A Level Maths. Bronze Set B, Paper 1 (Edexcel version) 2018 crashmaths Limited

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1 A Level Maths Bronze Set B, Paper (Edexcel version) 08 crashmaths Limited

2 A Level Maths CM Practice Paper (for Edexcel) / Bronze Set B Question Solution Partial Marks Guidance dy dx = x x e x oe Method to differentiate one of the x n terms, n 0 Uses the chain rule to differentiate the exponential term Any two terms differentiated correctly (unsimplified or better) All four terms differentiated correctly and simplified. Accept (a) 4 = p + p = 4 [4] [] equivalent forms e.g. instead of x x Sets up a correct equation using information about x and x Obtains the correct value of p (b) x = 4(4) + = 58 x 4 = 4(58) + = 4 Bft Bft Correct value of x ft their p Correct value of x 4 ft their p and their x 4 n= x n = x + x + x + x 4 = = 09 [4] Complete method to find the sum Correct value of the sum (a) (Since is acute,) cos = sin = p [] Complete method to find cos (allow ± here) Award method mark for complete method using right-triangle Correct expression of cos

3 (b) cosecθ = sinθ = sinθ cosθ = p p ft [] Complete method to find cosecθ using their (a) Correct expression of cosecθ (c) sin(θ 45) = sinθ cos45 cosθsin 45 = ( p p ) [] Complete method to find sin(θ 45) using their (a) and replacing some value for cos(45) and sin(45) Correct expression of sin(θ 45) 4 x + y = y = x, so gradient of l is B Correct gradient of l seen or implied 0 =, so gradient of l is Attempts to find the gradient of l (allow sign errors) Correct gradient of l Since the product of the gradients =, the lines l and l are perpendicular ft [4] Correct conclusion ft their gradients giving correct and clear reasoning 5 (a) f(x) > B [] Correct range of f Condone y in place of f(x) but do not accept x 5 (b) fg(x) = e ln(4 x) + = (4x) + = 8x + B Uses e ln(4x) = 4x at any stage Hence, fg(x) = 5 8x + = 5 Finds fg(x) in any form and sets it equal to 5 x = 4 [] Correct value of x 5 (c) y = e x + e x = y x = ln y, so f (x) = ln x for x > Bft [] Sets y equal to x and attempts to re-arrange for x (getting up to e x = is OK) Correct expression for f (x) Correct domain ft their 5(a)

4 6 (a) f() = () + = (> 0) Calculates values of f() and f() f() = () + = (< 0) since there has been a change of sign and f is continuous (on [, ]), f has a root between [,] [] Correctly calculated values and conclusion to complete the proof 6 (b) f(.5) = f '(x) = x 4x, so f '(.5) = 4(.5) = B * Correct value of f(.5) seen or implied anywhere Complete method to find f '(.5) Correct value of f '(.5) Applying NR process: α = α f(α ) f '(α ) =.5 ( ) ( ) = so the second approximation is.45 to sf (dep*) [5] Valid attempt at Newton-Raphson using their values Correct second approximation to sf 6 (c) Attempts to draw the graphs of y = x and y = x on the same axis Correctly drawn graphs and conclusion that states that the graphs intersect once and so there is only one root Alternative: draws the graph of y = f(x) states that the graph only intersects the x-axis once and so the equation f(x) = 0 only has one root since the graphs only intersect once, f(x) = 0 only has one root []

5 7 (a) = A(x +) + B( x) Complete and correct method to find one of the values When x =, = B B = When x =, = A A = So f(x) = ( x) + (x +) [] Correct value of A seen or implied by final answer Correct value of B seen or implied by final answer (values of A and B are enough, i.e. final line of ms not required) 7 (b) f(x) = ( x) + (x +) so f '(x) = ( x) ( ) + (x +) () f '(x) = ( x) (x +) Correct method to differentiate one of the terms, including intended use of the chain rule seen (this can be implied) Correct derivative At minimum point f '(x) = 0 ( x) = (x +) * Sets their derivative = 0 Cross-multiplying gives x + x + = 4 4x + x 6x = x = f = 8 9 (dep*) Carries out a complete process to solve their equation for x If they square root both sides, allow the for square rooting but must see consideration of ± for the final (otherwise A0^) so minimum point has coordinates, 8 9 [5] Correct coordinates

6 7 (c) f(x)dx = 0 ( x) + 0 (x +) dx = ln( x) + ln(x +) = ln + ln ln() + ln() = ln 0 * ft (dep*) [4] States integral of the form aln( x) + bln(x + ) Correct indefinite integration ft their a and b Substitutes in the correct limits (ft their b) in the correct order Their upper limit from (b) must make sense with respect to the picture and function, i.e. be positive and less than / Obtains correct result in the correct form ISW 8 d sin(θ + h) sinθ (sinθ) = lim dθ h 0 h sinθ cosh + sinhcosθ sinθ = lim h 0 h = lim h 0 sinθ(cosh ) h + cosθsinh h cosh sinh = (sinθ)lim + (cosθ)lim h 0 h h 0 h = (sinθ)(0) + (cosθ)() = cosθ AG B* * (dep*) [5] Correct expression for the derivative Expands the compound angle (allow a sign error in the formula) Correct expression Groups the sinθ terms and the cosθ terms and attempts to apply the limit Complete and convincing proof with no errors seen and correct limiting process seen No consideration of limits is A0

7 9 (a) p and q are odd numbers B Deduces that p and q must both be odd Let p = n + and q = m +, where n and m are (positive) integers Then p + q = (n +) + (m +) = 4n + 4n ++ 4m + 4m + = (n + m + n + n +) which is even. So the statement is true [] Attempts to characterise p and q as potentially distinct odd integers and find the sum of the squares Complete, convincing and technical proof (need to see p and q explicitly defined and n, m defined as integers) with no errors and conclusion Note : if you don t see p = n + and q = m +, but you do see (n + ) + (m + ) then give the by implication (but the is withheld unless characterisation is clear) Note : if p and q are given the same characterisation or n is used for both (for example), then M0 A0 Note : unproven statements such as sum of odd number is odd are not good enough for the 9 (b) Suppose for a contradiction that there are a finite number of primes. Let p, p,, p k be a collection of all the primes. * Attempts a proof by contradiction, assuming that there are finitely many primes Consider the number P = p p...p k + (dep*) Constructs the number P (P must have a prime factor but) none of the primes p, p,, p k divide P, so P must be prime [] Complete and convincing proof with clear reasoning for why the construction of P implies the existence of another prime 9 (b) ALT Suppose for a contradiction that there are a finite number of primes. Then the largest prime exists and let p k be this prime * Attempts a proof by contradiction, assuming that there are finitely many primes Consider P = p k! +. (dep*) Constructs the number P (P must have a prime factor but) none of p k or any of the smaller primes divide P, so P must be prime [] Complete and convincing proof with clear reasoning for why the construction of P implies the existence of another prime

8 0 (a) (i) y = (x x) + 8 = x = x Extracts a factor of and attempts to complete the square on their left-over expression Correct unsimplified expression = x + 5 so a =, b = and c = 5 [] Completes the square correctly, obtaining the answer in the required form or values of a, b and c stated 0 (a) (ii) Coordinates of the maximum point is, 5 B Bft [] Maximum point Correct coordinates ft their 0(a) 0 (b) B B B Correct shape of the graph Correct x intersections Correct y intersection [] 0 (c) x + 6x + 8 = k(x + ) x + 6x + 8 = kx + k x + kx 6x + k 8 = 0 x + (k 6)x + (k 8) = 0 AG [] Eliminates y from the two equations and attempts to move all the terms to one side Complete and convincing proof with no errors seen

9 0 (d) If the curve and line are tangent, then they only intersect once, so (k 6) 4()(k 8) = 0 * Sets discriminant of the equation equal to 0 k k + 6 8(k 8) = 0 k 8k +00 = 0 Obtains the correct TQ k = ( 8) ± ( 8) 4()(00) () k = 4 ± 4 6 (dep*) [4] Complete method to solve their quadratic for k Use of a calculator does not score the method mark if it leads to the wrong answer Correct values of k (a) 6 = a + d 7 = a + 6d = 5d d = 5 B B One equation correct A second correct equation Attempts to solve their simultaneous equations Correct value of a or d a = 6 d = 6 + 5, so a = 4 5 [5] Correct value of a and d (b) First term =.05, common difference = 0.0 In the nth week, she donates =.75 Hence.75 =.05 + (n )(0.0) n = 8 B B Correct first term and common difference of sequence seen or implied at any stage Correct amount of donation in the nth week seen or implied at any stage Sets up correct equation and attempts to find n Correct value of n So sum of donations over the n week period is S 8 = 8 [ (.05) + (8 )(0.0) ] = 67.0 [6] Uses their values of a and d and their n to find S n Correct sum of donations with units (accept 670p)

10 SUBSTITUTION (R) Let u = x, then x = u and dx = du When x =, u = and when x =, u = So 5x x dx = 5( u) u( du) = ( 0 u + 5u u )du = 0u + 5u 5 5 = 0 5 () + () + 0 () () 5 B * **(dep*) (dep**) Chooses to use substitution. This is an overall process mark. Award for: ) attempting use substitution u =, changing terms to u s ) integrating and using appropriate limits States substitution u = x and a correct dx in terms of du (or equivalent) Attempts to get all aspects of the integral in terms of u s Condone slips in signs and coefficients States or implies integral of the form au / + bu 5/ Correct integral Substitutes the correct limits into the integral in the correct order = = = 4 = (6 4) AG [7] Obtains the given result convincingly with no errors seen

11 ALT SUBSTITUTION (R) Chooses to use substitution. This is an overall process mark. Award for: ) attempting use substitution u =, changing terms to u s ) integrating and using appropriate limits Let u = x, then u du = dx, so dx = u du When x =, u = and when x =, u = So 5x x dx = 5( u )u( u du) = ( 0u +0u 4 )du B * States substitution u = x and a correct dx in terms of du (or equivalent) Attempts to get all aspects of the integral in terms of u s Condone slips in signs and coefficients = 0 u + u 5 **(dep*) States or implies integral of the form au + bu 5 Correct integral = 0 () + () ( ) ( ) 5 (dep**) Substitutes the correct limits into the integral in the correct order = (9) = 4 = (6 4) [7] Obtains the given result convincingly with no errors seen

12 ALT PARTS (R) Chooses to use parts. This is an overall process mark. Award for: ) Attempting to use parts the correct way around ) using limits 5x x dx = 5x( x ) () ( ) = 0 x( x) = 0 x( x) ( x) () 5( ) + 5 ( x) 5( x) dx B * **(dep*) States or implies x dx = ( x) Uses parts correctly once and obtains expression of the form Ax( x) + B ( x) dx Integrates a second time to obtain integral of the form Px( x) + Q( x) 5 Correct integral of 0 x( x) 4 ( x) 5 seen or implied This may be partitioned as in the mark scheme so you may see the integral in separate parts (the partitioned parts may be evaluated separately also) = 0 ()() + 5 ( )() + 0 ( )() + 5 ()() (dep**) Substitutes correct limits into their integral in the correct order = = = 4 = (6 4) AG [7] Obtains the given result convincingly with no errors seen

13 (a) e.g. The liquid s surface is a circle of radius m and so has area π B [] Any sensible explanation/illustration of why A = π (b) π dh dt = 0.06π h h h dh dt = 0.06 dh = 0.06dt h = 0.06t + c B* B(dep*) Separates variables correctly Attempted integration of one of the sides When t = 0, h = 4, so c = 4 = 4 h = t Uses initial conditions to find their c in an expression which contains the terms at and b h, a, b 0 h = 0.008t h = ( 0.008t) AG [4] Obtains the given result convincingly with no errors seen (c) Tank is empty when h = 0, i.e. ( 0.008t) = t = 0 t = 50 s so 4.7 minutes [] Sets h = 0 and attempts to re-arrange for t If they expand into a TQ, must see a valid attempt to solve this Correct value of t in seconds Correct value of t in minutes (d) e.g. The area of the liquid s surface now changes as the liquid drains B [] Correct explanation

MATHEMATICS A2/M/P1 A LEVEL PAPER 1

MATHEMATICS A2/M/P1 A LEVEL PAPER 1 Surname Other Names Candidate Signature Centre Number Candidate Number Examiner Comments Total Marks MATHEMATICS A LEVEL PAPER 1 Bronze Set B (Edexcel Version) CM Time allowed: 2 hours Instructions to