GCE Mathematics Advanced Subsidiary GCE Unit 7: Core Mathematics Mark Scheme for June 0 Oford Cambridge and RSA Eaminations
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7 Mark Scheme June 0 (i) ( ) ( ) 9 p ( ) seen or q q or q (their q) r Reasonably correct curve for quadrants only y in st and rd If p, q, r found correctly, then ISW slips in format. ( ) + M0 A0 ( ) (BOD) ( ) B0 A0 ( ) B0 A0 ( + ) B0 (BOD) ( ) B0 N.B. Ignore feathering now that answers are scanned. Reasonably correct shape, not touching aes more than twice. Very good curves for y in st and rd quadrants SC If 0, very good single curve in either st or rd quadrant and nothing in other three quadrants. Correct shape, not touching aes, asymptotes clearly the aes. Allow slight movement away from asymptote at one end but not more. Not finite. (ii) (i) Translation units parallel to y ais Must be translation/translated not shift, move etc. 0 Or For parallel to the y ais allow vertically, up, in the (positive) y direction. Do not accept in/on/across/up/along the y ais (ii) or or in final answer seen (Allow ) in final answer is M0 ± is A0
7 Mark Scheme June 0 8 8 8( 0) ( 9)( )( 0) 9, y, y (i) 0 Attempt to eliminate or y Correct term quadratic (not necessarily all in one side) Correct method to solve quadratic values correct y values correct SR If A0 A0, one correct pair of values, spotted or from correct factorisation www Attempt to epress both surds in terms of One term correct Must be a clear attempt to reduce to one variable. Condone poor algebra for first mark. If eliminated: y y ( ) Leading to y 89y 800 0 e.g. 00 (y )(y ) =0 etc. Fully correct (not ) (ii) ( 0) 0 Multiply numerator and denominator by or - or attempt to epress both terms of numerator in terms of (e.g. dividing both terms by ) One of a, b correctly obtained Both a = and b= correctly obtained Check both numerator and denominator have been multiplied
7 Mark Scheme June 0 k k 8k 0 (k )( k ) 0 k or k = or = or = 8 If candidates use k and rearrange: k - 8 k + = 0 8 k =k + k = 9k + k + 9k 0k + = 0 (9k )(k )=0 k or k = 9 9 or = or = 8 7 (i) (ii) 0 < 7 ( )( ) >, < - * D * D 8 Use a substitution to obtain a quadratic or factorise into brackets each containing Correct method to solve a quadratic Attempt to calculate k Substitute, rearrange and square both sides Correct method to solve quadratic Attempt to calculate k equations or inequalities both dealing with all terms resulting in a b, a 9, b 0 - and - seen www Accept as two separate inequalities provided not linked by or (must be ) Rearrange to collect all terms on one side Correct method to find roots, - seen Correct method to solve quadratic inequality i.e. > their higher root, < their lower root (not wrapped, strict inequalities, no and ) No marks unless evidence of substitution (quadratic seen or square rooting or squaring of roots found). = 0 may be implied. Allow as a substitution. No marks if straight to quadratic formula to get = = and no further working No marks if k then k 8k 0 SC If M0 Spotted solutions www each Justifies solutions eactly B Do not ISW after correct answer if contradictory inequality seen. Allow Do not ISW after correct answer if contradictory inequality seen. e.g. for last two marks, - > > scores A0
7 Mark Scheme June 0 8 (i) dy 0 y 7 (ii) d y d y When =, > 0 so minimum pt ft ft 7 Attempt to differentiate (one non-zero term correct) Completely correct Sets their d y 0 Correct value for - www Correct value of y for their value of Correct method e.g. substitutes their from (i) into d y their (must involve ) and considers sign. ft from their dy differentiated correctly and correct substitution of their value of and consistent final conclusion NB If second derivate evaluated, it must be correct (8 for = ). If more than one value of used, ma A0 NB = (and therefore possibly y = 7) can be found from equating the incorrect differential dy = + to 0. This could score A0 A0 ft If more than one value of found, allow ft for one correct value of y dy Allow comparing signs of their either side of their, comparing values of y to their 7 d y SC = a constant correctly obtained from their dy and correct conclusion (ft)
7 Mark Scheme June 0 9 (i) Gradient of AB = 7 9 Gradient of AC = Verte A OR: Length of AB = (7 ) ( ) 0 AC = ( ) ( 9 ) 0 BC = ( 7) ( 9 ) 00 Shows that AB + AC = BC Verte A * D * D y y Uses for any points One correct gradient (may be for gradient of BC =) Gradients for both AB and AC found correctly Attempts to show that m m oe, accept negative reciprocal Correct use of Pythagoras, square rooting not needed Any length or length squared correct All three correct Correct use of Pythagoras to show AB + AC = BC Do not allow final mark if verte A found from wrong working. (Dependent on st M ) Accept BÂC etc for verte A or between AB and AC Allow if marked on diagram. i.e must add squares of shorter two lengths 9 (ii) Midpoint of BC is 7 9, Length of BC = ( 7) ( 9 ) = (, -) Radius = ( ) ( y ) ( ) ( ) ( y ) 0 y 8y 0 0 00 0 * ** D* D** 7 Uses y, y AC ( out of subs correct) Correct centre (cao) o.e. for BC, AB or Correct method to find d or r or d or r o.e. for BC, AB or AC (must be consistent with their midpoint if found) ( a) + (y b) seen for their centre ( a) + (y b) = their r Correct equation Correct equation in required form Substitution method (into + y + a + by + c = 0) Substitutes all points to get equations in a,b,c At least equations correct Correct method to find one variable One of a, b, c correct Correct method to find other values All values correct Correct equation in required form Alternative markscheme for last marks with f,g, c method: y 8y for their centre D* c = (±) + 0 D** c = 0 Correct equation in required form Ends of diameter method (p, q) to (c, d): Attempts to use ( p) ( c) + (y q) (y d) = 0 for BC,AC or AB M ( 7) ( + ) + (y ) (y + 9) = 0 A for both brackets correct, A for both y brackets correct + y + 8y 0 = 0 SC If M A0 A0 then if both brackets correct and if both y brackets correct for AC or AB
0(i) (ii) (iii) (iv) 7 Mark Scheme June 0 0,, 0,0 +, +, + ( + )( ) + 8 + dy 8 When =, gradient = Gradient of l = On curve, when = -, y = y = ( + ) y = + Attempt to find gradient of curve when = - ( ) ( ) 8 So line is a tangent +ve cubic with distinct roots (0, ) labelled or indicated on y-ais (-, 0), (, 0) and (, 0) labelled or indicated on - ais and no other - intercepts Obtain one quadratic factor (can be unsimplified) Attempt to multiply a quadratic by a linear factor Attempt to differentiate (one non-zero term correct) Fully correct epression www Confirms gradient = at = **AG May be embedded in equation of line Correct y coordinate Correct equation of line using their values Correct answer in correct form Substitute = - into their d y Obtain gradient of CWO Correct conclusion Substitution method into ( p) + (y q) = their r Correct method to find d or r or d or r * Substitutes all points to get equations in p,q D At least equations correct Correct method to find one variable One of p, q correct Correct equation [ ( ) ( y ) 0 ] Correct equation in required form [ y 8y 0 0 ] For first, left end of curve must finish below ais and right end must end above ais. Allow slight wrong curvature at one end but not both ends. No cusp at either turning point. No straight lines drawn with a ruler. Condone (0, ) as maimum point. To gain second and third B marks, there must be an attempt at a curve, not just points on aes. Final can be awarded for a negative cubic. Alternative for first marks: Attempt to epand all brackets with an appropriate number of terms (including an term) Epansion with at most incorrect term Correct, answer (can be unsimplified) Allow if done in part(i) please check. M mark is for any equation of line with any non-zero numerical gradient through (-, their evaluated y) Alternatives ) Equates equation of l to equation of curve and attempts to divide resulting cubic by ( + ) Obtains ( + ) ( ) (=0) Concludes repeated root implies tangent at = - ) Equates their gradient function to and uses correct method to solve the resulting quadratic Obtains ( + )( ) = 0 oe Correctly concludes gradient = when = -
7 Mark Scheme June 0 Allocation of method mark for solving a quadratic e.g. 8 0 ) If the candidate attempts to solve by factorisation, their attempt when epanded must produce the correct quadratic term and one other correct term (with correct sign): ( )( 9) 0 and 8 obtained from epansion ( )( ) 0 and obtained from epansion ( 9)( ) 0 M0 only term correct ) If the candidate attempts to solve by using the formula a) If the formula is quoted incorrectly then M0. b) If the formula is quoted correctly then one sign slip is permitted. Substituting the wrong numerical value for a or b or c scores M0 ( ) 8 earns (minus sign incorrect at start of formula) ( ) 8 earns (8 for c instead of 8) ( ) 8 M0 ( sign errors: initial sign and c incorrect) ( ) 8 M0 (b on the denominator) Notes for equations such as 8 0, then b = would be condoned in the discriminant and would not be counted as a sign error. Repeating the sign error for a in both occurrences in the formula would be two sign errors and score M0. c) If the formula is not quoted at all, substitution must be completely correct to earn the ) If the candidate attempts to complete the square, they must get to the square root stage involving ±; we are looking for evidence that the candidate knows a quadratic has two solutions! 8 0 8 0 8 0 9 9 This is where the is awarded arithmetical errors may be condoned provided seen or implied If a candidate makes repeated attempts (e.g. fails to factorise and then tries the formula), mark only what you consider to be their last full attempt. 7
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