OCR MEI C4 (Core Mathematics 4) 2009 June

Express \(4\cos\theta - \sin\theta\) in the form \(R\cos(\theta + \alpha)\), where \(R > 0\) and \(0 < \alpha < \frac{1}{2}\pi\). Hence solve the equation \(4\cos\theta - \sin\theta = 3\), for \(0 \leq \theta \leq 2\pi\). [7]

Using partial fractions, find \(\int \frac{x}{(x+1)(2x+1)} \, dx\). [7]

A curve satisfies the differential equation \(\frac{dy}{dx} = 3x^2y\), and passes through the point \((1, 1)\). Find \(y\) in terms of \(x\). [4]

The part of the curve \(y = 4 - x^2\) that is above the \(x\)-axis is rotated about the \(y\)-axis. This is shown in Fig. 4. Find the volume of revolution produced, giving your answer in terms of \(\pi\). [5] \includegraphics{figure_4}

A curve has parametric equations $$x = at^3, \quad y = \frac{a}{1+t^2},$$ where \(a\) is a constant. Show that \(\frac{dy}{dx} = \frac{-2}{3t(1+t^2)^2}\). Hence find the gradient of the curve at the point \((a, \frac{1}{2}a)\). [7]

Given that \(\cos\text{ec}^2\theta - \cot\theta = 3\), show that \(\cot^2\theta - \cot\theta - 2 = 0\). Hence solve the equation \(\cos\text{ec}^2\theta - \cot\theta = 3\) for \(0° \leq \theta \leq 180°\). [6]

When a light ray passes from air to glass, it is deflected through an angle. The light ray ABC starts at point A \((1, 2, 2)\), and enters a glass object at point B \((0, 0, 2)\). The surface of the glass object is a plane with normal vector \(\mathbf{n}\). Fig. 7 shows a cross-section of the glass object in the plane of the light ray and \(\mathbf{n}\). \includegraphics{figure_7}

1. Find the vector \(\overrightarrow{AB}\) and a vector equation of the line AB. [2]

The surface of the glass object is a plane with equation \(x + z = 2\). AB makes an acute angle \(\theta\) with the normal to this plane.

1. Write down the normal vector \(\mathbf{n}\), and hence calculate \(\theta\), giving your answer in degrees. [5]

The line BC has vector equation \(\mathbf{r} = \begin{pmatrix} 0 \\ 0 \\ 2 \end{pmatrix} + \mu \begin{pmatrix} -2 \\ -2 \\ -1 \end{pmatrix}\). This line makes an acute angle \(\phi\) with the normal to the plane.

1. Show that \(\phi = 45°\). [3]
2. Snell's Law states that \(\sin\theta = k\sin\phi\), where \(k\) is a constant called the refractive index. Find \(k\). [2]

The light ray leaves the glass object through a plane with equation \(x + z = -1\). Units are centimetres.

1. Find the point of intersection of the line BC with the plane \(x + z = -1\). Hence find the distance the light ray travels through the glass object. [5]

Archimedes, about 2200 years ago, used regular polygons inside and outside circles to obtain approximations for \(\pi\).

1. Fig. 8.1 shows a regular 12-sided polygon inscribed in a circle of radius 1 unit, centre O. AB is one of the sides of the polygon. C is the midpoint of AB. Archimedes used the fact that the circumference of the circle is greater than the perimeter of this polygon. \includegraphics{figure_8.1}
   1. Show that AB = \(2\sin 15°\). [2]
   2. Use a double angle formula to express \(\cos 30°\) in terms of \(\sin 15°\). Using the exact value of \(\cos 30°\), show that \(\sin 15° = \frac{1}{4}\sqrt{2 - \sqrt{3}}\). [4]
   3. Use this result to find an exact expression for the perimeter of the polygon. Hence show that \(\pi > 6\sqrt{2 - \sqrt{3}}\). [2]
2. In Fig. 8.2, a regular 12-sided polygon lies outside the circle of radius 1 unit, which touches each side of the polygon. F is the midpoint of DE. Archimedes used the fact that the circumference of the circle is less than the perimeter of this polygon. \includegraphics{figure_8.2}
   1. Show that DE = \(2\tan 15°\). [2]
   2. Let \(t = \tan 15°\). Use a double angle formula to express \(\tan 30°\) in terms of \(t\). Hence show that \(t^2 + 2\sqrt{3}t - 1 = 0\). [3]
   3. Solve this equation, and hence show that \(\pi < 12(2 - \sqrt{3})\). [4]
3. Use the results in parts (i)(C) and (ii)(C) to establish upper and lower bounds for the value of \(\pi\), giving your answers in decimal form. [2]