Numerical solution of equations

The values generated by the iterative formula $x_{n+1} = \sqrt{\tfrac{1}{2}x_n^2 + 4x_n^{-3}}$, starting from $x_1 = 1.5$, approach $\alpha$.

Feb/March 2016

It is stated that $a$ is a positive constant for which $\int_0^a (1 + 2x + 3e^{3x})\,dx = 250$.

Feb/March 2017

Apply the trapezium rule with four intervals to estimate $\int_0^8 \ln(x + 2)\,dx$, giving the result to 3 significant figures.

Feb/March 2018

You are told that $\int_{-a}^{2a} 4e^{-2x}\,dx = 25$, with $a$ a positive constant.

Feb/March 2018

The curve is defined by $y = \dfrac{e^{2x}}{4x + 1}$, and $P$ is a point on it with $y$-coordinate $10$.

Feb/March 2019

The curve is given by $y = x^3 e^{0.2x}$ for $x \ge 0$. At the point $P$ on this curve, the gradient is $15$.

Feb/March 2020

If $2\ln(x+1)+\ln x=\ln(x+9)$, prove that $x=\sqrt{\frac{9}{x+2}}$.

Feb/March 2021

The curve is defined by $e^{2x}y - e^{y} = 100$.

Feb/March 2022

Given that $\int_1^a \left( \frac{4}{1+2x} + \frac{3}{x} \right)\, dx = \ln 10$, where $a$ is a constant greater than $1$.

Feb/March 2023

The diagram depicts a section of the curve with equation $y = \dfrac{x^3}{x + 2}$. At the point $P$, the gradient of the curve equals $6$.

Feb/March 2024

Let the x-coordinates of the intersection points be represented by $\alpha$ and $\beta$, with $\alpha < \beta$.

Feb/March 2025

Show by drawing a suitable pair of graphs that the equation $e^{2x} = 2 - x$ has only one root.

May/June 2010

By drawing a suitable pair of graphs, demonstrate that the equation $\ln x = 2 - x^2$ has only one root.

May/June 2010

By drawing an appropriate pair of graphs, show that the equation $\ln x = 2 - x^2$ has just one root.

May/June 2010

The variables $x$ and $y$ obey the equation $y = Kx^m$, with $K$ and $m$ as constants. The graph of $\ln y$ plotted against $\ln x$ is a straight line that passes through the points $(0, 2.0)$ and $(6, 10.2)$, as illustrated in the diagram.

May/June 2011

By using a suitable pair of graphs, show that the equation $e^{2x} = 14 - x^2$ has exactly two real roots.

May/June 2011

The sequence $x_1, x_2, x_3, \ldots$ defined by $x_1 = 1$, $x_{n+1} = \frac{1}{2}\sqrt[3]{x_n^2 + 6}$ tends to the value $\alpha$.

May/June 2011

The sequence $x_1, x_2, x_3, \ldots$ specified by $x_1 = 1$, $x_{n+1} = \frac{1}{2}\sqrt[3]{x_n^2 + 6}$ has limit $\alpha$.

May/June 2011

A curve is defined by the parametric equations $x = \dfrac{1}{(2t + 1)^2}$, $y = \sqrt{t + 2}$. The point $P$ on the curve is associated with parameter $p$, and the gradient at $P$ is stated to be $-1$.

May/June 2012

The diagram depicts the curve $y = \frac{\sin 2x}{x + 2}$ for $0 \leq x \leq \frac{1}{2}\pi$. The $x$-coordinate of the highest point $M$ is called $\alpha$.

May/June 2012

A curve is defined by parametric equations $x = \frac{1}{(2t + 1)^2}$ and $y = \sqrt{t + 2}$. The point $P$ on the curve is associated with parameter $p$, and the gradient at $P$ is stated to be $-1$.

May/June 2012

By sketching an appropriate pair of graphs, show that the equation $\cot x = 4x - 2$, where $x$ is in radians, has a single root for $0 \leq x \leq \frac{\pi}{2}$.

May/June 2013

Using a suitable pair of sketches, demonstrate that the equation $3e^x = 8 - 2x$ has a single root.

May/June 2013

By drawing an appropriate pair of graphs, show that the equation $\cot x = 4x - 2$, where $x$ is in radians, has only one root for $0 \le x \le \tfrac{1}{2}\pi$.

May/June 2013

By sketching an appropriate pair of graphs, demonstrate that the equation $3\ln x = 15 - x^3$ has exactly one real root.

May/June 2014

The variables $x$ and $y$ are related by the equation $y = K(2^{px})$, where $K$ and $p$ are constants. The graph of $\ln y$ plotted against $x$ is a straight line that goes through $(1.35, 1.87)$ and $(3.35, 3.81)$, as shown in the diagram.

May/June 2014

It is stated that $\int_0^a \left(\frac{1}{2}e^{3x} + x^2\right)\,dx = 10$, where $a$ is a positive constant.

May/June 2014

The variables $x$ and $y$ are related by $y = K 2^{px}$, where $K$ and $p$ are constants. The plot of $\ln y$ against $x$ forms a straight line that goes through the points $(1.35, 1.87)$ and $(3.35, 3.81)$, as the diagram shows.

May/June 2014

You are told that $\int_{0}^{a} \left( \frac{1}{2} e^{3x} + x^2 \right) \, dx = 10$, with $a$ a positive constant.

May/June 2014

If $\int_0^a (3e^{\frac{1}{2}x} + 1)\,dx = 10$, demonstrate that the positive constant $a$ obeys $a = 2 \ln\left(\frac{16 - a}{6}\right)$.

May/June 2015

Use sketches of two suitable graphs to show that the equation $|3x| = 16 - x^4$ has two real roots.

May/June 2015

By sketching an appropriate pair of graphs, show that the equation $|3x| = 16 - x^4$ has two real roots.

May/June 2015

The curve is defined by $y = \frac{3x^2}{x^2 + 4}$. At the point on the curve where the $x$-coordinate is positive and equal to $p$, the gradient is $\tfrac{1}{2}$.

May/June 2016

The curve is defined by $y = 6xe^{\frac{1}{3}x}$. At the point on it where the $x$-coordinate is $p$, the gradient is $40$.

May/June 2016

The curve is given by $y = 6x e^{\frac{1}{3}x}$. At the point on the curve where the $x$-coordinate is $p$, the gradient of the curve equals $40$.

May/June 2016

From $5^x = 3^{4y}$, apply logarithms to establish that $y = mx$.

May/June 2017

It is given that $\int_{0}^{a} 4e^{\frac{1}{2}x + 3} \, dx = 835$.

May/June 2017

The terms generated by the iterative rule $x_{n+1} = \frac{2x_n^2 + x_n + 9}{(x_n + 1)^2}$, beginning with $x_1 = 2$, settle to $\alpha$.

May/June 2017

By sketching an appropriate pair of graphs, show that the equation $x^3 = 11 - 2x$ has only one real root.

May/June 2017

The variables $x$ and $y$ are related by $y = \frac{K}{a^{2x}}$, where $K$ and $a$ are constants. A plot of $\ln y$ against $x$ is a straight line that passes through the points $(0.6, 1.81)$ and $(1.4, 1.39)$, as the diagram shows.

May/June 2017

By drawing appropriate graphs, demonstrate that the equation $x^3 = 11 - 2x$ has exactly one real root.

May/June 2017

The variables $x$ and $y$ are related by $y = \frac{K}{a^{2x}}$, with $K$ and $a$ as constants. The plot of $\ln y$ against $x$ is a straight line that goes through the points $(0.6, 1.81)$ and $(1.4, 1.39)$, as shown in the diagram.

May/June 2017

The diagram presents the curve whose equation is $y = \frac{5\ln x}{2x + 1}$. It meets the $x$-axis at $P$ and has a maximum at $M$.

May/June 2018

You are given that $\int_{0}^{a} (1 + e^{\frac{1}{2}x})^{2} \, dx = 10$, where $a$ is a positive constant.

May/June 2018

The information provided is that $\int_0^a (1 + e^{\frac{1}{2}x})^2\,dx = 10$, with $a$ a positive constant.

May/June 2018

The diagram depicts the curve whose equation is $y = \frac{8 + x^3}{2 - 5x}$. The maximum point is labelled $M$.

May/June 2019

The diagram depicts the curve given by the parametric equations $x = 3t - 6e^{-2t}$, $y = 4t^2 e^{-t}$, for $0 \le t \le 2$. At the point $P$ on this curve, the $y$-coordinate is $1$.

May/June 2019

The diagram represents the curve with parametric equations $x = 3t - 6e^{-2t}$, $y = 4t^2 e^{-t}$, for $0 \leq t \leq 2$. Point $P$ lies on the curve, and its $y$-coordinate is $1$.

May/June 2019

The diagram illustrates a section of the curve whose equation is $y = x^3 \cos 2x$. This curve reaches a maximum at the point $M$.

May/June 2020

It is stated that $\int_0^a \left(\frac{4}{2x + 1} + 8x\right) \, dx = 10$, with $a$ a positive constant.

May/June 2020

We are told that $\int_0^a \left( \frac{4}{2x + 1} + 8x \right) \, dx = 10$, where $a$ is a positive constant.

May/June 2020

The diagram illustrates the curve given by $y = \frac{3x + 2}{\ln x}$. This curve has a minimum point $M$.

May/June 2021

The diagram depicts the curve defined by the parametric equations $x = 4t + e^{2t}$, $y = 6t\sin 2t$, for $0 \leq t \leq 1$. The point $P$ on the curve is assigned parameter $p$ and has $y$-coordinate $3$.

May/June 2021

The diagram represents the curve whose parametric equations are $x = 4t + e^{2t}$, $y = 6t\sin 2t$, where $0 \leq t \leq 1$. Point $P$ lies on the curve and has parameter $p$ and $y$-coordinate $3$.

May/June 2021

Using sketches of $y = |5 - 2x|$ and $y = 3 \ln x$ on one diagram, show that the equation $|5 - 2x| = 3 \ln x$ has exactly two roots.

May/June 2022

Use the trapezium rule with three intervals to show that the value of $\int_{1}^{4} \ln x\,dx$ comes out as approximately $\ln 12$.

May/June 2022

The diagram illustrates the curve $y = 3e^{2x-1}$. The shaded region is enclosed by the curve and the lines $x = a$, $x = a + 1$ and $y = 0$, with $a$ a constant. The area of the shaded region is stated to be $120$ square units.

May/June 2022

The figure displays the curve $y = 3e^{2x-1}$. The shaded area is enclosed by the curve together with the lines $x = a$, $x = a + 1$ and $y = 0$, where $a$ is a constant. The area of this shaded region is given as $120$ square units.

May/June 2022

You are told that $\int_0^a (3e^{2x} - 1)\,dx = 12$, with $a$ a positive constant.

May/June 2023

The diagram depicts the graph of $y = 3 - e^{-\frac{1}{2}x}$. On the same diagram, sketch $y = |5x - 4|$, and demonstrate that $3 - e^{-\frac{1}{2}x} = |5x - 4|$ has precisely two real roots. The two roots of $3 - e^{-\frac{1}{2}x} = |5x - 4|$ are labelled $\alpha$ and $\beta$, with $\alpha < \beta$.

May/June 2023

The diagram gives the graph of $y = 3 - e^{-\frac{1}{2}x}$.

May/June 2023

A curve is defined by $y = \frac{1 + e^{2x}}{1 + 3x}$. It has one and only one stationary point $P$.

May/June 2024

The diagram represents the curve given by the equation $y = \frac{\ln(2x + 1)}{x + 3}$. This curve has a maximum point $M$.

May/June 2024

The graph depicts the curve with equation $y = \frac{\ln(2x + 1)}{x + 3}$. This curve includes a maximum point $M$.

May/June 2024

On one diagram, sketch the graphs of $y = 3e^{-2x}$ and $y = \sec x$ for values of $x$ such that $0 \leq x < \frac{1}{2}\pi$.

May/June 2025

The diagram displays sections of the curves given by $y = 4e^{-2x}$ and $y = 1 + 0.5\sin 3x$. $P$ marks one point where the curves intersect, and the shaded part is enclosed by the two curves together with the $y$-axis.

May/June 2025

The diagram displays sections of the curves $y = 4e^{-2x}$ and $y = 1 + 0.5\sin 3x$. Point $P$ is where the curves intersect, and the shaded area is enclosed by the two curves and the $y$-axis.

May/June 2025

The diagram displays portions of the curves with equations $y = 4e^{-2x}$ and $y = 1 + 0.5\sin 3x$. Point $P$ is where the curves meet, and the shaded part is enclosed by the two curves and the $y$-axis.

May/June 2025

The curve with equation $y = \frac{6}{x^2}$ meets the line $y = x + 1$ at the point $P$.

Oct/Nov 2010

The curve given by $y = \frac{6}{x^2}$ meets the line $y = x + 1$ at the point $P$.

Oct/Nov 2010

The values produced by the iterative relation $x_{n+1} = \frac{7x_n}{8} + \frac{5}{2x_n^4}$, starting from $x_1 = 1.7$, approach $\alpha$.

Oct/Nov 2010

Verify by calculation that the cubic equation $x^3 - 2x^2 + 5x - 3 = 0$ has a root between $x = 0.7$ and $x = 0.8$.

Oct/Nov 2011

The diagram presents the curve $y = x - 2\ln x$ together with its minimum point $M$.

Oct/Nov 2011

By sketching an appropriate pair of graphs, show that the equation $\frac{1}{x} = \sin x$, where $x$ is measured in radians, has just one root for $0 < x \leq \tfrac{1}{2}\pi$.

Oct/Nov 2011

The diagram depicts the curve $y = (x - 4)e^{\frac{1}{2}x}$. At the point $P$, the curve’s gradient is $3$.

Oct/Nov 2011

The diagram depicts the curve $y = \cos x$, for $0 \le x \le \frac{\pi}{2}$. A rectangle $OABC$ is shown, with $B$ lying on the curve and having $x$-coordinate $\theta$, while $A$ and $C$ lie on the axes, as illustrated. The shaded region $R$ is enclosed by the curve and by the lines $x = \theta$ and $y = 0$.

Oct/Nov 2012

The diagram illustrates the section of the curve $y = \sqrt{(2 - \sin x)}$ for $0 \le x \le \frac{1}{2}\pi$.

Oct/Nov 2012

The sketch displays the curve $y = \cos x$, for $0 \le x \le \frac{1}{2}\pi$. A rectangle $OABC$ has been drawn, with $B$ located on the curve at the point whose $x$-coordinate is $\theta$, and $A$ and $C$ lying on the axes, as shown. The shaded region $R$ is enclosed by the curve and by the lines $x = \theta$ and $y = 0$.

Oct/Nov 2012

The diagram presents the curve $y = x^4 + 2x - 9$. This curve crosses the positive $x$-axis at the point $P$.

Oct/Nov 2013

The diagram depicts part of the curve $y = 8x + \frac{1}{2}e^x$. The shaded region $R$ is enclosed by the curve and the lines $x = 0$, $y = 0$ and $x = a$, with $a$ positive. The area of $R$ is $\frac{1}{2}$.

Oct/Nov 2013

The diagram displays the curve $y = x^4 + 2x - 9$. This curve meets the positive $x$-axis at point $P$.

Oct/Nov 2013

The variables $x$ and $y$ are linked by $y = ab^{x}$, with $a$ and $b$ constant. A graph of $\ln y$ against $x$ is a straight line that goes through $(0.75, 1.70)$ and $(1.53, 2.18)$, as the diagram shows.

Oct/Nov 2014

The diagram displays a segment of the curve $y = \frac{x^{2}}{1 + e^{3x}}$ together with its maximum point $M$. Let the $x$-coordinate of $M$ be $m$.

Oct/Nov 2014

The polynomial $p(x)$ is given by $p(x) = x^4 - 3x^3 + 3x^2 - 25x + 48$. The diagram depicts the curve $y = p(x)$, which meets the $x$-axis at $(\alpha, 0)$ and $(3, 0)$.

Oct/Nov 2014

The variables $x$ and $y$ are related by the equation $y = a(b^x)$, where $a$ and $b$ are constants. A graph of $\ln y$ plotted against $x$ is a straight line that passes through the points $(0.75, 1.70)$ and $(1.53, 2.18)$, as shown in the diagram.

Oct/Nov 2014

The diagram depicts a section of the curve $y = \frac{x^2}{1 + e^{3x}}$ together with its maximum point $M$. The $x$-coordinate of $M$ is written as $m$.

Oct/Nov 2014

By drawing an appropriate pair of graphs, show that the equation $\ln x = 4 - \frac{1}{2}x$ has exactly one real root, $\alpha$.

Oct/Nov 2015

Starting from $x_1 = 2$, the values generated by the iterative formula $x_{n+1} = 2 + \frac{4}{x_n^2 + 2x_n + 4}$ approach $\alpha$.

Oct/Nov 2015

The variables $x$ and $y$ obey the relation $y = Kx^m$, where $K$ and $m$ are constants. The graph of $\ln y$ plotted against $\ln x$ is a straight line running through the points $(0.22, 3.96)$ and $(1.32, 2.43)$, as shown in the diagram.

Oct/Nov 2015

You are told that $\int_0^a (3e^{3x} + 5e^x)\,dx = 100$, with $a$ a positive constant.

Oct/Nov 2015

The positive constant $a$ is such that $\int_{-a}^{a} \left(4e^{2x} + 5\right)\,dx = 100$.

Oct/Nov 2016

The graph illustrates the curve $y = \frac{4 \ln x}{x^2 + 1}$ together with its stationary point $M$. Let the $x$-coordinate of $M$ be $m$.

Oct/Nov 2016

The values generated by the iterative relation $x_{n+1} = \frac{4}{x_n} + \frac{2x_n}{3}$, starting from $x_1 = 2$, approach $\alpha$.

Oct/Nov 2016

x and y are linked by $y = Kx^p$, where $K$ and $p$ are constants. As shown in the diagram, the graph of $\ln y$ against $\ln x$ is a straight line that passes through $(1.28,\,3.69)$ and $(2.11,\,4.81)$.

Oct/Nov 2016

The variable $x$ is such that $1.3^{2x} < 80$ and $|3x - 1| > |3x - 10|$.

Oct/Nov 2017

The sketch depicts the curve $y = x^2 + 3x + 1 + 5\cos\frac{1}{2}x$. It cuts the $y$-axis at the point $P$, and the gradient of the curve there is $m$. The point $Q$ lies on the curve and has $x$-coordinate $q$, with gradient $-m$ at $Q$.

Oct/Nov 2017

The curve is given by $y = \tan\left(\tfrac{1}{2}x\right) + 3\sin\left(\tfrac{1}{2}x\right)$. It has a stationary point $M$ for values of $x$ in the range $\pi < x < 2\pi$.

Oct/Nov 2017

The diagram depicts the curve $y = 4e^{-2x}$ together with a straight line. The curve intersects the $y$-axis at $P$. The straight line intersects the $y$-axis at the point $(0, 9)$, and its gradient matches the gradient of the curve at $P$. The line cuts the curve at two points, including $Q$, as indicated.

Oct/Nov 2017

The condition is that $x$ satisfies $1.3^{2x} < 80$ and $|3x - 1| > |3x - 10|$.

Oct/Nov 2017

The diagram depicts the curve $y = x^2 + 3x + 1 + 5\cos\frac{1}{2}x$. This curve meets the $y$-axis at $P$, where its gradient is $m$. Point $Q$ lies on the curve, has $x$-coordinate $q$, and the gradient there is $-m$.

Oct/Nov 2017