Pure: Vectors and Numerical Methods Depth: Fluency and Exam Drill - Worksheets, Questions and Revision

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A-Level · Pure Mathematics

P16D Pure: Vectors and Numerical Methods Depth: Fluency and Exam Drill

EDEXCEL 9MA0 · Calculator allowed · about 50 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

a = 3i - 4j + 12k. Find |a|.

(Total for Question 1 is 1 mark)

f(x) = x³ - 4x - 1. Evaluate f(2) and f(3).

(Total for Question 2 is 1 mark)

Using the iteration formula x_(n+1) = √3x_n + 1, with x0 = 2, find x1, giving your answer to 4 decimal places.

(Total for Question 3 is 1 mark)

State the Newton-Raphson iteration formula for finding an improved estimate x_(n+1) for a root of f(x) = 0.

(Total for Question 4 is 1 mark)

f(x) = x² - 7. Find f'(x).

(Total for Question 5 is 1 mark)

The trapezium rule is to be used with 2 strips of equal width to estimate the integral from x = 0 to x = 4. State the width of each strip.

(Total for Question 6 is 1 mark)

Relative to a fixed origin O, OP = 2i + 3j - k and OQ = 5i - j + 2k. Find the vector PQ.

(Total for Question 7 is 1 mark)

State the formula for the scalar (dot) product of vectors a = a1 i + a2 j + a3 k and b = b1 i + b2 j + b3 k.

(Total for Question 8 is 1 mark)

a = 2i + j - 3k, b = i - 2j + 4k. Find a . b.

(Total for Question 9 is 1 mark)

f(x) = x³ - 2x - 5. Given that f(2) = -1 and f(3) = 16, state whether f(x) = 0 has a root between x = 2 and x = 3, with a reason.

(Total for Question 10 is 1 mark)

Show that the equation x³ - 5x - 2 = 0 has a root between x = 2 and x = 3.

(Total for Question 11 is 2 marks)

f(x) = x³ - 5x - 2. Using x0 = 3, apply the Newton-Raphson method once to f(x) to find x1, giving your answer to 4 decimal places.

(Total for Question 12 is 2 marks)

Relative to a fixed origin O, points P and Q have position vectors OP = i + 2j - k and OQ = 4i - 2j + 5k. Find the vector PQ, and hence find |PQ|, giving your answer as an exact simplified surd.

(Total for Question 13 is 2 marks)

Using the trapezium rule with 4 strips of equal width, find an estimate for the integral from x = 0 to x = 2 of (x²+1) dx, giving your answer to 2 decimal places.

(Total for Question 14 is 2 marks)

f(x) = x³ - 4x² + 2.

(a)Show that the equation f(x) = 0 has a root between x = 3 and x = 4.(2)

(b)Show that the equation f(x) = 0 can be rearranged into the form x = 4 - 2/x².(2)

(Total for Question 15 is 4 marks)

Using the iteration formula x_(n+1) = 4 - 2/x_n², with x0 = 3.7, find the values of x1, x2 and x3, each to 4 decimal places, and hence state an estimate for the root of x³ - 4x² + 2 = 0, to 2 decimal places.

(Total for Question 16 is 4 marks)

g(x) = x³ + 2x - 6. The equation g(x) = 0 has a root α in the interval [1,2].

(a)Using x0 = 1.5, apply the Newton-Raphson method once to g(x) to find x1, giving your answer to 4 decimal places.(2)

(b)Apply the Newton-Raphson method a second time to find x2, giving your answer to 4 decimal places.(2)

(Total for Question 17 is 4 marks)

h(x) = e^x - 4x. Show that h(x) = 0 has a root between x = 0.3 and x = 0.4. By further decimal search, show that this root is x = 0.36, correct to 2 decimal places.

(Total for Question 18 is 3 marks)

Relative to a fixed origin O, the line l has vector equation r = (2i - j + k) + t(i + 2j - 2k), where t is a scalar parameter. The point A has position vector 9i + 3j - 5k.

(a)Show that the point C with position vector 4i + 3j - 3k lies on l, stating the corresponding value of t.(2)

(b)Find the vector AC, and hence find |AC|, giving your answer as an exact simplified surd.(2)

(Total for Question 19 is 4 marks)

J = the definite integral, from x = 0 to x = 1, of 1/(1+x) dx. The exact value of J is ln(2).

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

(b)Given that the exact value of J is ln(2), find the percentage error in your estimate from part (a), giving your answer to 2 significant figures.(1)

(Total for Question 20 is 3 marks)

Mark scheme · P16D Pure: Vectors and Numerical Methods Depth: Fluency and Exam Drill

Question 1

B1 13 cao
Answer: 13

Question 2

B1 f(2) = -1, f(3) = 14, both required
Answer: f(2) = -1, f(3) = 14

Question 3

B1 2.6458 cao
Answer: 2.6458

Question 4

B1 x_(n+1) = x_n - f(x_n)/f'(x_n)
Answer: x_(n+1) = x_n - f(x_n)/f'(x_n)

Question 5

B1 2x cao
Answer: 2x

Question 6

B1 2 cao
Answer: 2

Question 7

B1 3i - 4j + 3k cao
Answer: PQ = 3i - 4j + 3k

Question 8

B1 a.b = a1 b1 + a2 b2 + a3 b3
Answer: a.b = a1b1 + a2b2 + a3b3

Question 9

B1 -12 cao
Answer: -12

Question 10

B1 yes, since f changes sign (from negative to positive) and f is continuous
Answer: Yes, by the change of sign

Question 11

M1 evaluates f(2) = -4 and f(3) = 10
A1 cso: sign change (negative to positive) with f continuous confirms a root in (2,3)
Answer: f(2) = -4, f(3) = 10; sign change confirms a root

Question 12

M1 substitutes x0=3 into x1 = x0 - f(x0)/f'(x0), using f'(x)=3x²-5
A1 x1 = 2.5455 cao
Answer: x1 = 2.5455

Question 13

M1 PQ = OQ - OP = 3i - 4j + 6k
A1 |PQ| = √9+16+36 = √61 cao (exact)
Answer: PQ = 3i - 4j + 6k, |PQ| = √61

Question 14

M1 uses the five correct ordinates 1, 1.25, 2, 3.25, 5 with strip width 0.5
A1 estimate = 4.75 cao
Answer: 4.75

Question 15

(a) M1 evaluates f(3) = -7 and f(4) = 2
(a) A1 cso: sign change confirms a root in (3,4)
(a) Answer: f(3) = -7, f(4) = 2; sign change confirms a root
(b) M1 rearranges x³-4x²+2=0 to x²(x-4) = -2
(b) A1 cso: divides by x² to obtain x = 4 - 2/x²
(b) Answer: x = 4 - 2/x²

Question 16

M1 substitutes x0 = 3.7 into the iteration formula
A1 x1 = 3.8539
A1 x2 = 3.8653, x3 = 3.8661 (ft)
A1 root estimate = 3.87 (2dp, ft)
Answer: x1 = 3.8539, x2 = 3.8653, x3 = 3.8661; root awrt 3.87

Question 17

(a) M1 uses g'(x)=3x²+2 and substitutes x0=1.5 into the Newton-Raphson formula
(a) A1 x1 = 1.4571 cao
(a) Answer: x1 = 1.4571
(b) M1 applies the formula again using x1 from (a) (ft)
(b) A1 x2 = 1.4562 cao (ft)
(b) Answer: x2 = 1.4562

Question 18

B1 h(0.3) = awrt 0.150 (positive) and h(0.4) = awrt -0.108 (negative); sign change confirms a root in (0.3, 0.4)
M1 evaluates h(0.355) = awrt 0.006 (positive) and h(0.365) = awrt -0.019 (negative)
A1 cso: since h(0.355) > 0 and h(0.365) < 0, the root lies between 0.355 and 0.365, so x = 0.36 correct to 2dp
Answer: x = 0.36 (2dp)

Question 19

(a) M1 sets 2+t=4, -1+2t=3 and 1-2t=-3, and solves for t
(a) A1 cso: t=2 satisfies all three equations, so C lies on l
(a) Answer: t = 2, C lies on l
(b) M1 AC = OC - OA = -5i + 0j + 2k
(b) A1 |AC| = √25+0+4 = √29 cao
(b) Answer: AC = -5i + 2k, |AC| = √29

Question 20

(a) M1 uses the three correct ordinates 1, 2/3, 0.5 with strip width 0.5
(a) A1 estimate = 0.708 (3dp)
(a) Answer: 0.708
(b) B1 awrt 2.2% (ft from (a))
(b) Answer: awrt 2.2%

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Pure: Vectors and Numerical Methods Depth: Fluency and Exam Drill - Worksheets, Questions and Revision

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