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A-Level · Pure Mathematics

P16 Pure: Vectors and Numerical Methods Depth

EDEXCEL 9MA0 · Calculator allowed · about 130 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Relative to a fixed origin O, a = 2i - j + 2k and b = i + 2j - 2k.

(a)Find |a|.(2)

(b)Find the vector c = a + 2b, giving your answer in terms of i, j and k.(2)

(c)Find |c|, giving your answer to 2 decimal places.(2)

(d)Find the angle between a and b, giving your answer in degrees to 1 decimal place.(3)

(Total for Question 1 is 9 marks)

f(x) = x³ - 6x - 3.

(a)Show that the equation f(x) = 0 has a root between x = 2 and x = 3.(2)

(b)Show that the equation f(x) = 0 can be rearranged into the form x = (6x+3)^1/3.(2)

(c)Using the iteration x_(n+1) = (6x_n+3)^1/3 with x0 = 2, find the values of x1, x2 and x3, giving your answers to 4 decimal places.(3)

(d)By considering the change of sign of f(x) on a suitable interval, show that the root of f(x) = 0 in (2,3) is x = 2.669, correct to 3 decimal places.(3)

(Total for Question 2 is 10 marks)

f(x) = x³ - 5x - 3.

(a)Show that f(x) = 0 has a root α between x = 2 and x = 3.(2)

(b)Using x0 = 2.5, apply the Newton-Raphson method once to f(x) to find x1, giving your answer to 6 decimal places.(3)

(c)Apply the Newton-Raphson method a second time to find x2, giving your answer to 6 decimal places.(2)

(d)Given that x1 and x2 already agree to 4 decimal places (both round to 2.4909), comment on the rate of convergence of the Newton-Raphson method here compared with the fixed-point iteration used in Question 2.(2)

(Total for Question 3 is 9 marks)

Relative to a fixed origin O, the points P and Q have position vectors OP = 3i + 5j and OQ = 11i - 3j.

(a)Find the vector PQ.(2)

(b)The point R lies on PQ such that PR:RQ = 3:1. Find the position vector of R.(3)

(c)The point S has position vector 18i - 2j. Show that O, R and S are collinear.(2)

(Total for Question 4 is 7 marks)

Lines l1 and l2 have vector equations
l1: r = (i + 2j + 3k) + t(i - j + 2k)
l2: r = (5i - 2j + 5k) + s(-2i + 2j + 2k)
where s and t are scalar parameters.

(a)Show that l1 and l2 intersect, and find the position vector of their point of intersection.(6)

(b)Find the angle between l1 and l2, giving your answer in degrees to 1 decimal place.(4)

(Total for Question 5 is 10 marks)

The curve C has equation y = √x² + 16, for 0 ≤ x ≤ 4.

(a)Using the trapezium rule with 4 strips of equal width, find an estimate for the integral from x=0 to x=4 of √x²+16 dx, giving your answer to 3 decimal places. You should show the value of y at each of the 5 values of x used.(5)

(b)State, with a reason, whether your estimate in part (a) is an overestimate or an underestimate of the true value of the integral.(2)

(Total for Question 6 is 7 marks)

Two lines, l1 and l2, have direction vectors 2i + 3j - 6k and 3i - 6j + 2k respectively.

(Total for Question 7 is 5 marks)

g(x) = e^x - 3x - 2. The equation g(x) = 0 has a root α in the interval [2,3].

(a)Show that α lies between x = 2.1 and x = 2.2.(2)

(b)By further systematic decimal search (or otherwise), show that α = 2.13, correct to 2 decimal places.(3)

(Total for Question 8 is 5 marks)

f(x) = 2x³ - 9x² + 5.

(a)Show that x0 = 3 is not a suitable first approximation for solving f(x) = 0 using the Newton-Raphson method, explaining what goes wrong.(3)

(b)Using x0 = 4.4 instead, apply the Newton-Raphson method once to find x1, giving your answer to 4 decimal places.(3)

(Total for Question 9 is 6 marks)

The line l has vector equation r = t(i + 2j + 2k), where t is a scalar parameter. The point P has position vector 8i + j + 4k.

(a)Find the position vector of N, the foot of the perpendicular from P to l.(6)

(b)Find the distance PN, giving your answer in the form k*√5, where k is an integer.(3)

(c)The point P' is the reflection of P in the line l. Find the position vector of P'.(3)

(Total for Question 10 is 12 marks)

I = integral from x=0 to x=1 of 1/(1+x²) dx.

(a)Using the trapezium rule with 2 strips of equal width, find an estimate for I, giving your answer to 3 decimal places.(4)

(b)Using the trapezium rule with 4 strips of equal width, find a second estimate for I, giving your answer to 4 decimal places.(4)

(c)Given that the exact value of I is π/4, find the percentage error in each of your estimates from parts (a) and (b), giving your answers to 2 significant figures.(3)

(d)Comment on how the percentage error changes as the number of strips is doubled, with reference to your answers to part (c).(1)

(Total for Question 11 is 12 marks)

A drone follows a path such that its position vector, in km, relative to a fixed point O, at time t hours (t ≥ 0) is given by r(t) = t i + t² j + t k. A ground station is located at the point P with position vector 4i + 2j + 5k.

(a)Show that the square of the distance between the drone and the ground station at time t is given by D(t) = t⁴ - 2t² - 18t + 45.(4)

(b)Show that the value of t at which the distance is a minimum satisfies 2t³ - 2t - 9 = 0.(3)

(c)Show that this equation has a root between t = 1 and t = 2.(2)

(d)Taking t0 = 1.85 as a first approximation, use the Newton-Raphson method once to find an improved estimate t1 for the root, giving your answer to 4 decimal places.(4)

(Total for Question 12 is 13 marks)

Mark scheme · P16 Pure: Vectors and Numerical Methods Depth

Question 1

(a) M1 attempts 2² + (-1)² + 2²
(a) A1 |a| = 3 cao
(a) Answer: |a| = 3
(b) M1 attempts 2b = 2i + 4j - 4k, or the full component sum a + 2b
(b) A1 c = 4i + 3j - 2k oe
(b) Answer: c = 4i + 3j - 2k
(c) M1 attempts 4² + 3² + (-2)²
(c) A1 |c| = √29 = 5.39 awrt
(c) Answer: |c| = √29, awrt 5.39
(d) M1 attempts a.b = (2)(1) + (-1)(2) + (2)(-2)
(d) M1 attempts cos(θ) = (a.b)/(|a||b|) with correct |a| and |b|
(d) A1 θ = 116.4 degrees awrt
(d) Answer: θ = awrt 116.4 degrees

Question 2

(a) M1 evaluates f(2) and f(3): f(2) = -7, f(3) = 6
(a) A1 sign change (f(2) < 0, f(3) > 0) and f continuous, so a root lies in (2,3) cso
(a) Answer: f(2) = -7, f(3) = 6; sign change confirms a root in (2,3)
(b) M1 rearranges x³ - 6x - 3 = 0 to x³ = 6x + 3
(b) A1 takes cube roots of both sides to obtain x = (6x+3)^1/3 cso
(b) Answer: x = (6x+3)^1/3
(c) M1 correct method, x1 = (6(2)+3)^1/3 = 15^1/3, x1 = 2.4662 awrt
(c) A1 x2 = 2.6109 awrt
(c) A1 x3 = 2.6526 awrt
(c) Answer: x1 = 2.4662, x2 = 2.6109, x3 = 2.6526
(d) M1 evaluates f(2.6685) and f(2.6695), or an equivalent bounding interval for 2.669
(d) A1 f(2.6685) = -0.0089 (awrt, negative) and f(2.6695) = 0.0065 (awrt, positive), both correct
(d) A1 sign change on (2.6685, 2.6695), f continuous, hence root = 2.669 correct to 3dp cso
(d) Answer: Root = 2.669 (3dp)

Question 3

(a) M1 evaluates f(2) and f(3): f(2) = -5, f(3) = 9
(a) A1 sign change (f(2) < 0, f(3) > 0), f continuous, so a root lies in (2,3) cso
(a) Answer: f(2) = -5, f(3) = 9; sign change confirms a root in (2,3)
(b) M1 correct derivative f'(x) = 3x² - 5
(b) M1 correct Newton-Raphson substitution using f(2.5) = 0.125 and f'(2.5) = 13.75
(b) A1 x1 = 2.490909 awrt
(b) Answer: x1 = 2.490909
(c) M1 correct (ft) substitution using x1: f(x1) = 0.000619, f'(x1) = 13.6139
(c) A1 x2 = 2.490864 awrt
(c) Answer: x2 = 2.490864
(d) B1 recognises Newton-Raphson converges much faster here; only one iteration was needed to reach 4dp accuracy
(d) B1 valid comparison: the fixed-point iteration in Question 2 was still changing in the 2nd decimal place after three iterations (x3 = 2.6526 against the true root 2.669), so it converges much more slowly (linearly) than Newton-Raphson (which converges quadratically)
(d) Answer: Newton-Raphson converges much faster (quadratic convergence) than the fixed-point iteration of Question 2 (linear convergence): here x1 and x2 already agree to 4dp, whereas the Question 2 iteration was still 0.016 away from its root after 3 iterations.

Question 4

(a) M1 attempts OQ - OP
(a) A1 PQ = 8i - 8j oe
(a) Answer: PQ = 8i - 8j
(b) M1 correct method OR = OP + (3/4)PQ, or equivalent ratio approach
(b) A1 correct substitution (3i+5j) + (3/4)(8i-8j)
(b) A1 OR = 9i - j oe
(b) Answer: OR = 9i - j
(c) M1 shows OS = 2 OR, i.e. 18i - 2j = 2(9i - j)
(c) A1 valid conclusion: since OS is a scalar multiple of OR, OR and OS are parallel and share the common point O, so O, R, S are collinear cso
(c) Answer: OS = 2 OR, so O, R and S are collinear

Question 5

(a) M1 equates two components (e.g. i and k) to form a pair of simultaneous equations in s and t
(a) A1 correct pair of equations, e.g. t + 2s = 4 and t - s = 1 oe
(a) M1 solves the equations simultaneously
(a) A1 t = 2, s = 1 oe
(a) B1 verifies using the remaining (j) component: both sides equal 0, confirming consistency, so the lines intersect
(a) A1 position vector of intersection = 3i + 7k oe cso
(a) Answer: Lines intersect at the point with position vector 3i + 7k
(b) M1 attempts dot product of direction vectors (1)(-2) + (-1)(2) + (2)(2)
(b) A1 d1.d2 = 0
(b) M1 recognises cos(θ) = 0 implies the lines are perpendicular
(b) A1 θ = 90 degrees cao
(b) Answer: θ = 90 degrees (the lines are perpendicular)

Question 6

(a) M1 correct strip width h = 1
(a) M1 attempts at least 3 correct y-values (y0=4, y1=4.1231, y2=4.4721, y3=5, y4=5.6569, or ft)
(a) A1 all 5 y-values correct awrt 4dp
(a) M1 correct trapezium structure (h/2)[y0+y4+2(y1+y2+y3)]
(a) A1 estimate = 18.424 awrt
(a) Answer: Integral (approx) = 18.424
(b) B1 states overestimate
(b) B1 valid reason referencing the shape of the curve, e.g. y = √x²+16 is concave up (convex) on [0,4], so the chords used by the trapezium rule lie above the curve
(b) Answer: Overestimate, because the curve is concave up (convex) over [0,4], so the trapezium chords lie above the curve.

Question 7

M1 attempts dot product (2)(3) + (3)(-6) + (-6)(2)
A1 dot product = -24
M1 attempts |d1| and |d2| (both equal 7)
M1 attempts cos(θ) = (d1.d2)/(|d1||d2|)
A1 θ = 119.3 degrees awrt
Answer: θ = awrt 119.3 degrees

Question 8

(a) M1 evaluates g(2.1) and g(2.2)
(a) A1 g(2.1) = -0.1338 (<0), g(2.2) = 0.4250 (>0); sign change confirms root in (2.1,2.2) cso
(a) Answer: g(2.1) = -0.1338, g(2.2) = 0.4250; sign change confirms root in (2.1,2.2)
(b) M1 evaluates g(2.125) and g(2.135), or an equivalent bounding interval for 2.13
(b) A1 g(2.125) = -0.0021 (awrt, negative) and g(2.135) = 0.0520 (awrt, positive), both correct
(b) A1 sign change on (2.125,2.135), g continuous, hence α = 2.13 correct to 2dp cso
(b) Answer: α = 2.13 (2dp)

Question 9

(a) M1 finds f'(x) = 6x² - 18x
(a) M1 evaluates f'(3) = 54 - 54 = 0
(a) A1 valid explanation: since f'(3) = 0, the tangent to the curve at x0 = 3 is horizontal, so the Newton-Raphson formula requires dividing by f'(x0) = 0, which is undefined; x1 cannot be found (geometrically, the tangent never meets the x-axis)
(a) Answer: f'(3) = 0, so the tangent at x0=3 is horizontal and the Newton-Raphson formula fails (division by zero)
(b) M1 correct Newton-Raphson substitution using f(4.4) = 1.128 and f'(4.4) = 36.96
(b) A1 correct unrounded working x1 = 4.4 - 1.128/36.96
(b) A1 x1 = 4.3695 awrt
(b) Answer: x1 = 4.3695

Question 10

(a) M1 writes a general point on l as (t, 2t, 2t)
(a) M1 forms the vector from P to this general point (or vice versa) in terms of t
(a) M1 sets the dot product of this vector with the direction vector (1,2,2) equal to zero
(a) A1 correct equation 9t - 18 = 0 oe
(a) A1 t = 2
(a) A1 N = 2i + 4j + 4k
(a) Answer: N = 2i + 4j + 4k
(b) M1 forms vector PN = N - P
(b) M1 attempts magnitude √(-6)²+3²+0²
(b) A1 PN = 3sqrt(5) cso, exact form required
(b) Answer: PN = 3sqrt(5)
(c) M1 recognises that N is the midpoint of P and P'
(c) M1 correct method P' = 2N - P, or equivalent
(c) A1 P' = -4i + 7j + 4k
(c) Answer: P' = -4i + 7j + 4k

Question 11

(a) M1 correct strip width h = 0.5
(a) M1 correct y-values y0=1, y1=0.8 (at x=0.5), y2=0.5
(a) A1 all y-values correct
(a) A1 estimate = 0.775 using (h/2)[y0+y2+2y1]
(a) Answer: Estimate = 0.775
(b) M1 correct strip width h = 0.25
(b) M1 at least 3 correct y-values (1, 0.9412, 0.8, 0.64, 0.5)
(b) A1 all 5 y-values correct
(b) A1 estimate = 0.7828 awrt
(b) Answer: Estimate = 0.7828
(c) M1 correct method: percentage error = |estimate - π/4|/(π/4) x 100
(c) A1 2-strip percentage error = 1.3% awrt
(c) A1 4-strip percentage error = 0.33% awrt
(c) Answer: 2-strip error = awrt 1.3%; 4-strip error = awrt 0.33%
(d) B1 valid comment, e.g. doubling the number of strips (halving h) reduces the percentage error by a factor of about 4 (from 1.3% to 0.33%), consistent with the trapezium rule's error decreasing in proportion to h²
(d) Answer: The error falls by a factor of about 4 when the number of strips doubles (1.3% to 0.33%), consistent with trapezium-rule error being proportional to h^2.

Question 12

(a) M1 forms the vector from P to the drone (or vice versa): (t-4, t²-2, t-5)
(a) M1 squares and sums the three components
(a) A1 correct unsimplified expression (t-4)²+(t²-2)²+(t-5)²
(a) A1 simplifies to D(t) = t⁴-2t²-18t+45 cso
(a) Answer: D(t) = t⁴ - 2t² - 18t + 45
(b) M1 differentiates D(t): D'(t) = 4t³-4t-18
(b) M1 sets D'(t) = 0
(b) A1 divides by 2 to obtain 2t³-2t-9 = 0 cso
(b) Answer: 2t³ - 2t - 9 = 0
(c) M1 evaluates g(1) = 2(1)-2(1)-9 and g(2) = 2(8)-2(2)-9
(c) A1 sign change (g(1) = -9 < 0, g(2) = 3 > 0), g continuous, hence root in (1,2) cso
(c) Answer: g(1) = -9, g(2) = 3; sign change confirms a root between t=1 and t=2
(d) M1 correct derivative g'(t) = 6t²-2
(d) M1 correct Newton-Raphson substitution using g(1.85) = -0.0368 and g'(1.85) = 18.535
(d) A1 correct unrounded value t1 = 1.85198...
(d) A1 t1 = 1.8520 awrt
(d) Answer: t1 = 1.8520

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