Note If the candidate believes that e x = 0 solves to x = 0 or gives an extra solution of x = 0, then withhold the final accuracy mark.

Size: px

Start display at page:

Download "Note If the candidate believes that e x = 0 solves to x = 0 or gives an extra solution of x = 0, then withhold the final accuracy mark."

Berniece Farmer
5 years ago
Views:

1 . (a) Eithr y = or ( 0, ) (b) Whn =, y = ( 0 + ) = 0 = 0 ( + ) = 0 ( )( ) = 0 Eithr = (for possibly abov) or = A 3. Not If th candidat blivs that = 0 solvs to = 0 or givs an tra solution of = 0, thn withhold th final accuracy mark. (c) dy = (4 ) ( + ) A A 3 Not : (thir u ) + ( + )(thir v ) A: Any on trm corrct. A: Both trms corrct. (d) (4 ) ( ) 0 + = = ( 7)( ) = 0 = 7, A Whn 7 7 y =, = 9, whn, y = = dd A Not dy st : For stting thir found in part (c) qual to 0. nd : Factoris or liminat out corrctly and an attmpt to factoris a 3-trm quadratic or apply th formula to candidat s a + b + c. S ruls for solving a thr trm quadratic quation on pag of this Appndi. 3 rd dd: An attmpt to us at last on -coordinat on y = ( + ). Not that this mthod mark is dpndnt on th award of th two prvious mthod marks in this part. Som candidats writ down corrsponding y-coordinats without any working. It may b ncssary on som occasions to us your calculator to chck that at last on of th two y-coordinats found is corrct to awrt sf. Final A: Both{ = }, y = 7 and { = }, y = 9. cao Not that both act valus of y ar rquird. 7 []

2 . (a) ( + )( ) ( + )( 3) = ( ) ( 3) A af 3 Not : An attmpt to factoris th numrator. : Corrct factorisation of dnominator to giv ( + )( 3). Can b sn anywhr. (b) + 9 ln = = + = 3 = ( ) 3 = 3 d A af cso 4 Not : Uss a corrct law of logarithms to combin at last two trms. This usually is achivd by th subtraction law of logarithms to giv + 9 ln = + Th product law of logarithms can b usd to achiv ln ( + 9 ) = ln (( + )). Th product and quotint law could also b usd to achiv + 9 In = 0 ( ) + d: Rmoving ln s corrctly by th ralisation that th anti-ln of is. Not that this mark is dpndnt on th prvious mthod mark bing awardd. : Collct trms togthr and factoris. Not that this is not a dpndnt mthod mark A: or or. af Not that th answr nds to b in trms of. Th dcimal answr is Not that th solution must b corrct in ordr for you to award this final accuracy mark. [7]

3 3. (i) (a) ln(3 7) = ln(3 7) = Taks of both sids of th quation. This can b implid by 3 7 =. Thn rarrangs to mak th subjct. d 3 7 = + 7 = { = } + 7 Eact answr of. 3 3 A 3 (b) 3 7+ = ln (3 7+ ) = ln Taks ln (or logs) of both sids of th quation. ln 3 + ln 7+ = ln Applis th addition law of logarithms. ln = ln ln = ln A o (ln 3 + 7) = +ln Factorising out at last two trms on on sid and collcting numbr trms on th othr sid. dd + ln = { = } Eact answr of 7 + ln 3 + ln 7 + ln3 A o

4 (ii) (a) f() = +3, y = +3 y 3 = ln (y 3)= Attmpt to mak (or swappd y) th subjct ln y 3 = Maks th subjct and taks ln of both sids Hnc ( ) = ln( 3) ln( 3) or ln ( 3 ) f or ( ) = ln( y 3) f (s appndi) A cao f (): Domain: > 3 or (3, ) Eithr > 3 or (3, ) or Domain > 3. 4 (b) g()=ln( ),, > fg() = ln( ) +3 {=( ) + 3} fg() : Rang: y > 3 or (3, ) An attmpt to put function g into function f. ln( ) +3 or ( ) +3 or + 4. A isw Eithr y > 3 or (3, ) or Rang>3 or fg()>3. 3 [] 4. (a) P = 0 t t = 0 P = 0 o = 0() = 0 0

5 (b) t P = = 0 t 0 t rarrangs quation to mak th subjct. 000 t = ln 0 t = awrt.6 or 3 yars A Not t = or t = awrt.6 t = will scor A0 t (c) dp = 6 dt k and k 0. l 6 t t A (d) t 0 = 6 t 0 = ln 6 { } Using 0 = d P and d t an attmpt to solv t to find th valu of t or P = 0 l 0 ln 6 or P = 0 ( ) Substituts thir valu of t back into th quation for P. d 0(0) P = = 0 0 or awrt 0 A 3 6 []

6 . (a) Curv rtains shap whn > k ln Curv rflcts through th -ais whn > k ln (0, k ) and ( ln, 0 3 in th corrct positions. (b) Corrct shap of curv. Th curv should b containd in quadrants, and 3 (Ignor asymptot) ( k, 0) and ( 0, ln k ) (c) Rang of f: f() > k or y > k or ( k, ) Eithr f() > k or y> k or ( k, )or f > k or Rang > k.

7 (d) y = k y + k = Attmpt to mak ln(y + k) = (or swappd y) th subjct ln( y + k) = Maks th subjct and taks ln of both sids Hnc ( ) = ln( + k) ln( + k) or ln ( + k) A cao 3 f () f (): Domain: > k or ( k, ) Eithr > k or ( k, ) or Domain > k or ft on sidd ft inquality thir part (c) RANGE answr [0] 6. (a) f() = + ( + 4) ( )( + 4) R, 4,. ( )( + 4) ( ) + f() = ( ) ( + 4) An attmpt to combin to on fraction Corrct rsult of combining all thr fractions A = ( )( + 4) + = Simplifis to giv th corrct [( + 4)( )] numrator. Ignor omission of dnominator ( + 4)( 3) = [( + 4)( )] An attmpt to factoris th d numrator. ( 3) = ( ) Corrct rsult A cso AG A

8 (b) g() = 3 R, ln. Apply quotint rul: u = 3 du = v = dv = g () = ( ) ( ( ) 3) vu' uv' Applying v Corrct diffrntiation A = ( ) + 3 = Corrct rsult A AG ( ) cso 3 (c) g () = = = ( ) =( ) Puts thir diffrntiatd numrator qual to thir dnominator. = = A ( 4)( ) = 0 Attmpt to factoris or solv quadratic in = 4 or = = ln 4 or = 0 both = 0, ln 4 A 4 [] 7. (a) g() (b) fg() = f ( ) =3 + ln = + 3 A (fg : + 3 ) (c) fg () 3

9 (d) d ( + 3 ) = + 6 A + 6 = + (6 ) = 0 0, 6 = 0 A = 0, 6 A A 6 [0]. (a) f () = A = 3 (+) = 0 = A f( ) = 3 (b) = 0.96 = = (c) Choosing (0.7, 0.7 6) or an appropriat tightr intrval. f(0.7 ) = f(0.7 6)= A Chang of sign (and continuity) root (0.7, 0.7 6) cso A 3 ( = 0.76, is corrct to 4 dcimal placs) Not: = is accurat [] 9. (a) + = + = ln = (ln ) A

10 (b) dy = + dy = (ln ) = 6 y = 6 (ln ) y = ln A 4 [6] 0. (a) 6 3 ln 3 = ln 6 or ln = ln or ln 3 6 = 0 = (only this answr) Acso Answr = with no working or no incorrct working sn: A ln 6 Not: = from ln = ln 3 = ln M0A0 ln = ln 6 ln 3 = (ln 6 ln 3) allow, = (no wrong working) A (b) ( ) = 0 (any 3 trm form) ( 3)( ) = 0 = 3 or = Solving quadratic dp = ln 3, = 0 (or ln ) A 4 st for attmpting to multiply through by : Allow y, X, vn, for nd is for solving quadratic as far as gtting two valus for or y or X tc 3 rd is for convrting thir answr(s) of th form = k to = lnk (must b act) A is for ln3 and ln or 0 (Both rquird and no furthr solutions) [6]. (a) D = 0, t =, = 0 =.33 awrt A (b) D = 0 + 0, t =, = = 3.49 (*) Acso

11 Alt. (b) = = 3.49 (*) Acso (Main schm is for (0 + 0 ), or {0 + thir(a)} N.B. Th answr is givn. Thr ar many corrct answrs sn which dsrv M0A0 or A0 T (c) = 3 3 T = = T = ln T = or 3. or 3 A 3 T st M is for ( ) = 3 o.. nd T M is for convrting = k (k > 0) to T k = ln indpndnt of st M.. This is Trial and improvmnt: as schm, corrct procss for thir quation (two qual to 3 s.f.) A as schm [7]. (a) 4 C (b) 300 = t t = 7 sub. T = 300 and attmpt to rarrang to 0.0t = a, whr a Q 0.0t 7 = 400 A corrct application of logs t = 7.49 A 4 dt (c) = 0 0.0t ( for k 0.0t ) A dt At t = 0, rat of dcras = (±).64 C / min A 3

12 (d) T >, (sinc 0.0t 0 as t ) [9] 3. dy = A At = 3, gradint of normal = = 3 3 y ln l = 3( 3) y = A [] 00a 4. (a) Stting p 300 at t = = + a (300 = 00a); a = 0. (c.s.o.) (*) d A 3 0.t (b) 00(0.) 0 = 0.t + 0. ; 0.t = 6. A Corrctly taking logs to 0. t = ln k t = 4 (3.9..) A 4 (c) Corrct drivation: (Showing division of num. and dn. by 0.t ; using a) (d) Using t, 0.t 0, p 336 = 00 A 0. [0]. (a) log3 = log taking logs = log log 3 or log3 = log =.46 cao A 3 A

13 + (b) = log + = 4 + = 4 = or quivalnt; 4 multiplying by to gt a linar quation A 4 (c) sc = / cos sin = cos tan = = 4, A 3 us of tan [0] 6. (a) I = 3 + Using limits corrctly to giv +. (c.a.o.) A 3 must subst 0 and and subtract (b) A = (0, ); y = dy = Equation of tangnt: y = +; c =. ; A 4 attmpting to find q. of tangnt and subst in y = 0, must b linar quation (c) + y = y + 4y = + 4 y = y + 4 ; A putting y = and att. to rarrang to find. g 4 ( ) = or quivalnt A 3 must b in trms of (d) gf(0) = g(); =3 ; A att to put 0 into f and thn thir answr into g []

14 7. (a) (i) = a y (ii) In both sids of (i) i. ln = ln a y ln or ( y =) log a = ln a = y ln a * y ln a = ln cso = ylna is BO Must s ln a y or us of chang of bas formula. (b) y = ln a ln dy,, = * ln a, A cso dy dy ALT. or = ln a, = ln a, A cso nds som corrct attmpt at diffrntiating. (c) log 0 0 = A is (0, ) y A = from(b) m = or or (or bttr) 0 ln a 0 ln0 qu of targt y = m ( 0) i. y = ( 0) or y = + (o.) 0ln0 0 ln0 ln0 A 4 Allow ithr ft thir y A and m (d) y = 0 in (c) 0 = +, = 0 ln 0 0ln0 ln0 ln0 = 0 0 ln 0 or 0( ln0) or 0 ln0( ) ln0 A Attmpt to solv corrct quation. Allow if a not = 0. [0] y. (a) O Shap 3 p = 3 or { 3, 0} sn

15 (b) Gradint of tangnt at Q = q Gradint of normal = q Attmpt at quation of OQ [y = q] and substituting = q, y = ln 3q or attmpt at quation of tangnt [y 3 ln q = q( q)] with = 0, y = 0 or quating gradint of normal to (ln 3q)/q q + ln 3q = 0 (*) A 4 (c) ln 3 = 3 = ; = 3 ; A (d) = 0.90; = , 3 = , 4 = ; A Root = (3 dcimal placs) A 3 [] 9. dy = 6 A At =, dy = ; y = 4 ln A; Tangnt is y 4 + ln = ( ) At y = 0, = + ln = ln + ln = ln A [7] 0. (a) A is (, 0); B is (0, ) ; (b) y = Chang ovr and y, = y y = ln ( + ) y = + ln ( + ) A f : / + ln ( + ), > A A

16 (c) f() = 0 is quivalnt to = 0 Lt g() = g(3) =. g(4) =.3 Sign chang root α A (d) n + = + ln( n + ), = 3. = A 3 = A 4 = 3.04 = 3.0 Nds convincing argumnt on 3 d.p. accuracy Tak 3.03 and nt itration is rducing 3.0 Answr: 3.0 (3 d.p.) A [4]. (i) + 3 = = ln 6 = (ln 6 3) A 3 (ii) ln (3 + ) = = 4 = ( 4 ) 3 A 3 [6]. (a) y y = ln O shap -intrcpt lablld

17 (b) d y = so tangnt lin to (, ) is y = + C th lin passs through (, ) so = + C and C = 0 Th lin passs through th origin. A 3 y y = y= ln O (c) All lins y = m passing through th origin and having a gradint > 0 li abov th -ais. Thos having a gradint < will li blow th lin. y = so it cuts y = ln btwn = and =. (d) 0 =.6 n = 3 =.9 =. A 3 =. 4 =. =.7 A 3 () Whn =.7, ln = > 0 3 Whn =.6, ln = < 0 A Chang of sign implis thr is a root btwn. A 3 [3]

4037 ADDITIONAL MATHEMATICS

4037 ADDITIONAL MATHEMATICS CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Ordinary Lvl MARK SCHEME for th Octobr/Novmbr 0 sris 40 ADDITIONAL MATHEMATICS 40/ Papr, maimum raw mark 80 This mark schm is publishd as an aid to tachrs and candidats,