Pure: Numerical Methods - Worksheets, Questions and Revision

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A-Level · Pure Mathematics

P9 Pure: Numerical Methods

EDEXCEL 9MA0 · Calculator allowed · about 145 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

The equation x³ + 2x - 5 = 0 has a single real root, α.

(a)Show that α lies between x = 1 and x = 2.(2)

(b)Show that the equation x³ + 2x - 5 = 0 can be rearranged into the iterative form x_(n+1) = (5 - 2x_n)^1/3.(2)

(c)Using x₀ = 1.5, apply the iteration formula from part (b) to find x₁, x₂ and x₃, giving each answer to 4 decimal places.(4)

(d)Show that α = 1.328 correct to 3 decimal places.(2)

(Total for Question 1 is 10 marks)

The curve y = e^-x is shown for 0 ≤ x ≤ 2. The table gives values of y at x = 0, 0.5, 1, 1.5 and 2.
x : 0 0.5 1 1.5 2
y : 1 p 0.3679 q 0.1353

(a)Find the values of p and q, each correct to 4 decimal places.(2)

(b)Use the trapezium rule with all 4 strips to find an estimate for the integral of e^-x from 0 to 2, giving your answer to 4 decimal places.(3)

(c)Given that the exact value of the integral of e^-x from 0 to 2 is 1 - e^-2, calculate the percentage error in the trapezium rule estimate found in part (b), giving your answer to 1 decimal place.(3)

(d)Explain, with reference to the shape of the curve y = e^-x, why the trapezium rule gives an overestimate of the true value of the integral on this interval.(2)

(Total for Question 2 is 10 marks)

Fahima records the velocity, v m/s, of her car t seconds after moving away from a junction.
t : 0 10 20 30 40
v : 0 4.2 7.8 10.1 11.5

(a)Use the trapezium rule with all 4 strips to estimate the total distance travelled by the car in the first 40 seconds.(3)

(b)State, giving a reason, whether your answer to part (a) is likely to be an overestimate or an underestimate of the true distance travelled.(2)

(c)Explain why the trapezium rule would give the exact distance travelled if the velocity had increased at a constant rate throughout the 40 seconds.(2)

(Total for Question 3 is 7 marks)

Let f(x) = x³ - 7x + 3.

(a)Show that f(x) = 0 has a root between x = 0 and x = 1.(2)

(b)Show that, using the Newton-Raphson method with x₀ = 0.5, x₁ = 0.44.(2)

(c)Find x₂ and x₃ using the Newton-Raphson method, giving each answer to 6 decimal places, and hence state the root of f(x) = 0 correct to 4 decimal places.(3)

(d)Show that the root found in part (c) is x = 0.441 correct to 3 decimal places.(2)

(Total for Question 4 is 9 marks)

The curve y = 1/x is defined for 1 ≤ x ≤ 3. The table gives values of y.
x : 1 1.5 2 2.5 3
y : 1 ? ? 0.4 0.3333

(a)Find the missing values of y at x = 1.5 and x = 2, each to 4 decimal places.(2)

(b)Use the trapezium rule with all 4 strips to find an estimate for the integral of 1/x from 1 to 3, giving your answer to 4 decimal places.(3)

(c)Given that the exact value of the integral of 1/x from 1 to 3 is ln(3), calculate the percentage error in the estimate found in part (b), giving your answer to 2 decimal places.(3)

(d)By considering the second derivative of y = 1/x, explain why the trapezium rule overestimates the integral on this interval.(2)

(Total for Question 5 is 10 marks)

Let f(x) = ln(x) + x - 4, for x > 0.

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

(b)Show that f'(x) = 1/x + 1, and hence write down the Newton-Raphson iteration formula for solving f(x) = 0.(2)

(c)Using x₀ = 3, find x₁ and x₂ using the Newton-Raphson method, giving each answer to 6 decimal places.(3)

(d)Prove that the root of f(x) = 0 is x = 2.93 correct to 2 decimal places.(2)

(Total for Question 6 is 9 marks)

This question tests your understanding of key ideas used in numerical methods.

(a)State the condition on f(x) that is required for the change-of-sign (sign change) test for a root to be valid on an interval [a,b].(1)

(b)Write down the general iteration formula used in the Newton-Raphson method for estimating a root of f(x) = 0.(1)

(c)The function f(x) = 1/(x-2) satisfies f(1) < 0 and f(3) > 0. Explain why the change-of-sign method alone does not guarantee a root of f(x) = 0 between x = 1 and x = 3.(2)

(d)Give one reason why the Newton-Raphson method may fail to converge to a root if the starting value x₀ is chosen close to a stationary point of y = f(x).(2)

(e)The iteration x_(n+1) = g(x_n) converges to a root α. State, with a reason, the condition on g'(α) required for this convergence.(2)

(Total for Question 7 is 8 marks)

The equation x² + 8x - 10 = 0 has a positive root, α. The iteration x_(n+1) = (10 - x_n²)/8 is used to estimate α, starting from x₀ = 1.

(a)Use the quadratic formula to show that the exact positive root of x² + 8x - 10 = 0 is α = -4 + √26.(3)

(b)Show that x² + 8x - 10 = 0 can be rearranged to give x = (10 - x²)/8.(1)

(c)Find g'(x), and show that |g'(α)| < 1.(2)

(d)By finding x₁, x₂, x₃ and x₄ (each to 4dp), show that the iteration converges to α by oscillating alternately above and below the true value (a cobweb pattern), rather than approaching α from one side only (a staircase pattern).(3)

(e)State how the resulting diagram would differ if g'(α) were positive instead of negative, with |g'(α)| still less than 1.(1)

(Total for Question 8 is 10 marks)

Let f(x) = x³ - x - 3.

(a)Show that f(x) = 0 has a root between x = 1 and x = 2.(2)

(b)Using the iteration x_(n+1) = (x_n + 3)^1/3 with x₀ = 1.5, find x₁, x₂, x₃ and x₄, giving each to 4 decimal places, and state the root correct to 4 decimal places.(4)

(c)Show that the alternative rearrangement x_(n+1) = x_n³ - 3, using the same x₀ = 1.5, does not converge to the root found in part (b) by finding x₁ and x₂.(3)

(d)Use the Newton-Raphson method with x₀ = 1.7 to find the root of f(x) = 0 correct to 4 decimal places, showing at least two iterations.(4)

(e)By comparing your answers to parts (b) and (d), state which method reached the root to 4 decimal places using fewer iterations.(1)

(Total for Question 9 is 14 marks)

Let f(x) = x³ - 3x + 1.

(a)Show that f'(x) = 0 when x = 1, and explain why the Newton-Raphson formula cannot be used with x₀ = 1 to estimate a root of f(x) = 0.(2)

(b)Show that f(x) = 0 has a root between x = 1.4 and x = 1.6.(2)

(c)Using x₀ = 1.5, apply the Newton-Raphson method to find this root correct to 3 decimal places, showing your iterations.(4)

(d)Prove that this root is correct to 3 decimal places by considering a suitable sign change.(2)

(Total for Question 10 is 10 marks)

Let f(x) = 2sin(x) - x + 1, where x is measured in radians.

(a)Show that f(x) = 0 has a root between x = 2.3 and x = 2.4.(2)

(b)Find f'(x), and hence write down the Newton-Raphson iteration formula for solving f(x) = 0.(2)

(c)Using x₀ = 2.3, find x₁ and x₂, giving each answer to 4 decimal places, and state the root of f(x) = 0 correct to 4 decimal places.(4)

(d)Show that this root is x = 2.38 correct to 2 decimal places.(2)

(Total for Question 11 is 10 marks)

Let f(x) = xe^x - 5. This question compares two numerical methods for solving f(x) = 0.

(a)Show that f(x) = 0 has a root between x = 1.3 and x = 1.4.(2)

(b)Show that the equation xe^x = 5 can be rearranged into the iterative form x_(n+1) = ln(5/x_n).(2)

(c)Using x₀ = 1.3, find x₁, x₂, x₃ and x₄, giving each to 4 decimal places, and comment on the pattern of convergence.(4)

(d)By first finding f'(x) using the product rule, use the Newton-Raphson method with x₀ = 1.3 to find x₁ and x₂, giving each to 6 decimal places, and state the root of f(x) = 0 correct to 4 decimal places.(5)

(e)Comment on the relative rates of convergence of the two methods used in parts (c) and (d), referring to the number of iterations required to obtain the root correct to 4 decimal places.(2)

(Total for Question 12 is 15 marks)

Mark scheme · P9 Pure: Numerical Methods

Question 1

(a) M1 evaluates f(1) and f(2)
(a) A1 f(1) = -2, f(2) = 7; opposite signs and f continuous on [1,2], so by the change of sign rule a root lies in (1,2) cso
(a) Answer: f(1) = -2, f(2) = 7, sign change confirms a root in (1,2)
(b) M1 rearranges to x³ = 5 - 2x
(b) A1 takes the cube root of both sides to give x = (5 - 2x)^1/3 cso
(b) Answer: x_(n+1) = (5 - 2x_n)^1/3
(c) M1 correct substitution of x₀ = 1.5 into the iteration formula
(c) A1 x₁ = 1.2599 (awrt 4dp)
(c) A1 x₂ = 1.3536 (awrt 4dp)
(c) A1 x₃ = 1.3186 (awrt 4dp)
(c) Answer: x₁ = 1.2599, x₂ = 1.3536, x₃ = 1.3186
(d) M1 evaluates f(1.3275) and f(1.3285)
(d) A1 f(1.3275) = -0.0056 (<0) and f(1.3285) = +0.0017 (>0); change of sign in (1.3275, 1.3285) confirms α = 1.328 to 3dp cso
(d) Answer: α = 1.328 (3dp)

Question 2

(a) B1 p = 0.6065 (awrt 4dp)
(a) B1 q = 0.2231 (awrt 4dp)
(a) Answer: p = 0.6065, q = 0.2231
(b) M1 correct trapezium structure with h = 0.5: (h/2)[y₀ + y₄ + 2(y₁+y₂+y₃)]
(b) A1 correct value inside the bracket, 6.5904... (or equivalent unsimplified form)
(b) A1 0.8826 (awrt 4dp)
(b) Answer: Estimate = 0.8826
(c) B1 exact value = 0.8647 (awrt 4dp)
(c) M1 percentage error = |estimate - exact| / exact x 100
(c) A1 2.1% (awrt 1dp)
(c) Answer: Percentage error = 2.1%
(d) B1 states that y = e^-x is convex (curves upwards / concave up) on [0,2]
(d) B1 hence each chord joining consecutive points lies above the curve, so the trapezium rule overestimates the area/integral
(d) Answer: The curve is convex, so the chords lie above the curve and the trapezium rule overestimates.

Question 3

(a) M1 correct trapezium structure with h = 10: (h/2)[v₀ + v₄ + 2(v₁+v₂+v₃)]
(a) A1 correct value inside the bracket, 55.7
(a) A1 278.5 (m) cao
(a) Answer: Distance approx 278.5 m
(b) B1 states underestimate
(b) B1 valid reason, e.g. the rate of increase of v is slowing (successive differences 4.2, 3.6, 2.3, 1.4 are decreasing) so the v-t graph is concave, meaning the curve lies above the chords, giving an underestimate
(b) Answer: Underestimate, since the v-t graph is concave (the increases in v are getting smaller).
(c) B1 a constant rate of increase means the v-t graph is a straight line between recorded points
(c) B1 the trapezium rule fits a straight line exactly between each pair of points, so there is no approximation error and the area under a straight line segment is exact
(c) Answer: With constant acceleration the v-t graph is linear, and the trapezium rule is exact for a straight line, so no error arises.

Question 4

(a) M1 evaluates f(0) and f(1)
(a) A1 f(0) = 3, f(1) = -3; opposite signs and f continuous, so by the change of sign rule a root lies in (0,1) cso
(a) Answer: f(0) = 3, f(1) = -3, sign change confirms a root in (0,1)
(b) M1 states f'(x) = 3x² - 7 and substitutes x₀ = 0.5 into x₁ = x₀ - f(x₀)/f'(x₀), using f(0.5) = -0.375 and f'(0.5) = -6.25
(b) A1 x₁ = 0.5 - (-0.375/-6.25) = 0.44 cso
(b) Answer: x₁ = 0.44
(c) M1 correct Newton-Raphson substitution using x₁ = 0.44
(c) A1 x₂ = 0.440808 (awrt 6dp)
(c) A1 root = 0.4408 (4dp) cao (x₃ = 0.440808, confirming stability)
(c) Answer: x₂ = 0.440808, root = 0.4408 (4dp)
(d) M1 evaluates f(0.4405) and f(0.4415)
(d) A1 f(0.4405) = +0.0020 (>0) and f(0.4415) = -0.0044 (<0); change of sign confirms root = 0.441 to 3dp cso
(d) Answer: root = 0.441 (3dp)

Question 5

(a) B1 y(1.5) = 0.6667 (awrt 4dp)
(a) B1 y(2) = 0.5
(a) Answer: y(1.5) = 0.6667, y(2) = 0.5
(b) M1 correct trapezium structure with h = 0.5: (h/2)[y₀+y₄+2(y₁+y₂+y₃)]
(b) A1 correct value inside the bracket, 4.4667
(b) A1 1.1167 (awrt 4dp)
(b) Answer: Estimate = 1.1167
(c) M1 attempts percentage error = |estimate - exact|/exact x 100
(c) A1 uses ln(3) = 1.0986 (awrt 4dp)
(c) A1 1.64% (awrt 2dp)
(c) Answer: Percentage error = 1.64%
(d) B1 d2y/dx2 = 2/x³ > 0 for 1 ≤ x ≤ 3, so the curve is convex on this interval
(d) B1 hence the chords lie above the curve, so the trapezium rule overestimates the true area
(d) Answer: y = 1/x is convex on [1,3] (d2y/dx2 > 0), so chords lie above the curve, giving an overestimate.

Question 6

(a) M1 evaluates f(2) and f(3)
(a) A1 f(2) = -1.3069 (<0), f(3) = 0.0986 (>0); opposite signs and f continuous on [2,3], so a root lies in (2,3) cso
(a) Answer: f(2) = -1.3069, f(3) = 0.0986, sign change confirms a root in (2,3)
(b) B1 f'(x) = 1/x + 1, obtained by differentiating ln(x) + x - 4 term by term cso
(b) B1 x_(n+1) = x_n - (ln(x_n)+x_n-4)/(1/x_n + 1)
(b) Answer: f'(x) = 1/x + 1; x_(n+1) = x_n - (ln(x_n)+x_n-4)/(1/x_n+1)
(c) M1 correct Newton-Raphson substitution with x₀ = 3
(c) A1 x₁ = 2.926041 (awrt 6dp)
(c) A1 x₂ = 2.926271 (awrt 6dp)
(c) Answer: x₁ = 2.926041, x₂ = 2.926271
(d) M1 evaluates f(2.925) and f(2.935)
(d) A1 f(2.925) = -0.0017 (<0) and f(2.935) = +0.0117 (>0); change of sign confirms root = 2.93 to 2dp cso
(d) Answer: root = 2.93 (2dp)

Question 7

(a) B1 f must be continuous on [a,b]
(a) Answer: f(x) must be continuous on the interval [a,b].
(b) B1 x_(n+1) = x_n - f(x_n)/f'(x_n)
(b) Answer: x_(n+1) = x_n - f(x_n)/f'(x_n)
(c) B1 f(x) = 1/(x-2) is discontinuous (has a vertical asymptote) at x = 2, which lies inside the interval [1,3]
(c) B1 the change of sign is caused by the discontinuity, not by a root; f(x) = 0 has no solution at all here, so the method's condition (continuity) is not met, and its conclusion is invalid cso
(c) Answer: f is discontinuous at x=2 in the interval, so the sign change is due to the asymptote, not a root; the method requires continuity.
(d) B1 near a stationary point f'(x) is close to zero, so the tangent to the curve is nearly horizontal
(d) B1 the tangent then meets the x-axis a long way from x₀ (or, if f'(x₀)=0 exactly, the formula requires division by zero and cannot be used), so the next iterate can be far from the required root
(d) Answer: f'(x₀) is close to (or equal to) zero, so the near-horizontal tangent gives a next estimate far from the root, or the formula is undefined.
(e) B1 |g'(α)| < 1
(e) B1 reason: this ensures the size of the error reduces at each step, so successive iterates get closer to α
(e) Answer: |g'(α)| < 1, so that the error decreases at each iteration.

Question 8

(a) M1 correct application of the quadratic formula with a=1, b=8, c=-10
(a) A1 x = (-8 ± √64+40)/2 = -4 ± √26
(a) A1 takes the positive root, α = -4 + √26 cso
(a) Answer: α = -4 + √26
(b) B1 correct rearrangement: 8x = 10 - x², so x = (10-x²)/8 cso
(b) Answer: x = (10-x²)/8
(c) M1 g'(x) = -2x/8 = -x/4
(c) A1 g'(α) = -1.0990/4 = -0.2748, so |g'(α)| = 0.2748 < 1 cso
(c) Answer: g'(x) = -x/4; |g'(α)| = 0.2748 < 1
(d) M1 correct repeated substitution into the iteration formula
(d) A1 x₁ = 1.1250, x₂ = 1.0918, x₃ = 1.1010, x₄ = 1.0985 (awrt 4dp)
(d) A1 values alternate above (1.1250, 1.1010) and below (1.0918, 1.0985) α = 1.0990, consistent with g'(α) < 0, giving a cobweb (not staircase) diagram cso
(d) Answer: x₁=1.1250, x₂=1.0918, x₃=1.1010, x₄=1.0985; values oscillate either side of α, confirming a cobweb pattern.
(e) B1 the iteration would converge monotonically from one side only (a staircase diagram) rather than oscillating
(e) Answer: It would give a staircase diagram, converging from one side only.

Question 9

(a) M1 evaluates f(1) and f(2)
(a) A1 f(1) = -3, f(2) = 3; opposite signs and f continuous on [1,2], so a root lies in (1,2) cso
(a) Answer: f(1) = -3, f(2) = 3, sign change confirms a root in (1,2)
(b) M1 correct repeated substitution into the iteration formula
(b) A1 x₁ = 1.6510 (awrt 4dp)
(b) A1 x₂ = 1.6692, x₃ = 1.6714 (awrt 4dp each)
(b) A1 x₄ = 1.6717 and hence root = 1.6717 (4dp) cao
(b) Answer: x₁=1.6510, x₂=1.6692, x₃=1.6714, x₄=1.6717; root = 1.6717 (4dp)
(c) M1 correct substitution x₁ = 1.5³ - 3
(c) A1 x₁ = 0.375, x₂ = 0.375³ - 3 = -2.9473 (awrt 4dp)
(c) A1 the values are moving rapidly away from 1.6717, so this rearrangement diverges rather than converging cso
(c) Answer: x₁ = 0.375, x₂ = -2.9473; values diverge away from the root.
(d) M1 states f'(x) = 3x² - 1
(d) M1 correct Newton-Raphson substitution with x₀ = 1.7
(d) A1 x₁ = 1.6722 (awrt 4dp)
(d) A1 x₂ = 1.6717 and root = 1.6717 (4dp) cao
(d) Answer: x₁ = 1.672229, x₂ = 1.671700, root = 1.6717 (4dp)
(e) B1 the Newton-Raphson method (part (d)), which reached 1.6717 after 2 iterations, compared with 4 iterations needed for the rearrangement in part (b)
(e) Answer: Newton-Raphson converged faster (2 iterations versus 4).

Question 10

(a) M1 f'(x) = 3x² - 3, so f'(1) = 3-3 = 0 cso
(a) A1 the Newton-Raphson formula requires dividing by f'(x₀); since f'(1) = 0 this division is undefined, so x₁ cannot be found (geometrically, the tangent at x=1 is horizontal and never meets the x-axis)
(a) Answer: f'(1) = 0, so the formula requires division by zero and cannot be used.
(b) M1 evaluates f(1.4) and f(1.6)
(b) A1 f(1.4) = -0.456 (<0), f(1.6) = 0.296 (>0); sign change confirms root in (1.4,1.6) cso
(b) Answer: f(1.4) = -0.456, f(1.6) = 0.296, sign change confirms a root in (1.4,1.6)
(c) M1 correct Newton-Raphson substitution with x₀ = 1.5, using f'(x) = 3x²-3
(c) A1 x₁ = 1.5333 (awrt 4dp)
(c) A1 x₂ = 1.5321 (awrt 4dp)
(c) A1 root = 1.532 (3dp) cao
(c) Answer: x₁ = 1.533333, x₂ = 1.532091, root = 1.532 (3dp)
(d) M1 evaluates f(1.5315) and f(1.5325)
(d) A1 f(1.5315) = -0.0024 (<0), f(1.5325) = +0.0017 (>0); sign change confirms root = 1.532 to 3dp cso
(d) Answer: root = 1.532 (3dp)

Question 11

(a) M1 evaluates f(2.3) and f(2.4) in radians
(a) A1 f(2.3) = +0.1914 (>0), f(2.4) = -0.0491 (<0); sign change confirms root in (2.3,2.4) cso
(a) Answer: f(2.3) = 0.1914, f(2.4) = -0.0491, sign change confirms a root in (2.3,2.4)
(b) B1 f'(x) = 2cos(x) - 1
(b) B1 x_(n+1) = x_n - (2sin(x_n)-x_n+1)/(2cos(x_n)-1)
(b) Answer: f'(x) = 2cos(x) - 1; x_(n+1) = x_n - (2sin(x_n)-x_n+1)/(2cos(x_n)-1)
(c) M1 correct Newton-Raphson substitution with x₀ = 2.3
(c) A1 x₁ = 2.3821 (awrt 4dp)
(c) A1 x₂ = 2.3801 (awrt 4dp)
(c) A1 root = 2.3801 (4dp) cao
(c) Answer: x₁ = 2.3821, x₂ = 2.3801, root = 2.3801 (4dp)
(d) M1 evaluates f(2.375) and f(2.385)
(d) A1 f(2.375) = +0.0124 (>0), f(2.385) = -0.0121 (<0); sign change confirms root = 2.38 to 2dp cso
(d) Answer: root = 2.38 (2dp)