SPS SPS SM Pure (SPS SM Pure) 2021 May

1. For a small angle \(\theta\), where \(\theta\) is in radians, show that \(2\cos\theta + (1 - \tan\theta)^2 \approx 3 - 2\theta\). [3]
2. Hence determine an approximate solution to \(2\cos\theta + (1 - \tan\theta)^2 = 28\sin\theta\). [2]

Solve the equation \(|2x - 1| = |x + 3|\). [3]

Solve the equation \(2^{4x-1} = 3^{5-2x}\), giving your answer in the form \(x = \frac{\log_{10} a}{\log_{10} b}\). [6]

A sequence of transformations maps the curve \(y = e^x\) to the curve \(y = e^{2x+3}\). Give details of these transformations. [3]

A curve has equation \(x^3 - 3x^2y + y^2 + 1 = 0\).

1. Show that \(\frac{dy}{dx} = \frac{6xy - 3x^2}{2y - 3x^2}\). [4]
2. Find the equation of the normal to the curve at the point \((1, 2)\). [4]

In this question you must show detailed reasoning. A circle touches the lines \(y = \frac{1}{2}x\) and \(y = 2x\) at \((6, 3)\) and \((3, 6)\) respectively. \includegraphics{figure_6} Find the equation of the circle. [7]

It is given that there is exactly one value of \(x\), where \(0 < x < \pi\), that satisfies the equation $$3\tan 2x - 8\tan x = 4.$$

1. Show that \(t = \sqrt[3]{\frac{1}{2} + \frac{1}{3}t - \frac{1}{3}t^2}\), where \(t = \tan x\). [3]
2. Show by calculation that the value of \(t\) satisfying the equation in part (i) lies between 0.7 and 0.8. [2]
3. Use an iterative process based on the equation in part (i) to find the value of \(t\) correct to 4 significant figures. Use a starting value of 0.75 and show the result of each iteration. [3]
4. Solve the equation \(3\tan 4y - 8\tan 2y = 4\) for \(0 < y < \frac{1}{4}\pi\). [2]

Find the general solution of the differential equation $$(2x^3 - 3x^2 - 11x + 6)\frac{dy}{dx} = y(20x - 35).$$ Give your answer in the form \(y = f(x)\). [9]

1. Show that the two non-stationary points of inflection on the curve \(y = \ln(1 + 4x^2)\) are at \(x = \pm\frac{1}{2}\). [6]

\includegraphics{figure_9} The diagram shows the curve \(y = \ln(1 + 4x^2)\). The shaded region is bounded by the curve and a line parallel to the \(x\)-axis which meets the curve where \(x = \frac{1}{2}\) and \(x = -\frac{1}{2}\).

1. Show that the area of the shaded region is given by $$\int_0^{\ln 2} \sqrt{e^y - 1} \, dy.$$ [3]
2. Show that the substitution \(e^y = \sec^2\theta\) transforms the integral in part (ii) to \(\int_0^{\frac{\pi}{4}} 2\tan^2\theta \, d\theta\). [2]
3. Hence find the exact area of the shaded region. [3]