Question Answer Marks Guidance x x x mult throughout by (x + )(x ) or combining fractions and mult up oe (can retain denominator throughout). Condone a single computational error provided that there xx) ( (x) (x)(x) is no conceptual error. Do not condone omission of brackets unless it is clear from subsequent work that they were assumed. x 3x x any fully correct multiplied out form (including say, if s correctly cancelled) soi x 3x = 0 = x(x 3) D dependent on first.for any method leading to both solutions. Collecting like terms and forming quadratic (= 0) and attempting to solve *(provided that it is a quadratic and b² 4ac 0).Using either correct quadratic equation formula (can be error when substituting), factorising (giving correct x² and constant terms when factors multiplied out), completing the square oe soi.* x = 0 or 3 for both solutions www. [4] SC B for x = 0 (or x = 3) without any working SC B for x = 0 (or x = 3) without above algebra but showing that they satisfy equation SC M0 SCB for first two stages followed by stating x = 0 SC M0 SCB for first two stages and cancelling x to obtain x = 3 only.
Question r Marks Guidance x A B C x (x) x x x = Ax(x ) B(x ) Cx x (x) x Ax( x ) B( x ) Cx x = 0, = B B = x = ½, ½ = C/4 C = 6 x coeffs: 0 = A + C A = 3 x 3 6 x (x) x x x B correct partial fractions Using a correct method to find a coefficient (equating numerators and substituting oe or using cover-up) Condone omission of brackets only if brackets are implied by subsequent work. Must go as far as finding a coefficient. Not dependent on B B = www C = 6 www A = 3 www isw for incorrect assembly of partial fractions following correct A,B,C A B SC x can get /5 max from B0 (for B=6) x Ax B C SC x can get B (C=6) and can continue x for full marks if the first fraction is then split. SC A B C Dx can get B (C=6, D=0) x x x [5]
3 3 x + A Bx+ C = + xx ( +) x ( x +) 3 x + = A(x + ) + (Bx + C)x x = 0 = A coefft of x : 0 = A + B B = coefft of x: 3 = C B correct partial fractions equating coefficients at least one of B,C correct 3 x + 3 x = + xx ( +) x ( x +) [6] 4 4 xx ( +4 A Bx + C = + ) x x +4 Ax ( +4)+(Bx+Cx ) = xx ( + 4) 4 = A( x +4) +( Bx+ C) x x = 0 4 = 4A A = coefft of x : 0 = A + B B = coeffts of x: 0 = C 4 x = xx ( +4) x x + 4 B D [6] correct partial fractions A= Substitution or equating coeffts B= C= 0 5 x 4x = 3 x x + x(x + ) 4x(x ) = 3(x )(x + ) x + x 4x + 8x = 3x 3x 6 0 = 5x 3x 6 = (5x + )(x 3) x = /5 or 3. cao [5] Clearing fractions expanding brackets oe factorising or formula
Question Answer Marks Guidance 6 (i) x (x)( x) A B x x x = A( x) + B( + x) x = ½ ½ = B( + ½ ) B = /3 x = = 3A A = /3 [3] expressing in partial fractions of correct form (at any stage) and attempting to use cover up, substitution or equating coefficients Condone a single sign error for only. www cao www cao (accept A/(+x) +B/(-x), A = /3, B = /3 as sufficient for full marks without needing to reassemble fractions with numerical numerators)
Question Answer Marks Guidance 6 (ii) x /3 /3 ( x)( x) x x ( ) 3 x ( x) correct binomial coefficients throughout for first three terms of ( ) ( ) ( ) ( ) [( )( x ) ( x)...(( ) x x...)] either (x) or (+x) oe ie,( ),( )( )/, not ncr form. 3 Or correct simplified coefficients seen. + x + 4x [ x4x... ( xx 3...)] x + x (or /3/-/3 of each expression, ft their A/B) If k(-x+x²) () not clearly stated separately, condone absence of inner brackets (ie +x+4x²x + x²) only if subsequently it is clear that brackets were assumed, otherwise A0. [ie x+x² is A0 unless it is followed by the correct answer] Ignore any subsequent incorrect terms (3x3x...) x x... so a = and b = or from expansion of x( x) (x) 3 www cao OR x ( x x ²) = x ( ( x + x ²)) correct binomial coefficients throughout for (-(x+x²)) oe (ie,-), at least as far as necessary terms (+x) (NB third term = x ( + x + x ² + (-)(-)( x + x ²)²/ + ) of expansion unnecessary and can be ignored) = x ( + x + x ² + x ² ) A x(+x) www = x + x².. so a = and b = ww cao Valid for ½ < x < ½ or x < ½ B independent of expansion. Must combine as one overall range. condone s (although incorrect) or a combination. Condone also, say ½ < x < ½ but not x < ½ or < x < or ½ > x > ½ [5]
7(i) v = 0e t t d t = 0e + c when t = 0, v = 0 0 = 0 + c c = 0 so v= 0 0e t [4] separate variables and intend to integrate t e 0 finding c cao (ii) As t e / t 0 v 0 So long term speed is 0 m s [] ft (for their c>0, found) (iii) ( w 4)( w+5) = A B w 4 + w+5 A( w+ 5)+ B( w 4) = ( w 4)( w+5) A(w + 5) + B(w 4) w = 4: = 9A A = /9 w = 5: = 9B B = /9 /9 /9 = ( w 4)( w+5) w 4 w+5 = 9( w 4) 9( w+5) [4] cover up, substitution or equating coeffs /9 /9 (iv) dw ( w 4)( w+5) dt = dw = dt ( w 4)( w+ 5) [ ] dw = dt 9( w 4) 9( w+5) ln( w 4) ln( w + 5) = t + c 9 9 w 4 ln = t+ c 9 w + 5 When t = 0, w = 0 6 c = ln = ln 9 5 9 5 w 4 9 ln = t ln + w + 5 5 9 w 4 9 + t 4.5t e t ln = 5 = e = 0.4e * w + 5 5 ft E [6] separating variables substituting their partial fractions integrating correctly (condone absence of c) correctly evaluating c (at any stage) combining lns (at any stage) www (v) As t e 4 5 t 0 w 4 0 So long term speed is 4 m s. []