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Locating roots by change of sign

39 questions

LessonNot started
The diagram shows part of the graph of y=f(x)y = \text{f}(x), where f(x)\text{f}(x) is a cubic polynomial in xx.123xyExplain why one of the roots of the equation f(x)=0\text{f}(x) = 0 cannot be found by the sign change method.

(2 marks)
●●●●●Level 22 marksStart
Show that the equation x3−2x−5=0x^3 - 2x - 5 = 0 has a root between x=2x = 2 and x=3x = 3.
●●●●●Level 22 marksStart
Show that x3−7x+2=0x^3 - 7x + 2 = 0 has a root between x=0x = 0 and x=1x = 1.
●●●●●Level 22 marksStart
Show that the equation x3+2x−7=0x^{3} + 2x - 7 = 0 has a root between x=1x = 1 and x=2x = 2. (2 marks)
●●●●●Level 22 marksStart
The equation x3+3ln⁡x=5x^3 + 3\ln x = 5 has exactly one real root.
(a) Show by calculation that the root lies between 1.541.54 and 1.551.55.

(2 marks)

In order to find the root, the iterative formula

xn+1=(p+qln⁡xn)r,x_{n+1} = \left(p + q\ln x_n\right)^r,

with initial value x0=1.5x_0 = 1.5, is to be used.
(b) Determine the values of the constants pp, qq and rr.

(2 marks)
(c) Hence find the root correct to 4 significant figures. Show the result of each step of the iteration process.

(2 marks)
●●●●●Level 36 marksStart
Show that the equation ex=3−xe^x = 3 - x has a root between x=0x = 0 and x=1x = 1.
●●●●●Level 32 marksStart
f(x)=x3+x−1f(x) = x^3 + x - 1.
(a) Show that f(x)=0f(x) = 0 has a root in the interval [0,1][0, 1].
(b) Show that this root is 0.70.7 correct to 1 decimal place.
●●●●●Level 34 marksStart
The function f(x)=1x−3f(x) = \dfrac{1}{x - 3} satisfies f(2)=−1f(2) = -1 and f(4)=1f(4) = 1. Explain why this does not show that f(x)=0f(x) = 0 has a root between x=2x = 2 and x=4x = 4.
●●●●●Level 32 marksStart
Show that the equation ln⁡x+x=2\ln x + x = 2 has a root between x=1x = 1 and x=2x = 2.
●●●●●Level 32 marksStart
f(x)=x3−3x+1f(x) = x^3 - 3x + 1. Show that f(x)=0f(x) = 0 has three roots, one in each of the intervals [−2,−1][-2, -1], [0,1][0, 1] and [1,2][1, 2].
●●●●●Level 33 marksStart
(a) Show that 2x+x=52^x + x = 5 has a root between x=1x = 1 and x=2x = 2.
(b) Show that this root is 1.71.7 correct to 1 decimal place.
●●●●●Level 34 marksStart
Show that the equation e−x=x\mathrm{e}^{-x} = x has a root between x=0.5x = 0.5 and x=0.6x = 0.6. (3 marks)
●●●●●Level 33 marksStart
f(x)=x3−x−4f(x) = x^{3} - x - 4.
(a) Show that f(x)=0f(x) = 0 has a root in the interval [1,2][1, 2]. (2 marks)
(b) Show further that the root lies in the interval [1.7,1.8][1.7, 1.8]. (2 marks)
●●●●●Level 34 marksStart
A student notes that tan⁡1=1.56>0\tan 1 = 1.56 > 0 and tan⁡2=−2.19<0\tan 2 = -2.19 < 0, and concludes that tan⁡x=0\tan x = 0 has a root between x=1x = 1 and x=2x = 2.

Explain why this conclusion is wrong. (3 marks)
●●●●●Level 33 marksStart
f(x)=2x3−3x−4f(x) = 2x^{3} - 3x - 4. The equation f(x)=0f(x) = 0 has a single real root α\alpha.

Show that α=1.65\alpha = 1.65 correct to 2 decimal places. (4 marks)
●●●●●Level 34 marksStart
The equation x+ln⁡x=3x + \ln x = 3 has a single root α\alpha.
(a) Show that α\alpha lies between 11 and 33. (2 marks)
(b) Show further that 2<α<2.52 < \alpha < 2.5. (2 marks)
●●●●●Level 34 marksStart
f(x)=x3−4x+1f(x) = x^{3} - 4x + 1.

Show that the equation f(x)=0f(x) = 0 has three real roots, locating each in an interval of integers. (5 marks)
●●●●●Level 35 marksStart
1.61.822.22.42.62.833.2510xyFigure 1

Figure 1 shows part of the curve y=x3−4x−2y = x^{3} - 4x - 2. The curve crosses the xx-axis at the point where x=αx = \alpha.
(a) Show by calculation that α\alpha lies between 22 and 33. (2 marks)
(b) Show that the equation x3−4x−2=0x^{3} - 4x - 2 = 0 can be rearranged into the form x=4x+23x = \sqrt[3]{4x+2}. (1 mark)
(c) Use the iterative formula xn+1=4xn+23x_{n+1} = \sqrt[3]{4x_{n}+2}, with x1=2x_{1} = 2, to find α\alpha correct to 33 decimal places. Show the result of each iteration. (3 marks)
●●●●●Level 36 marksStart
f(x)=x3−5x+1f(x) = x^3 - 5x + 1.
(a) Show that the equation f(x)=0f(x) = 0 has a root α\alpha between x=2x = 2 and x=2.5x = 2.5. (2 marks)
(b) Taking x0=2x_0 = 2 as a first approximation to α\alpha, apply the Newton–Raphson method twice to find x2x_2. Give your answer to 4 decimal places. (3 marks)
(c) Explain why the Newton–Raphson method fails if x0=1.3x_0 = 1.3 is used instead. (1 mark)
●●●●●Level 36 marksStart
f(x)=x3−5x+1\mathrm{f}(x) = x^3 - 5x + 1.

The equation f(x)=0\mathrm{f}(x) = 0 has a root α\alpha.
(a) Show that α\alpha lies between 0.20.2 and 0.210.21. (2 marks)
(b) Using the Newton-Raphson method with first approximation x1=0.2x_1 = 0.2, find x2x_2 and x3x_3, and hence state the value of α\alpha correct to 5 decimal places. (4 marks)
(c) Explain why the Newton-Raphson method fails if x1=53x_1 = \sqrt{\dfrac{5}{3}} is used as the first approximation. (2 marks)
●●●●●Level 38 marksStart
f(x)=x3−3x+ln⁡xx>0\text{f}(x) = x^3 - 3x + \ln x \qquad x > 0
(a) Show that the equation f(x)=0\text{f}(x) = 0 has a root α\alpha in the interval [1.6, 1.7][1.6,\ 1.7]

(2 marks)
(b) Using the iteration formula

xn+1=3xn−ln⁡xn3x_{n+1} = \sqrt[3]{3x_n - \ln x_n}

with x1=1.6x_1 = 1.6, find the values of x2x_2, x3x_3 and x4x_4, giving each answer to 3 decimal places.

(2 marks)
●●●●●Level 34 marksStart
The diagram shows part of the curve y=f(x)y = \text{f}(x), where f(x)=3ln⁡x+5−2x\text{f}(x) = 3\ln x + 5 - 2x. The equation f(x)=0\text{f}(x) = 0 has a positive root α\alpha close to x=0.2x = 0.2.12346xy(a) Explain why using the sign change method with x=0.1x = 0.1 and x=6x = 6 will fail to locate α\alpha.

(1 mark)
(b) Show that the equation f(x)=0\text{f}(x) = 0 can be written as x=e13(2x−5)x = \text{e}^{\frac{1}{3}(2x - 5)}.

(2 marks)
(c) Use the iterative formula xn+1=e13(2xn−5)x_{n+1} = \text{e}^{\frac{1}{3}(2x_n - 5)} with a starting value of x1=0.2x_1 = 0.2 to find the value of α\alpha correct to 4 significant figures, showing the result of each iteration.

(3 marks)
(d) An alternative iterative formula is xn+1=F(xn)x_{n+1} = \text{F}(x_n), where F(x)=1.5ln⁡x+2.5\text{F}(x) = 1.5\ln x + 2.5.

By considering F′(0.2)\text{F}'(0.2) explain why this iterative formula will not find α\alpha.

(3 marks)
●●●●●Level 49 marksStart
A continuous curve has equation

y=x3cos⁡2x0⩽x⩽π4y = x^3\cos 2x \qquad 0 \leqslant x \leqslant \dfrac{\pi}{4}

The curve has a stationary point at the point QQ, where x>0x > 0
(a) Show, using calculus, that the xx coordinate of QQ is a solution of the equation

x=12arctan⁡(32x)x = \dfrac{1}{2}\arctan\left(\dfrac{3}{2x}\right)

(4 marks)

Using the iteration formula

xn+1=12arctan⁡(32xn)x1=0.6x_{n+1} = \dfrac{1}{2}\arctan\left(\dfrac{3}{2x_n}\right) \qquad x_1 = 0.6
(b) find, to 4 decimal places, the value of

(i) x2x_2

(ii) x3x_3

(3 marks)
(c) Using a suitable interval and a suitable function which should be stated, show that the xx coordinate of QQ is 0.5960.596 correct to 3 decimal places.

(2 marks)
●●●●●Level 49 marksStart
f(x)=x3+3x−7f(x) = x^3 + 3x - 7. The equation f(x)=0f(x) = 0 has a single real root α\alpha.
(a) Show that α\alpha lies between 1.41.4 and 1.451.45. (2 marks)
(b) Show that the equation f(x)=0f(x) = 0 can be rearranged as x=7−3x3x = \sqrt[3]{7 - 3x}. (1 mark)
(c) Using the iterative formula xn+1=7−3xn3x_{n+1} = \sqrt[3]{7 - 3x_n} with x0=1.5x_0 = 1.5, find x1x_1, x2x_2 and x3x_3, giving each to 4 decimal places. (3 marks)
(d) By choosing a suitable interval, show that α=1.406\alpha = 1.406 correct to 3 decimal places. (2 marks)
●●●●●Level 48 marksStart
f(x)=x3−4x+1f(x) = x^3 - 4x + 1
(a) Show that the equation f(x)=0f(x) = 0 has a root α\alpha such that 1.8<α<1.91.8 < \alpha < 1.9. (2 marks)
(b) Show that the Newton–Raphson iteration for this equation can be written as

xn+1=2xn3−13xn2−4x_{n+1} = \dfrac{2x_n^3 - 1}{3x_n^2 - 4} (2 marks)
(c) Using x0=2x_0 = 2, find α\alpha correct to 4 decimal places. Show the result of each iteration. (3 marks)
(d) Explain why the Newton–Raphson method cannot be used to find α\alpha starting from x0=23x_0 = \dfrac{2}{\sqrt{3}}. (1 mark)
●●●●●Level 48 marksStart
f(x)=x3+x−5\mathrm{f}(x) = x^3 + x - 5
(a) Show that the equation f(x)=0\mathrm{f}(x) = 0 has a root α\alpha between 1.51.5 and 1.61.6. (2 marks)
(b) Show that the equation f(x)=0\mathrm{f}(x) = 0 can be rearranged into the form x=5−x3x = \sqrt[3]{5 - x}. (1 mark)
(c) Use the iterative formula

xn+1=5−xn3x_{n+1} = \sqrt[3]{5 - x_n}

with x0=1.5x_0 = 1.5 to find x1x_1, x2x_2 and x3x_3, giving each answer to 44 decimal places. (3 marks)
(d) Show that α=1.516\alpha = 1.516 correct to 33 decimal places. (2 marks)
●●●●●Level 48 marksStart
f(x)=x3+4x−7\mathrm{f}(x) = x^3 + 4x - 7
(a) Show that the equation f(x)=0\mathrm{f}(x) = 0 has a root α\alpha between 1.21.2 and 1.31.3. (2 marks)
(b) Taking x0=1.3x_0 = 1.3, apply the Newton–Raphson procedure twice to f(x)\mathrm{f}(x), giving x2x_2 to 55 decimal places. (4 marks)
(c) Write down the value of α\alpha correct to 44 significant figures. (1 mark)
●●●●●Level 47 marksStart
The equation

x2=e7−xx^2 = \mathrm{e}^{7 - x}

has a single solution, x=αx = \alpha
(a) By considering a suitable change of sign, show that α\alpha lies between 44 and 55

(2 marks)
(b) Show that the equation x2=e7−xx^2 = \mathrm{e}^{7 - x} can be rearranged to give

x=7−2ln⁡xx = 7 - 2\ln x

(3 marks)
(c) Use the iterative formula

xn+1=7−2ln⁡xnx_{n+1} = 7 - 2\ln x_n

with x1=5x_1 = 5, to find x2x_2, x3x_3 and x4x_4

Give your answers to three decimal places.

(2 marks)
●●●●●Level 47 marksStart
The diagram below shows a sector of a circle OABOAB.

The chord ABAB divides the sector into a triangle and a shaded segment.

Angle AOBAOB is θ\theta radians.OABθ(a) Given that the area of the shaded segment is one third of the area of the sector OABOAB, show that

2θ=3sin⁡θ2\theta = 3\sin\theta

(4 marks)
(b) Use a suitable change of sign to show that a solution to the equation

2θ=3sin⁡θ2\theta = 3\sin\theta

lies in the interval given by θ∈[1.4,1.6]\theta \in [1.4, 1.6]

(2 marks)
●●●●●Level 46 marksStart
In a model of a sprint race, the speed of an athlete, vv m s−1^{-1}, tt seconds after the start of the race, is modelled by the equation

v=12(18−t)ln⁡(t+1)0⩽t⩽18v = \frac{1}{2}(18 - t)\ln(t + 1) \qquad 0 \leqslant t \leqslant 18

Using the model,
(a) show that the speed of the athlete is a maximum when tt satisfies the equation

(t+1)ln⁡(t+1)+t−18=0(t + 1)\ln(t + 1) + t - 18 = 0

(4 marks)

The equation in part (a) has exactly one root, α\alpha, in the interval 0⩽t⩽180 \leqslant t \leqslant 18
(b) Show that α\alpha lies in the interval [5,6][5, 6]

(2 marks)
(c) Using t0=6t_0 = 6 as a first approximation for α\alpha, apply the Newton–Raphson method once to find a second approximation for α\alpha, giving your answer to 3 decimal places.

(2 marks)
●●●●●Level 48 marksStart
f(x)=xcos⁡x+20⩽x⩽π\text{f}(x) = x\cos x + 2 \qquad 0 \leqslant x \leqslant \pi

where xx is measured in radians.
(a) Find f′(x)\text{f}'(x)

(2 marks)

The curve with equation y=f(x)y = \text{f}(x) has a maximum turning point at x=αx = \alpha
(b) Show that α\alpha lies in the interval [0.8,0.9][0.8, 0.9]

(2 marks)
(c) Using x0=0.8x_0 = 0.8 as a first approximation for α\alpha, apply the Newton–Raphson method once to f′(x)\text{f}'(x) to find a second approximation for α\alpha, giving your answer to 3 decimal places.

(3 marks)
●●●●●Level 47 marksStart
f(x)=x−3ln⁡xx>0\text{f}(x) = x - 3\ln x \qquad x > 0
(a) Show that the equation f(x)=0\text{f}(x) = 0 has a root α\alpha in the interval [4.5, 5][4.5,\ 5]

(2 marks)
(b) Find f′(x)\text{f}'(x)

(2 marks)
(c) Using x0=5x_0 = 5 as a first approximation for α\alpha, apply the Newton–Raphson method once to obtain a second approximation for α\alpha.

Give your answer to 3 decimal places.

(2 marks)

A student attempts to use the Newton–Raphson method with x0=3x_0 = 3 as a first approximation for α\alpha.
(d) Explain why the method fails.

(1 mark)
●●●●●Level 47 marksStart
f(x)=5sin⁡2x−3x+20≤x≤π\text{f}(x) = 5\sin 2x - 3x + 2 \qquad 0 \le x \le \pi

where xx is measured in radians.

The curve with equation y=f(x)y = \text{f}(x) has a single maximum turning point in the interval 0≤x≤π0 \le x \le \pi
(a) Using calculus, find the xx coordinate of this maximum turning point, giving your answer to 3 decimal places.

(4 marks)

The equation f(x)=0\text{f}(x) = 0 has a single root α\alpha in the interval 0≤x≤π0 \le x \le \pi

Given that f(1.3)=0.678\text{f}(1.3) = 0.678 and f(1.4)=−0.525\text{f}(1.4) = -0.525, to 3 decimal places,
(b) explain why α\alpha must lie in the interval [1.3, 1.4][1.3,\ 1.4]

(1 mark)
(c) Using x0=1.4x_0 = 1.4 as a first approximation for α\alpha, apply the Newton–Raphson method once to obtain a second approximation for α\alpha.

Show your method and give your answer to 3 significant figures.

(2 marks)
●●●●●Level 47 marksStart
In this question you must show detailed reasoning.

The diagram shows the curve with equation y=2x−5(x+1)2y = \dfrac{2x - 5}{(x + 1)^2} for x>−1x > -1. The tangent to the curve at the point PP has gradient 11.2.5Pxy(a) Show that the xx-coordinate of PP satisfies the equation x3+3x2+5x−11=0x^3 + 3x^2 + 5x - 11 = 0.

(5 marks)
(b) Show by calculation that the xx-coordinate of PP lies between 11 and 1.21.2.

(2 marks)
(c) Show that the iteration

xn+1=11−xn3−3xn25x_{n+1} = \frac{11 - x_n^3 - 3x_n^2}{5}

cannot converge to the xx-coordinate of PP whatever starting value is used.

(2 marks)
(d) Use the Newton-Raphson method, with initial value 11, to determine the coordinates of PP correct to 5 decimal places.

(5 marks)
●●●●●Level 514 marksStart
A curve has equation y=kcos⁡x+x−x2y = k\cos x + x - x^2, where xx is in radians and kk is a positive constant.
(a) Given that the curve has a point of inflection where x=23πx = \tfrac23\pi, show that k=4k = 4.

(5 marks)

It is also given that the curve crosses the positive xx-axis at exactly one point.
(b) Show by calculation that the xx-coordinate of this point lies between 1.21.2 and 1.61.6.

(2 marks)
(c) Starting with x0=1.4x_0 = 1.4, use the Newton-Raphson method to find this xx-coordinate correct to 5 decimal places. Show the result of each iteration to 6 decimal places.

(3 marks)
(d) By choosing suitable bounds, verify that your answer to part (c) is correct to 5 decimal places.

(1 mark)
●●●●●Level 511 marksStart
The diagram shows the cross-section of a horizontal pipe, which is a circle with centre OO and radius 5050 cm. The pipe contains some water. In the cross-section, the surface of the water is the chord ABAB and the angle AOBAOB is θ\theta radians. The water occupies the shaded segment bounded by the chord ABAB and the arc ABAB, which has area 900900 cm2^2.OABθ rad50 cm(a) Show that θ=sin⁡θ+0.72\theta = \sin\theta + 0.72.

(3 marks)
(b) Let F(θ)=sin⁡θ+0.72\text{F}(\theta) = \sin\theta + 0.72.

By considering the value of F′(θ)\text{F}'(\theta) where θ=1.7\theta = 1.7, explain why using an iterative method based on the equation in part (a) will converge to the root, assuming that 1.71.7 is sufficiently close to the root.

(2 marks)
(c) Use the iterative formula θn+1=sin⁡θn+0.72\theta_{n+1} = \sin\theta_n + 0.72 with a starting value of 1.71.7 to find the value of θ\theta correct to 4 significant figures.

You should show the result of each iteration.

(3 marks)
(d) Use a change of sign method to show that the value of θ\theta found in part (c) is correct to 4 significant figures.

(3 marks)
●●●●●Level 511 marksStart
f(x)=x3−4x+1\mathrm{f}(x) = x^3 - 4x + 1

Figure 4 shows part of the curve with equation y=f(x)y = \mathrm{f}(x).-2-112-6-4-2246xyFigure 4
(a) Show that the equation f(x)=0\mathrm{f}(x) = 0 has a root α\alpha in the interval [1.8,1.9][1.8, 1.9]. (2 marks)
(b) Taking x0=1.9x_0 = 1.9, apply the Newton–Raphson procedure twice to f(x)\mathrm{f}(x) to obtain an approximation for α\alpha, giving your answer to 55 decimal places. (4 marks)
(c) Explain, with reference to the gradient of the curve, why x0=1.15x_0 = 1.15 would be a poor choice of first approximation for the Newton–Raphson procedure. (2 marks)
(d) An alternative approach uses the iterative formula

xn+1=4xn−13x_{n+1} = \sqrt[3]{4x_n - 1}

Using this formula with x0=1.9x_0 = 1.9, find the value of x2x_2, giving your answer to 44 decimal places. (2 marks)
●●●●●Level 510 marksStart
(a) The equation

x4=e9−2xx^4 = \mathrm{e}^{9 - 2x}

has a single solution, x=αx = \alpha

By considering a suitable change of sign, show that α\alpha lies between 22 and 33

(2 marks)
(b) Show that the equation x4=e9−2xx^4 = \mathrm{e}^{9 - 2x} can be rearranged to give

x=4.5−2ln⁡xx = 4.5 - 2\ln x

(3 marks)
(c) (i) Use the iterative formula

xn+1=4.5−2ln⁡xnx_{n+1} = 4.5 - 2\ln x_n

with x1=3x_1 = 3, to find x2x_2, x3x_3 and x4x_4

Give your answers to three decimal places.

(ii) Figure 1 below shows a sketch of parts of the graphs of y=4.5−2ln⁡xy = 4.5 - 2\ln x and y=xy = x12341234Oy = xy = 4.5 − 2 ln xxyOn Figure 1, draw a staircase or cobweb diagram to show how convergence takes place.

Label, on the xx-axis, the positions of x2x_2, x3x_3 and x4x_4

(iii) Explain why the iterative formula xn+1=4.5−2ln⁡xnx_{n+1} = 4.5 - 2\ln x_n fails to converge to α\alpha when the starting value is x1=e2.25x_1 = \mathrm{e}^{2.25}

(5 marks)
●●●●●Level 510 marksStart
The diagram shows a sector of a circle OABOAB.

The point CC lies on OBOB such that ACAC is perpendicular to OBOB.

Angle AOBAOB is θ\theta radians.OBAθC(a) Given the area of the sector OABOAB is 2\sqrt{2} times the area of the triangle OACOAC, show that

sin⁡2θ=2 θ\sin 2\theta = \sqrt{2}\,\theta

(4 marks)
(b) Use a suitable change of sign to show that a solution to the equation

sin⁡2θ=2 θ\sin 2\theta = \sqrt{2}\,\theta

lies in the interval given by θ∈[π5,π4]\theta \in \left[\dfrac{\pi}{5}, \dfrac{\pi}{4}\right]

(2 marks)
(c) The Newton-Raphson method is used to find an approximate solution to the equation sin⁡2θ=2 θ\sin 2\theta = \sqrt{2}\,\theta
(c) (i) Using θ1=π4\theta_1 = \dfrac{\pi}{4} as a first approximation for θ\theta, apply the Newton-Raphson method twice to find the value of θ3\theta_3

Give your answer to three decimal places.

(ii) Explain how a more accurate approximation for θ\theta can be found using the Newton-Raphson method.

(iii) Explain why using θ1=π8\theta_1 = \dfrac{\pi}{8} as a first approximation in the Newton-Raphson method does not lead to a solution for θ\theta.

(6 marks)
●●●●●Level 512 marksStart

Iterative methods (x = g(x))

33 questions

LessonNot started
(a) Show that x3−7x+2=0x^3 - 7x + 2 = 0 can be written as x=x3+27x = \dfrac{x^3 + 2}{7}.
(b) Using xn+1=xn3+27x_{n+1} = \dfrac{x_n^3 + 2}{7} with x0=0x_0 = 0, find the root near 00 to 3 decimal places.
●●●●●Level 25 marksStart
The iteration xn+1=2xn+53x_{n+1} = \sqrt[3]{2x_n + 5} is used with x0=2x_0 = 2 to find a root of x3−2x−5=0x^3 - 2x - 5 = 0. Find x1x_1, x2x_2 and x3x_3, giving each to 4 decimal places.
●●●●●Level 23 marksStart
Show that the equation x2−5x+3=0x^{2} - 5x + 3 = 0 can be rearranged into the form

x=x2+35x = \dfrac{x^{2} + 3}{5} (2 marks)
●●●●●Level 22 marksStart
The iteration xn+1=5xn−2x_{n+1} = \sqrt{5x_{n} - 2} is used with x0=2x_{0} = 2.

Find x1x_{1}, x2x_{2} and x3x_{3}, giving each to 4 decimal places. (3 marks)
●●●●●Level 23 marksStart
The equation x3+3ln⁡x=5x^3 + 3\ln x = 5 has exactly one real root.
(a) Show by calculation that the root lies between 1.541.54 and 1.551.55.

(2 marks)

In order to find the root, the iterative formula

xn+1=(p+qln⁡xn)r,x_{n+1} = \left(p + q\ln x_n\right)^r,

with initial value x0=1.5x_0 = 1.5, is to be used.
(b) Determine the values of the constants pp, qq and rr.

(2 marks)
(c) Hence find the root correct to 4 significant figures. Show the result of each step of the iteration process.

(2 marks)
●●●●●Level 36 marksStart
The iteration xn+1=3xn+4x_{n+1} = \sqrt{3x_n + 4} converges to a limit α\alpha. Find, with integer coefficients, the equation that α\alpha satisfies, and hence state the value of α\alpha.
●●●●●Level 34 marksStart
Explain why the iteration xn+1=xn3−52x_{n+1} = \dfrac{x_n^3 - 5}{2} does not converge to the root of x3−2x−5=0x^3 - 2x - 5 = 0 near x=2.09x = 2.09.
●●●●●Level 33 marksStart
(a) Show that x3+4x−1=0x^3 + 4x - 1 = 0 can be written as x=1x2+4x = \dfrac{1}{x^2 + 4}.
(b) Use xn+1=1xn2+4x_{n+1} = \dfrac{1}{x_n^2 + 4} with x0=0x_0 = 0 to find the root to 4 decimal places.
●●●●●Level 35 marksStart
An iteration xn+1=g(xn)x_{n+1} = g(x_n) converges to a root α\alpha at which g′(α)=−0.7g'(\alpha) = -0.7. State, with a reason, whether the staircase/cobweb diagram is a staircase or a cobweb.
●●●●●Level 32 marksStart
The iteration xn+1=12(xn+3xn)x_{n+1} = \dfrac12\left(x_n + \dfrac{3}{x_n}\right) is used with x0=2x_0 = 2.
(a) Find x1x_1, x2x_2 and x3x_3, giving each to 44 decimal places.
(b) State the exact value the sequence converges to.
●●●●●Level 34 marksStart
The iteration xn+1=5xn−2x_{n+1} = \sqrt{5x_{n} - 2} converges to a limit α\alpha.

Find the exact value of α\alpha. (4 marks)
●●●●●Level 34 marksStart
An iteration xn+1=g(xn)x_{n+1} = g(x_{n}) converges to a root α\alpha, and 0<g′(α)<10 < g'(\alpha) < 1.

Describe the appearance of the convergence on a staircase/cobweb diagram, and state how the behaviour would differ if −1<g′(α)<0-1 < g'(\alpha) < 0. (3 marks)
●●●●●Level 33 marksStart
The iteration xn+1=xn2−1x_{n+1} = x_{n}^{2} - 1 is started with x0=2x_{0} = 2.

Calculate x1x_{1}, x2x_{2} and x3x_{3}, and explain what the values show about this iteration. (3 marks)
●●●●●Level 33 marksStart
The iteration xn+1=12(xn+7xn)x_{n+1} = \dfrac{1}{2}\left(x_{n} + \dfrac{7}{x_{n}}\right) is used with x0=3x_{0} = 3.
(a) Find x1x_{1} and x2x_{2}, giving each to 6 decimal places. (2 marks)
(b) State the exact value the iteration converges to, and verify that x2x_{2} agrees with it to 4 decimal places. (2 marks)
●●●●●Level 34 marksStart
1.61.822.22.42.62.833.2510xyFigure 1

Figure 1 shows part of the curve y=x3−4x−2y = x^{3} - 4x - 2. The curve crosses the xx-axis at the point where x=αx = \alpha.
(a) Show by calculation that α\alpha lies between 22 and 33. (2 marks)
(b) Show that the equation x3−4x−2=0x^{3} - 4x - 2 = 0 can be rearranged into the form x=4x+23x = \sqrt[3]{4x+2}. (1 mark)
(c) Use the iterative formula xn+1=4xn+23x_{n+1} = \sqrt[3]{4x_{n}+2}, with x1=2x_{1} = 2, to find α\alpha correct to 33 decimal places. Show the result of each iteration. (3 marks)
●●●●●Level 36 marksStart
f(x)=x3−3x+ln⁡xx>0\text{f}(x) = x^3 - 3x + \ln x \qquad x > 0
(a) Show that the equation f(x)=0\text{f}(x) = 0 has a root α\alpha in the interval [1.6, 1.7][1.6,\ 1.7]

(2 marks)
(b) Using the iteration formula

xn+1=3xn−ln⁡xn3x_{n+1} = \sqrt[3]{3x_n - \ln x_n}

with x1=1.6x_1 = 1.6, find the values of x2x_2, x3x_3 and x4x_4, giving each answer to 3 decimal places.

(2 marks)
●●●●●Level 34 marksStart
The diagram shows part of the curve y=f(x)y = \text{f}(x), where f(x)=3ln⁡x+5−2x\text{f}(x) = 3\ln x + 5 - 2x. The equation f(x)=0\text{f}(x) = 0 has a positive root α\alpha close to x=0.2x = 0.2.12346xy(a) Explain why using the sign change method with x=0.1x = 0.1 and x=6x = 6 will fail to locate α\alpha.

(1 mark)
(b) Show that the equation f(x)=0\text{f}(x) = 0 can be written as x=e13(2x−5)x = \text{e}^{\frac{1}{3}(2x - 5)}.

(2 marks)
(c) Use the iterative formula xn+1=e13(2xn−5)x_{n+1} = \text{e}^{\frac{1}{3}(2x_n - 5)} with a starting value of x1=0.2x_1 = 0.2 to find the value of α\alpha correct to 4 significant figures, showing the result of each iteration.

(3 marks)
(d) An alternative iterative formula is xn+1=F(xn)x_{n+1} = \text{F}(x_n), where F(x)=1.5ln⁡x+2.5\text{F}(x) = 1.5\ln x + 2.5.

By considering F′(0.2)\text{F}'(0.2) explain why this iterative formula will not find α\alpha.

(3 marks)
●●●●●Level 49 marksStart
The diagram shows a sector OABOAB of a circle with centre OO and radius 55 cm. The angle AOBAOB is θ\theta radians. The shaded segment is bounded by the chord ABAB and the arc ABAB, and the perimeter of the shaded segment is 2121 cm.OABθ rad5 cm(a) Show that θ=4.2−2sin⁡(12θ)\theta = 4.2 - 2\sin\left(\tfrac12\theta\right).

(4 marks)

The equation θ=4.2−2sin⁡(12θ)\theta = 4.2 - 2\sin\left(\tfrac12\theta\right) has only one root for 0<θ<π0 < \theta < \pi.
(b) This root can be found by using the iterative formula θn+1=4.2−2sin⁡(12θn)\theta_{n+1} = 4.2 - 2\sin\left(\tfrac12\theta_n\right) with a starting value of θ1=2\theta_1 = 2.
  • Write down the values of θ2\theta_2, θ3\theta_3 and θ4\theta_4.
  • Hence find the value of this root correct to 3 significant figures.
(3 marks)

The diagram shows the graph of y=4.2−2sin⁡(12θ)y = 4.2 - 2\sin\left(\tfrac12\theta\right) and the line y=θy = \theta, for 0≤θ≤π0 \le \theta \le \pi.1231234θy(c) • Use this diagram to show how the iterative process used in (b) converges to this root.
  • State the type of convergence.
(3 marks)
●●●●●Level 410 marksStart
f(x)=x3−6x+2f(x) = x^3 - 6x + 2.
(a) Show f(x)=0f(x) = 0 has a root α\alpha in [0,1][0, 1].
(b) Using xn+1=xn3+26x_{n+1} = \dfrac{x_n^3 + 2}{6} with x0=0x_0 = 0, find α\alpha to 3 decimal places.
●●●●●Level 45 marksStart
f(x)=x3−5x+1f(x) = x^{3} - 5x + 1.
(a) Show that f(x)=0f(x) = 0 has a root α\alpha in the interval [0,1][0, 1]. (2 marks)
(b) Show that the equation can be rearranged as x=x3+15x = \dfrac{x^{3} + 1}{5}. (1 mark)
(c) Using x0=0.2x_{0} = 0.2, apply the iteration xn+1=xn3+15x_{n+1} = \dfrac{x_{n}^{3} + 1}{5} to find α\alpha correct to 3 decimal places. (3 marks)
●●●●●Level 46 marksStart
A continuous curve has equation

y=x3cos⁡2x0⩽x⩽π4y = x^3\cos 2x \qquad 0 \leqslant x \leqslant \dfrac{\pi}{4}

The curve has a stationary point at the point QQ, where x>0x > 0
(a) Show, using calculus, that the xx coordinate of QQ is a solution of the equation

x=12arctan⁡(32x)x = \dfrac{1}{2}\arctan\left(\dfrac{3}{2x}\right)

(4 marks)

Using the iteration formula

xn+1=12arctan⁡(32xn)x1=0.6x_{n+1} = \dfrac{1}{2}\arctan\left(\dfrac{3}{2x_n}\right) \qquad x_1 = 0.6
(b) find, to 4 decimal places, the value of

(i) x2x_2

(ii) x3x_3

(3 marks)
(c) Using a suitable interval and a suitable function which should be stated, show that the xx coordinate of QQ is 0.5960.596 correct to 3 decimal places.

(2 marks)
●●●●●Level 49 marksStart
f(x)=x3+3x−7f(x) = x^3 + 3x - 7. The equation f(x)=0f(x) = 0 has a single real root α\alpha.
(a) Show that α\alpha lies between 1.41.4 and 1.451.45. (2 marks)
(b) Show that the equation f(x)=0f(x) = 0 can be rearranged as x=7−3x3x = \sqrt[3]{7 - 3x}. (1 mark)
(c) Using the iterative formula xn+1=7−3xn3x_{n+1} = \sqrt[3]{7 - 3x_n} with x0=1.5x_0 = 1.5, find x1x_1, x2x_2 and x3x_3, giving each to 4 decimal places. (3 marks)
(d) By choosing a suitable interval, show that α=1.406\alpha = 1.406 correct to 3 decimal places. (2 marks)
●●●●●Level 48 marksStart
f(x)=x3+x−5\mathrm{f}(x) = x^3 + x - 5
(a) Show that the equation f(x)=0\mathrm{f}(x) = 0 has a root α\alpha between 1.51.5 and 1.61.6. (2 marks)
(b) Show that the equation f(x)=0\mathrm{f}(x) = 0 can be rearranged into the form x=5−x3x = \sqrt[3]{5 - x}. (1 mark)
(c) Use the iterative formula

xn+1=5−xn3x_{n+1} = \sqrt[3]{5 - x_n}

with x0=1.5x_0 = 1.5 to find x1x_1, x2x_2 and x3x_3, giving each answer to 44 decimal places. (3 marks)
(d) Show that α=1.516\alpha = 1.516 correct to 33 decimal places. (2 marks)
●●●●●Level 48 marksStart
The equation

x2=e7−xx^2 = \mathrm{e}^{7 - x}

has a single solution, x=αx = \alpha
(a) By considering a suitable change of sign, show that α\alpha lies between 44 and 55

(2 marks)
(b) Show that the equation x2=e7−xx^2 = \mathrm{e}^{7 - x} can be rearranged to give

x=7−2ln⁡xx = 7 - 2\ln x

(3 marks)
(c) Use the iterative formula

xn+1=7−2ln⁡xnx_{n+1} = 7 - 2\ln x_n

with x1=5x_1 = 5, to find x2x_2, x3x_3 and x4x_4

Give your answers to three decimal places.

(2 marks)
●●●●●Level 47 marksStart
f(x)=e12xx>0\text{f}(x) = e^{\frac{1}{2}x} \qquad x > 0

g(x)=40ln⁡xx>0\text{g}(x) = 40\ln x \qquad x > 0
(a) Find

(i) f′(x)\text{f}'(x)

(ii) g′(x)\text{g}'(x)

(2 marks)

Given that f′(x)=g′(x)\text{f}'(x) = \text{g}'(x) at x=αx = \alpha
(b) show that α\alpha satisfies the equation

xe12x−80=0xe^{\frac{1}{2}x} - 80 = 0

(2 marks)

The iterative formula

xn+1=2ln⁡(80xn)x_{n+1} = 2\ln\left(\frac{80}{x_n}\right)

is used with x1=5x_1 = 5 to find an approximate value for α\alpha.
(c) Calculate, giving each answer to 4 decimal places,

(i) the value of x2x_2

(ii) the value of α\alpha

(3 marks)
●●●●●Level 47 marksStart
In this question you must show detailed reasoning.

The diagram shows the curve with equation y=2x−5(x+1)2y = \dfrac{2x - 5}{(x + 1)^2} for x>−1x > -1. The tangent to the curve at the point PP has gradient 11.2.5Pxy(a) Show that the xx-coordinate of PP satisfies the equation x3+3x2+5x−11=0x^3 + 3x^2 + 5x - 11 = 0.

(5 marks)
(b) Show by calculation that the xx-coordinate of PP lies between 11 and 1.21.2.

(2 marks)
(c) Show that the iteration

xn+1=11−xn3−3xn25x_{n+1} = \frac{11 - x_n^3 - 3x_n^2}{5}

cannot converge to the xx-coordinate of PP whatever starting value is used.

(2 marks)
(d) Use the Newton-Raphson method, with initial value 11, to determine the coordinates of PP correct to 5 decimal places.

(5 marks)
●●●●●Level 514 marksStart
The diagram shows the cross-section of a horizontal pipe, which is a circle with centre OO and radius 5050 cm. The pipe contains some water. In the cross-section, the surface of the water is the chord ABAB and the angle AOBAOB is θ\theta radians. The water occupies the shaded segment bounded by the chord ABAB and the arc ABAB, which has area 900900 cm2^2.OABθ rad50 cm(a) Show that θ=sin⁡θ+0.72\theta = \sin\theta + 0.72.

(3 marks)
(b) Let F(θ)=sin⁡θ+0.72\text{F}(\theta) = \sin\theta + 0.72.

By considering the value of F′(θ)\text{F}'(\theta) where θ=1.7\theta = 1.7, explain why using an iterative method based on the equation in part (a) will converge to the root, assuming that 1.71.7 is sufficiently close to the root.

(2 marks)
(c) Use the iterative formula θn+1=sin⁡θn+0.72\theta_{n+1} = \sin\theta_n + 0.72 with a starting value of 1.71.7 to find the value of θ\theta correct to 4 significant figures.

You should show the result of each iteration.

(3 marks)
(d) Use a change of sign method to show that the value of θ\theta found in part (c) is correct to 4 significant figures.

(3 marks)
●●●●●Level 511 marksStart
The diagram shows a sector OABOAB of a circle with centre OO and radius rr. The angle AOBAOB is θ\theta radians, where 0<θ<12π0 < \theta < \tfrac12\pi. The tangent to the circle at AA meets the line OBOB extended at TT. The region RR is bounded by the arc ABAB and the lines BTBT and ATAT. The areas of region RR and sector OABOAB are in the ratio 6:56 : 5.Oθ radrTRAB(a) Show that θ=tan⁡−1(2.2θ)\theta = \tan^{-1}(2.2\theta).

(4 marks)

The equation θ=tan⁡−1(2.2θ)\theta = \tan^{-1}(2.2\theta) has only one root for 0<θ<12π0 < \theta < \tfrac12\pi.
(b) This root can be found by using the iterative formula θn+1=tan⁡−1(2.2θn)\theta_{n+1} = \tan^{-1}(2.2\theta_n) with a starting value of θ1=1\theta_1 = 1.
  • Write down the values of θ2\theta_2, θ3\theta_3 and θ4\theta_4.
  • Hence find the value of this root correct to 3 significant figures.
(3 marks)

The diagram shows the graph of y=tan⁡−1(2.2θ)y = \tan^{-1}(2.2\theta) and the line y=θy = \theta, for 0≤θ≤12π0 \le \theta \le \tfrac12\pi.0.511.50.511.5θy(c) • Use this diagram to show how the iterative process used in (b) converges to this root.
  • State the type of convergence.
(3 marks)
(d) Draw a suitable diagram to show why using an iterative process with the formula θn+1=511tan⁡θn\theta_{n+1} = \tfrac{5}{11}\tan\theta_n does not converge to the root found in (b).

(2 marks)
●●●●●Level 512 marksStart
Two new podcasts, AA and BB, are launched on the same day. Each podcast is downloaded 500500 times in Week 1. The number of downloads of podcast AA will increase by 6060 for every subsequent week. The number of downloads of podcast BB will increase by 7%7\% for every subsequent week.
(a) Verify that in Week 10 podcast AA is downloaded 121121 more times than podcast BB, correct to the nearest whole number.

(3 marks)

Let Week XX be the first week in which podcast BB is downloaded more times than podcast AA.
(b) Show that XX satisfies the inequality X>1+log⁡1.07(0.12X+0.88)X > 1 + \log_{1.07}(0.12X + 0.88).

(3 marks)
(c) Use the iterative formula xn+1=1+log⁡1.07(0.12xn+0.88)x_{n+1} = 1 + \log_{1.07}(0.12x_n + 0.88) with x1=20x_1 = 20 to find the value of XX.

You should show the result of each iteration.

(3 marks)
(d) (i) Give a reason why the model for podcast AA may not be realistic.

(ii) Give a different reason why the model for podcast BB may not be realistic.

(2 marks)
●●●●●Level 511 marksStart
tNFigure 1

A group of red squirrels is released into a forest and the number of red squirrels in the forest is then monitored.

The number of red squirrels, NN, in the forest is modelled by the equation

N=180+300e−0.1t−420e−0.3tN = 180 + 300\mathrm{e}^{-0.1t} - 420\mathrm{e}^{-0.3t}

where tt months is the time after the squirrels are released.

Figure 1 is a sketch of NN against tt.

Use the equation of the model to answer parts (a) to (e).
(a) State the number of red squirrels released into the forest.

(1 mark)

In the long term, the number of red squirrels in the forest approaches LL.
(b) State the value of LL.

(1 mark)

The number of red squirrels in the forest reaches its maximum value after TT months.
(c) Find the value of TT, giving your answer to 3 decimal places.

(Solutions based entirely on calculator technology are not acceptable.)

(5 marks)

The number of red squirrels in the forest is 250250 for the second time after MM months.
(d) Show that MM is a solution of the equation

t=10ln⁡(307+42e−0.3t)t = 10\ln\left(\frac{30}{7 + 42\mathrm{e}^{-0.3t}}\right)

(2 marks)

Using the iteration formula

tn+1=10ln⁡(307+42e−0.3tn)with t1=13t_{n+1} = 10\ln\left(\frac{30}{7 + 42\mathrm{e}^{-0.3t_n}}\right) \qquad \text{with } t_1 = 13
(e) (i) find, to 4 decimal places, the value of t2t_2

(ii) find, to 4 decimal places, the value of MM

(3 marks)
●●●●●Level 512 marksStart
Figure 1dOxhA frog jumps from a point OO on horizontal ground. Figure 1 shows the path of the frog from OO until it lands, a horizontal distance of dd cm from OO.

When the frog has travelled a horizontal distance of xx cm, its height above the ground is hh cm.

The frog is modelled as a particle travelling in a vertical plane above horizontal ground.

The path of the frog is modelled by the equation

h=x−0.25xe0.04x0⩽x⩽dh = x - 0.25x\text{e}^{0.04x} \qquad 0 \leqslant x \leqslant d

Use the model to answer parts (a),
(b) and (c).
(a) Find the value of dd, giving your answer to 2 decimal places.

(Solutions relying entirely on calculator technology are not acceptable.)

(3 marks)
(b) Show that the maximum value of hh occurs when

x=25ln⁡(10025+x)x = 25\ln\left(\dfrac{100}{25 + x}\right)

(4 marks)

Using the iteration formula

xn+1=25ln⁡(10025+xn)x1=15x_{n+1} = 25\ln\left(\dfrac{100}{25 + x_n}\right) \qquad x_1 = 15
(c) (i) find the value of x2x_2 to 2 decimal places,

(ii) find, by repeated iteration, the horizontal distance travelled by the frog before it reaches its maximum height. Give your answer to 2 decimal places.

(3 marks)
●●●●●Level 510 marksStart
f(x)=x3−4x+1\mathrm{f}(x) = x^3 - 4x + 1

Figure 4 shows part of the curve with equation y=f(x)y = \mathrm{f}(x).-2-112-6-4-2246xyFigure 4
(a) Show that the equation f(x)=0\mathrm{f}(x) = 0 has a root α\alpha in the interval [1.8,1.9][1.8, 1.9]. (2 marks)
(b) Taking x0=1.9x_0 = 1.9, apply the Newton–Raphson procedure twice to f(x)\mathrm{f}(x) to obtain an approximation for α\alpha, giving your answer to 55 decimal places. (4 marks)
(c) Explain, with reference to the gradient of the curve, why x0=1.15x_0 = 1.15 would be a poor choice of first approximation for the Newton–Raphson procedure. (2 marks)
(d) An alternative approach uses the iterative formula

xn+1=4xn−13x_{n+1} = \sqrt[3]{4x_n - 1}

Using this formula with x0=1.9x_0 = 1.9, find the value of x2x_2, giving your answer to 44 decimal places. (2 marks)
●●●●●Level 510 marksStart
(a) The equation

x4=e9−2xx^4 = \mathrm{e}^{9 - 2x}

has a single solution, x=αx = \alpha

By considering a suitable change of sign, show that α\alpha lies between 22 and 33

(2 marks)
(b) Show that the equation x4=e9−2xx^4 = \mathrm{e}^{9 - 2x} can be rearranged to give

x=4.5−2ln⁡xx = 4.5 - 2\ln x

(3 marks)
(c) (i) Use the iterative formula

xn+1=4.5−2ln⁡xnx_{n+1} = 4.5 - 2\ln x_n

with x1=3x_1 = 3, to find x2x_2, x3x_3 and x4x_4

Give your answers to three decimal places.

(ii) Figure 1 below shows a sketch of parts of the graphs of y=4.5−2ln⁡xy = 4.5 - 2\ln x and y=xy = x12341234Oy = xy = 4.5 − 2 ln xxyOn Figure 1, draw a staircase or cobweb diagram to show how convergence takes place.

Label, on the xx-axis, the positions of x2x_2, x3x_3 and x4x_4

(iii) Explain why the iterative formula xn+1=4.5−2ln⁡xnx_{n+1} = 4.5 - 2\ln x_n fails to converge to α\alpha when the starting value is x1=e2.25x_1 = \mathrm{e}^{2.25}

(5 marks)
●●●●●Level 510 marksStart

The Newton–Raphson method

29 questions

LessonNot started
f(x)=x3−2x−5f(x) = x^3 - 2x - 5. Using the Newton–Raphson method with x0=2x_0 = 2, find x1x_1.
●●●●●Level 22 marksStart
The positive root of x2−7=0x^2 - 7 = 0 is 7\sqrt7. Using Newton–Raphson with x0=3x_0 = 3, find x1x_1 and x2x_2 to 4 decimal places.
●●●●●Level 23 marksStart
Using the Newton–Raphson method on f(x)=x3−5f(x) = x^3 - 5 with x0=2x_0 = 2, find x1x_1 and x2x_2 to 4 decimal places. (This estimates 53\sqrt[3]{5}.)
●●●●●Level 33 marksStart
The equation ex+x=4e^x + x = 4 has a root near x=1x = 1. Use the Newton–Raphson method with x0=1x_0 = 1 to find x1x_1 and x2x_2 to 4 decimal places.
●●●●●Level 33 marksStart
f(x)=x3+x−1f(x) = x^3 + x - 1. Using Newton–Raphson with x0=1x_0 = 1, find x1x_1 and x2x_2 to 4 decimal places.
●●●●●Level 33 marksStart
A student applies the Newton–Raphson method to f(x)=x3−2x−5f(x) = x^3 - 2x - 5, whose only real root is near 2.092.09, but starts at x0=0x_0 = 0. Find x1x_1 and explain why this starting value is a poor choice.
●●●●●Level 33 marksStart
(a) Show that applying the Newton–Raphson method to f(x)=x2−Nf(x) = x^2 - N gives the iteration xn+1=12(xn+Nxn)x_{n+1} = \dfrac12\left(x_n + \dfrac{N}{x_n}\right).
(b) Hence, with N=10N = 10 and x0=3x_0 = 3, estimate 10\sqrt{10} to 4 decimal places.
●●●●●Level 34 marksStart
f(x)=x3−4x+1f(x) = x^{3} - 4x + 1.

Using the Newton–Raphson method with x0=2x_{0} = 2, find x1x_{1} and x2x_{2}, giving x2x_{2} to 5 decimal places. (4 marks)
●●●●●Level 34 marksStart
Using the Newton–Raphson method on f(x)=x2−10f(x) = x^{2} - 10 with x0=3x_{0} = 3, find x1x_{1} and x2x_{2}, and comment on how close x2x_{2} is to 10\sqrt{10}. (4 marks)
●●●●●Level 34 marksStart
The equation cos⁡x=x\cos x = x has a single root.

Using the Newton–Raphson method on f(x)=cos⁡x−xf(x) = \cos x - x with x0=0.7x_{0} = 0.7, find x1x_{1} and x2x_{2}, giving each to 5 decimal places. (5 marks)
●●●●●Level 35 marksStart
(a) Show that applying the Newton–Raphson method to f(x)=x3−20f(x) = x^{3} - 20 gives the iteration

xn+1=2xn3+203xn2x_{n+1} = \dfrac{2x_{n}^{3} + 20}{3x_{n}^{2}} (3 marks)
(b) With x0=3x_{0} = 3, find x1x_{1} and x2x_{2} to 5 decimal places, and state what the iteration converges to. (3 marks)
●●●●●Level 36 marksStart
f(x)=x3−4x+1f(x) = x^{3} - 4x + 1.

Explain why the Newton–Raphson method fails if it is started at x0=23x_{0} = \dfrac{2}{\sqrt{3}}. (3 marks)
●●●●●Level 33 marksStart
The equation ex=3x\mathrm{e}^{x} = 3x has a root near x=0.6x = 0.6.

Using the Newton–Raphson method on f(x)=ex−3xf(x) = \mathrm{e}^{x} - 3x with x0=0.6x_{0} = 0.6, find x1x_{1} and x2x_{2}, giving each to 5 decimal places. (5 marks)
●●●●●Level 35 marksStart
Use the Newton–Raphson method on f(x)=x5−30f(x) = x^{5} - 30 with x0=2x_{0} = 2 to find x1x_{1} and x2x_{2}, giving each to 5 decimal places, and state the exact value the sequence approaches. (5 marks)
●●●●●Level 35 marksStart
f(x)=x3−5x+1f(x) = x^3 - 5x + 1.
(a) Show that the equation f(x)=0f(x) = 0 has a root α\alpha between x=2x = 2 and x=2.5x = 2.5. (2 marks)
(b) Taking x0=2x_0 = 2 as a first approximation to α\alpha, apply the Newton–Raphson method twice to find x2x_2. Give your answer to 4 decimal places. (3 marks)
(c) Explain why the Newton–Raphson method fails if x0=1.3x_0 = 1.3 is used instead. (1 mark)
●●●●●Level 36 marksStart
f(x)=x3−5x+1\mathrm{f}(x) = x^3 - 5x + 1.

The equation f(x)=0\mathrm{f}(x) = 0 has a root α\alpha.
(a) Show that α\alpha lies between 0.20.2 and 0.210.21. (2 marks)
(b) Using the Newton-Raphson method with first approximation x1=0.2x_1 = 0.2, find x2x_2 and x3x_3, and hence state the value of α\alpha correct to 5 decimal places. (4 marks)
(c) Explain why the Newton-Raphson method fails if x1=53x_1 = \sqrt{\dfrac{5}{3}} is used as the first approximation. (2 marks)
●●●●●Level 38 marksStart
The curve y=(x+3)e−x+x2y = (x + 3)\text{e}^{-x} + x^2 has one stationary point for x>0x > 0, which is close to x=0.7x = 0.7.
(a) Show that the xx-coordinate of this stationary point satisfies the equation 2xex−x−2=02x\text{e}^{x} - x - 2 = 0.

(2 marks)
(b) Show that the Newton-Raphson iterative formula for finding the root of the equation in part (a) can be written in the form

xn+1=2xn2exn+22(xn+1)exn−1.x_{n+1} = \frac{2x_n^2\text{e}^{x_n} + 2}{2(x_n + 1)\text{e}^{x_n} - 1}.

(4 marks)
(c) Apply the Newton-Raphson formula with initial value x1=0.7x_1 = 0.7 to find x2x_2 and x3x_3.

(1 mark)
(d) Find the coordinates of this stationary point, giving each coordinate correct to 3 decimal places.

(2 marks)
●●●●●Level 49 marksStart
The equation ln⁡x+x=2\ln x + x = 2 has a root near x=1.5x = 1.5. Use the Newton–Raphson method with x0=1.5x_0 = 1.5 to find x1x_1 to 4 decimal places.
●●●●●Level 42 marksStart
f(x)=x3−4x+1f(x) = x^3 - 4x + 1
(a) Show that the equation f(x)=0f(x) = 0 has a root α\alpha such that 1.8<α<1.91.8 < \alpha < 1.9. (2 marks)
(b) Show that the Newton–Raphson iteration for this equation can be written as

xn+1=2xn3−13xn2−4x_{n+1} = \dfrac{2x_n^3 - 1}{3x_n^2 - 4} (2 marks)
(c) Using x0=2x_0 = 2, find α\alpha correct to 4 decimal places. Show the result of each iteration. (3 marks)
(d) Explain why the Newton–Raphson method cannot be used to find α\alpha starting from x0=23x_0 = \dfrac{2}{\sqrt{3}}. (1 mark)
●●●●●Level 48 marksStart
f(x)=x3+4x−7\mathrm{f}(x) = x^3 + 4x - 7
(a) Show that the equation f(x)=0\mathrm{f}(x) = 0 has a root α\alpha between 1.21.2 and 1.31.3. (2 marks)
(b) Taking x0=1.3x_0 = 1.3, apply the Newton–Raphson procedure twice to f(x)\mathrm{f}(x), giving x2x_2 to 55 decimal places. (4 marks)
(c) Write down the value of α\alpha correct to 44 significant figures. (1 mark)
●●●●●Level 47 marksStart
The function f is defined by

f(x)=arccos⁡(x2)for 0≤x≤a\text{f}(x) = \arccos\left(\frac{x}{2}\right) \quad \text{for } 0 \le x \le a

The curve with equation y=f(x)y = \text{f}(x) is shown below.π/2aOxy(a) State the value of aa

(1 mark)
(b) (i) On the diagram above, sketch the curve with equation

y=2cos⁡xfor 0≤x≤π2y = 2\cos x \quad \text{for } 0 \le x \le \frac{\pi}{2}

and sketch the line with equation

y=xfor 0≤x≤π2y = x \quad \text{for } 0 \le x \le \frac{\pi}{2}

(ii) Explain why the solution to the equation

x−2cos⁡x=0x - 2\cos x = 0

must also be a solution to the equation

2cos⁡x=arccos⁡(x2)2\cos x = \arccos\left(\frac{x}{2}\right)

(5 marks)
(c) Use the Newton-Raphson method with x0=0.5x_0 = 0.5 to find an approximate solution, x3x_3, to the equation

x−2cos⁡x=0x - 2\cos x = 0

Give your answer to four decimal places.

(3 marks)
●●●●●Level 49 marksStart
In a model of a sprint race, the speed of an athlete, vv m s−1^{-1}, tt seconds after the start of the race, is modelled by the equation

v=12(18−t)ln⁡(t+1)0⩽t⩽18v = \frac{1}{2}(18 - t)\ln(t + 1) \qquad 0 \leqslant t \leqslant 18

Using the model,
(a) show that the speed of the athlete is a maximum when tt satisfies the equation

(t+1)ln⁡(t+1)+t−18=0(t + 1)\ln(t + 1) + t - 18 = 0

(4 marks)

The equation in part (a) has exactly one root, α\alpha, in the interval 0⩽t⩽180 \leqslant t \leqslant 18
(b) Show that α\alpha lies in the interval [5,6][5, 6]

(2 marks)
(c) Using t0=6t_0 = 6 as a first approximation for α\alpha, apply the Newton–Raphson method once to find a second approximation for α\alpha, giving your answer to 3 decimal places.

(2 marks)
●●●●●Level 48 marksStart
f(x)=xcos⁡x+20⩽x⩽π\text{f}(x) = x\cos x + 2 \qquad 0 \leqslant x \leqslant \pi

where xx is measured in radians.
(a) Find f′(x)\text{f}'(x)

(2 marks)

The curve with equation y=f(x)y = \text{f}(x) has a maximum turning point at x=αx = \alpha
(b) Show that α\alpha lies in the interval [0.8,0.9][0.8, 0.9]

(2 marks)
(c) Using x0=0.8x_0 = 0.8 as a first approximation for α\alpha, apply the Newton–Raphson method once to f′(x)\text{f}'(x) to find a second approximation for α\alpha, giving your answer to 3 decimal places.

(3 marks)
●●●●●Level 47 marksStart
f(x)=x−3ln⁡xx>0\text{f}(x) = x - 3\ln x \qquad x > 0
(a) Show that the equation f(x)=0\text{f}(x) = 0 has a root α\alpha in the interval [4.5, 5][4.5,\ 5]

(2 marks)
(b) Find f′(x)\text{f}'(x)

(2 marks)
(c) Using x0=5x_0 = 5 as a first approximation for α\alpha, apply the Newton–Raphson method once to obtain a second approximation for α\alpha.

Give your answer to 3 decimal places.

(2 marks)

A student attempts to use the Newton–Raphson method with x0=3x_0 = 3 as a first approximation for α\alpha.
(d) Explain why the method fails.

(1 mark)
●●●●●Level 47 marksStart
f(x)=5sin⁡2x−3x+20≤x≤π\text{f}(x) = 5\sin 2x - 3x + 2 \qquad 0 \le x \le \pi

where xx is measured in radians.

The curve with equation y=f(x)y = \text{f}(x) has a single maximum turning point in the interval 0≤x≤π0 \le x \le \pi
(a) Using calculus, find the xx coordinate of this maximum turning point, giving your answer to 3 decimal places.

(4 marks)

The equation f(x)=0\text{f}(x) = 0 has a single root α\alpha in the interval 0≤x≤π0 \le x \le \pi

Given that f(1.3)=0.678\text{f}(1.3) = 0.678 and f(1.4)=−0.525\text{f}(1.4) = -0.525, to 3 decimal places,
(b) explain why α\alpha must lie in the interval [1.3, 1.4][1.3,\ 1.4]

(1 mark)
(c) Using x0=1.4x_0 = 1.4 as a first approximation for α\alpha, apply the Newton–Raphson method once to obtain a second approximation for α\alpha.

Show your method and give your answer to 3 significant figures.

(2 marks)
●●●●●Level 47 marksStart
In this question you must show detailed reasoning.

The diagram shows the curve with equation y=2x−5(x+1)2y = \dfrac{2x - 5}{(x + 1)^2} for x>−1x > -1. The tangent to the curve at the point PP has gradient 11.2.5Pxy(a) Show that the xx-coordinate of PP satisfies the equation x3+3x2+5x−11=0x^3 + 3x^2 + 5x - 11 = 0.

(5 marks)
(b) Show by calculation that the xx-coordinate of PP lies between 11 and 1.21.2.

(2 marks)
(c) Show that the iteration

xn+1=11−xn3−3xn25x_{n+1} = \frac{11 - x_n^3 - 3x_n^2}{5}

cannot converge to the xx-coordinate of PP whatever starting value is used.

(2 marks)
(d) Use the Newton-Raphson method, with initial value 11, to determine the coordinates of PP correct to 5 decimal places.

(5 marks)
●●●●●Level 514 marksStart
A curve has equation y=kcos⁡x+x−x2y = k\cos x + x - x^2, where xx is in radians and kk is a positive constant.
(a) Given that the curve has a point of inflection where x=23πx = \tfrac23\pi, show that k=4k = 4.

(5 marks)

It is also given that the curve crosses the positive xx-axis at exactly one point.
(b) Show by calculation that the xx-coordinate of this point lies between 1.21.2 and 1.61.6.

(2 marks)
(c) Starting with x0=1.4x_0 = 1.4, use the Newton-Raphson method to find this xx-coordinate correct to 5 decimal places. Show the result of each iteration to 6 decimal places.

(3 marks)
(d) By choosing suitable bounds, verify that your answer to part (c) is correct to 5 decimal places.

(1 mark)
●●●●●Level 511 marksStart
f(x)=x3−4x+1\mathrm{f}(x) = x^3 - 4x + 1

Figure 4 shows part of the curve with equation y=f(x)y = \mathrm{f}(x).-2-112-6-4-2246xyFigure 4
(a) Show that the equation f(x)=0\mathrm{f}(x) = 0 has a root α\alpha in the interval [1.8,1.9][1.8, 1.9]. (2 marks)
(b) Taking x0=1.9x_0 = 1.9, apply the Newton–Raphson procedure twice to f(x)\mathrm{f}(x) to obtain an approximation for α\alpha, giving your answer to 55 decimal places. (4 marks)
(c) Explain, with reference to the gradient of the curve, why x0=1.15x_0 = 1.15 would be a poor choice of first approximation for the Newton–Raphson procedure. (2 marks)
(d) An alternative approach uses the iterative formula

xn+1=4xn−13x_{n+1} = \sqrt[3]{4x_n - 1}

Using this formula with x0=1.9x_0 = 1.9, find the value of x2x_2, giving your answer to 44 decimal places. (2 marks)
●●●●●Level 510 marksStart
The diagram shows a sector of a circle OABOAB.

The point CC lies on OBOB such that ACAC is perpendicular to OBOB.

Angle AOBAOB is θ\theta radians.OBAθC(a) Given the area of the sector OABOAB is 2\sqrt{2} times the area of the triangle OACOAC, show that

sin⁡2θ=2 θ\sin 2\theta = \sqrt{2}\,\theta

(4 marks)
(b) Use a suitable change of sign to show that a solution to the equation

sin⁡2θ=2 θ\sin 2\theta = \sqrt{2}\,\theta

lies in the interval given by θ∈[π5,π4]\theta \in \left[\dfrac{\pi}{5}, \dfrac{\pi}{4}\right]

(2 marks)
(c) The Newton-Raphson method is used to find an approximate solution to the equation sin⁡2θ=2 θ\sin 2\theta = \sqrt{2}\,\theta
(c) (i) Using θ1=π4\theta_1 = \dfrac{\pi}{4} as a first approximation for θ\theta, apply the Newton-Raphson method twice to find the value of θ3\theta_3

Give your answer to three decimal places.

(ii) Explain how a more accurate approximation for θ\theta can be found using the Newton-Raphson method.

(iii) Explain why using θ1=π8\theta_1 = \dfrac{\pi}{8} as a first approximation in the Newton-Raphson method does not lead to a solution for θ\theta.

(6 marks)
●●●●●Level 512 marksStart

The trapezium rule

29 questions

LessonNot started
Use the trapezium rule with 4 strips to estimate ∫04x2+1 dx\displaystyle\int_0^4 \sqrt{x^2 + 1}\,dx, giving your answer to 3 decimal places.
●●●●●Level 23 marksStart
Use the trapezium rule with 5 ordinates to estimate ∫132x dx\displaystyle\int_1^3 2^x\,dx, giving your answer to 3 decimal places.
●●●●●Level 23 marksStart
The table below shows corresponding values of xx and yy for y=5+x4y = \sqrt{5 + x^4}

The values of yy are given to 3 decimal places as appropriate.

x00.40.81.21.6y2.2362.2422.3262.6603.399\begin{array}{c|ccccc} x & 0 & 0.4 & 0.8 & 1.2 & 1.6 \\ \hline y & 2.236 & 2.242 & 2.326 & 2.660 & 3.399 \end{array}
(a) Use the trapezium rule with all the values of yy in the table to find an estimate for

∫01.65+x4 dx\int_0^{1.6} \sqrt{5 + x^4}\,dx

giving your answer to 3 significant figures.

(3 marks)
(b) Use your answer to part (a) to find an estimate for

∫01.6(1+45+x4)dx\int_0^{1.6} \left(1 + 4\sqrt{5 + x^4}\right)dx

(2 marks)
●●●●●Level 25 marksStart
The diagram shows part of the curve y=ln⁡(1+x2)y = \ln(1 + x^2).2xy(a) Use the trapezium rule with 4 intervals of equal width to find an estimate for ∫02ln⁡(1+x2) dx\displaystyle\int_0^2 \ln(1 + x^2)\,dx. Give your answer correct to 3 significant figures.

(4 marks)
(b) Explain how the trapezium rule could be used to obtain a more accurate estimate for ∫02ln⁡(1+x2) dx\displaystyle\int_0^2 \ln(1 + x^2)\,dx.

(1 mark)
(c) Explain why it is not clear from the diagram whether the value from part (a) is an under-estimate or an over-estimate for ∫02ln⁡(1+x2) dx\displaystyle\int_0^2 \ln(1 + x^2)\,dx.

(2 marks)
●●●●●Level 37 marksStart
The diagram shows part of the curve y=11+xy = \dfrac{1}{1 + \sqrt{x}}.4xy(a) Use the trapezium rule with 4 intervals to find an estimate for ∫0411+x dx\displaystyle\int_0^4 \frac{1}{1 + \sqrt{x}}\,dx. Give your answer correct to 3 significant figures.

(4 marks)
(b) State whether the value from part (a) is an under-estimate or an over-estimate, giving a reason for your answer.

(1 mark)
(c) Explain how the trapezium rule could be used to obtain a more accurate estimate.

(1 mark)
●●●●●Level 36 marksStart
The table below shows corresponding values of xx and yy for y=x2+53y = \sqrt[3]{x^2 + 5}

The values of yy are given to 3 decimal places.

x11.522.533.5y1.8171.9352.0802.2412.4102.584\begin{array}{c|cccccc} x & 1 & 1.5 & 2 & 2.5 & 3 & 3.5 \\ \hline y & 1.817 & 1.935 & 2.080 & 2.241 & 2.410 & 2.584 \end{array}
(a) Using the trapezium rule with all the values of yy in the given table, obtain an estimate for

∫13.5x2+53 dx\int_{1}^{3.5} \sqrt[3]{x^2 + 5}\,\mathrm{d}x

giving your answer to 3 significant figures.

(3 marks)

Using your answer to part (a) and making your method clear, estimate
(b) (i) ∫13.5(4+x2+53)dx\displaystyle\int_{1}^{3.5} \left(4 + \sqrt[3]{x^2 + 5}\right)\mathrm{d}x

(ii) ∫13.58x2+403 dx\displaystyle\int_{1}^{3.5} \sqrt[3]{8x^2 + 40}\,\mathrm{d}x

(3 marks)
●●●●●Level 36 marksStart
A continuous curve has equation y=log⁡3(2x+5)y = \log_3(2x + 5)

The table below shows corresponding values of xx and yy for this curve, with the values of yy given to 4 decimal places as appropriate.

x00.40.81.21.62y1.46501.60011.71771.82181.91532\begin{array}{c|cccccc} x & 0 & 0.4 & 0.8 & 1.2 & 1.6 & 2 \\ \hline y & 1.4650 & 1.6001 & 1.7177 & 1.8218 & 1.9153 & 2 \end{array}

Using the trapezium rule with all the values of yy in the given table,
(a) find an estimate for

∫02log⁡3(2x+5) dx\int_{0}^{2} \log_3(2x + 5)\,\mathrm{d}x

giving your answer to 3 decimal places.

(3 marks)
(b) Using your answer to part (a) and making your method clear, estimate

(i) ∫02log⁡32x+5 dx\displaystyle\int_{0}^{2} \log_3\sqrt{2x + 5}\,\mathrm{d}x

(ii) ∫02log⁡3(18x+45) dx\displaystyle\int_{0}^{2} \log_3(18x + 45)\,\mathrm{d}x

(2 marks)
●●●●●Level 35 marksStart
Use the trapezium rule with 4 strips to estimate ∫021+x3 dx\displaystyle\int_0^2 \sqrt{1 + x^3}\,dx, giving your answer to 3 decimal places.
●●●●●Level 33 marksStart
(a) Use the trapezium rule with 4 strips to estimate ∫01ex2 dx\displaystyle\int_0^1 e^{x^2}\,dx to 3 decimal places.
(b) State, with a reason, whether this is an over- or under-estimate.
●●●●●Level 35 marksStart
(a) Use the trapezium rule with 4 strips to estimate ∫151x dx\displaystyle\int_1^5 \dfrac1x\,dx to 4 decimal places.
(b) The exact value is ln⁡5=1.6094\ln 5 = 1.6094. Explain why the estimate is larger.
●●●●●Level 35 marksStart
Use the trapezium rule with 6 strips to estimate ∫039−x2 dx\displaystyle\int_0^3 \sqrt{9 - x^2}\,dx, giving your answer to 3 decimal places.
●●●●●Level 33 marksStart
(a) Use the trapezium rule with 44 strips to estimate ∫0211+x2 dx\displaystyle\int_{0}^{2} \dfrac{1}{1 + x^{2}}\,dx, giving your answer to 4 decimal places. (4 marks)
(b) Given that the exact value is arctan⁡2=1.1071\arctan 2 = 1.1071, state with a reason whether the rule gives an underestimate here. (1 mark)
●●●●●Level 35 marksStart
Use the trapezium rule with 55 ordinates to estimate ∫13ln⁡x dx\displaystyle\int_{1}^{3} \ln x\,dx, giving your answer to 4 decimal places. (4 marks)
●●●●●Level 34 marksStart
Use the trapezium rule with 55 strips to estimate ∫01e−x2 dx\displaystyle\int_{0}^{1} \mathrm{e}^{-x^{2}}\,dx, giving your answer to 4 decimal places. (4 marks)
●●●●●Level 34 marksStart
(a) Use the trapezium rule with 44 strips to estimate ∫02e−x dx\displaystyle\int_{0}^{2} \mathrm{e}^{-x}\,dx, giving your answer to 4 decimal places. (4 marks)
(b) The exact value is 1−e−2=0.86471 - \mathrm{e}^{-2} = 0.8647. Explain, with reference to the shape of the curve, why the rule OVERestimates here. (2 marks)
●●●●●Level 36 marksStart
The table gives values of y=f(x)y = f(x).

x00.511.52y22.242.833.614.47\begin{array}{c|ccccc} x & 0 & 0.5 & 1 & 1.5 & 2 \\ \hline y & 2 & 2.24 & 2.83 & 3.61 & 4.47 \end{array}

Use the trapezium rule with all five ordinates to estimate ∫02f(x) dx\displaystyle\int_{0}^{2} f(x)\,dx, giving your answer to 3 significant figures. (3 marks)
●●●●●Level 33 marksStart
(a) Use the trapezium rule with 22 strips, and then with 44 strips, to estimate ∫121x dx\displaystyle\int_{1}^{2} \dfrac{1}{x}\,dx, giving each answer to 4 decimal places. (5 marks)
(b) The exact value is ln⁡2=0.6931\ln 2 = 0.6931. Comment on the effect of doubling the number of strips. (1 mark)
●●●●●Level 36 marksStart
(a) Use the trapezium rule with 44 strips to estimate ∫04x dx\displaystyle\int_{0}^{4} \sqrt{x}\,dx, giving your answer to 4 decimal places. (4 marks)
(b) Compare your estimate with the exact value 163\dfrac{16}{3}, and explain the direction of the error. (2 marks)
●●●●●Level 36 marksStart
The diagram shows the curve y=4−x2y = \sqrt{4 - x^2} for 0≤x≤20 \le x \le 2, together with the region under it.0.511.520.511.52xy(a) Use the trapezium rule with 44 strips of equal width to estimate

∫024−x2 dx\displaystyle\int_0^2 \sqrt{4 - x^2}\ \mathrm{d}x,

giving your answer to 3 significant figures. (4 marks)
(b) Explain, with reference to the shape of the curve, whether the trapezium rule gives an overestimate or an underestimate of the true area. (1 mark)
(c) By considering the shape of the region, write down the exact value of the integral. (1 mark)
●●●●●Level 36 marksStart
A carpenter is making a 22 metre length of wooden handrail.

The cross-section of the handrail is symmetrical about the xx-axis and is shown in the diagram below, where xx and yy are measured in centimetres.OxyThe table below gives values of yy for the lower half of the cross-section.
xx001.21.22.42.43.63.64.84.86.06.0
yy00−2.53-2.53−3.01-3.01−2.89-2.89−2.20-2.2000
Use the trapezium rule, with the values shown in the table above, to find the best estimate for the volume of wood in the handrail.

(5 marks)
●●●●●Level 35 marksStart
(a) Sketch the curve with equation

y=ax−5y = a^{x} - 5

where aa is a positive constant greater than 1

On your sketch show
  • the coordinates of the point of intersection of the curve with the yy-axis
  • the equation of the asymptote of the curve
(3 marks)

x00.40.81.21.62y11.39181.76822.29723.23955\begin{array}{c|cccccc} x & 0 & 0.4 & 0.8 & 1.2 & 1.6 & 2 \\ \hline y & 1 & 1.3918 & 1.7682 & 2.2972 & 3.2395 & 5 \end{array}

The table above shows corresponding values of xx and yy for the curve with equation y=3x−x2y = 3^{x} - x^2

The values of yy are given to 4 decimal places where appropriate.
(b) Use the trapezium rule with all the values of yy in the table to obtain an estimate for

∫02(3x−x2)dx\int_{0}^{2} \left(3^{x} - x^2\right)\mathrm{d}x

giving your answer to 2 decimal places.

(3 marks)
(c) Using your answer to part (b) and making your method clear, estimate

(i) ∫02(3x+x2)dx\displaystyle\int_{0}^{2} \left(3^{x} + x^2\right)\mathrm{d}x

(ii) ∫02(7−3x+x2)dx\displaystyle\int_{0}^{2} \left(7 - 3^{x} + x^2\right)\mathrm{d}x

(3 marks)
●●●●●Level 49 marksStart
O(1, 1.86)(2, 2.36)(3, 2.5)(4, 2.36)(5, 1.86)(6, 0)xyFigure 1

A gardener has a polytunnel, a long greenhouse with a curved plastic roof. At one end, the polytunnel is 6 m wide.

The height, yy metres, of the roof is measured at a point xx metres from one side of the polytunnel.

Figure 1 shows a plot of the cross-section of the polytunnel and the coordinate values (x,y)(x, y).
(a) Use the trapezium rule with all the yy values given in Figure 1 to estimate the cross-sectional area of the polytunnel.

(3 marks)

The polytunnel is 15 m long and is modelled as having the same cross-section along its whole length.
(b) Use the model and the answer to part (a) to estimate the volume of air inside the polytunnel, giving your answer in m3\text{m}^3 to 2 significant figures.

(2 marks)

Assuming the model,
(c) state, giving a reason for your answer, whether your answer for part (b) is an overestimate or an underestimate of the volume of air inside the polytunnel.

(1 mark)
●●●●●Level 46 marksStart
The table gives values of y=f(x)y = f(x).

x00.61.21.82.4y3.24.14.64.02.5\begin{array}{c|ccccc} x & 0 & 0.6 & 1.2 & 1.8 & 2.4 \\\\ \hline y & 3.2 & 4.1 & 4.6 & 4.0 & 2.5 \end{array}

Use the trapezium rule with all five values to estimate ∫02.4y dx\displaystyle\int_0^{2.4} y\,dx.
●●●●●Level 43 marksStart
(a) Use the trapezium rule with 4 strips to estimate ∫26ln⁡x dx\displaystyle\int_2^6 \ln x\,dx to 3 decimal places.
(b) State, with a reason, whether this under- or over-estimates the integral.
●●●●●Level 45 marksStart
I=∫131+x3  dxI = \int_1^3 \sqrt{1 + x^3}\;\mathrm{d}x

Figure 1 shows the region whose area is given by II.11.522.53123456xyFigure 1
(a) Use the trapezium rule with four strips of equal width to find an estimate for II, giving your answer to 44 decimal places. (4 marks)
(b) State, with a reason, whether your answer to part (a) is an overestimate or an underestimate of II. (2 marks)
●●●●●Level 46 marksStart
The table below shows corresponding values of xx and yy for y=log⁡5(4x)y = \log_5 (4x)

The values of yy are given to 3 decimal places as appropriate.

x0.50.7511.251.5y0.4310.6830.86111.113\begin{array}{c|ccccc} x & 0.5 & 0.75 & 1 & 1.25 & 1.5 \\ \hline y & 0.431 & 0.683 & 0.861 & 1 & 1.113 \end{array}
(a) Use the trapezium rule with all the values of yy in the table to find an estimate for

∫0.51.5log⁡5(4x) dx\int_{0.5}^{1.5} \log_5 (4x)\,dx

(3 marks)

Using your answer to part (a), and making your method clear, estimate
(b) (i) ∫0.51.5log⁡5(4x)2 dx\displaystyle\int_{0.5}^{1.5} \log_5 (4x)^2\,dx

(ii) ∫0.51.5log⁡5(100x) dx\displaystyle\int_{0.5}^{1.5} \log_5 (100x)\,dx

(3 marks)
●●●●●Level 46 marksStart
The table shows corresponding values of xx and yy for a continuous curve with equation y=f(x)y = \text{f}(x), where aa and bb are constants.

x00.511.522.5ya3.6b4.75.46.2\begin{array}{c|cccccc} x & 0 & 0.5 & 1 & 1.5 & 2 & 2.5 \\ \hline y & a & 3.6 & b & 4.7 & 5.4 & 6.2 \end{array}

An estimate for the area under the curve between x=0x = 0 and x=2.5x = 2.5 is found using the trapezium rule with all the yy values in the table.

Given that this estimate is 11.211.2
(a) show that

a+2b=11.2a + 2b = 11.2

(3 marks)

Given also that the sum of all the yy values in the table is 27.127.1
(b) find the value of aa and the value of bb.

(3 marks)
●●●●●Level 46 marksStart
(a) Use the substitution u=2x−3u = 2x - 3 to find

∫x2x−3  dx\displaystyle\int x\sqrt{2x - 3}\; dx,

giving your answer in terms of xx. (4 marks)

The finite region RR is bounded by the curve y=x2x−3y = x\sqrt{2x - 3}, the xx-axis and the lines x=2x = 2 and x=6x = 6.1234567510152025Rxy(b) Hence find the exact area of RR. (3 marks)
(c) Use the trapezium rule, with four strips of equal width, to estimate the area of RR. Explain, with reference to the shape of the curve, whether your estimate is an over-estimate or an under-estimate, and confirm that this is consistent with your answer to (b). (3 marks)
●●●●●Level 510 marksStart
The region bounded by the curve y=(x−6)ln⁡xy = (x - 6)\ln x and the xx-axis is shaded in the diagram below.Oxy(a) Use the trapezium rule with 55 ordinates to find an estimate for the area of the shaded region.

Give your answer correct to three significant figures.

(3 marks)
(b) Show that the exact area is given by

18ln⁡6−85418\ln 6 - \frac{85}{4}

Fully justify your answer.

(6 marks)
●●●●●Level 59 marksStart