Core Maths C3. Revision Notes


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1 Core Maths C Revision Notes October 0
2 Core Maths C Algebraic fractions... Cancelling common factors... Multipling and dividing fractions... Adding and subtracting fractions... Equations... 4 Functions... 4 Notation... 4 Domain, range and graph... 4 Defining functions... 5 Composite functions... 6 Inverse functions and their graphs... 6 Modulus functions... 7 Modulus functions f ()... 7 Modulus functions f ( )... 8 Standard graphs... 8 Combinations of transformations of graphs... 9 Trigonometr... 0 Sec, cosec and cot... 0 Graphs... 0 Inverse trigonometrical functions... Graphs... Trigonometrical identities... Finding eact values... Proving identities.... Eliminating a variable between two equations... Solving equations... R cos( + α)... 4 Eponentials and logarithms... 4 Natural logarithms... 4 Definition and graph... 4 Equations of the form e a + b p Differentiation... 5 Chain rule... 5 Product rule... 6 Quotient rule... 6 Derivatives of e and log e ln d... 8 d d d Trigonometric differentiation... 8 Chain rule further eamples... 9 Trigonometr and the product and quotient rules... 9 C 4/04/ SDB
3 6 Numerical methods... 0 Locating the roots of f() The iteration n + g( n )... 0 Conditions for convergence... Inde... C 4/04/ SDB
4 Algebraic fractions Cancelling common factors Eample: Simplif. 4 6 Solution: First factorise top and bottom full ( ) ( )( + ) 4 6 4( 9) 4( )( + ) and now cancel all common factors, in this case ( ) to give ( + ) Answer. 4( + ) Multipling and dividing fractions This is just like multipling and dividing fractions with numbers and then cancelling common factors as above. Eample: Simplif Solution: First turn the second fraction upside down and multipl 9 4 +, and then factorise full ( )( + ) ( + ), and cancel all common factors ( ) ( + )( ) Answer ( + ). Adding and subtracting fractions Again this is like adding and subtracting fractions with numbers; but finding the Lowest Common Denominator can save a lot of trouble later. Eample: Solution: Simplif 7 + First factorise the denominators ; we see that the L.C.D. is ( )( 4)( ) ( )( 4) ( )( ) ( ) 5( 4) ( )( 4)( ) ( )( )( 4) ( )( )( 4) ( )( )( 4) which cannot be simplified further. C 4/04/ SDB
5 Equations Eample: Solution: Solve + First multipl both sides b the Lowest Common Denominator which in this case is ( + ) ( ) ( + ) ( + ) , ( + )( ) 0 or Functions A function is an epression (often in terms of ) which has onl one value for each value of. Notation + 7, f () + 7 and f : + 7 are all was of writing the same function. Domain, range and graph The domain is the set of values which can take: this is sometimes specified in the definition and sometimes is evident from the function: e.g. can onl take positive or zero. The range is that part of the ais which is used. Eample: Find the range of the function f : with domain is a real number: <. Solution: First sketch the graph for values of between and and we can see that we are onl using the ais from 7 to, and so the range is is a real number: 7 < C 4/04/ SDB
6 Eample: Find the largest possible domain and the range for the function f : +. Solution: First notice that we cannot have the square root of a negative number and so cannot be negative 0 largest domain is real:. To find the range we first sketch the graph and we see that the graph will cover all of the ais from upwards and so the range is real: Eample: Find the largest possible domain and the range for the function f :. + Solution: The onl problem occurs when the denominator is 0, and so cannot be. Thus the largest domain is real:. To find the range we sketch the graph and we see that can take an value ecept, so the range is real: Defining functions Some mappings can be made into functions b restricting the domain. Eamples: ) The mapping where R is not a function as 9 is not defined, but if we restrict the domain to positive or zero real numbers then f : where R, 0 is a function. ) but f : where R is not a function as the image of is not defined, where R, is a function. C 4/04/ SDB 5
7 Composite functions To find the composite function fg we must do g first. Eample: f : and g : +. Find fg and gf. Solution: f is Think of f and g as rules times b subtract g is square add fg is square add times b subtract giving ( + ) + fg : + or fg() +. gf is times b subtract square add giving ( ) gf : or gf () Inverse functions and their graphs The inverse of f is the opposite of f: thus the inverse of multipl b is divide b and the inverse of square is square root. The inverse of f is written as f : note that this does not mean over f. The graph of f () is the reflection in of the graph of f () To find the inverse of a function (i) interchange and (ii) find in terms of. Eample: Find the inverse of f :. Solution: We have (i) interchanging and 6 (ii) solving for f : f ()  f () C 4/04/ SDB
8 Eample: Find the inverse of g : +. 5 Solution: We have + 5 (i) (ii) interchanging and + 5 solving for ( 5) g () ( ) g () g : 5 + Note that f f () f f (). Modulus functions Modulus functions f () f () is the positive value of f (), so to sketch the graph of f () first sketch the graph of f () and then reflect the part(s) below the ais to above the ais. Eample: Sketch the graph of. Solution: First sketch the graph of. 5 Then reflect the portion between 0 and in the ais 5 to give 5 5 C 4/04/ SDB 7
9 Modulus functions f ( ) In this case f ( ) f (), f ( 5) f (5), f ( 8.7) f (8.7) etc. and so the graph on the left of the ais must be the same as the graph on the right of the ais, so to sketch the graph, first sketch the graph for positive values of onl, then reflect the graph sketched in the ais. Eample: Sketch the graph of. Solution: First sketch the graph of for onl positive values of Then reflect the graph sketched in the ais to complete the sketch Standard graphs and and C 4/04/ SDB
10 Combinations of transformations of graphs We know the following transformations of graphs: f () a translated through b becomes f (  a) + b stretched factor a in the direction becomes a f () stretched factor a in the direction becomes f a reflected in the ais becomes f () reflected in the ais becomes f ( ) We can combine these transformations: Eamples: ) f ( ) is the image of f () under a stretch in the ais of factor followed b a translation, or the translation followed b the stretch. 0 ) + 6 is the image of f () under a stretch in the ais of factor 0 followed b a translation, 6 BUT these transformations cannot be done in the reverse order. To do a translation before a stretch we have to notice that + 6 ( 0 + ) which is the image of f () under a translation of followed b a stretch in the ais of factor. ) sin( + π) is the image of f () under a reflection in the ais followed π b a translation of, or the translation followed b the reflection. 0 C 4/04/ SDB 9
11 Trigonometr Sec, cosec and cot Secant is written sec ; cosecant is written cosec cos cos cotangent is written cot tan sin Graphs sin and sec cosec cot Notice that our calculator does not have sec, cosec and cot buttons so to solve equations involving sec, cosec and cot, change them into equations involving sin, cos and tan and then use our calculator as usual. Eample: Find cosec 5 o. Solution: cosec 5 o o. 74 to 4 S.F. sin Eample: Solve sec. for 0 π c Solution: sec.. cos 0. 5 cos..5 or π radians 0 C 4/04/ SDB
12 Inverse trigonometrical functions The inverse of sin is written as arcsin or sin and in order that there should onl be one value of the function for one value of we restrict the domain to π / π /. Similarl for the inverses of cos and tan, as shown below. Graphs arcsin arccos arctan R π / arcsin π / 0 arccos π π / < arctan < π / Trigonometrical identities You should learn these sin θ + cos θ tan θ + sec θ + cot θ cosec θ sin (A + B) sin A cos B + cos A sin B sin (A B) sin A cos B cos A sin B sin A sin A cos A cos (A + B) cos A cos B sin A sin B cos (A B) cos A cos B + sin A sin B cos A cos A sin A cos A sin A sin A ½ ( cos A) cos A ½ ( + cos A) sin ½ θ ½ ( cos θ) cos ½ θ ½ ( + cos θ) sin A sin A 4 sin A cos A 4 cos A cos A tan A + tan B tan( A + B)  tan A tan B tan A  tan B tan( A  B) + tan A tan B tan tan A tan  tan A sin P + sin Q sin P sin Q cos P + cos Q A A tan A tan A tan A sin cos cos P + Q P + Q P + Q P  Q cos P  Q sin P  Q cos P + Q P  Q cos P cos Q sin sin sin A cos B sin(a + B) + sin(a B) cos A sin B sin(a + B) sin(a B) cos A cos B cos(a + B) + cos(a B) sin A sin B cos(a + B) cos(a B) C 4/04/ SDB
13 Finding eact values When finding eact values ou ma not use calculators. Eample: Find the eact value of cos 5 o Solution: We know the eact values of sin 45 o, cos 45 o and sin 0 o, cos 0 o so we consider cos 5 cos (45 0) cos 45 cos0 + sin 45 sin Eample: Given that A is obtuse and that B is acute, and sin A / 5 and cos B 5 / find the eact value of sin (A + B). Solution: We know that sin (A + B) sin A cos B + cos A sin B so we must first find cos A and sin B. Using sin θ + cos θ cos A 9 / 5 6 / 5 and sin B 5 / / 69 cos A ± 4 / 5 and sin B ± / But A is obtuse so cos A is negative and B is acute so sin B is positive cos A 4 / 5 and sin B / sin (A + B) / 5 5 / + 4 / 5 / / 65 Proving identities. Start with one side, usuall the L.H.S., and fiddle with it until it equals the other side. Do not fiddle with both sides at the same time. cos Eample: Prove that A + cot A.  cos A cos A + cos A  + cos A Solution: L.H.S. cot A. Q.E.D.  cos A  ( sin A) sin A Eliminating a variable between two equations Eample: Eliminate θ from sec θ, tan θ. Solution: We remember that tan θ + sec θ sec θ tan θ. secθ + and tanθ ( + ) ( + ) +. Solving equations Here ou have to select the best identit to help ou solve the equation. Eample: Solve the equation sec A tan A, for 0 A 60 o. Solution: We know that tan θ + sec θ tan A + tan A C 4/04/ SDB
14 tan A + tan A 0, factorising gives (tan A )(tan A + ) 0 tan A or tan A A 45, 5, or 6.6, R cos( + α) An alternative wa of writing a cos ± b sin using one of the formulae listed below (I) R cos ( + α) R cos cos α R sin sin α (II) R cos ( α) R cos cos α + R sin sin α (III) R sin ( + α) R sin cos α + R cos sin α (IV) R sin ( α) R sin cos α R cos sin α To keep R positive and α acute we select the formula with corresponding + and signs. The technique is the same which ever formula we choose. Eample: Solve the equation sin 5 cos 6 for 0 o Solution: First re write in the above form: notice that the sin is +ve and the cos is ve so we need formula (IV). R sin ( α) R sin cos α R cos sin α sin 5 cos Equating coefficients of sin, R cos α Equating coefficients of cos, R sin α 5 [i] [ii] Squaring and adding [i] and [ii] R cos α + R sin α + 5 R (cos α + sin α) but cos α + sin α R 69 R ± But choosing the correct formula means that R is positive R + Substitute in [i] cos α / α.60 o or... But choosing the correct formula means that α is acute α.60 o. sin 5 cos sin(.60) and so to solve sin 5 cos 6 we need to solve sin(.60) 6 sin(.60) 6 / or o or 75. o. C 4/04/ SDB
15 Eample: Find the maimum value of sin 5 cos and the value(s) of for which it occurs. Solution: From the above eample sin 5 cos sin(.6). The maimum value of sin(anthing) is and occurs when the angle is 90, 450, 80 etc. i.e n the ma value of sin(.6) is when n n o. 4 Eponentials and logarithms Natural logarithms Definition and graph e.78 and logs to base e are called natural logarithms. log e is usuall written ln. e ln Note that e and ln are inverse functions and that the graph of one is the reflection of the other in the line. 4 Graph of e (a + b) + c. e e The graph of e is the graph of 6 e stretched b a factor of / in 4 the direction of the ais.  4 C 4/04/ SDB
16 The graph of e ( + ) is the graph of e translated through since 0 + ( ) and the graph of e ( +) + 4 is that of e ( + ) 0 translated through 4 e e + Equations of the form e a + b p Eample: Solve e + 5. Solution: Take the natural logarithm of each side, remembering that ln is the inverse of e. ln(e + ) ln 5 + ln 5 ln5 Eample: Solve ln( 5) 4. Solution: ln to S.F. Raise both sides to the power of e, remembering that e is the inverse of e ln( 5) e 4 ( 5) e 4 e to S.F. 5 Differentiation Chain rule If is a composite function like (5 7) 9 think of as u 9, where u 5 7 then the chain rule gives d d d d d d d du du d 8 du 9 u d 8 9(5 7) (0 ) 90 (5 The rule is ver simple, just differentiate the function of u and multipl b 7) 8. du. d C 4/04/ SDB 5
17 Eample: ( d ). Find. d Solution: ( ) ( ). Put u u d d du d du d d du u ( ) ( ) d d d d ( ). Product rule If is the product of two functions, u and v, then d dv du uv u + v. d d d Eample: Differentiate ( 5). Solution: ( 5) ( 5) so put u and v ( 5) d d dv du u + v d d ) ( 5 + ( 5) Quotient rule If is the quotient of two functions, u and v, then du dv v u u d d d. v d v Eample: Differentiate + 5 Solution:, so put u and v du dv v u d d d d v 6 C 4/04/ SDB
18 ( + 5) ( ( ) ( + 5) + 5) + 0 ( (4 + 5) + 4 5) ( ). Derivatives of e and log e ln. e ln d d e d d ln k ln k + ln ln k k ln or use the chain rule. d 0 + d d d k k Eample: Find the derivative of f () 5e at the point where. Solution: f () 5e f () 5e f () 5e 4.9 Eample: Differentiate the function f () ln ln 5 Solution: f () ln ln 5 ln + ln 5 ln ln 4 ln f () 4 or we can use the chain rule f () Eample: Find the derivative of f () log 0. Solution: f () log 0 ln + ln0 ln ln ln0 ln ln 0 + ln ln 0 (using change of base formula) f () 0 +. ln0 ln0 C 4/04/ SDB 7
19 Eample: d e. Find. d Solution: e u, where u d d du d du d d du e d e u d e. Eample: ln 7 d. Find. d Solution: ln u, where u 7 d d d d du d du d du u d 7. d d d d Eample: sin d. Find. d Solution: d First find d as this is easier. d d sin sin cos 6 sin cos d cosec 6 d 6 sincos sin 6 d d Trigonometric differentiation must be in RADIANS when differentiating trigonometric functions. f () f () important formulae sin cos cos sin sin + cos tan sec sec sec tan tan + sec cot cosec cosec cosec cot + cot cosec 8 C 4/04/ SDB
20 Chain rule further eamples Eample: sin 4. d Find. d Solution: sin 4. Put u sin u 4 d d du d du d d du 4u 4sin d d cos. Eample: d e sin. Find. d Solution: e u, where u sin d d du d du d d du e d d e sin cos cos e sin. Eample: d ln (sec ). Find. d Solution: ln u, where u sec d d du d du d d du sec tan d u d sec tan. Trigonometr and the product and quotient rules Eample: Differentiate cosec. Solution: cosec so put u and v cosec d dv du u + v d d d ( cosec cot ) + cosec cosec cot + cosec. C 4/04/ SDB 9
21 tan Eample: Differentiate 7 tan Solution:, so put u tan and v 7 7 d 7 sec tan d (7 ) d d d d 4 sec 6 sec tan tan 6 Numerical methods Locating the roots of f() 0 A quick sketch of the graph of f () can help to give a rough idea of the roots of f () 0. If f () changes sign between a and b and f () is continuous in this region then a root of f () 0 lies between a and b. Eample: Show that a root of the equation f () lies between and 4. Solution: f () , and f (4) Thus f () changes sign and f () is continuous there is a root between and 4. The iteration n + g( n ) Man equations of the form f () 0 can be re arranged as g () which leads to the iteration n + g ( n ). Eample: Show that the equation can be re arranged as and find a solution correct to four decimal places. Solution: From the previous eample we know that there is a root between and 4, so starting with we use the iteration n to give n C 4/04/ SDB
22 As 8 and 9 are both equal to.96 to 4 D.P. we conclude that a root of the equation is.96 to 4 D.P. Note that we have onl found one root: the sketch shows that there are more roots. Conditions for convergence If an equation is rearranged as g() and if there is a root α then the iteration n+ g( n ), starting with an approimation near α (i) will converge if  < g (α) < 4 5 (a) will converge without oscillating (b) will oscillate and converge if 0 < g (α) <, if < g (α) < 0, (ii) will diverge if g (α) <  or g (α) >. 0 C 4/04/ SDB
23 Inde algebraic fractions adding and subtracting, equations, 4 multipling and dividing, chain rule, 5 further eamples, 9 cosecant, 0 cotangent, 0 derivative e, 7 ln, 7 differentiation d, 8 d d d trigonometric functions, 8 equations graphical solutions, 0 eponential e a + b p, 5 graph of e (a + b) + c, 4 graph of e, 4 functions, 4 combining functions, 6 defining functions, 5 domain, 4 inverse functions, 6 modulus functions, 7 range, 4 graphs combining transformations, 9 standard functions, 8 iteration conditions for convergence, n + g( n ), 0 logarithm graph of ln, 4 natural logarithm, 4 product rule, 6 further eamples, 9 quotient rule, 6 further eamples, 9 R cos( + α), secant, 0 trigonometrical identities, trigonometr finding eact values, harder equations, inverse functions, proving identities, C 4/04/ SDB
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