OCR MEI C4 (Core Mathematics 4) 2012 January

Express \(\frac{x+1}{x^2(2x-1)}\) in partial fractions. [5]

Solve, correct to 2 decimal places, the equation \(\cot 2\theta = 3\) for \(0° < \theta < 180°\). [4]

Express \(3\sin x + 2\cos x\) in the form \(R\sin(x + \alpha)\), where \(R > 0\) and \(0 < \alpha < \frac{\pi}{2}\). Hence find, correct to 2 decimal places, the coordinates of the maximum point on the curve \(y = f(x)\), where $$f(x) = 3\sin x + 2\cos x, \quad 0 < x < \pi.$$ [7]

1. Complete the table of values for the curve \(y = \sqrt{\cos x}\).

   \(x\)	0	\(\frac{\pi}{6}\)	\(\frac{\pi}{4}\)	\(\frac{3\pi}{8}\)	\(\frac{\pi}{2}\)
   \(y\)		0.9612	0.8409

   Hence use the trapezium rule with strip width \(h = \frac{\pi}{8}\) to estimate the value of the integral \(\int_0^{\frac{\pi}{2}} \sqrt{\cos x} \, dx\), giving your answer to 3 decimal places. [3] Fig. 4 shows the curve \(y = \sqrt{\cos x}\) for \(0 \leq x \leq \frac{\pi}{2}\). \includegraphics{figure_4}
2. State, with a reason, whether the trapezium rule with a strip width of \(\frac{\pi}{16}\) would give a larger or smaller estimate of the integral. [1]

Verify that the vector \(2\mathbf{i} - \mathbf{j} + 4\mathbf{k}\) is perpendicular to the plane through the points A(2, 0, 1), B(1, 2, 2) and C(0, -4, 1). Hence find the cartesian equation of the plane. [5]

Given the binomial expansion \((1 + qx)^p = 1 - x + 2x^2 + \ldots\), find the values of \(p\) and \(q\). Hence state the set of values of \(x\) for which the expansion is valid. [6]

Show that the straight lines with equations \(\mathbf{r} = \begin{pmatrix} 4 \\ 2 \\ 4 \end{pmatrix} + \lambda \begin{pmatrix} 3 \\ 0 \\ 1 \end{pmatrix}\) and \(\mathbf{r} = \begin{pmatrix} -1 \\ 4 \\ 9 \end{pmatrix} + \mu \begin{pmatrix} -1 \\ 1 \\ 3 \end{pmatrix}\) meet. Find their point of intersection. [5]

Fig. 8 shows a cross-section of a car headlight whose inside reflective surface is modelled, in suitable units, by the curve $$x = 2t^2, y = 4t, \quad -\sqrt{2} < t < \sqrt{2}.$$ P\((2t^2, 4t)\) is a point on the curve with parameter \(t\). TS is the tangent to the curve at P, and PR is the line through P parallel to the \(x\)-axis. Q is the point (2, 0). The angles that PS and QP make with the positive \(x\)-direction are \(\theta\) and \(\phi\) respectively. \includegraphics{figure_8}

1. By considering the gradient of the tangent TS, show that \(\tan \theta = \frac{1}{t}\). [3]
2. Find the gradient of the line QP in terms of \(t\). Hence show that \(\phi = 2\theta\), and that angle TPQ is equal to \(\theta\). [8]

[The above result shows that if a lamp bulb is placed at Q, then the light from the bulb is reflected to produce a parallel beam of light.] The inside surface of the headlight has the shape produced by rotating the curve about the \(x\)-axis.

1. Show that the curve has cartesian equation \(y^2 = 8x\). Hence find the volume of revolution of the curve, giving your answer as a multiple of \(\pi\). [7]

\includegraphics{figure_9} Fig. 9 shows a hemispherical bowl, of radius 10 cm, filled with water to a depth of \(x\) cm. It can be shown that the volume of water, \(V\) cm\(^3\), is given by $$V = \pi(10x^2 - \frac{1}{3}x^3).$$ Water is poured into a leaking hemispherical bowl of radius 10 cm. Initially, the bowl is empty. After \(t\) seconds, the volume of water is changing at a rate, in cm\(^3\) s\(^{-1}\), given by the equation $$\frac{dV}{dt} = k(20 - x),$$ where \(k\) is a constant.

1. Find \(\frac{dV}{dx}\), and hence show that \(\pi x \frac{dx}{dt} = k\). [4]
2. Solve this differential equation, and hence show that the bowl fills completely after \(T\) seconds, where \(T = \frac{50\pi}{k}\). [5]

Once the bowl is full, the supply of water to the bowl is switched off, and water then leaks out at a rate of \(k\) cm\(^3\) s\(^{-1}\).

1. Show that, \(t\) seconds later, \(\pi(20 - x) \frac{dx}{dt} = -k\). [3]
2. Solve this differential equation. Hence show that the bowl empties in \(3T\) seconds. [6]