AS Level / Year 1 Edexcel Further Maths / CP1

Transcription

1 AS Level / Year 1 Edexcel Further Maths / CP1 March 2018 Mocks 2018 crashmaths Limited

2 1 (a) n k(2k 3) = 2 k 2 k=1 k=1 n n Use linearity 3 k k=1 n k=1 k(2k 3) = 2 n 6 (n +1)(2n +1) 3 n 2 (n +1) Uses standard results d = 1 n(n +1) [ 2(2n +1) 9 ] 6 = 1 6 n(n +1)(4n 7) Extracts a factor of 1 n(n +1) 6 d (b) 3n 3n k(2k 3) = k(2k 3) k=n k=1 Complete and convincing proof AO2.1 [4] n 1 k(2k 3) Correct partitioning k=1 = 1 6 (3n)(3n +1)(4(3n) 7) 1 6 (n 1)(n 1+1)(4(n 1) 7) Uses their (a) to find an expression for the sum. No need to simplify. ISW after any correct form has been reached = n 2 (3n +1)(12n 7) n 6 (n 1)(4n 11) [For illustration only, simplification not necessary] ft [2] 6

3 2 (a) Enlargement Description AO1.2 of scale factor 3 about the origin / (0,0) Description AO1.2 [2] (b) (c) Method 1 Q = 1 0 Correct matrix Q AO R = = 3 0 Correct R ft their Q ft 0 3 [2] Area of T = 1 2 (8)(6) = 24 Correct area of T (units2 ) detr = 9 Attempts to find the determinant of their R (c) Method 2 So Area of T * = 24 9 = 216 (units 2 ) Correct area of T* D = (0,0), E = ( 12, 18), F = ( 24,0) Attempts to find the coordinates of D, E and F under their R [3] So area of T * = 1 2 (24)(18) = 216 (units2 ) Attempts to find area of T* Correct area [3] 7

4 3 Method 1 (α + 2),(β + 2) and (γ + 2) satisfy (w 2) 3 + p(w 2) 2 + 6(w 2) +1 = 0 Substitutes x = w 2 AO3.1a w 3 6w 2 +12w 8 + pw 2 4 pw + 4 p + 6w = 0 Expands the brackets d w 2 + (p 6)w 2 + (18 4 p)w + (4 p 19) = 0 Correct simplified expression Comparing coefficients gives Compares coefficients AO2.2 p 6 = 8 p = 2 Correct values of p and q q = 18 4 p = = 26 [5] Method 2 α + β + γ = p, aβ + βγ + αγ = 6, αβγ = 1 (*) Seen or implied AO2.2 8 = (α + 2) + (β + 2) + (γ + 2) = (α + β + γ) = p + 6 p = 2 Considers sum of the roots and forms equation in p ft their (*) Correct value of p AO3.1a d q = (α + 2)(β + 2) + (α + 2)(γ + 2) + (β + 2)(γ + 2) = αβ + αγ + βγ + 4(α + β + γ) +12 q = 6 + 4( 2) +12 q = 26 Considers pair sum and forms equation in terms of q ft their (*) and their p Correct value of q d [5] 5

5 4 (a) i = 6(2 3i) 12 18i = (2 + 3i)(2 3i) 13 Multiplies top and bottom by 2 3i Correct denominator of 13 = i Correct simplification x + iy i = (x iy)(4 i) 13 Uses z* = x iy and good attempt to combine real and imaginary parts AO1.2 x i y = (4x y) i(x + 4y) So x = 4x y and y = x 4y Compares coefficients correctly ft their 6 and attempts to solve for x or y 2 + 3i d x = 3 8, y = Correct values of x and y [6] (b) r = Correct value of r ft their (a). See notes AO1.2 21/104 θ = tan 1 3 / 8 = Method to find the argument. See notes So z = 0.43(cos(28.3) + isin(28.3)) Correct expression [3]

6 (c) One point plotted correctly on Argand diagram All points plotted correctly AO1.2 AO1.2 (Points should ft their (a)) [2] 11 Question 4 Notes (b) for the correct value of r ft their (a). Note that you can accept any degree of accuracy, exact answers or approximated answers (these should be seen to at least two significant figures), but there must be some computation, so r = alone is B0. correct method to find the argument ft their (a). If their complex number is in another quadrant, then this does not matter, we only y need to see tan 1 x here. correct expression. Note the argument must be given to at least two significant figures and you should accept other valid arguments. (c) one point plotted correctly on the Argand diagram, coordinates ft their (a). Accept complex numbers plotted as points or vectors. all three points plotted correctly on Argand diagram, coordinates ft their (a). Accept complex numbers plotted as points or vectors. The positions of all points should be relatively correct.

7 5 (a) Let the two roots be α and β, then Correctly constrains the problem AO3.1a α + β = b a, αβ = c a, α β = m n α 2 = cm an and β 2 = cn am Finds expressions for α 2 and β 2 AO2.1 α 2 + β 2 = (α + β) 2 2αβ cm an + cn am = b2 a 2c 2 a Attempts to use relation α 2 + β 2 = (α + β) 2 2αβ AO2.1 d (b) acm2 + acn 2 = b2 2ac a 2 mn a 2 acm 2 + acn 2 = mnb 2 2acmn acm 2 + 2acmn + acn 2 = mnb 2 ac(m 2 + 2mn + n 2 ) = mnb 2 ac(m + n) 2 = mnb 2 * k( 10)(3+ 2) 2 = 3(2)8 2 k = Complete and convincing proof AO2.1 Uses condition in (a) with correct values Correct value of k AO2.2 [4] [2] 6

8 Question 5 Notes (a) Note there may be alternatives to this question. In general, you should apply the following scheme: correctly constrains the problem, correctly expresses the roots in terms of a, b, c, m and n, d uses their expressions and a suitable relation to obtain an expression only in terms of a, b, c, m and n, complete and convincing proof

9 6 (a) a = 3 Cao AO2.2 (b) Volume = 3 7 (x 3)dx = x 2 2 3x 7 Correct expression for the volume (condone omission of here and ignore limits) 3 Correct indefinite integration (condone omission of here) [1] = (7) (3) Substitutes limits into the expression the correct way around d = 8 = 25.1( ) Correct answer oe (c/i) = 16 cm Cao AO3.4 [1] [4] (c/ii) 2 x 6 = 12 cm Answer is 12 cm, but condone 6 cm as q not clear AO3.4 [1]

10 (c/iii) 3f( 1 4 x) = x 3 Correct expression, seen or implied AO ( ) 1 Volume = 9π 4 x 3 dx 12 Sets up correct integral with correct limits ft their 3f(1/4x) and (c/i) AO3.1b ft = 9π 1 8 (28)2 3(28) 1 8 (12)2 + 3(12) Correct method to integrate their volume expression definitely (d) = 288π = ( ) Correct volume oe AO3.4 idea that Bowl of the glass curves outwards and then inwards, while solid of revolution doesn t change direction of curvature / dy/dx is never zero on the curve Any reasonable description of the curvature issue AO3.5b [4] [1] 12

11 7 (a) = Considers correct 3 x 3 determinant Expands the determinant AO3.1a d = 2( 3 3) 1(6 + 6) = = 0 therefore the system has no solutions Expands at least one 2 x 2 determinant correctly Convincingly obtains 0 and gives a conclusion AO2.1 d [4] (b) 3x 4 = 1 3x = 5 3x + 4 = 5 3x = 1 so the system is not consistent Attempts to eliminate one or two variables to show inconsistency Convincing proof and conclusion AO2.1 AO2.1 [2] (c) (The planes defined by the equations) form a prism Description AO1.2 Question 7 Notes (b) 1st this is for an attempt to form two equations which are inconsistent, i.e. replacing 2y z in the 2 nd and 3 rd equation (shown in scheme), subtracting 2 nd and 3 rd equations (to obtain 2y z = 1), etc. [1] 7

12 8 (a) 4 2 r = 2 + t Correct expression oe. Accept row vectors in particular AO1.2 [1] (b) e.g. Let t = 1, then r = = 3,so ( 6, 3, 6) lies on the line Convincing proof AO2.1 [1] (c) Method = 2 2 = Considers dot product of v with their direction vector of the line = 2 4 = 8 8 = so v is perpendicular to the plane Dots v with a second vector in the plane Shows v is perpendicular to two vectors in the plane AO2.1 [3]

13 (c) Method 2 i j k = i j k Considers cross product between two vectors in the plane and expands the determinant correctly into 2x2 matrices i j k = 14i 28j Correctly does a cross product between two vectors in the plane (d) 14i 28j = 14(i 2j), so v is perpendicular to the plane Convincing proof with conclusion AO2.1 [3] 1 LHS correct AO1.2 r 2 = 0 RHS correct AO1.2 0 [2] (e/i) = 8 0 (, so C does not lie in the plane) 2 0 (e/ii) Method 1 ˆn = Shows dot product isn t 0. Condone no conclusion here AO2.1 Correct unit normal, seen or implied [1] Shortest distance = = = Method to work out shortest distance Correct shortest distance [3]

14 (e/ii) Method 2 5 Correct normalisation factor, seen or implied Shortest distance = (8)(1) + (0)(2) + ( 2)(0) 5 = 8 5 Uses shortest distance formula with their normalisation factor and equation of the plane Correct shortest distance [3] 7

15 9 (a) Let n = 1, then LHS = 0 2 = RHS = = So true for n = 1 Shows the statement is true for n = 1 AO2.1 Assume true for n = k, i.e k = 2k k2 k 0 2 k Assumes true for n = k. Either can be shown explicitly or implied AO2.1 Then k+1 = k Begins inductive stage clearly using the assumption AO2.1 d = 2k k2 k k k+1 + k2 k k+1 = 2k+1 (k +1)2 k k+1 = 2k+1 Obtains the correct matrix for the case where n = k + 1 in the required form AO2.1 If true for n = k, then it has been shown that is is also true for n = k + 1. Since true for n = 1, it is true for all positive integers n. Complete and convincing proof, with no errors seen and a complete conclusion AO2.4 [5]

16 (b) Let n = 2 m {, then for m 0,} m 2 m 2 2m 0 2 = 22m 0 2 m 0 2 m m = 22 2 m+2m Convincing proof. Need to use (a) and clearly show what they are doing AO2.1 [1] 6

17 10 (a/i) Circle in the correct quadrant AO1.2 and Circle touches real axis AO1.2 (a/ii) Straight vertical line AO1.2 Passes through 3 AO1.2 [4] (b) = 3 Uses Pythagoras AO3.1a z = 3+ i(2 + 3), z = 3+ i(2 3) Correct complex solutions, one mark for each correct solution. Deduct 1 A mark (max 2) for any additional solutions given (c) k = 2, k = 6 Correct values of k, one mark for each correct value of k. Deduct 1 A mark (max 2) for any additional solutions given AO2.2 AO2.2 AO2.2 AO2.2 [3] [2] 9

18 Marks breakdown by AO AO Number of marks % AO AO AO3 9 15