Paper Reference(s) 666/0 Edecel GCE Core Mathematics C Advanced Level Wednesda 0 Januar 00 Afternoon Time: hour 0 minutes Materials required for eamination Mathematical Formulae (Pink or Green) Items included with question papers Nil Candidates ma use an calculator allowed b the regulations of the Joint Council for Qualifications. Calculators must not have the facilit for smbolic algebra manipulation, differentiation or integration, or have retrievable mathematical formulae stored in them. Instructions to Candidates Write the name of the eamining bo (Edecel), our centre number, candidate number, the unit title (Core Mathematics C), the paper reference (666), our surname, initials and signature. Information for Candidates A booklet Mathematical Formulae and Statistical Tables is provided. Full marks ma be obtained for answers to ALL questions. The marks for the parts of questions are shown in round brackets, e.g. (). There are 9 questions in this question paper. The total mark for this paper is 7. Advice to Candidates You must ensure that our answers to parts of questions are clearl labelled. You must show sufficient working to make our methods clear to the Eaminer. Answers without working ma not gain full credit. N8A This publication ma onl be reproduced in accordance with Edecel Limited copright polic. 00 Edecel Limited.
. Epress as a single fraction in its simplest form.. f() = + (a) Show that f() = 0 can be rearranged as =,. () The equation f() = 0 has one positive root α. The iterative formula n + = n n is used to find an approimation to α. (b) Taking = 0, find, to decimal places, the values of, and. (c) Show that α =.07 correct to decimal places.. (a) Epress cos sin in the form R cos( + α), where R > 0 and 0 < α <. (b) Hence, or otherwise, solve the equation cos sin = for 0 <, giving our answers to decimal places.. (i) Given that = ln( ), find d. d () (ii) Given that = tan, show that d = d. () N8A
. Sketch the graph of = ln, stating the coordinates of an points of intersection with the aes. 6. Figure Figure shows a sketch of the graph of = f (). The graph intersects the -ais at the point (0, ) and the point A(, ) is the maimum turning point. Sketch, on separate aes, the graphs of (i) = f( ) +, (ii) = f( + ) +, (iii) = f(). On each sketch, show the coordinates of the point at which our graph intersects the -ais and the coordinates of the point to which A is transformed. (9) N8A
7. (a) B writing sec as d(sec ), show that cos d = sec tan. Given that = e sec, (b) find d. d The curve with equation = e sec, 6 < < 6, has a minimum turning point at (a, b). (c) Find the values of the constants a and b, giving our answers to significant figures. 8. Solve for 0 80. cosec cot = (7) 9. (i) Find the eact solutions to the equations (a) ln ( 7) =, (b) e 7 + =. () (ii) The functions f and g are defined b f () = e +, R, g() = ln ( ), R, >. (a) Find f and state its domain. (b) Find fg and state its range. END TOTAL FOR PAPER: 7 MARKS N8A
+ Q + Januar 00 666 Core Mathematics C Mark + = ( ) + + = ( + )( ) + = ( ) + ( + )( ) or ( )( ) + or ( + )( ) seen or implied anwhere in candidate s working. Award below + ( ) = ( ) ( + ) + ( ) or ( ) ( + ) ( ) ( + ) Attempt to combine. Correct result. A Decide to award here!! = ( ) ( + ) or ( )( + ) Either or ( )( + ) ( )( + ) or 9 6 A aef [] GCE Core Mathematics C (666) Januar 00
Q f( ) = + (a) f( ) = 0 + = 0 + = ( ) 0 Sets f( ) = 0(can be implied) and takes out a factor of from +, or from + (slip). + = + ( ) = + + + = + then rearranges to give the quoted result on the question paper. A AG () n + (b) Iterative formula: n + =, = 0 n + (0) + = (0) + An attempt to substitute = 0 into the iterative formula. Can be implied b =. or. or awrt. =.0788... =.079... Both = awrt. and = awrt.07 =.0878... = awrt.09 A A (c) Let f( ) = + = 0 f (.06) = 0.07867... f (.07) = 0.00009... Sign change (and f ( ) is continuous) therefore a root α is α α =.07 ( dp) such that (.06,.07) Choose suitable interval for, e.g. [.06,.07] or tighter an one value awrt sf d both values correct awrt sf, sign change and conclusion A As a minimum, both values must be correct to sf, candidate states change of sign, hence root. [8] Core Mathematics C (666) Januar 00
π cos sin = Rcos + α, R > 0, 0 < < Q (a) ( ) cos sin = Rcoscosα Rsinsinα Equate cos : = R cosα Equate sin : = Rsinα R = + = = { } ;.809.. R = + ; or awrt.8 A tanα = α = 0.0900... c Hence, cos sin = cos( + 0.0) (b) cos sin = tanα = ± or tanα =± or sinα = ± or cosα =± their R their R α = awrt 0. or π α = awrt 0.7π or α = A awrt.8 cos( + 0.0) = cos 0.0 0.6899... ( + ) = { = } cos( their α ) c ( + 0.0) = 0.88696... For appling ± = their R cos their R c = 0.7... awrt 0.7 c A ( ) c c 0.0 π 0.88696... {.688... } + = = π their 0.88 dd = c.979... c awrt.9 A Hence, = { 0.7,.9} () [9] Part (b): If there are an EXTRA solutions inside the range 0 < π, then withhold the final accurac mark if the candidate would otherwise score all marks. Also ignore EXTRA solutions outside the range 0 < π. Core Mathematics C (666) Januar 00
Q (i) = ln( + ) u du = ln( + ) = something + + ln( + ) A + ln( ) d + Appl quotient rule: u = ln( + ) v = du dv = = d + d + = d ( ) ln( + ) u ln( + ) v Appling correctl. Correct differentiation with correct bracketing but allow recover. A d ( + ) d = ln( + ) {Ignore subsequent working.} (ii) = tan tan differentiate sin using either the d sec cos = quotient rule or product rule. d sec = d sec { cos } = = Finding d d sec or an attempt to b reciprocating d. * A d* = d + tan For writing down or appling the identit sec = + tan, which must be applied/stated completel in. d* Hence, = d +, (as required) For the correct proof, leading on from the previous line of working. A AG () [9] Core Mathematics C (666) Januar 00 6
Q = ln Right-hand branch in quadrants and. Correct shape. B (,0) O ( ),0 Left-hand branch in quadrants and. Correct shape. Completel correct sketch and both, 0 and, 0 ( { }) ( { }) B B [] Core Mathematics C (666) Januar 00 7
Q6 (i) = f( ) + A' (, ) ( ) { 0, } Shape of and must have a maimum in quadrant and a minimum in quadrant or on the positive -ais. Either ({ } ) Both ({ } ) 0, or '(, ) B A B 0, and '(, ) A B (ii) = f( + ) + A '({} 0, 6) An translation of the original curve. The translated maimum has either -coordinate of 0 (can be implied) or -coordinate of 6. The translated curve has maimum ({ 0,6 } ) and is in the correct position on the Cartesian aes. B B B O (iii) = f ( ) ( 0, ) ( ) A ', 6 Shape of with a minimum in quadrant and a maimum in quadrant. Either ({ } ) Both ({ } ) 0, or '(, 6) B A B 0, and '(, 6) A B O [9] Core Mathematics C (666) Januar 00 8
Q7 (a) = sec = = (cos ) cos = d (cos ) ( sin ) = ± ((cos ) (sin ) ) d (cos ) ( sin ) or (cos ) (sin ) A sin sin = = = d cos cos cos sec tan Convincing proof. Must see both underlined steps. A AG (b) = e sec du dv = = d d u = e v = sec e sec tan Either e e or Seen sec sec tan or implied Both e e and sec sec tan A Applies vu + uv correctl for their e sec e sectan u, u, v, v d = + e sec + e sec tan A isw (c) Turning point = 0 d Hence, e sec ( + tan ) = 0 {Note e 0, sec 0, so + tan = 0, } Sets their d = 0 and factorises (or cancels) d out at least e from at least two terms. giving tan = tan =± k ; k 0 { } = 0.8800 = a = 0.9600... Either = b = e sec( 0.96) ( 0.96) Hence, { } c awrt 0.96 or awrt. A = 0.809... = 0.8 (sf ) 0.8 A cao π Part (c): If there are an EXTRA solutions for (or a) inside the range π < <, ie. 0. < < 0. or ANY 6 6 EXTRA solutions for (or b), (for these values of ) then withhold the final accurac mark. π Also ignore EXTRA solutions outside the range π < <, ie. 0. < < 0.. 6 6 [] Core Mathematics C (666) Januar 00 9
Q8 cosec cot, ( eqn *) = 0 80 Using cosec = + cot gives + cot cot = o Writing down or using cosec =± ± cot or cosec θ =± ± cot θ. cot cot 0 = or cot = cot For either cot cot { = 0} or cot = cot A cot ( cot ) = 0 or cot = Attempt to factorise or solve a quadratic (See rules for factorising quadratics) or cancelling out cot from both sides. d cot = 0 or cot = Both cot = 0 and cot =. A cot = 0 (tan ) = 90, 70 =, cot = tan = =, =.,. Candidate attempts to divide at least one of their principal angles b. This will be usuall implied b seeing =. resulting from cot =. dd Overall, = {.,,., } Both =. and =. A Both = and = B [7] If there are an EXTRA solutions inside the range 0 80 and the candidate would otherwise score FULL MARKS then withhold the final accurac mark (the sith mark in this question). Also ignore EXTRA solutions outside the range 0 80. Core Mathematics C (666) Januar 00 0
Q9 (i)(a) ln( 7) = e = e ln( 7) Takes e of both sides of the equation. This can be implied b 7 = e. e + 7 { } 7 = e = =.80... Then rearranges to make the subject. Eact answer of e + 7. d A (b) 7 + e = 7 + ( ) ln e = ln Takes ln (or logs) of both sides of the equation. 7+ ln ln e ln + = Applies the addition law of logarithms. ln + 7 + = ln ln + 7+ = ln A oe (ln + 7) = + ln Factorising out at least two terms on one side and collecting number terms on the other side. dd (ii) (a) + ln = = 7 + ln f( ) e, = + { 0.087... } Eact answer of + ln 7 + ln A oe () = e + = e Attempt to make (or swapped ) the subject ln ( ) = ln ( ) = Makes e the subject and takes ln of both sides Hence f ( ) f ( ) = ln( ) or ln( ) or ln ( ) f ( ) = ln( ) (see appendi) : Domain: > or (, ) Either > or (, ) or Domain >. B (b) g( ) = ln( ),, > A cao { } ln( ) fg( ) = e + = ( ) + An attempt to put function g into function f. ln( ) e + or ( ) + or +. A isw fg( ) : Range: > or (, ) Either > or (, ) or Range > or fg( ) >. B [] Core Mathematics C (666) Januar 00