For use oly i Badmito School November 0 C Note C Notes (Edecel) Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets
For use oly i Badmito School November 0 C Note Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets
For use oly i Badmito School November 0 C Note Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets
For use oly i Badmito School November 0 C Note Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets 4
For use oly i Badmito School November 0 C Note Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets 5
For use oly i Badmito School November 0 C Note Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets 6
For use oly i Badmito School November 0 C Note Factor ad Remaider Theorem Simple algebraic divisio; use of the Factor Theorem ad the Remaider Theorem. Oly divisio by ( + a) or ( a) will be required. Studets should kow that if f() = 0 whe = a, the ( a) is a factor of f(). Studets may be required to factorise cubic epressios such as 4 ad 6 6. Studets should be familiar with the terms quotiet ad remaider ad be able to determie the remaider whe the polyomial f() is divided by (a + b). We eed to be able to fid a, b ad c i the followig: 58 a b c. We do this by comparig coefficiets. Whe we multiply out a b c we ca see that the Hece a. Whe we multiply out b c Hece b ad so b. we ca see that the coefficiet is a. coefficiet isb. Whe we multiply out c Hece c 5 ad so c 8. we ca see that the coefficiet is c. Hece we see that 5 8 8 8 is 8 (ote that the costat term i which acts as a check that the above is correct). If we used the otatio f( ) 5 8 We say that is a factor of f ( ). It follows from this that we could the write f( ) 8 f() 8 0 0. Factor Theorem So we see that if is a factor of f( ) the f() 0. I fact the followig is true: We ca use this to fid factors of polyomials. ( a) is a factor of f( ) if ad oly if f( a) 0. If we have to fid a factor of f( ) p q r s we are lookig for a value a such that f( a) 0. SHORT CUT : The oly values of a that we eed to try are factors of s. Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets 7
For use oly i Badmito School November 0 C Note We simply calculate f (), f () etc. util we get a value of zero. For eample if we were asked to factorise f( ) 66 0. f( ) 6 6 we could calculate We deduce, from the Factor Theorem, that ( ) is a factor of f ( ). We the use the above method to write f( ) 5 6. We the eed to factorise the quadratic (if possible). I this case it does factorise to give f( ). Eample Fid a if ( ) is a factor of f( ) a 5 6. ( ) is a factor so f () 0 (by Factor Theorem) So f () 8 4a0 6 4a 4 0, hece a 6. Eample Fid a ad b i f( ) a b give that ad are factors. is a factor so f ( ) 0. Hece f ( ) 84ab 0. From this we see that 4a b 0 ad so a b 0 (). is a factor so f () 0. Hece f () 7 9ab 0. From this we see that 9a b 5 ad so a b 5 (). Addig () ad () gives 5a 5. So a. Pluggig this back ito () gives b 8. Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets 8
For use oly i Badmito School November 0 C Note Remaider Theorem Whe the fuctio f ( ) is divided by a the remaider is f a. b Whe the fuctio f ( ) is divided by a b the remaider is f a. NB: If ( a b) is a factor of f ( ) the the remaider is zero, that is, from the above result, b f 0. We kow this result already from the factor theorem. a Whe we write f( ) i the form f( ) ( a) g( ) r we say that g( ) is the quotiet ad r is the remaider. e.g. Fid the remaider whe is divided by 7. If we let f( ) the the remaider we wat is We should proceed as follows: 7 f. Type i the followig (o the Casio) The type i X X X to get So 7 7 7 7 f 9. So the remaider whe is divided by 7 is 9. 7 9 We ca also see that Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets 9
For use oly i Badmito School November 0 C Note Eample (a) I the fuctio f( ) a b 6, a ad b are itegers. Fid a ad b give that: (i) the remaider whe f( ) is divided by is 4. (ii) the remaider whe f ( ) is divided by is 60. (b) Show that is a factor ad factorise f ( ) completely. The remaider theorem shows us that: (i) ca be rewritte as f () 4. (ii) ca be rewritte as f () 60. a b ad so a b 7 () f() 6 4 a b ad so 4a b 46. That is a b () f () 6 60 Solvig () ad () gives a 6 ad b. Hece. f( ) 6 6 (b) f( ) 6 6 0 so f( ) 6 6 5 6 is a factor. Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets 0
For use oly i Badmito School November 0 C Note Coordiate geometry i the, y plae Coordiate geometry of the circle usig the equatio of a circle i the form a yb r Studets should be able to fid the radius ad the coordiates of the cetre of the circle give the equatio of the circle, ad vice versa. y r P O Cosider the circle show above with radius r. If we look at the poit P we see that, from Pythagoras y r. This is the equatio of a circle cetre the origi, radius r. If we moved the cetre of the circle to Eample Fid the cetre ad radius of the circle give by ab, the the equatio would be y y 4 6 0 We eed to complete the square o the terms ad the y terms. y 46y0 y y 4 90 6 So the cetre is (, -) ad the radius is 4. a yb r. Eample The poits A (, 6) ad B (9, 4) lie o a circle i such a way that AB is a diameter of the circle. Fid the equatio of the circle. As AB is a diameter of the circle, the midpoit of AB must be the cetre of the circle. Hece the 6 64 cetre of the circle is,, that is (6, 0). AB 6 46 0 so the diameter is 0 ad hece the radius is 5. Hece the equatio of the circle is y 6 0 5 5 Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets
For use oly i Badmito School November 0 C Note ad icludig use of the followig circle properties: (i) the agle i a semicircle is a right agle; Q R O P (ii) the perpedicular from the cetre to a chord bisects the chord If ay chord is draw o the circle the the perpedicular from the cetre of the circle to the chord bisects the chord y A P O X B This meas that if we draw the chord AB o the circle ad the draw the lie through O perpedicular to AB, such that it hits AB at X. It follows that AX=XB. (iii) the perpedicularity of radius ad taget. y r P O I the diagram the taget to the circle at P has bee draw. The lie OP is at right agles to this taget. Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets
For use oly i Badmito School November 0 C Note Geometric Series The sum of a fiite geometric series; the sum to ifiity of a coverget geometric series, icludig the use of r.geometric series. The geeral term ad the sum to terms are required. We have see that a arithmetic sequece is a sequece i which the differece betwee cosecutive terms is costat. A geometric sequece is a sequece i which the ratio betwee cosecutive terms is costat. (e.g., 6,, 4, 48, ). As before the first term is represeted by a ad the commo ratio is represeted by r. I the eample, 6,, 4, 48, we could write it as I fact we could write it as 0 4,,,,... 4,,,,.... Whe we write it i this was we see that the th term is u. I geeral if the first term is a ad the commo ratio is r the the first few terms of the sequece will be a, ar, ar, ar... We ca see from this that the th term is u ar. Suppose that S is the sum of the first terms of the sequece whose th term is We ca write S a ar ar... ar u ar. If we multiply both sides by r we get rs ar ar ar... ar If we ow subtract rs from S we get... S rs a ar ar ar ar ar ar... ar aar Hece we see that S r a r The sum of the first terms is ad so S a r r S a r r.. Alteratively we could have got S a r r Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets
For use oly i Badmito School November 0 C Note Eample The third ad fifth terms of a geometric series are 40 ad 0. Fid the first term ad the commo ratio, give that it is positive. Third term of 40 ca be writte as ar 40 () 4 Fifth term of 0 ca be writte as ar 0 () Dividig () by () gives Pluggig this back ito () gives a 60. r ad so, give that r is positive, 4 r. Eample Fid the sum of the first 0 terms of the sequece : (a) 5, 5, 45, 5,. (b) 40, 0, 0, 5,. (a) The first term, a, is 5 ad the commo ratio, r, is. a r 5 So, usig the formula S we have S r 0 4760. (b) The first term, a, is 40 ad the commo ratio, r, is ½. a r So, usig the formula S r 40 we have S 0 80 (to sf). What would have happeed if we had looked at the first 00 terms istead of the first 0 terms i the two eamples we have just looked at? I eample (a) the value of S would have bee very large. I eample (b) the value of S would have got much closer to 80. Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets 4
For use oly i Badmito School November 0 C Note The sum to ifiity of a coverget geometric series. I eample (a) we could ot have foud the sum of a ifiite umber of terms sice this would itself have bee ifiite. However i eample (b) we could have foud the sum of a ifiite umber of terms ad the aswer would have bee 80. What determies whether or ot the sum of a ifiite umber of terms ca be foud or ot? Look a r agai at the formula S. r The key part of this formula whe we are cosiderig a ifiite umber of terms is r. If r (as i eample ) the If r (as i eample ) the The rule is as follows.. r gets very large as gets large. r teds towards zero as gets large. If r the r 0 as, so the sum to ifiity of the sequece is simply a S. r If r or r the the sum to ifiity of the sequece caot be foud. Eample Fid the ifiite sum of 08, 6,, 4,. The first term, a, is 08 ad the commo ratio, r, is. a 08 So, usig the formula S we have S 6. r Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets 5
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For use oly i Badmito School November 0 C Note Sequeces ad Series Biomial epasio of ( ) for a positive iteger. The otatios! ad ( a b) may be required.. Epasio of r If is a positive iteger the!.... I how may ways ca we arrage the letters p, q, r, s, t? The first could be ay oe of 5 letters, the secod ay oe of 4 ad so o. Hece we ca arrage the letters p, q, r, s, t i 5! ways. If we ow replaced p, q ad s with the letter a the this will reduce the umber of arragemets by a factor of! (sice p, q ad r ca be arraged i! ways). If we the replaced s ad t with the letter b the this will further reduce the umber of arragemets by a factor of! (sice s ad t ca be arraged i! ways). Hece we see that if we the letters a a a b b, they ca be arraged i I geeral if we had the letters a a a a a a a a a a a b b b b b b 5!!! ways. r r these could be arraged i! r! r! ways. We use the otatio r or C r to represet!. r! r! We ca see from this that!,!!!!! ad!!! ad i geeral...! r r r! r!... r NB : 0! ad so! ad 0 0!!!!0! Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets 7
For use oly i Badmito School November 0 C Note Now cosider the epasio of 5 a b. We ca write this as ababababa b. Whe multiplyig out we have to choose a letter from each of these 5 brackets. We could choose a from 4 of the brackets ad b from the other bracket. Each of these will 4 cotribute a term of the form ab. Or we could choose a from of the brackets ad b from the other brackets. Each of these will cotribute a term of the form ab. Ad so o How may terms of the form Each ab are there? ab term came from choosig a from of the brackets ad b from the other brackets. The umber of ab terms is, therefore, the same umber as the umber of arragemets of the letters a a a b b. We saw that these could be arraged i 5!!!, or 5, ways. So the ab term i the epasio of a b 5 is 5 ab Similarly the 4 ab term i the epasio of a b 5 is 5 4 4 ab. 5 5 5 5 5 5 ab a a b a b a b a b b. 5 4 0 So we see that 5 5 4 4 5 I geeral we see that, as is stated o the C formulae sheet : ad ab a a b a b... a b... b 0 r r r... r r......... r Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets 8
For use oly i Badmito School November 0 C Note Eample (a) Fid the first three terms i the epasio of 7. From the above we see that 7 76 765 7... 484 80... (b) Hece fid 7.00 to 6dp. If we let 0.00.00. the 7 7 7 So from (a) we see that.00 0.04 0.000084 0.00000080. 7 Hece see that.00.04084 (to 6dp). Eample Fid the first three terms i the epasio of 5. From the above we see that 5 5 4 54 54 5... 480 9080 70 Eample Fid the term i the epasio of 7 5. We ca get a term by multiplyig the i multiplyig the i with the with the term i 5 7. term i 5 7 or by The The term i 7 5 term i 7 7 5 5 5 55. is 5 7 4 5 5 5 475. is So 5 7... 55 475... Hece the term i the epasio of 5 7 is 55 475 05.. Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets 9
For use oly i Badmito School November 0 C Note Eample 4 If the first three terms of p 5 are, 0 ad Usig the same method as before we see that: 5 5 5 5 p C C p C p 0 5p 0p q the fid p ad q. We are told that the first three terms are, 0 ad Hece 5p 0 ad so p 4. Also 0 p q ad so q 60. q. Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets 0
For use oly i Badmito School November 0 C Note Trigoometry The sie ad cosie rules, ad the area of a triagle i the form absi C Sie Rule The Sie rule is writte as a A si b c or si B si C si A a si B b sic c e.g. Fid ad i the followig: si8 8 54 7. 7. si54 7.si 8. cm (tosf) si 54 You must use brackets o the calculator 0 5 si si0 5 si0 si 5 si0 si 5 d SIN 57.6 (to dp) Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets
For use oly i Badmito School November 0 C Note Cosie Rule This is a b c bccos A ad cos A b c a bc Eample Fid ad i the followig: 4 54 9 4 4 54 6789 8.4 cm(to sf) 454cos 9 4 cos 9 cos 0.665... 48. (todp) d COS Use Cosie rule if you kow either (a) three sides or (b) two sides ad the eclosed agle. Otherwise use Sie rule Area of triagle is si ab C Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets
For use oly i Badmito School November 0 C Note Radia measure, icludig use for arc legth ad area of sector of a circle. Use of the formulae s r ad A r for a circle. We are familiar with measurig oe revolutio i degrees ad kowig it to be we could measure a revolutio i radias, where oe revolutio is radias. 60. Alteratively So we see that to covert radias to degrees we multiply by 80 we multiply by. 80 ad to covert degrees to radias It follows from the above that 0, 45, 60 6 4 Arc legth of sector r Cosider the sector show above. The arc legth is 60 of the circumferece of the circle, that is Arc legth of sector r r 60 80. The area of the sector is of the area of the circle, that is 60 Area of sector r r 60 80 It follows from the above that if we measure i radias the sice we have: Arc legth of sector, s r Area of sector, A r is equal to radias 80 Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets
For use oly i Badmito School November 0 C Note Eam Questio Figure A R B r r O Figure shows the sector OAB of a circle of radius r cm. The area of the sector is 5 cm ad AOB =.5 radias. (a) Prove that r = 5. (b) Fid, i cm, the perimeter of the sector OAB. () () The segmet R, shaded i Fig, is eclosed by the arc AB ad the straight lie AB. (c) Calculate, to decimal places, the area of R. () Edecel GCE Pure Mathematics P Jue 00 (a) r. Area of sector 5.5 so.5 5 r. Hece we have r 0 ad so r 0 5 (b) Arc legth of sector r 5.5 5 So total perimeter is 5 5 5 7 5 (c) Area of triagle AOB is 5 5 si.5 0si.5 9.975 cm (to dp). Area of R is 5 9.975 5.05 cm (to dp). Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets 4
For use oly i Badmito School November 0 C Note Trigoometry Sie, cosie ad taget fuctios. We eed to defie si, cos ad ta for agles outside of the rage of 0 to 80 which we have used i triagles. y 0.8 P 0.6 0.4 0. O - -0.75-0.5-0.5 0.5 0.5 0.75-0. -0.4-0.6-0.8 - Cosider the poit P show above which is away from the origi O ad is such that OP makes a agle of (measured aticlockwise) with the positive -ais. cos, si are defied to be the coordiates of P ad ta is defied to be the gradiet of OP. Kowledge ad use of si ta ad si cos. cos Sice cos, si are defied to be the coordiates of P we have the followig triagle: P si O cos It follows from Pythagoras theorem o this triagle that We use the otatio ad si si cos cos. si cos. si cos It follows from our kowledge of gradiets that, sice ta is the gradiet of OP ad sice si cos, si are the coordiates of P that ta. cos Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets 5
For use oly i Badmito School November 0 C Note Their graphs, symmetries ad periodicity. Kowledge of graphs of curves with equatios such as y si, y si, y si is epected. 6 The graph of y si The graph of y cos 80 60 Suppose we have to solve si k. Suppose we have to solve cos k. We draw y si ad y k (as above) We draw y cos ad y k (as above) ad we look for the -coordiates of ad we look for the -coordiates of the poits of itersectio. the poits of itersectio. It is clear from the graph that if is the It is clear from the graph that if is the -coordiate (from calculator) of oe of the -coordiate (from calculator) of oe of the poits of itersectio the 80 is the poits of itersectio the 60 is the -coordiate of aother poit of itersectio. -coordiate of aother poit of itersectio. The -coordiates of all further poits of itersectio ca be obtaied from addig or subtractig multiples of 60 to either of these solutios. The -coordiates of all further poits of itersectio ca be obtaied from addig or subtractig multiples of 60 to either of these solutios. Eample Solve si 0.7, 0 60 Eample Solve cos 0., 0 60 Our calculators give us oe value, Our calculators give us oe value, amely si (0.7) 44.4. amely cos ( 0.) 0.5 The other value, that appears o The other value, that appears o a differet part of the curve is, a differet part of the curve is, from above, 80 44.4 5.6 (to dp). from above, 60 0.5 58.5 (to dp). All other solutios are obtaied by addig o or subtractig multiples of to either of these aswers. All other solutios are obtaied 60 by addig o or subtractig multiples of 60 to either of these aswers. Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets 6
For use oly i Badmito School November 0 C Note The graph of y ta 80 Suppose we have to solve ta k. We draw y ta ad y k (as above) ad we look for the -coordiates of the poits of itersectio. It is clear from the graph that if is the -coordiates (from calculator) of oe of the poits of itersectio the 80 is the -coordiate of aother poit of itersectio. As before, all other solutios are obtaied by addig o or subtractig multiples of these aswers. However we ca see that whe we do this it is the same as 60 to either of Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets 7
For use oly i Badmito School November 0 C Note Eample Solve ta Our calculators give us oe value, amely ta () 6.4. All other values are obtaied by addig o multiples of 80 to this aswer, to give 4.4 I summary If is a solutio to si k the so is 80. If is a solutio to cos If is a solutio to ta k the so is 60. k the so is 80. I all three cases all other solutios are foud by addig or subtractig multiples of 60 to these two solutios. From these graphs we see that y si, for eample, is as follows: The period of the curve is the value of p such that f p f for all values of. So the period of y si is 60 (or i radias) ad the period of y cos is 60 (or i radias) ad the period of y ta is 80 (or i radias) Solutio of simple trigoometric equatios i a give iterval. Studets should be able to solve equatios such as si 4 for 0 cos 0 for 80 80 ta for 90 70 6cos si si 5 0 6 for for 90 70 Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets 8
For use oly i Badmito School November 0 C Note See eample sheet for these. A way of rememberig si, cos ad ta for certai agles Agle, 0 0 45 60 90 Itegers 0 4 Square root 0 si Divide by 0 cos Flip order aroud si ta 0 cos 0 Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets 9
For use oly i Badmito School November 0 C Note Eamples. Solve the followig trigoometric equatios i the give itervals : 4 (a) cos for 0 60 (to dp) 5 4. (from calc) or 60 (4. ) 6.9 4. or 6.9 (b) ta for 0 60 (to dp) 7.6 (from calc) or 80 7.6 5.6 7.6 or 5.6 (c) si for 0 (eact values) 0 (from calc) or 80 ( 0 ) 0 We have to add 60 to 0 because 0 is ot i rage. 0 or 0 7 or 6 6 (d) si for 80 80 (to dp) 4.8 (from calc) or 80 (4.8 ) 8. 4.8 or 8. (e) cos for 0 (eact values) 45 (from calc) or 60 (45 ) 5 7 or 4 4 (f ) si cos for ( to dp) si cos cos cos ta 60 (from calc) or 80 (60 ) 40 We have to subtract 60 from 40 because 40 is ot i rage. 0 or 60 or Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets 0
For use oly i Badmito School November 0 C Note. Solve the followig trigoometric equatios i the give itervals: (a) cos 5 for 80 80 (eact values) 45 5 5 5 0 (from calc) or 60 (0 ) 40 We have to subtract 60 from 40 because 40 is ot i rage. 5 0 or 0 85 or 55 (b) si for 0 (eact values) 5 6 6 4 6 0 (from calc) or 80 (0 ) 50 66 or 86 or 0 0 (c) ta 40 for 0 60 (to dp) 40 40 0 40 6.4 0.4 or 8.4 (from calc) or 80 (6.4 ) 4.4 Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets
For use oly i Badmito School November 0 C Note. Solve the followig trigoometric equatios i the give itervals: (a) ta for 0 80 0 540 45 (from calc) or 80 (45 ) 5 We have to add 60 to get all values i rage. 45 or 5 or 405 5 or 75 or 5 (b) cos 4 for 0 80 0 4 70 4 0 (from calc) or 60 (0 ) 0 We have to add 60 to get all values i rage. 4 0 or 0 or 90 or 690 7.5 or 8.5 or 97.5 or 7.5 (c) si for 0 0 60 0 (from calc) or 80 (-0 ) 0 0 is ot i rage so we have to add 60 to get 0. 0 or 0 05 or 65 7 or Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets
For use oly i Badmito School November 0 C Note 4. Solve the followig trigoometric equatios i the give itervals : (a) cos si 0 for 0 (eact values) Use the fact that cos si Replace cos with si si si0 Replace si with s s s0 s s0 s s 0 s or s Solve si 0 (from calc) or 80 (-0 ) 0 0 is ot i rage so we have to add 60 to get 0. 0 or 0 Solve si 90 (from calc) or 80 (90 ) 90 So 90 or 0 or 0 Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets
For use oly i Badmito School November 0 C Note (b) si cos 6 0 for 0 60 (to dp) Use the fact that cos si Replace si with cos c c cos cos 6 0 Replace cos with c c c6 0 c c6 0 4 0 c or c 4 Solve cos.8 (from calc) or 60 (.8 ) 8. Solve cos 4 4.4 (from calc) or 60 (4.4 ) 8.6 So.8 or 8. or 4.4 or 8.6 Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets 4
For use oly i Badmito School November 0 C Note 5. Solve the followig trigoometric equatios i the give itervals: (a) si for 0 5 4 6 6 96 Either (i) si 6 or (ii) si 6 4 4 (i) 6 60 (from calc) or 80 (60 ) 0 (ii) 6 60 (from calc) or 80 (-60 ) 40 60 is outside of rage so we add 60 to get 00 So 6 60 or 0 or 40 or 00 4 or 84 or 04 or 64 7 7 or or or 5 5 5 5 50 50 0 (b) cos 50 for 0 60 Either (i) cos 50 or ( ii) cos 50 (i) 50 45 or 60 (45 ) 5 5 is outside of rage so we subtract 60 to get 45 (ii) 50 5 or 5 50 45 or 45 or 5 or 5 5 or 95 or 85 or 75 Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets 5
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For use oly i Badmito School November 0 C Note y a ad its graph. Logarithms ad Epoetial Fuctios We have already met equatios of the form 8 ad we have oly bee able to solve these equatios by ispectio where is either a iteger or a fractio. Now cosider the equatio 7. The ide,, is the power to which must be raised to give 7. As preset we do ot have ay otatio for what we call. There is o fuctio that we kow of which eables us to epress i terms of ad 7. It is clearly useful for there to be such a fuctio ad this is where we use logarithms. Aother word for this ide,, is logarithm. We write log 7, or (i words) is the log (base ) of 7. NB : log0 is more simply kow as log. The key thig to remember about logarithms is this: See how the a slides: log b a b a a b log a b EXAM TIP: If you forget this, type i log00 ito your calculators. This gives us. So we ca write log0 00 ad we kow that this ca be rewritte as 0 00. Hece we ca see that log a b ca be rewritte as a b. So, for eample, if log 8 the usig the above we see that 8 ad so 4. We kow that the graph of y a is a epoetial graph ad it follows that y log a is a reflectio of this i the lie y. y a y y log a NB y log a does ted to ifiity but it does so very slowly. For eample log0 000000 6. Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets 7
For use oly i Badmito School November 0 C Note Laws of logarithms. To iclude log y log log y log log log y log a a a a a a a k klog loga log log a a a a y We have three laws of idices: y y ( I) a a a ( II) a a a y y ( III) a a y y If a m ad y a the from the above result log a m ad y log a From (I) we have that From (II) we have that m y a ad so log a m y Hece log m log m log a a a m a y ad so log m a y Hece log m log m log a a a y From (III) we have that y y y y m a a ad so loga m y ad so y loga m logam y Hece loga m ylog m a Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets 8
For use oly i Badmito School November 0 C Note Studets may use the chage of base formula. The formula book has that log a logb. log a b log0 b So, for eample, loga b log a 0 The solutio of equatios of the form b a. To solve b a We saw earlier that 0 00 ca be rewritte as log0 00. I the same way b a ca be rewritte as log a b. Some calculators calculate log a b directly, others do ot. log0 b If they do t, the use the above, that loga b. log0 a So, i geeral: The solutio to b log b log. a 0 a is a b log0 Eample Solve 7. We saw earlier that 0 00 ca be rewritte as log0 00. I the same way 7 ca be rewritte as log 7. Some calculators calculate log 7 directly to give.77 (to sf). log0 7 If they do t, the use the above, that log 7.77 (to sf). log 0 Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets 9
For use oly i Badmito School November 0 C Note Eample If k log a the fid the followig i terms of k: (a) log (b) log 7a (c) log 9a a (a) log a log a k (b) log 7alog 7 log ak (c) log 9a log 9 log a log 9 log a k Eample If k log the fid log9 i terms of. log9 k Usig the above we see that log9 log 9 Eample 4 log log. Solve Usig first law of logarithm, we ca rewrite the above as log. We saw earlier that log0 00 ca be rewritte as 0 00. I the same way log ca be rewritte as. Hece 8. So 7 ad hece. 7 Eample 5 Solve loglog log4. Hece loglog log4 0. Usig first law of logarithm, we ca rewrite the above as log 0. 4 We saw earlier that log0 00 ca be rewritte as 0 00. I the same way log 0 0 ca be rewritte as. 4 4 Hece 4 ad so Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets 40
For use oly i Badmito School November 0 C Note Eample 6 5 7 5 0. Solve Puttig y 5 gives that y 5. Hece the above equatio ca be rewritte as y 7y 0. Factorisig gives y y Hece y or y 4. So 5 or 5 4 4 0. Usig the method of earlier questios, we see that 0.68 (to sf) or 0.86 (to sf). Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets 4
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For use oly i Badmito School November 0 C Note Differetiatio Applicatios of differetiatio to maima ad miima ad statioary poits, icreasig ad decreasig fuctios. The otatio f ( ) may be used for the secod order derivative. To iclude applicatios to curve sketchig. Maima ad miima problems may be set i the cotet of a practical problem. Cosider the curve y 9 show above. The curve has positive gradiet whe ad whe. It has a egative gradiet whe. It has zero gradiet whe ad whe. A curve is said to be icreasig at ay poit if its gradiet is positive at that poit, i.e. d y 0 d. A curve is said to be decreasig at ay poit if its gradiet is egative at that poit, i.e. d y 0 d. A poit o a curve is said to be a statioary poit if d y 0 at that poit. d So i the above eample, the curve is icreasig whe ad whe. It is decreasig whe. It has statioary poits at ad. Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets 4
For use oly i Badmito School November 0 C Note We ca fid the statioary poits without the graph by calculatig d y d ad solvig d y 0 d. dy y 9 so 6 9 d. Solvig d y 0 d gives us This gives or. 6 9 0 The gradiet of the gradiet is deoted by d y d. We see that i this eample d y d 6 6 It is clear from the graph that the y-value at is the largest value i that part of the curve. It is d y called a (local) maimum. We see that, 0 d. d y If is egative at a statioary poit the the statioary poit is a local maimum. d d y It is also true that if is positive at a statioary poit the the statioary poit is a local d miimum. Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets 44
For use oly i Badmito School November 0 C Note Eample A factory produces cartos for ice cream. Each carto is i the shape of a closed cuboid with base dimesios cm by cm ad height h cm. h Give that the capacity of a carto has to be 0 cm, (a) epress h i terms of, (b) fid the surface area, A cm i terms of. The factory wats to miimise the surface area of a carto. (c) Fid the value of for which A is a miimum (d) Prove that this value of A is a miimum. (e) Calculate the miimum value of A.. (a) The volume of the cuboid is h h. Sice the volume is 0, it follows that 70 h 0 ad so h. (b) The surface area of all si sides is A h h h h 6 8h. h 70 Usig h gives 70 960 A6 8 6. (c) A is a miimum whe d A 0 d. A6 960 da 960 960 d If d A 0 d the 960. Hece 960 ad so 6.7 cm (to sf). (d) d A 590 590. d d A Whe 6.7, 0 so it is a miimum. d (e) Whe 6.7, A 708 cm (to sf) Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets 45
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For use oly i Badmito School November 0 C Note Defiite Itegratio Evaluatio of defiite itegrals. Iterpretatio of the defiite itegral as the area uder a curve. The area eclosed betwee the curve 6 d. This area is shaded i the diagram show below. y, the -ais, the lies ad 6 is deoted by 6 6 d 796 Studets will be epected to be able to evaluate the area of a regio bouded by a curve ad give straight lies. E.g. fid the fiite area bouded by the curve y 6 ad the lie y. dy will ot be required. We eed to fid first of all the -coordiates of the poits of itersectio of y. Solvig these gives 0 ad 4. 6 ad y Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets 47
For use oly i Badmito School November 0 C Note The area is the the differece betwee the two areas show below: So the area is 4 4 6 d d. 0 0 This is equivalet to 4 4 6 d 4 d. 0 0 Thus we see that the area is 4 4 4 d 0 64 0 So if the curves y f( ) ad y g( ) itersect at a ad b the the area eclosed betwee b b b them is f ( ) d g( ) d f ( ) g( ) d a a a Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets 48
For use oly i Badmito School November 0 C Note NB Use your calculators effectively o these. Suppose we had to fid Type i the followig: 4 4 4 9 46d 6 4 4 SHIFT RCL ) OR 4 STO X, T You should ow see the followig display: 4 X The type i The type i 4 4 6 to get 5 8 SHIFT RCL ). OR STO X, T You should ow see the followig display: X The type i Hece 4 4 6 to get 7 4. 4 4 4 = 9 46d 6 4 5 7 9 8 4 4 Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets 49
For use oly i Badmito School November 0 C Note Approimatio of area uder a curve usig the trapezium rule. E.g. evaluate values of at = 0, 0.5, 0.5, 0.75 ad. d usig the 0 The area eclosed betwee the curve y, the -ais, the lies 0 ad ca be estimated usig 4 trapezia. We ca calculate the y coordiates as follows: 0 0.5 0.5 0.75.00..4.58.7 The area of the 4 trapezia is therefore: 0.5 0.5 0.5 0.5.00...4.4.58.58.7 0.5.00...4.4.58.58.7 0.5.00..4.58.7.40 Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets 50
For use oly i Badmito School November 0 C Note Cosider the area uder the curve show below where is the umber of trapezia, h is the width of each trapezium ad y 0, y,... y, y are the heights as show below: y y0 y y y h The trapezium rule shows us that the area is approimately h y 0 y y... y As stated o the formula sheet, b h y y0 y y... y a d where h b a Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets 5
For use oly i Badmito School November 0 C Note Eample The table below shows the values of y.6 for differet values of : 0 0.5.5.5 y.6 0 0.775.49.760 (a) (b) Fill i the blaks Use the trapezium rule to estimate.6 d. 0 (a) Use calculator to fill i the followig: 0 0.5.5.5 y.6 0 0.55 0.775.0.49.496.760 (b) b h b a yd y0 y y... y where h. a Formula book states that How to use the formula. a 0 0.5.5.5 y.6 0 0.55 0.775.0.49.496.760 b y 0 y y y y 4 y 5 y 6 So we see that a 0, b. The largest value of y is y 6 so 6. ba 0 Hece h 0.5 More simply, h is the gap betwee the values. 6 0 0.5.6 d 0.760 0.55 0.775.0.49.496 0.5.6 d 0.760 0.55 0.775.0.49.496.96 (to sf). So 0 Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets 5