Locating roots by change of sign, iteration x = g(x) and cobweb/staircase diagrams, the Newton-Raphson method, the trapezium rule and whether it over or under estimates.
12 original questions, 79 marks. Type your answers below for an instant mark and a score (algebra in any equivalent form, e.g. 2sqrt(3) or 2√3), or print the worksheet, mark it with the answers sheet, and use the solutions to see every step written out by hand.
Show that the equation f(x) = 0 has a root α in the interval [1.5, 1.6]
By choosing a suitable interval, show that α = 1.57 correct to 2 decimal places.
4 marks
Show answer
f(1.5) = -0.625 0, f continuous, so a root lies in [1.5, 1.6]; f(1.565) = -0.0370 0, so 1.565 < α < 1.575 and α = 1.57 to 2 dp
2
The equation x3 − 6x + 2 = 0 has a root α between 0 and 1
Show that the equation can be written in the form x = x3 + 26
Use the iteration formula xn+1 = xn3 + 26 with x1 = 0.5 to find the values of x2, x3 and x4, giving each answer to 4 decimal places.
Write down the value of α correct to 3 decimal places.
5 marks
Show answer
a) 6x = x3 + 2 so x = (x3 + 2)6b) x₂ = 0.3542, x₃ = 0.3407, x₄ = 0.3399c) α = 0.340
3
The equation x4 − 3x − 5 = 0 has a root β close to x = 2
Taking x1 = 2 as a first approximation, apply the Newton-Raphson method twice to find x2 and x3. Give each answer to 4 decimal places.
4 marks
Show answer
x2 = 1.8276, x3 = 1.7961
4
The table shows values of y = e−x2 for 0 ≤ x ≤ 1, with values of y given to 4 decimal places.
x
0
0.25
0.5
0.75
1
y
1
0.9394
0.5698
0.3679
Complete the table, giving the missing value of y to 4 decimal places.
Use the trapezium rule with all the values of y in the completed table to find an estimate for 1∫0 e−x2 dx, giving your answer to 3 significant figures.
4 marks
Show answer
a) y = 0.7788b) 0.743
5
The curve C has equation y = e0.5x − x2
C has a turning point P whose x coordinate α lies between 0 and 1
Show that α is a solution of the equation x = 0.25e0.5x
Using the iteration formula xn+1 = 0.25e0.5xn with x1 = 0.3, find the values of x2, x3 and x4, giving each answer to 4 decimal places.
Using a suitable interval and a suitable function that should be stated, show that α = 0.2888 to 4 decimal places.
A student finds f(1.9) and f(2.1) and says that, because the signs are different, the equation f(x) = 0 has a root between 1.9 and 2.1
Find f(1.9) and f(2.1)
Explain why the student's conclusion is not correct.
(ii) g(x) = 4x2 − 16x + 15
Show that g(1) and g(3) have the same sign.
By considering g(2), explain why the equation g(x) = 0 has at least two roots between 1 and 3, even though g(1) and g(3) have the same sign.
6 marks
Show answer
a) f(1.9) = -28.1, f(2.1) = 32.1b) f is not continuous: there is an asymptote at x = 2, so the sign change is not caused by a rootc) g(1) = 3 and g(3) = 3, both positived) g(2) = -1 < 0: sign changes on [1, 2] and on [2, 3], g continuous, so at least two roots
7
The diagram shows part of the curve with equation y = 2 + ln x and the line y = x. They meet at the point where x = α
Show that α lies between 3 and 3.5
Use the iteration formula xn+1 = 2 + ln xn with x1 = 1.5 to find x2 and x3, giving each answer to 4 decimal places.
On the diagram, draw a staircase diagram to show how the iteration in part (b) moves towards α. Mark x2 and x3 on the x-axis.
A different rearrangement gives the iteration formula xn+1 = exn − 2. Using x1 = 3, find x2, x3 and x4 to 3 decimal places, and explain why this formula cannot be used to find α.
9 marks
Show answer
a) f(3) = -0.0986 0, so 3 < α < 3.5b) x₂ = 2.4055, x₃ = 2.8777c) staircase (see diagram) climbing up towards αd) x₂ = 2.718, x₃ = 2.051, x₄ = 1.052: the values move away from α (diverge from it)
8
f(x) = x3 − 3x2 + 5
The equation f(x) = 0 has exactly one real root γ
Show that, for this equation, the Newton-Raphson formula can be written xn+1 = 2xn3 − 3xn2 − 53xn2 − 6xn
Using this formula with x1 = −1, find x2 and x3, giving each answer to 4 decimal places.
Explain why the Newton-Raphson method cannot be used with x1 = 2
6 marks
Show answer
a) xₙ₊₁ = xₙ - (xₙ3 - 3xₙ2 + 5)(3xₙ2 - 6xₙ) = (2xₙ3 - 3xₙ2 - 5)(3xₙ2 - 6xₙ)b) x₂ = -1.1111, x₃ = -1.1038c) f'(2) = 0: the tangent is horizontal and never meets the x-axis (division by zero)
9
The diagram shows part of the curve with equation y = √(x3 + 1). The region R, shown shaded, is bounded by the curve, the x-axis, the y-axis and the line x = 2
x
0
0.5
1
1.5
2
y
1
1.4142
3
Complete the table, giving the values of y to 4 decimal places.
Use the trapezium rule with all the values in the table to find an estimate for the area of R, giving your answer to 3 significant figures.
Using the diagram, state with a reason whether your answer to part (b) is an overestimate or an underestimate.
Explain how the trapezium rule could be used to give a more accurate estimate.
7 marks
Show answer
a) 1.0607 and 2.0917b) 3.28c) overestimate: the curve bends upwards (convex), so each trapezium top lies above the curved) use more strips of smaller width
10
The table shows some values of x and y for a continuous curve with equation y = f(x)
x
1
1.5
2
2.5
3
y
2.4
3.1
3.5
3.2
2.6
Use the trapezium rule with all the values in the table to find an estimate for 3∫1 f(x) dx
Using your answer to part (a), find an estimate for 3∫1 (3f(x) − 2x) dx
Using your answer to part (a), write down an estimate for 2∫0 f(x + 1) dx, giving a reason.
6 marks
Show answer
a) 6.15b) 10.45c) 6.15: the curve is translated 1 unit left and the limits move with it
11
The region R is bounded by the curve with equation y = x ln x, the x-axis and the line x = 3
Show that the curve meets the x-axis at x = 1 for x > 0
Use the trapezium rule with four strips of equal width to find an estimate for the area of R, giving your answer to 4 decimal places.
Use integration by parts to find the exact area of R, giving your answer in the form a ln 3 + b, where a and b are rational.
Find the percentage error in the estimate from part (b), and explain, by considering d2ydx2, why the trapezium rule gives an overestimate here.
10 marks
Show answer
a) x ln x = 0, x > 0, so ln x = 0 and x = 1b) 2.9666c)92 ln 3 - 2d) 0.776%; y'' = 1/x > 0, convex, so trapezia lie above the curve
12
The concentration of a medicine in a patient's blood, C mg per litre, t hours after it is taken, is modelled by
C = 8te−0.5t, t ≥ 0
Find the maximum concentration according to the model, giving your answer to 3 significant figures.
The medicine is effective while C ≥ 4. The concentration first reaches 4 mg per litre when t = 0.715 (3 decimal places). It falls back to 4 mg per litre when t = T
Show that T is a root of the equation h(t) = 0, where h(t) = 2te−0.5t − 1, and show that 4 < T < 5
Taking t1 = 4.5 as a first approximation, apply the Newton-Raphson method twice to find an approximation for T, giving your answer to 3 decimal places.
Hence find, giving your answer in hours to 2 decimal places, how long the medicine is effective for.
10 marks
Show answer
a) 5.89 mg per litre (at t = 2)b) 8te-0.5t = 4 gives 2te-0.5t - 1 = 0; h(4) = 0.0827, h(5) = -0.179c) T = 4.307d) 3.59 hours
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Every question here is original and written by Robinson Tuition. It follows the Edexcel AS specification and suits AQA and OCR AS maths too; this is not an exam-board paper.