AS Level / Year 1 Edexcel Maths / Paper 1

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Hector Perkins
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1 AS Level / Year Edexcel Maths / Paper March 8 Mocks 8 crashmaths Limited

2 4x + 4x + 3 = 4( x + x) + 3 Takes out a factor of 4 from first two terms or whole expression = 4 x Completes the square correctly on their x + x Correct unsimplified expression oe = 4 x + + Cao 4 st for 4x + 4x + 3 = 4 x + x ( ) + 3 or 4x + 4x + 3 = 4 x + x Question Notes nd completes the square correctly on their x + x, must have a leading coefficient = st for = 4 x + nd cao (or equivalent, i.e. = 4 x etc.) 4 7//8 v8final

3 x dx = 3 x 3 dx = x {+c} Writes 3 x = 3 x and attempts to integrate Correct unsimplified indefinite integration (ignore constants) x 3 dx = x = x {+c} 3 dx = = x 3 x 3 3 Substitutes limits into their expression the correct way around (no simplifying necessary) = 3 3 = 3 ( 3 ) = ( ) * Rationalises the denominator (expression must be correct) Convincing proof, all steps shown, with answer in given form or correct value of k stated d 5 Question Notes st writes the integrand in index form and attempts to integrate (adds to power, divides by resulting power) st correct unsimplified integration (ignore constants) nd substitutes the limits into their indefinite integral the correct way around 3 rd rationalises the denominator correctly. This is dependent on the st and nd and requires the expression to be correct. nd complete and convincing proof, with all steps shown and no errors seen. Answer should be given in required form or k stated. 7//8 v8final

4 3!!! " a AB = = a 3!!! " AB = 5 5 ( a 3) + = 5 5!!! " Attempts to find expression for AB Forms the correct equation oe AO3.a a 6a = 5 a 6a + 5 = Attempts to solve their 3TQ for a Correct maximum value of a stated d ( )( a ) = a 5 a max = 5 4!!!"!!!" st considers OB OA nd correct method used to solve their 3TQ for a Question 3 Notes nd correct maximum value of a stated. Condone a = 5 stated (can assume they are referring to the maximum). 7//8 v8final

5 4 (a) x = 4, Correct values of x. One mark for each correct value. See notes for guidance if extra values given (b) (, tan( 4) ) Correct coordinates oe. Accept exact value of y or decimal equivalents (c) x = 3, x = 3 Correct asymptotes. One mark for each correct value. See notes for guidance if extra equations given B B B [] B B (d) Correct shape AO. B x, y intercepts and asymptotes ft their (a), (b) and (c) shown on sketch correctly BFT 7 7//8 v8final

6 Question 4 Notes (a) One B mark for each correct x intercept. If additional correct values (ignoring range) are given, ignore these values and give the B marks. If additional incorrect values are given (in or out of range), deduct B mark for each incorrect value given (max B marks). (b) Cao, accept decimal or exact value of tan(4) =.839. Condone just y = tan(4), i.e. coordinate form given. (c) One B mark for each correct asymptote. If additional correct asymptotes (ignoring range) are given, ignore these values and give the B marks. If additional incorrect asymptotes are given deduct B mark for each incorrect asymptote given (max B marks). (d) st B for the correct shape of the graph nd B x, y intercepts and asymptotes ft their (a), (b) and (c) shown on their sketch correctly 7//8 v8final

7 5 (a) Method x + 3x 9 x x 3 x 5x x 3 4x Attempts long division (all steps correct up to * ) Correct remainder 3x 5x (*) 3x 6x 9x 9x so the remainder is 5 (a) Method f() = () 3 () 5() = 5 so the remainder is 5 Substitutes into f(x) Correct remainder (b) Method x 7x 4 x + 3 x 3 x 5x x 3 + 6x Attempts long division (all steps correct up to * ) Completely correct long division, with conclusion at the end 7x 5x (*) 7x x 4x 4x Therefore, (x + 3) is a factor of f(x) 7//8 v8final

8 (b) Method f( 3) = ( 3) 3 ( 3) 5( 3) = = Therefore, (x + 3) is a factor of f(x) Substitutes 3 into f(x) Complete and convincing proof with no errors seen. Must see the terms/groups of terms evaluated (c) Other factor is x 7x 4 Method to find other factor of f(x) f(x) = (x + 3)(x 7x 4) = (x + 3)(x +)(x 4) If candidates use Method in (b), award for correct workings in (b) Solutions are x = 3,, 4 Correct solutions Question 5 Notes (a) Method and : no statements are necessary, but the final answer must be clearly identified, i.e. underlined etc. (b) Method : completely correct long division with conclusive statement, i.e. therefore (x + 3) is a factor of f(x) Method : complete and convincing proof. This is a show that question, so we need to see some explicit evaluation of the terms before they conclude f( 3) is equal to. (c) st uses a correct method to find the other factor, i.e. inspection (implied by obtaining one correct coefficient) or long division. st obtains the correct factor nd correct solutions. Cao Note: If candidates use Method in (b), then they can score for correct workings in (b) 7 7//8 v8final

9 6 (a) Method (a) Method lim h ( m(x + h) + c) mx + c h ( ) mh = lim h h = lim m = m h so the gradient of the line y = mx + c is m lim x' x ( mx'+ c) mx + c x' x ( ) m( x' x) = lim x' x x' x so the gradient of the line y = mx + c is m = lim m = m x' x Considers ( m(x + h) + c) mx + c h ( ) Complete and convincing proof with correct limiting process seen Considers ( mx'+ c) mx + c x' x ( ) Complete and convincing proof with correct limiting process seen AO. AO. AO. AO. (b) Method 6(x + h) 3 6x 3 6x 3 +8x h +8xh + 6h 3 6x 3 lim = lim h h h h 8x h +8xh + 6h 3 = lim h h ( ) = lim h 8x +8xh + 6h = 8x Considers 6(x + h)3 6x 3 Attempts to simplify numerator Complete and convincing proof with correct limiting process seen h AO. AO. d (b) Method 6x' 3 6x 3 6(x' x)(x' + xx'+ x ) lim = lim x' x x' x x' x (x' x) = 6lim x' x ( x' + xx'+ x ) = 6(x + x + x ) = 8x Considers 6x'3 6x 3 x' x Obtains a factor of x' + xx'+ x Complete and convincing proof with correct limiting process seen AO. AO. d 5 7//8 v8final

10 7 (a) ()()sin6 = 5. Attempts to find area cm Correct area to dp. Condone 5 cm (b) a = + ()()cos6 = 4 Uses cosine rule a = 4 =.(355...) cm Correct value of a. Accept any degree of accuracy (c) (d/i) e.g. sin x = sin6 sin x = sin6 a 4 = Uses the sine rule with correct combinations of sides 5 93 Identifies correct value of x oe AO3.a B x = sin 6 [] B [] (d/ii) sin 6 = Convincing illustration about why the other angle fails AO3.a B > 8, so this angle would violates the property that angles in a triangle add together to give 8. [] 7 7//8 v8final

11 8 (a) 5 x = 4 k 6 x Equates 5 x k 6 x = 4 AO3.a log( 5 ) x = log( 4 k 6 x ) Takes logs to both sides (accept any base or ln) d x log5 = ( k 6x)log 4 Uses power rule for logs correctly d x log5 = k log 4 ( 6log 4)x x log5 + ( 6log 4)x k log 4 = * Complete and convincing proof with no errors seen AO. [4] (b) ( 6log 4) 4( log5) ( k log 4) < Uses the discriminant (only interested in LHS here, ignore their choice of inequality/equality symbol) AO3.a 4k log5log 4 > 36log 4 k < 36log 4 4 log5log 4 k < 9log 4 log5 k < 9log 4 log5 = 9log.5 log5 * Attempts to re-arrange to get k on one side (ft using their chosen inequality/equality symbol). Ignore sign and inequality preservation errors also Complete and convincing proof with no errors seen. All stages must be correct and clearly shown. NB: Look out for fudged workings that appear to lead to the right answer but come from mishandlings of negatives and inequality signs AO. d 7 7//8 v8final

12 9 (a/i) m AB = ( ) 3 = 3 Attempts to find gradient of AB (a/ii) gradient of perpendicular bisector = 3 Midpoint of AB = 3, 3 Correct gradient of perp bisector ft their gradient of AB ft Correct coordinates of midpoint of AB B y + 3 = 3 x + 3 6x + y + = Attempts to use their midpoint and (a/i) to find equation of perp. bisector Correct equation oe (b) Method Gradient of normal to C at A is Correct gradient of normal to C at A. Seen or implied AO3.a So equation of normal is x + y + 7 = Correct equation of normal to C at A AO3.a x coordinate of centre given by solution to 5x + 5 = x = Attempts to solve simultaneously their equation of the normal and their (a/i) Obtains one coordinate correctly AO3.a AO. d + y + 7 = y = 6 y = 3 coordinates of C are (, 3) Obtains the second coordinate correctly (must show that x = y = 3) AO. [5] 7//8 v8final

13 (b) Method Gradient of normal to C at A is Correct gradient of normal to C at A. Seen or implied AO3.a So equation of normal is x + y + 7 = Correct equation of normal to C at A AO3.a Now + ( 3) + 7 = =, so (, 3) lies on the normal Substitutes (, 3) into their normal and their (a/ii) AO3.a d 6( ) + ( 3) + = + =, so (, 3) lies the perpendicular bisector of AB centre of C must have the coordinates (, 3) Shows that (, 3) lies on their normal OR their (a/ii) Shows that (, 3) lies on the normal and perp bisector and concludes that therefore the centre must be at (, 3) AO. AO. [5] (c) e.g. r = ( 3) + ( 3 ) = 5 r = 5 Correct radius oe (d) ( x +) + ( y + 3) = 5 st B : LHS correct nd Bft : RHS correct ft their (d) AO. AO. B Bft 5 Question 9 Notes (c) attempts to use Pythagoras (or equivalently, distance between two points formula) to find an expression for r or r. Condone one sign error. 7//8 v8final

14 (a) Correct shape AO. B Correct y intercept at (,) B (b) Asymptotes to x axis as x ± Passes through the origin Maximum and minimum points shown clearly and in the correct places AO. AO. AO. B B B 5 7//8 v8final

15 (a) C 3 =! 3! ( 3)! Uses the definition. Accept 7! instead of ( 3)! explicitly shown AO. = = 7 6 = * Convincing proof AO. (b) Method n n m m r m = n! m! n m! ( )! ( )!( n r)! ( ) n m r m Uses the definition. Accept equivalent expressions for n r, i.e. n m r + m. Condone n m r m AO. = = n! m! n m! ( )! ( )!( n r)! ( ) n m r m = n! m! ( r m)! ( n r)! r! r! Introduces r! r! AO. d = n! r! ( n r)! r! m! ( r m)! = n r r m Complete and convincing proof with no errors seen AO. 7//8 v8final

16 (b) Method n r r m = n! r! n r! ( ) r! m! ( r m)! Uses the definition AO. n! = r! ( n r! ) r! m! ( r m)! = n! m! ( n m)! ( n m)! ( r m)! ( n m r + m)! Introduces ( n m)! ( n m)! AO. d = n! m! ( n m)! ( n m)! ( r m)! ( n m r + m)! = n n m m r m Complete and convincing proof with no errors seen AO. (c) Let m = in (b), then Substitutes m = into (b) AO. n r r = n n r n r r = n n r r Clearly uses = r, n = n to give a complete and convincing proof AO. n r = n n r r 7 Question Notes (b) Special case: combinatorial proofs should be sent to review. 7//8 v8final

17 (i) dy dx = x3 x x x = x x 3 Attempts to write the gradient function in index form AO3.a y = (x x 3 )dx = x + x + c Attempts to integrate indefinitely Correct indefinite integration, including constant AO3.a d When x =, y = 4 4 = + + c c = 3 Substitutes initial conditions into their expression for y to find c d y = x + x + 3 States y in terms of x. NB: value of c alone is not enough [6] (ii) e.g. let f(x) = x, g(x) = x, then f(x)g(x)dx = x 3 dx = 4 Choose two functions for f and g and attempts to compute f g or f g AO. f(x)dx g(x)dx = x dx x dx = 3 = 6 4, so therefore Jessie is wrong. 6 Attempts to compute the other integral Both integrals computed correctly and shown not to be equal + conclusion, i.e. therefore Jessie is wrong or therefore f g fg AO. AO. 9 7//8 v8final

18 Question Notes (a) st attempts to write the gradient function in index form. M mark should be awarded for clearly separating the fraction into two terms and attempting to use relevant index law 3 rd we need to see y expressed in terms of x. In other words, candidates who find the value of c and leave their answer there cannot access the final A mark. Special case: st M and A mark as per scheme. Then: y dy' = (x' x' )dx' 4 x y 4 = x + x y = x + x No primes needed. nd for integrating both sides, nd for correct integration, 3 rd for correct limits, 3 rd correct answer oe (b) st chooses two functions for f and g (need not be distinct) and attempts to compute one of the integrals nd attempts to compute the other integral both integrals correctly evaluated, shown to be different and a conclusion. 7//8 v8final

19 3 (a) f '(x) = x 3 3 x = x 9 x Attempts to differentiate f wrt x AO3.a f '(4) = (4) 9 (4) = 8 9 = Substitutes 4 into their gradient function d So gradient of l is Correct gradient of the line l At x = 4, f(4) = = 4 Correct y coordinate when x = 4. Seen or implied B So equation of l is y = x 8 Cao [5] (b) g'(x) = x + qx Differentiates g wrt x correctly AO3.a g'( ) = q = Forms correct equation using their g AO3.a d q = 9 Correct value of q AO. 8 7//8 v8final

20 4 ** Scheme change: (a/i) has changed to mark (from ), while (a/ii) is now marks (from ) ** (a/i) = r h r 3 r h = 4 3 r 3 h r 4r 3 * Convincing proof AO. B [] (a/ii) A = 4 r + rh Writes down correct formula for surface area of the solid AO. A = 4 r + r r 4r 3 Correct expression oe (no need to simplify) A = 4 3 r + 4 r (b) da dr = 8 3 r 4 r Attempts to differentiate their A wrt r Correct differentiation AO3.b da dr = 8 3 r = 4 r =... r Sets their derivative equal to and attempts to re-arrange for r AO3.b d r 3 = 9 r = 9 3, so A is minimised when r is 9 3 Convincing proof AO. [4] 7//8 v8final

21 (c) d A dr = 8 3 π + 48 Attempts to differentiate their da r 3 dr again wrt r d A dr r= 9 3 π = 8 3 π π { = 8π} > Considers d A dr r= 9 3 π and shows that it is greater than (see notes). AO. NB: this mark requires their second derivative to be correct Since d A dr r= 9 3 π >, the surface area is minimised when 9 π 3 Question 4 Notes Conclusion AO.4 (b) nd no conclusion is OK here. (c) st 9 3 d A substitutes π into (must be correct) and shows that it is greater than. Candidates can either show the substitution dr or state the outcome they don t need to do both, as the positivity is fairly obvious, but they must state that it is positive. Can be implied in the conclusion. 7//8 v8final

22 Marks breakdown by AO AO Number of marks % AO 6 6 AO 4 4 AO //8 v8final

Paper 1 (Edexcel Version)

AS Level / Year 1 Paper 1 (Edexcel Version) Set A / Version 1 017 crashmaths Limited 1 y = 3x 4 + x x +1, x > 0 (a) ydx = 3x 3 3 3 + x 3 / x + x {+c} Attempts to integrate, correct unsimplified integration