Past Exam Questions

The equation of a curve is \(y = x \ln(8 - x)\). The gradient of the curve is equal to 1 at only one point, when \(x = a\).

(i) Show that \(a\) satisfies the equation \(x = 8 - \frac{8}{\ln(8 - x)}\).

(ii) Verify by calculation that \(a\) lies between 2.9 and 3.1.

(iii) Use an iterative formula based on the equation in part (i) to determine \(a\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

The curve with equation \(y = x^2 \cos \frac{1}{2}x\) has a stationary point at \(x = p\) in the interval \(0 < x < \pi\).

1. Show that \(p\) satisfies the equation \(\tan \frac{1}{2}p = \frac{4}{p}\).
2. Verify by calculation that \(p\) lies between 2 and 2.5.
3. Use the iterative formula \(p_{n+1} = 2 \arctan \left( \frac{4}{p_n} \right)\) to determine the value of \(p\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

The diagram shows the curve \(y = \csc x\) for \(0 < x < \pi\) and part of the curve \(y = e^{-x}\). When \(x = a\), the tangents to the curves are parallel.

(i) By differentiating \(\frac{1}{\sin x}\), show that if \(y = \csc x\) then \(\frac{dy}{dx} = -\csc x \cot x\). [3]

(ii) By equating the gradients of the curves at \(x = a\), show that \(a = \arctan \left( \frac{e^a}{\sin a} \right)\). [2]

(iii) Verify by calculation that \(a\) lies between 1 and 1.5. [2]

(iv) Use an iterative formula based on the equation in part (ii) to determine \(a\) correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3]

A curve has parametric equations

\(x = t^2 + 3t + 1, \quad y = t^4 + 1.\)

The point \(P\) on the curve has parameter \(p\). It is given that the gradient of the curve at \(P\) is 4.

1. Show that \(p = \sqrt[3]{2p + 3}\).
2. Verify by calculation that the value of \(p\) lies between 1.8 and 2.0.
3. Use an iterative formula based on the equation in part (i) to find the value of \(p\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

The diagram shows part of the curve with parametric equations

\(x = 2 \ln(t + 2)\), \(y = t^3 + 2t + 3\).

1. Find the gradient of the curve at the origin. [5]
2. At the point \(P\) on the curve, the value of the parameter is \(p\). It is given that the gradient of the curve at \(P\) is \(\frac{1}{2}\).
1. Show that \(p = \frac{1}{3p^2 + 2} - 2\). [1]
2. By first using an iterative formula based on the equation in part (a), determine the coordinates of the point \(P\). Give the result of each iteration to 5 decimal places and each coordinate of \(P\) correct to 2 decimal places. [4]

The diagram shows the curves \(y = e^{2x-3}\) and \(y = 2 \ln x\). When \(x = a\) the tangents to the curves are parallel.

1. Show that \(a\) satisfies the equation \(a = \frac{1}{2}(3 - \ln a)\). [3]
2. Verify by calculation that this equation has a root between 1 and 2. [2]
3. Use the iterative formula \(a_{n+1} = \frac{1}{2}(3 - \ln a_n)\) to calculate \(a\) correct to 2 decimal places, showing the result of each iteration to 4 decimal places. [3]

The diagram shows the curve \(y = e^{-\frac{1}{2}x^2} \sqrt{(1 + 2x^2)}\) for \(x \geq 0\), and its maximum point \(M\).

(i) Find the exact value of the \(x\)-coordinate of \(M\). [4]

(ii) The sequence of values given by the iterative formula \(x_{n+1} = \sqrt{(\ln(4 + 8x_n^2))}\), with initial value \(x_1 = 2\), converges to a certain value \(\alpha\). State an equation satisfied by \(\alpha\) and hence show that \(\alpha\) is the \(x\)-coordinate of a point on the curve where \(y = 0.5\). [3]

(iii) Use the iterative formula to determine \(\alpha\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

The curve \(y = \frac{\ln x}{x + 1}\) has one stationary point.

(i) Show that the \(x\)-coordinate of this point satisfies the equation \(x = \frac{x + 1}{\ln x}\), and that this \(x\)-coordinate lies between 3 and 4.

(ii) Use the iterative formula \(x_{n+1} = \frac{x_n + 1}{\ln x_n}\) to determine the \(x\)-coordinate correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

The diagram shows the curve \(y = \frac{\sin x}{x}\) for \(0 < x \leq 2\pi\), and its minimum point \(M\).

(i) Show that the \(x\)-coordinate of \(M\) satisfies the equation \(x = \tan x\).

(ii) The iterative formula \(x_{n+1} = \arctan(x_n) + \pi\) can be used to determine the \(x\)-coordinate of \(M\). Use this formula to determine the \(x\)-coordinate of \(M\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

The equation of a curve is \(y = \ln x + \frac{2}{x}\), where \(x > 0\).

(i) Find the coordinates of the stationary point of the curve and determine whether it is a maximum or a minimum point.

(ii) The sequence of values given by the iterative formula \(x_{n+1} = \frac{2}{3 - \ln x_n}\), with initial value \(x_1 = 1\), converges to \(\alpha\). State an equation satisfied by \(\alpha\), and hence show that \(\alpha\) is the \(x\)-coordinate of a point on the curve where \(y = 3\).

(iii) Use this iterative formula to find \(\alpha\) correct to 2 decimal places, showing the result of each iteration.

The curve with equation \(y = \frac{x^3}{e^x - 1}\) has a stationary point at \(x = p\), where \(p > 0\).

(a) Show that \(p = 3(1 - e^{-p})\).

(b) Verify by calculation that \(p\) lies between 2.5 and 3.

(c) Use an iterative formula based on the equation in part (a) to determine \(p\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

The equation of a curve is \(y = \frac{x}{\cos^2 x}\), for \(0 \leq x < \frac{1}{2}\pi\). At the point where \(x = a\), the tangent to the curve has gradient equal to 12.

(a) Show that \(a = \cos^{-1} \left( \sqrt[3]{\frac{\cos a + 2a \sin a}{12}} \right)\).

(b) Verify by calculation that \(a\) lies between 0.9 and 1.

(c) Use an iterative formula based on the equation in part (a) to determine \(a\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

The curve \(y = x \sqrt{\sin x}\) has one stationary point in the interval \(0 < x < \pi\), where \(x = a\) (see diagram).

(a) Show that \(\tan a = -\frac{1}{2}a\).

(b) Verify by calculation that \(a\) lies between 2 and 2.5.

(c) Show that if a sequence of values in the interval \(0 < x < \pi\) given by the iterative formula \(x_{n+1} = \pi - \arctan\left(\frac{1}{2}x_n\right)\) converges, then it converges to \(a\), the root of the equation in part (a).

(d) Use the iterative formula given in part (c) to determine \(a\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

The equation of a curve is \(y = \sqrt{\tan x}\), for \(0 \leq x < \frac{1}{2}\pi\).

(a) Express \(\frac{dy}{dx}\) in terms of \(\tan x\), and verify that \(\frac{dy}{dx} = 1\) when \(x = \frac{1}{4}\pi\).

The value of \(\frac{dy}{dx}\) is also 1 at another point on the curve where \(x = a\), as shown in the diagram.

(b) Show that \(t^3 + t^2 + 3t - 1 = 0\), where \(t = \tan a\).

(c) Use the iterative formula \(a_{n+1} = \arctan \left( \frac{1}{3} (1 - \tan^2 a_n - \tan^3 a_n) \right)\) to determine \(a\) correct to 2 decimal places, giving the result of each iteration to 4 decimal places.

The diagram shows the curve \(y = \frac{\arctan x}{\sqrt{x}}\) and its maximum point \(M\) where \(x = a\).

(a) Show that \(a\) satisfies the equation \(a = \tan \left( \frac{2a}{1 + a^2} \right)\).

(b) Verify by calculation that \(a\) lies between 1.3 and 1.5.

(c) Use an iterative formula based on the equation in part (a) to determine \(a\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

Let \(f(x) = \frac{e^{2x} + 1}{e^{2x} - 1}\), for \(x > 0\).

(a) The equation \(x = f(x)\) has one root, denoted by \(a\). Verify by calculation that \(a\) lies between 1 and 1.5. [2]

(b) Use an iterative formula based on the equation in part (a) to determine \(a\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

(c) Find \(f'(x)\). Hence find the exact value of \(x\) for which \(f'(x) = -8\). [6]

The curve with equation \(y = e^{-2x} \ln(x-1)\) has a stationary point when \(x = p\).

1. Show that \(p\) satisfies the equation \(x = 1 + \exp\left( \frac{1}{2(x-1)} \right)\), where \(\exp(x)\) denotes \(e^x\).
2. Verify by calculation that \(p\) lies between 2.2 and 2.6.
3. Use an iterative formula based on the equation in part (i) to determine \(p\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

The diagram shows the curves \(y = 4 \, \cos \frac{1}{2} x\) and \(y = \frac{1}{4-x}\), for \(0 \leq x < 4\). When \(x = a\), the tangents to the curves are perpendicular.

1. Show that \(a = 4 - \sqrt{2 \sin \frac{1}{2} a}\).
2. Verify by calculation that \(a\) lies between 2 and 3.
3. Use an iterative formula based on the equation in part (i) to determine \(a\) correct to 3 decimal places. Give the result of each iteration to 5 decimal places.

The constant a is such that \(\int_0^a xe^{-2x} \, dx = \frac{1}{8}\).

(a) Show that \(a = \frac{1}{2} \ln(4a + 2)\).

(b) Verify by calculation that a lies between 0.5 and 1.

(c) Use an iterative formula based on the equation in (a) to determine a correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

It is given that \(\int_{1}^{a} \ln(2x) \, dx = 1\), where \(a > 1\).

(i) Show that \(a = \frac{1}{2} \exp \left( 1 + \frac{\ln 2}{a} \right)\), where \(\exp(x)\) denotes \(e^x\).

(ii) Use the iterative formula \(a_{n+1} = \frac{1}{2} \exp \left( 1 + \frac{\ln 2}{a_n} \right)\) to determine the value of \(a\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.