PARTIAL FRACTIONS AND BINOMIAL EXPANSION 1., Study notes of Pre-Calculus

Download PARTIAL FRACTIONS AND BINOMIAL EXPANSION 1. and more Study notes Pre-Calculus in PDF only on Docsity! City of London Academy 1 C4: QUESTIONS FROM PAST EXAM PAPERS – PARTIAL FRACTIONS AND BINOMIAL EXPANSION 1. (a) Find the values of the constants A, B and C. (4) (b) Hence, or otherwise, expand in ascending powers of x, as far as the term in x 2 . Give each coefficient as a simplified fraction. (7) (Total 11 marks) 2. Given that f(x) can be expressed in the form (a) find the values of B and C and show that A = 0. (4) (b) Hence, or otherwise, find the series expansion of f(x), in ascending powers of x, up to and including the term in x 2 . Simplify each term. (6) (c) Find the percentage error made in using the series expansion in part (b) to estimate the value of f (0.2). Give your answer to 2 significant figures. (4) (Total 14 marks) 3. Given that, for , where A and B are constants, (a) find the values of A and B. (3) (b) Hence, or otherwise, find the series expansion of f(x), in ascending powers of x, up to and including the term in x 3 , simplifying each term. (6) (Total 9 marks)    2121 1052 2       x C x B A xx xx   21 1052 2   xx xx 3 2 , )1()23( 163227 )(f 2 2     x xx xx x , )1()23()23( )(f 2 x C x B x A x       . )21( 13 )(f 2 1 2     x x x x 222 1 )21()21()21( 13 , x B x A x x x        City of London Academy 2 4. (a) Find the binomial expansion of in ascending powers of x up to and including the term in x 3 , simplifying each term. (4) (b) Show that, when x = the exact value of √(1 – 8x) is (2) (c) Substitute into the binomial expansion in part (a) and hence obtain an approximation to √23. Give your answer to 5 decimal places. (3) (Total 9 marks) 5. f(x) = , │x│ < 4 Find the binomial expansion of f (x) in ascending powers of x, up to and including the term in x 3 . Give each coefficient as a simplified fraction. (Total 6 marks) 6. (a) Expand , in ascending powers of x up to and including the term in x 2 . Simplify each term. (5) (b) Hence, or otherwise, find the first 3 terms in the expansion of as a series in ascending powers of x. (4) (Total 9 marks) 7. (a) Use the binomial theorem to expand , in ascending powers of x, up to and including the term in x 3 , giving each term as a simplified fraction. (5) (b) Use your expansion, with a suitable value of x, to obtain an approximation to Give your answers to 7 decimal places. (2) (Total 7 marks) , 8 1 ,)81(  xx , 100 1 . 5 23 100 1 x )4( 1 x 3 4 where, )34( 1   x x )34( 8 x x   3 8 ,)38( 3 1  xx 3 )7.7( City of London Academy 5 (c) State, with a reason, whether your series expansion in part (b) is valid for x = . (2) (Total 10 marks) 17. The function f is given by f(x) = , x  , x  2, x  1. (a) Express f(x) in partial fractions. (3) (b) Hence, or otherwise, prove that f(x) < 0 for all values of x in the domain. (3) (Total 6 marks) ==================================================================================== MARK SCHEME 1. (a) B1 M1 A1 A14 (b) M1 B1 B1 M1 ft their A1 ft stated or implied A1 A17 [11] 2. (a) 27x 2 + 32x + 16 ≡ A(3x + 2)(1 – x) + B(1 – x) + C(3x + 2) 2 Forming this identity Substitutes either or M1 x = 1 into 2 1 )1)(2( )1(3   xx x 2A        22 5 10 1 2 2 1x x A x x B x C x         1x  3 3 1B B     2x  12 3 4C C          12 12 5 10 2 1 2 1 1 2 2 x x x x x x                  1 21 1 ... x x x       1 2 1 1 ... 2 2 4 x x x                   2 22 5 10 1 2 1 2 1 1 1 ... 1 2 2 x x x x x x                  5 ...  1 2 A B C  23 ... ... 2 x   0x 3 2–x     416–12,– 3 5 3 20 3 5 3 64 3 2  BBBx City of London Academy 6 their identity or equates 3 terms or M1 substitutes in values to write down three x = 1, 27 + 32 + 16 = 25C 75 = 25C C =3 simultaneous equations. Both B = 4 and C = 3 A1 (Note the A1 is dependent on both method marks in this part.) 27 = – 3A + 9C 27 = – 3A + 27 0 = –3A Equate x 2 : A = 0 Compares coefficients or substitutes in a third x-value or uses simultaneous x = 0, 16 = 2A + B + 4C equations to show A = 0. B14 16 = 2A + 4 + 12 0 = 2A A = 0 (b) f(x) = = 4(3x + 2) –2 + 3(1 – x) –1 Moving powers to top on any one of the two expressions M1 = = = Either or 1 ± (–1)(–x) from either first or second expansions respectively Ignoring 1 and 3, any one correct {............} A1 expansion. Both {............} correct. A1 = = 4 + (0x) ; A1; A16 (c) Actual = f (0.2) = Attempt to find the actual value of f(0.2) = or seeing awrt 4.3 and believing it is candidate‟s actual f(0.2). M1 Or         )–1( 3 )23( 4 2 xx     1–2– 2 3 )–1(3)1(24 xx    1–2– 2 3 )–1(31 xx         ...)( 2! )3)(–2(– );)(2(–11 2 2 3 2 3 xx );)(2(–1 2 3x        ...)(– 2! )2)(–1(– );)(–1(–13 2xx    ...13...3–1 22 4 27  xxxx 2 4 39;04 xx  2 4 39 x )8.0)(76.6( 164.608.1  676 2935 ...64.34171597 408.5 48.23  City of London Academy 7 Actual = f (0.2) = Candidates can also attempt to find the actual value by using = with their A, B and C. Estimate = f(0.2) = Attempt to find an estimate for M1ft f(0.2) using their answer to (b) = 4 + 0.39 = 4.39 %age error = M1 = 1.112095408... = 1.1%(2sf) 1.1% A1 cao4 [14] 3. (a) 3x  1  A (1  2x) + B Let x = = B  B M1 Considers this identity and either substitutes x = , equates coefficients or solves simultaneous equations Equate x terms; 3 = 2A  A = A1;A1 3 A = (No working seen, but A and B correctly stated  award all three marks. If one of A or B correctly stated give two out of the three marks available for this part.) (b) M1 Moving powers to top on any one of the two expressions dM1 Either 1 ± 2x or 1 ± 4x from either first or second expansions respectively )2.0–1( 3 )2)2.0(3( 4 2   )–1()23()23( 2 x C x B x A     676 2935 ...64.3417159775.3 76.6 4  2 4 39 )2.0(4 100 6...4.34171597 ....3417159764–4.39  100 actual actual - estimatetheir  ; 2 1 1 2 3  2 1 2 1 2 3 2 1 2 3 ;  B 2 2 11 2 3 )21()21()(f   xxx            ...)2( !3 )3)(2)(1( )2( !2 )2)(1( );2()1(1 32 2 3 xxx City of London Academy 10 0.959168 M1 5×0.959168 M1 = 4.795 84 cao A13 [9] 5. f (x) = M1 = B1 M1 A1ft ft their A1, A16 Alternative f (x) = M1 = B1 M1 A1 = A1, A16 [6] 6. ** represents a constant (which must be consistent for first accuracy mark) (a) with **  1 23 2 1 – )4( )4( 1 x x   ...2 1 – ....)1()4(  ...)1(2 1 or....)1( 2 1 ...                              ... 4!3 ))(–)(–(– 42 ))(–(– 4 )(–1... 3 2 5 2 3 2 12 2 3 2 1 2 1 xxx       4 x ... 2048 5 – 256 3 , 16 1 – 2 1 32  xxx 2 1 – )4( )4( 1 x x          ....4 3.2.1 ––– 4 2.1 –– 4)(–4 32 7 – 2 5 2 3 2 1 22 5 – 2 3 2 1 2 3 – 2 12 1 –  xxx ... 2048 5 – 256 3 , 16 1 – 2 1 32  xxx 2 1 2 1 2 1 2 1 4 3 1 2 1 4 3 1)4()34( )34( 1                 xx x x                     ...)*(* !2 );*(* 2 1 1 2 1 22 3 2 1 xx City of London Academy 11 Ignore subsequent working outside brackets B1 Expands to give a simplified or an un-simplified ; M1 A correct simplified or an un-simplified [...........] expansion with candidate‟s followed through (**x) A1ft Award SC M1 if you see SC: A1 isw A1 isw 5 (b) (x + 8) Writing (x + 8) multiplied by candidate‟s part (a) expansion. M1 Multiply out brackets to find a constant term, two x terms and two x 2 terms. M1 Anything that cancels to 4 + 2x; A1; A1 4 [9]                                        ... 128 27 ; 8 3 1 2 1 ... 4 3 !24 3 2 1 1 2 1 2 2 2 3 2 1 xx xx        ... 256 27 ; 16 3 2 1 2xx 2 1 or )4( 2 1  2 1 )**1(   x )*(* 2 1 1 x          22 3 2 1 )*(* !2 )*(* 2 1 xx                ;... 8 3 1 2 1 x        ... 128 27 8 3 1 2xxK       2 128 27 ...;......... 2 1 x        ... 256 27 16 3 2 1 2xx ... 32 33 ;24 ... 32 27 2 3 4 .... 16 3 2 1 2 2 2    xx xx xx 2 32 33 x City of London Academy 12 7. ** represents a constant (which must be consistent for first accuracy mark) (a) with **  1 Award SC M1 if you see Takes 8 outside the bracket to give any of or 2. B1 Expands to give a simplified or an un-simplified ; M1; A correct simplified or an un-simplified {............} expansion with candidate‟s followed through (**x) A1ft Either or anything that cancels to ; A1; 5 (b) = 2 – 0.025 – 0.0003125 – 0.0000065104166... = 1.97468099... Attempt to substitute x = 0.1 into a candidate‟s binomial expansion. M1 awrt 1.9746810 A1 2 You would award B1M1A0 for 3 1 3 1 3 1 3 1 8 3 12 8 3 1)8()38(              xx x                                                      ...)*(* !3 3 5 3 2 3 1 )*(* !2 3 2 3 1 );*(* 3 1 12 32 xxx 32 )*(* !3 3 5 3 2 3 1 )*(* !2 3 2 3 1 xx                                                                                                       ... 8 3 !3 3 5 3 2 3 1 8 3 !2 3 2 3 1 ; 8 3 3 1 12 32 xxx        ... 1536 5 64 1 ; 8 1 12 32 xxx ... 768 5 32 1 ; 4 1 2 32  xxx 3 1 )8( 3 1 )**1( x )*(* 3 1 1 x              .............. 8 1 12 x x 4 1 2  ...)1.0( 768 5 )1.0( 32 1 )1.0( 4 1 2)7.7( 323 1  City of London Academy 15 Way 2 f(x) = (3 + 2x) –3 = with **  1 B1 or (3) –3 (See note below) M1 Expands (3 + 2x) –3 to give and un-simplified (3) –3 + (–3)(3) –4 (**x); A1ft A correct un-simplified or simplified {..........} expansion with candidate‟s followed thro‟ (**x) = = = A1; Anything that cancels to : A1 Simplified Attempts using Maclaurin expansions need to be escalated up to your team leader. If you feel the mark scheme does not apply fairly to a candidate please escalate the response up to your team leader. Special Case: If you see the constant in a candidate‟s final binomial expression, then you can award B1 [5] 9. ** represents a constant                    ...)*(*)3( !3 )5)(4)(3( )*(*)3( !2 )4)(3( );*(*)3)(3()3( 36 2543 x xx 27 1                    ...)2()3( !3 )5)(4)(3( )2()3( !2 )4)(3( );2()3)(3()3( 36 2543 x xx                                   ...)8( 729 1 )10( )4( 243 1 )6();2( 81 1 )3( 27 1 3 2 x xx ... 729 80 81 8 ; 27 2 27 1 32  xxx 27 2 27 1 x  729 80 81 8 32 xx  27 1 City of London Academy 16 Takes 2 outside the bracket to give any of (2) –2 or B1 Expands (1 + * * x) –2 to give an unsimplified 1 + (–2)(**x); M1 A correct unsimplified {..........} expansion with candidate‟s (**x) A1 Anything that cancels to ; A1; Simplified A1 [5] Aliter Way 2 f(x) = (2 – 5x) –2 or (2) –2 B1 Expands (2 – 5x) –2 to give an unsimplified (2) –2 + (–2)(2) –3 (**x); M1 A correct unsimplified {..........} expansion with candidate‟s (**x) A1 22 22- 2 5 1 4 1 2 5 1)2(5x)-(2)f(                xx x            ...)*(* !3 )4)(3)(2( )*(* !2 )3)(2( );*)(*2(1 4 1 32 xxx 4 1 ... 8 5 15 16 11 4; 4 1 1 4 1 ... 8 125 16 75 ; 4 5 4 1 ... 2 125 4 75 ;51 4 1 ... 2 5 !3 )4)(3)(2( 2 5 !2 )3)(2( ; 2 5 )2(1 4 1 32 32 32 32                                        xxx xxx xx x xxx 4 5 4 1 x  8 125 16 75 32 xx                        ...)*(*)2( !3 )4)(3)(2( )*(*)2( !2 )3)(2( );*(*)2)(2()2( 35 2432 x xx 4 1 City of London Academy 17 Anything that cancels to ; A1; Simplified A1 Attempts using Maclaurin expansions need to be referred to your team leader. [5] 10. (a) (1+ M1 (corr bin coeffs) M1 (powers of –2x) = 1 + x, + x 2 + x 3 + ... A1 A1 4 Alternative May use McLaurin f(0) = 1 and f(0) = 1 to obtain 1 st two terms 1 + x M1 A1 Differentiates two further times and uses formula with correct factorials to give M1 A1 4 (b) . So series is (previous series) M1A1 ft 2 [6] 11. (a) Considers 3x 2 + 16 = A(2 + x) 2 + B(1 – 3x)(2 + x) + C(1 – 3x) M1 and substitutes x = –2 , or x = 1/3 , or compares coefficients and solves simultaneous equations To obtain A = 3, and C = 4 A1 A1 ... 8 5 15 16 11 4; 4 1 1 4 1 ... 8 125 16 75 ; 4 5 4 1 ...)125( 16 1 )4( )25( 16 1 )3();5( 8 1 )2( 4 1 ...)5()2( !3 )4)(3)(2( )5()2( !2 )3)(2( );5()2)(2()2( 32 32 3 2 35 2432                                                           xxx xxx x xx x xx 4 5 4 1 x  8 125 16 75 32 xx  ....))2( 3.2.1 ) 2 5)( 2 3)( 2 1( )2( 2.1 ) 2 3)( 2 1 )2( 2 1 32 (           xxx 2 3 2 5 32 2 5 2 3 xx  2 1 2 1 2 1 )21(100)200100(   xx 10 1 City of London Academy 20 (x = – )  = A A = 1 A1 3 (b) 2(1 – x) –1 = 2[1 + x + x 2 + … ] M1 [A1] Use of binomial with n = –1 scores M1(×2) (1 + 3x) –1 = [1 – 3x + (3x) 2 + …] M1 [A1]  = 2 + 2x + 2x 2 + 1 – 3x + 9x 2 = 3 – x + 11x 2 A1 5 (c) (1 + 3x) –1 requires |x| < , so expansion is not valid. M1, A1 2 [10] 17. (a) , and correct method for finding A or B M1 A = 1, B = 2 A1, A1 3 (b) f(x) = M1 A1 Argument for negative, including statement that square terms are positive for all values of x. (f.t. on wrong values of A and B) A1 ft 3 [6] EXAMINERS‟ REPORTS 1. The first part of question 5 was generally well done. Those who had difficulty generally tried to solve sets of relatively complicated simultaneous equations or did long division obtaining an incorrect remainder. A few candidates found B and C correctly but either overlooked finding A or did not know how to find it. Part (b) proved very testing. Nearly all were able to make the connection between the parts but there were many errors in expanding both (x – 1) –1 and (2 + x) –1 . Few were able to write (x – 1) –1 as –(1 – x) –1 and the resulting expansions were incorrect in the majority of cases, both 1 + x – x 2 and 1 – x – x 2 being common. (2 + x) –1 was handled better but the constant in was frequently incorrect. Most recognised that they should collect together the terms of the two expansions but a few omitted their value of A when collecting the terms. 2. Part (a) was tackled well by many candidates. The majority of candidates were able to write down the correct identity. The most popular strategy at this stage (and the best!) was for candidates to substitute x = 1 and x = into their identity to find the values of the constants B and C. The substitution of x = caused problems for a few candidates which led them to find an incorrect value for B. Many candidates demonstrated that constant A was zero by use of a further value of x or by comparing coefficients in their identity. A significant minority of candidates manipulated their original identity and then compared coefficients to produce three equations in order to solve them simultaneously. In part (b), most candidates were able to rewrite their partial fractions with negative powers and apply the two binomial 3 1 3 4 3 4 !2 )2)(1(  )31)(1( 53 xx x   3 1 12)1)(2( )1(3       x B x A xx x 22 )1( 2 )2( 1     xx 2 1 1 2 1 2 1         x 3 2– 3 2– City of London Academy 21 expansions correctly, usually leading to the correct answer. A significant minority of candidates found the process of manipulating 4(3x + 2) –2 to challenging. A significant number of candidates were unsure of what to do in part (c). Some candidates found the actual value only. Other candidates found the estimated value only. Of those who progressed further, the most common error was to find the difference between these values and then divide by their estimate rather than the actual value. Some candidates did not follow the instruction to give their final answer correct to 2 significant figures and thus lost the final accuracy mark. 3. In part (a) candidates needed to start with the correct identity; although correct solutions were seen from a good proportion of the candidature, a significant number of candidates started with the wrong identity and thus gained no marks. The most common wrong starting point was to use 3x –≡A(1 –2x) 2 +B(1 –2x) , but 3x–1≡A(1–2x) 2 +B and 3x– 1≡A(1–2x) 2 +Bx also occasionally appeared. Candidates using the first identity often produced answers A= , B= – ; the same values but for the wrong constants. Candidates using the second identity could produce the „correct‟ answers B= ,A=– (eg by setting x = 0 and x = ) but this is fortuitous and clearly gains no marks. Generally candidates showed a good understanding of the work on expanding series in part (b) and most were able to gain some credit. The mark scheme allowed four marks to be gained for the correct unsimplified expansions, as far as the term in x 3 , of (1–2x) –1 and (1–2x) –2 This helped some candidates who went on to make numerical or sign errors when simplifying their expansions and errors in part (a) only affected the final two accuracy marks. Candidates who multiplied (3x–1) by the expansion of (1–2x) –2 gave solutions that were not dependent on their answers in part (a) and it was not uncommon to see a score of zero marks in part (a) followed by a score of six marks in part (b). 4. This proved a suitable starting question and there were many completely correct solutions. The majority of candidates could complete part (a) successfully. In part (b), those who realised that working in common (vulgar) fractions was needed usually gained the method mark but, as noted in the introduction, the working needed to establish the printed result was frequently incomplete. It is insufficient to write down . The examiners accepted, for example, = = . In part (c), most candidates realised that they had to evaluate their answer to part (a) with x = 0.01. However many failed to recognise the implication of part (b), that this evaluation needed to be multiplied by 5. It was not uncommon for candidates to confuse parts (b) and (c) with the expansion and decimal calculation appearing in (b) and fraction work leading to √ 23 appearing in (c). 5. This proved a suitable starting question and the majority of candidates gained 5 or 6 of the available 6 marks. Nearly all could obtain the index as but there were a minority of candidates who had difficulty in factorising out 4 from the brackets and obtaining the correct multiplying constant of . Candidates‟ knowledge of the binomial expansion itself was good and, even if they had an incorrect index, they could gain the method mark here. An unexpected number of candidates seemed to lose the thread of the question and, having earlier obtained the correct multiplying factor and expanded correctly, forgot to multiply their expansion by . 6. This question was also generally well tackled with about 50% of candidates obtaining at least 8 of the 9 marks available. A substantial minority of candidates were unable to carry out the first step of writing as , with the outside the brackets usually written incorrectly as either 2 or 4. Many candidates were ( ) 2 2 3+1 – x 2 1 2 3 2 1 2 3 2 1 5 23 100 8 –1  100 92 100 8 –1  25 23 5 23 2 1– 2 1 2 1 2 1 – 4 1        x 2 1 )34( 1 x 2 1 4 3 1 2 1         x 2 1 City of London Academy 22 able to use a correct method for expanding a binomial expression of the form (1 + ax) n . A variety of incorrect values of a and n, however, were seen by examiners with the most common being a as , 3 and –3 and n as , –1 and –2. Some candidates, having correctly expanded , forgot to multiply their expansion by . As expected, sign errors, bracketing errors and simplification errors were also seen in this part. A significant minority of candidates expanded as far as x 3 , and were not penalised on this occasion. In part (b), most candidates realised that they needed to multiply (x + 8) by their expansion from part (a) although a small minority attempted to divide (x + 8) by their expansion. A surprising number of candidates attempted to expand (x + 8) to obtain a power series. Other candidates omitted the brackets around x+ 8 although they progressed as if “invisible” brackets were there. The mark scheme allowed candidates to score 2 marks out of 4 even if their answer in (a) was incorrect and many candidates were able to achieve this. 7. In part (a), a majority of candidates produced correct solutions, but a minority of candidates were unable to carry out the first step of writing . Those who did so were able to complete the remainder of this part but some bracketing errors, sign errors and manipulation errors were seen. In part (b), many candidates realised that they were required to substitute x = 0.1into their binomial expansion. About half of the candidates were able to offer the correct answer to 7 decimal places, but some candidates made calculation errors even after finding the correct binomial expansion in part (a). A few candidates used their calculator to evaluate the cube root of 7.7 and received no credit. 8. The majority of candidates produced correct solutions to this question, but a substantial minority of candidates were unable to carry out the first step of writing (3 + 2 x) –3 as . Those who were able to do this could usually complete the remainder of the question but some sign errors and manipulation errors were seen. Another common error was for candidates to apply and/or in the third and fourth terms of their expansion. 9. The majority of candidates produced correct solutions to this question, but a substantial minority of candidates were unable to carry out the first step of writing (2 – 5x) as . Those who were able to do this could usually complete the remainder of the question but some sign errors were seen. 10. Many candidates were successful with part (a). The most frequent error involved the use of powers of 2x rather than (– 2x). The careless use of brackets led both to sign errors in part (a) and also a significant number of incorrect answers in (b), the most common being either 100, 10 or × [answer to part (a)]. 11. This was a high scoring question for many candidates, with even weaker candidates often picking up half marks. A problem in part (a) was that some candidates used the fact that B = 0 to find A or C or both, especially by candidates who compared coefficients and found themselves getting “bogged down” with manipulating the equations. Candidates 4 3 2 1 2 1 4 3 1         x 2 1 3 1 3 1 8 3 12 as )38(        x x 2 3 2 1 27 1         x !2 )1( nn !3 )2)(1(  nnn 2 2 5 1 4 1         x 1000 1