AS Level Further Maths

Transcription

1 AS Level Further Maths Bronze Set B, Core Pure (Edexcel version) 2018 crashmaths Limited

2 AS Further Maths CM Core Pure 1 Practice Paper (for Edexcel) / Bronze Set B Question Solution Partial Marks Guidance 1 n = r 3 4 r r=1 n r=1 = n2 4 (n +1)2 4 n 2 (n +1) *(dep*) Attempts to use linearity of the sum Substitutes in correct formula for sum of cubes and sum of natural numbers in their expression = 1 n(n +1) [ n(n +1) 8 ] 4 (dep**) n(n + 1) = 1 4 n(n +1)(n2 + n 8) [4] Extracts a factor of 1 4 Obtains answer in correct form or states the values of a, b and c with sufficient working and no errors seen. ISW 2 2(w 1) 3 (w 1) 2 + (w 1) = 0 Substitutes z = w ± 1 into equation 2(w 3 3w 2 + 3w 1) (w 2 2w +1) + (w 1) = 0 (dep*) Method to expand terms in their expression. Condone sign errors, but we need to see correct binomial coefficients for each expansion. Correct unsimplified expression 2w 2 7w 2 + 9w 4 = 0 So w w w 2 = 0 [4] Correct equation in w in correct form. ISW once correct form has been reached. 3 (a) k 1 k 3 0 k k = 0 4( 2(k 1) 3) (k 2 k) + 2(k) = 0 8k k 2 + k + 2k = 0 k 2 + 5k + 4 = 0 so k = 1, k = 4 (dep*) Equates determinant of matrix of coefficients to 0 and correctly expands the determinant into 2 2 determinants. Condone omission of sign on middle term Expands determinant fully to form a 3TQ and uses a complete method to solve their 3TQ Correct values of k and no others. ISW 3 (b) (i) For k = 1, eliminating x from first two equations gives y 2z = 44/7 2, and so is not consistent with the last eq. Method to show system is inconsistent for one of their values of k Convincing proof for k = 1

3 For k = 4, eliminating x from first two equations gives y 5z = 14 2, and so is not consistent with the last eq. Convincing proof for k = 4 [Candidates do not need to mention that the planes are not parallel] 3 (b) (ii) (The planes form a triangular) prism 4 α = 7 6 β [1] Correct description. The word prism alone is sufficient Correct relationship between α and β, seen or implied αβ = 14 3, α + β = k 3 Correctly relates coefficients to roots, seen or implied 7β 2 6 = 14 3 β 2 = 4 β = 2 Attempts to solve for one of the roots so α = 7 3 Both roots correctly found Then k = , so k = 13 [5] Correct value of k 4 ALT Let the roots be 7x and 6x Then product of roots is 14/3, so 42x 2 = 42 3 x2 = 1 9 Forms an expression relating the roots by the ratio, seen or implied Correctly relates coefficients to roots, seen or implied Attempts to solve for x (or equivalent) So x = 1 3 Sum of roots is 13x = k/3, so k = 13 [5] Correct value of x Correct value of k

4 5 (a) (i) ( i) ( 7 + i 3) = i + 28i 4 3 Expands brackets, generating four terms and uses i 2 = 1 5 (a) (ii) = i i 7 + i 3 = = ( )( 7 i 3) ( )( 7 i 3) i = 7 + i i + 28i i 52 = 3 + i [2] Correct final answer Multiplies numerator and denominator by conjugate of z 2 Obtain correct simplified numerator or denominator Correct final answer 5 (b) [1] Both points A and B correctly plotted on an Argand diagram in relatively correct positions. Condone no labels and even the wrong labels. Detailed scale is not required. 5 (c) AOB = arg(z 1 ) arg(z 2 ) In this question, a calculator is not allowed. = arg z 1 z 2 ( ) = arg 3 + i = π π 6 = 5π AG 6 Carries out complete non-calculator based method to find AOB: uses arg(u/v). Must state they are using arg(u/v) or M0 use of cosine rule, leading to cosθ = k (middle step oe i necessary): cosθ = = = ! 4 ## 13 " ## 2 13 $ 2 step necessary oe [2] Complete and convincing proof. 4 [ AOB = arg(z 1 ) arg(z 2 ) = π tan tan 1 7 = 5π 6 M0 A0 unless justified without use of a calculator.] is

5 6 (a) A = [1] Correct matrix A 6 (b) Enlargement matrix is k 0 0 k Correct enlargement matrix, seen or implied B = k k 1 0 = 0 k k 0 Uses B = (their enlargement matrix) A Correct matrix B. ISW 6 (c) (0,0) [1] Correct coordinates. Accept x = 0, y = 0. Accept O or the origin. 6 (d) Matrix representing T 1 followed by T 2 is k 1 4 k 0 = 3k 2k 4k k 3k 2k 1 4k k 4 = k 8k = k = 4 *(dep*) Attempts to compute CB or BC Sets up correct equation using their CB or BC Correct value of k = 0 44, so image of (2, 3) is (0, 44) (dep**) [5] Uses their k to find the image of (2, 3) Correct image as a coordinate 6 (e) = 0 0, so yes, P is invariant under T [2] Method to determine whether their P is invariant under their T. Identifies that P is still invariant under T with convincing working [SC: Yes, P is invariant since the origin is invariant under all linear transformations is SC B2]

6 7 (a) f(16) = (16)2 = = = 0, as required [2] Substitutes 16 into f Complete and convincing proof, showing one intermediate step between substitution and final answer and a brief conclusion, i.e. as required, qed, therefore, a = 16 7 (a) ALT a a2 = 0 a 3 = 1 16 a4 a = 16 [2] Sets f(a) = 0 and attempts to solve for a. Allow working in x Correct value of a. If working in terms of x, must give a statement that refers to a such as therefore, a = (b) f '(x) = 3 2 x 1 2 x Attempts to differentiate f 3 2 x 1 2 x = 0 3 x 2 x 1 2 = 0 x = 9 (x 0) so b = 9 (dep*) Sets their derivative equal to 0 and solves for x Correct value of b Then c = (9)2 = = 27 4 AG [4] Convincingly shows c = 27/4. Condone omission of intermediate steps here, but they must show the substitution at least.

7 7 (c) Volume of revolution of curve is π 16 0 x x2 dx 16 = π x x x4 dx 0 Attempts to find the integral of y 2. Ignore limits here. = π 1 4 x x x Correct indefinite integral of y 2 = π 1 4 (16) (16) (16)5 0 = π (dep*) Substitutes correct limits into their indefinite integral correctly Correct volume of revolution as an exact value or to at least 2 sf Volume of cylinder = π (16) = 729π Volume formed by shaded region is 729π π = which is 1150 units3 to 3sf [7] Correct volume of cylinder formed by the revolution of the rectangle For volume formed by rectangle volume formed by curve Correct answer to 3 significant figures. ISW 8 (a) The lines have different direction vectors so they are not parallel See guidance for rest of scheme States that the lines are not parallel λ = 1 λ = 5 λ = µ = 5 µ = 7 µ = [4] Equate at least two components and solve for λ or µ One correct value for λ or µ Verify that all three equations are not satisfied and the lines fail to intersect

8 8 (b) i 1 = = Evaluates dot product between i 4j + 2z and one direction vector Shows i 4j + 2z is perpendicular to one line (i.e. dot product = 0) i 1 = = so the vector i 4j + 2z is perpendicular to both lines Convincing proof 8 (b) ALT 2a + b + 3c = 0 2a + b + c = 0 States or implies one correct equation giving conditions on the normal vector 2b + 4c = 0 so b : c = 2 : 1 Solves correctly for the ratio of two of the components oe If c = 2, b = 4 and a = 1. So i 4j + 2z is perpendicular to both lines Convincing proof 8 (b) ALT Normal vector given by = 2i + 8j 4k 3 1 Attempts to compute cross-product of direction vectors Two components of the cross-product correct so i 4j + 2z is perpendicular to both lines (as it is a multiple of this) Convincing proof 8 (c) Midpoint of line segment from two lines is 1 2, 1 2, 1 2 States or implies correct position vector of the midpoint of a line segment connecting the two lines So d = = 3 2 Hence equation of plane is x 4y + 2z = 3 2 Use result of part (ii) and their midpoint to find d. They must identify their point as a midpoint. Correct equation of plane oe Beware of fortuitous answers

9 8 (c) ALT 1(1) 4(1) + 2(0) d 21 1( 2) 4(0) + 2(1) d = 21 Forms an equation in d by equation perpendicular distances from both lines to the plane 3 d = d States a correct equation d = 3 2 Hence equation of plane is x 4y + 2z = 3 2 Correct equation of plane oe Beware of fortuitous answers 9 5 p + q = 200 States or implies correct equation Let the two roots be x + iy and x iy Then sum of roots is 5 + 2x = 13 x = 4 Since roots lie on a circle of radius 5, y 2 = 25 y = 3 Then q = product of roots = 5(x 2 + y 2 ) = 5(25) = 125 So p = ( )/5 = 65, [8] Attempts to use sum of roots property to find x Correct value of x Attempts to use the fact that the roots lie on a circle of radius 5 Correct value of y Attempts to use q = ± product of roots Correct value of p, q 10 (a) f(1) = 1(2) = 2, so true for n = 1 Assume statement is true for n = k, i.e. k(k + 1) is divisible by 2 Shows that the statement is true for n = 1 with conclusion (which may be seen at a later stage) Then f(k +1) f(k) = (k +1)(k + 2) k(k +1) = (k +1)[k + 2 k] Makes the assumption and attempts to compute f(k + 1) f(k) or just f(k + 1) = 2(k +1) 2(k + 1) is divisible by 2 Hence f(k +1) = f(k) + 2(k +1) is divisible by 2 Convincingly shows the statement is true for n = k + 1 Therefore, if true for n = k, it has been shown that by induction, it is also true for n = k + 1. Since true for n = 1, it follows (by induction) that it is true for all positive integers n [4] Conclusion with all underlined elements

10 10 (b) g(1) = = 0, which is divisible by 6 Assume the statement is true for n = k, i.e. k 3 k is divisible by 6 Shows that the statement is true for n = 1 with conclusion (which may be seen at a later stage) Then g(k +1) g(k) = (k +1) 3 (k +1) (k 3 k) = k 3 + 3k 2 + 3k +1 k 1 k 3 + k Makes the assumption and attempts to compute g(k + 1) g(k) or just g(k + 1) = 3k 2 + 3k = 3k(k +1) 3k(k + 1) is a multiple of 6 since it is divisible by 3 and by part (a), it is also divisible by 2 Explains why 3k(k + 1) is divisible by 6. NB: no need to mention part (a) 3k(k + 1) is divisible by 3 and 2, and so it is also divisible by 6 is Hence g(k + 1) = g(k) + 3k(k + 1) is divisible by 6 Obtains g(k + 1) = g(k) + 3k(k + 1) and states it is divisible by 6 Therefore, if true for n = k, it has been shown that by induction, it is also true for n = k + 1. Since true for n = 1, it follows (by induction) that it is true for all positive integers n [5] Convincing conclusion with all underlined elements