn 4 the total fre5uency/m 4 the lower boundary of the class median8 4 the cumulative fre5uency before classmedianf 4 the fre5uency of the class medianf m4 the lower boundary of the class medianB4 the class width
Q1=LQ1+(n4−FfQ1)CQ3=LQ3+EF3n4−FfQ3GHC
E3amp4e % 8ind the median and inter5uartile range of below data.
Q2=20.5+EF502−2312GH10=22.167
Q1=10.5+EF504−1010GH10=13
Q3=30.5+EF3(50)4−358GH10=33.625
(e) calculate the mean and standard deviation of ungrouped and grouped data, from rawdata and from given totals such as
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∑ i=1n(xi−a)and ∑ i=1n(xi−a)2
Mean and Standard "e6iation o Un!rouped and rouped "ata
Un!rouped "ata
Oean
x̄=∑xn
Standard 6eviation
s=∑(x−x̄)2n−−−−−−−−−√ s=∑x2n−(∑xn)2−−−−−−−−−−−−−− ⎷ ̂ ^
rouped "ata
Oean
x̄=∑fx∑f
Standard 6eviation
s=∑f(x−x̄)2n−−−−−−−−−−√ s=∑fx2∑f−(∑fx∑f)2−−−−−−−−−−−−−−−− ⎷ ̂ ^
(f) select and use the appropriate measures of central tendency and measures ofdispersion;
Measures o Centra4 Tendenc5
Oeasure of central tendency is an average of a set of measurements.
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• Mode 0 the number that occurs most fre5uently.
• Median 0 the value of the middle item in a set of observations.
• Mean 0 average value of the distribution.
Measures o "ispersion
Oeasures of 6ispersion is group of analytical tools that describes the spread or variability ofa data set.
• $an!e - 6ifference between the largest and smallest sample values.
• OarianceD Standard "e6iation - Oeasures the dispersion around an average.
• Coeicient o 6ariation - Axpressed in a relative value.
• (g) calculate the 9earson coefficient of s"ewness;
Pearson coeicient o se
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Positi6e se
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Addition Princip4e
0f we want to find the probability that event # happens or event ? happens, we shouldadd the probability that # happens to the probability that ? happens.
Addition $u4e#
P&A or B+ 7 P&A+ 8 P&B+
E3amp4e# # single D-sided die is rolled.
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7omp&ementar": C'haustive and ?utua&&" C'c&usive Cvents
7omp&ementar" Cvent
The comp&ement of an" event A: AL is the event that A does not occur.
C'haustive Cvent
A set of events is co&&ective&" e'haustive if at &east one of the events mustoccur. 9or e'amp&e: when ro&&ing a si'Psided die: the outcomes ,: K: O: : :and Q are co&&ective&" e'haustive: because the" inc&ude the entire range of possib&e outcomes. Thus: a&& samp&e spaces are co&&ective&" e'haustive.
P(AB)=1
-utually E#clusive Event
Two events are Lmutua&&" e'c&usiveL if the" cannot occur at the same time.C'amp&e: tossing a coin once: which can resu&t in either heads or tai&s: but notboth.
P(A∩B)=0
P(AB)=P(A)+P(B)
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(e) use the formula
Independent Cvents 7onditiona& =robabi&ities
Formu4a
P(AB)=P(A)+P(B)−P(A∩B)
Independent E6ents
Two events are independent means that the occurrence o one does not aect thepro:a:i4it5 o the other. Two events, # and ?, are independent if and only if
P(A∩B)=P(A).P(B)
Conditiona4 Pro:a:i4it5
Conditiona4 Pro:a:i4it5 is the probability that an event will occur, when another event is"nown to occur or to have occurred.
=iven two events # and ?, , the conditional probability of # given ? is defined as the5uotient of the Hoint probability of # and ?, and the probability of ?
P(A|B)=P(A∩B)P(B)
P(A∩B)=P(A|B).P(B)
P&A B+ 7 P&A+ 8 P&B+ 0 P&A ∩ B+W
• (f) calculate conditional probabilities, and identify independent events;
• (g) use the formulae
P&A ∩ B+ 7 P&A+ 3 P&BA+ 7 P&B+ 3 P&A B+W
(h) use the rule of total probability.
http://coolmathsolutions.blogspot.com/2013/03/independent-events-conditional.htmlhttp://coolmathsolutions.blogspot.com/2013/03/independent-events-conditional.htmlhttp://coolmathsolutions.blogspot.com/2013/03/independent-events-conditional.htmlhttp://coolmathsolutions.blogspot.com/2013/03/the-fundamental-laws-of-set-algebra.htmlhttp://coolmathsolutions.blogspot.com/2013/03/the-fundamental-laws-of-set-algebra.htmlhttp://coolmathsolutions.blogspot.com/2013/03/independent-events-conditional.htmlhttp://coolmathsolutions.blogspot.com/2013/03/independent-events-conditional.htmlhttp://coolmathsolutions.blogspot.com/2013/03/independent-events-conditional.htmlhttp://coolmathsolutions.blogspot.com/2013/03/the-fundamental-laws-of-set-algebra.html
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The 9undamenta& Daws of 8et A&gebra
Commutative la+s
AB=BA
A∩B=B∩A
2ssociative La+s
(AB)C=A(BC)
(A∩B)∩C=A∩(B∩C)
%istributive La+s
A(B∩C)=(AB)∩(AC)
A∩(BC)=(A∩B)(A∩C)
3mpotent La+s
A∩A=A
AA=A
%omination La+s
AU=U
A∩=
2bsorption La+s
A(A∩B)=A
A∩(AB)=A
3nverse La+s
AA′=U
A∩A′=
• La+ of Complement
(A′)′=A
U′=
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′=U
• %e-or&an4s La+
(AB)′=A′∩B′
(A∩B)′=A′B′
• elative Complement of 5 in 2
A−B=A∩B′
Chapter 1, Pro:a:i4it5 "istri:utions
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1,.1 "iscrete random 6aria:4es
Var(X)=E(X2)−[E(X)]2
Var(X)=σ2x=∑ [x2i6P(xi)]−µ2x
• (a) identify discrete random variables;
• (b) construct a probability distribution table for a discrete random variable;
• (c) use the probability function and cumulative distribution function of a discrete randomvariable;
• (d) calculate the mean and variance of a discrete random variable;
"iscrete random 6aria:4es
# discrete 6aria:4e is a variable which can only ta"e a counta:4e num:er o 6a4ues.
# discrete random variable _ is uni5uely determined by
•
0ts set of possible values _
• 0ts probability density function (pdf)% # real-valued function f (`) deQned for each x$_ as the probability that _ has the value x.
• Pro:a:i4it5 "ensit5 Function
f(x)=Pr(X=x)
∑ x=inf(xi)=1• Cumu4ati6e "istri:ution Function 8(x) is defined to be
F(x)=P(X≤x)
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"iscrete Pro:a:i4it5 "istri:ution
The probability that random variable _ will ta"e the value xi is denoted by p(xi) whereP(xi)=P(X=xi)
Mean and Oariance of a Discrete Random Variable
Oean
E(X)=µx=∑ [xi6P(xi)]ariance
1,.* Continuous random 6aria:4es
• (e) identify continuous random variables;
• (f) relate the probability density function and cumulative distribution function of acontinuous random variable;
• (g) use the probability density function and cumulative distribution function of acontinuous random variable;
• (h) calculate the mean and variance of a continuous random variable;
Continuous $andom Oaria:4es
# random variable _ is continuous if its set of possible values is an entire interval ofnumbers
Pro:a:i4it5 "ensit5 Function
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/et _ be a continuous random variables. Then a pro:a:i4it5 distri:ution or pro:a:i4it5densit5 unction &pd+ o is a function f (x) such that for any two numbers a and b,
P(a≤X≤b)=
∫baf(x)dx
The graph of f is the density curve.
#rea of the region between the graph of f and the x U axis is e5ual to $.
The Cumu4ati6e "istri:ution Function
The cumulative distribution function, 8(x) for a continuous random variables _ is definedfor every number x by
F(x)=P(X≤x)=∫x−∞f(y)dy
P(a≤X≤b)=F(b)−F(a)8or each x, 8(x) is the area under the density curve to the left of x.
E3pected Oa4ue
The expected or mean value of a continuous random variables _ with pdf f(x) is
E(X)=µx=∫∞−∞x.f(x)dx
Oariance
The variance of continuous random variables _ with pdf f(x) is
σ2x=Var(X)=∫∞−∞(x−µ)2.f(x)dxE[(x−µ)2]
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Var(X)=E(X2)−[E(x)]2
1,.' Binomia4 distri:ution
• (i) use the probability function of a binomial distribution, and find its mean and variance;
• (H) use the binomial distribution as a model for solving problems related to science andtechnology;
Binomia4 "istri:ution
The :inomia4 distri:ution is used when there are exactly two mutually exclusive outcomes
of a trial. These outcomes are appropriately labeled CsuccessC and CfailureC.
0f the random variable _ follows the binomial distribution with parameters n and p, _ (?(n,9), the probability of getting exactly k successes in n trials is given by the probability massfunction
f(k;n, p)=Pr(X=k)=(nk) pk(1− p)n−k0f _ ?(n, p), _ is a binomially distributed random variable, then the expected value of_ is
Mean =npand the variance is
Variance =np(1− p)
n number of successes p is the probability of success in ?inomial 6istribution, assumes that p is fixed for alltrials.
1,.- Poisson distri:ution
• (") use the probability function of a 9oisson distribution, and identify its mean andvariance;
• (l) use the 9oisson distribution as a model for solving problems related to science andtechnology;
http://coolmathsolutions.blogspot.com/2013/03/binomial-distribution.htmlhttp://coolmathsolutions.blogspot.com/2013/03/binomial-distribution.htmlhttp://coolmathsolutions.blogspot.com/2013/03/binomial-distribution.htmlhttp://coolmathsolutions.blogspot.com/2013/03/poisson-distribution.htmlhttp://coolmathsolutions.blogspot.com/2013/03/poisson-distribution.htmlhttp://coolmathsolutions.blogspot.com/2013/03/poisson-distribution.htmlhttp://coolmathsolutions.blogspot.com/2013/03/poisson-distribution.htmlhttp://coolmathsolutions.blogspot.com/2013/03/poisson-distribution.htmlhttp://coolmathsolutions.blogspot.com/2013/03/binomial-distribution.htmlhttp://coolmathsolutions.blogspot.com/2013/03/binomial-distribution.htmlhttp://coolmathsolutions.blogspot.com/2013/03/binomial-distribution.htmlhttp://coolmathsolutions.blogspot.com/2013/03/poisson-distribution.htmlhttp://coolmathsolutions.blogspot.com/2013/03/poisson-distribution.htmlhttp://coolmathsolutions.blogspot.com/2013/03/poisson-distribution.htmlhttp://coolmathsolutions.blogspot.com/2013/03/poisson-distribution.html
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=oisson \istribution
Poisson distribution is a discrete probabi&it" distribution that e'pressesthe probabi&it" of a given number of events occurring in a #'ed interva& of time andUor space if these events occur with a Mnown average rate and
independent&" of the time since the &ast event.A discrete stochastic variab&e is said to have a =oisson distribution withparameter S: if for M N S: ,: K: ... the probabi&it" mass function of isgiven b"
f(k;λ)=Pr(X=k)=λke−λk!
is the number of occurrences of an eventj the probabi&it" of which is givenb" the function is a positive rea& number.
?ean and
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The norma4 distri:ution is a continuous probability distribution, defined by the formula
f(x)=1σ2π−−√e−(x−µ)22σ2The normal distribution is also often denoted
Xk N(µ,σ2)
Standard orma4 "istri:ution
0f 4 and 4 $, the distribution is called the standard norma4 distri:ution.
8ormula for -score%
z=x−µσ
• is the C-scoreC (Standard Score)
• x is the value to be standardied
• is the mean
• is the standard deviation
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orma4 Appro3imations
Binomia4 Appro3imation
The normal distribution can be used as an approximation to the binomial distribution,under certain circumstances, namely%
0f _ ?(n, p) and if n is 4ar!e and&or p is c4ose to X, then _ is approximately 1(np,np5)
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Poisson Appro3imation
The normal distribution can also be used to approximate the 9oisson distribution for4ar!e 6a4ues o (the mean of the 9oisson distribution).
0f _ 9o() then for large values of l, _ 1(, ) approximately.
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Chapter 1
Samp4in! and Estimation
1.1 Samp4in!
• (a) distinguish between a population and a sample, and between a parameter and astatistic;
• (b) identify a random sample;
• (c) identify the sampling distribution of a statistic;
• (d) determine the mean and standard deviation of the sample mean;
• (e) use the result that _ has a normal distribution if _ has a normal distribution;
• (f) use the central limit theorem;
• (g) determine the mean and standard deviation of the sample proportion;
• (h) use the approximate normality of the sample