OCR MEI C3 (Core Mathematics 3)

Fig. 9 shows the curve \(y = \frac{x^2}{3x - 1}\). P is a turning point, and the curve has a vertical asymptote \(x = a\). \includegraphics{figure_1}

1. Write down the value of \(a\). [1]
2. Show that \(\frac{dy}{dx} = \frac{x(3x - 2)}{(3x - 1)^2}\) [3]
3. Find the exact coordinates of the turning point P. Calculate the gradient of the curve when \(x = 0.6\) and \(x = 0.8\), and hence verify that P is a minimum point. [7]
4. Using the substitution \(u = 3x - 1\), show that \(\int \frac{x^2}{3x - 1} dx = \frac{1}{27} \int \left( u + 2 + \frac{1}{u} \right) du\). Hence find the exact area of the region enclosed by the curve, the \(x\)-axis and the lines \(x = \frac{2}{3}\) and \(x = 1\). [7]

Differentiate \(\sqrt{1 + 6x^2}\). [4]

Show that the curve \(y = x^2 \ln x\) has a stationary point when \(x = \frac{1}{\sqrt{e}}\). [6]

The equation of a curve is \(y = \frac{x^2}{2x + 1}\).

1. Show that \(\frac{dy}{dx} = \frac{2x(x + 1)}{(2x + 1)^2}\). [4]
2. Find the coordinates of the stationary points of the curve. You need not determine their nature. [4]

1. Differentiate \(\sqrt{1 + 2x}\).
2. Show that the derivative of \(\ln(1 - e^{-x})\) is \(\frac{1}{e^x - 1}\). [4]

The function \(\text{f}(x) = \frac{\sin x}{2 - \cos x}\) has domain \(-\pi \leqslant x \leqslant \pi\). Fig. 8 shows the graph of \(y = \text{f}(x)\) for \(0 \leqslant x \leqslant \pi\). \includegraphics{figure_6}

1. Find \(\text{f}(-x)\) in terms of \(\text{f}(x)\). Hence sketch the graph of \(y = \text{f}(x)\) for the complete domain \(-\pi \leqslant x \leqslant \pi\). [3]
2. Show that \(\text{f}'(x) = \frac{2\cos x - 1}{(2 - \cos x)^2}\). Hence find the exact coordinates of the turning point P. State the range of the function \(\text{f}(x)\), giving your answer exactly. [8]
3. Using the substitution \(u = 2 - \cos x\) or otherwise, find the exact value of \(\int_0^\pi \frac{\sin x}{2 - \cos x} dx\). [4]
4. Sketch the graph of \(y = \text{f}(2x)\). [1]
5. Using your answers to parts (iii) and (iv), write down the exact value of \(\int_0^{\frac{\pi}{2}} \frac{\sin 2x}{2 - \cos 2x} dx\). [2]

Fig. 3 shows the curve defined by the equation \(y = \arcsin(x - 1)\), for \(0 \leqslant x \leqslant 2\). \includegraphics{figure_7}

1. Find \(x\) in terms of \(y\), and show that \(\frac{dx}{dy} = \cos y\). [3]
2. Hence find the exact gradient of the curve at the point where \(x = 1.5\). [4]

A curve has equation \(y = \frac{x}{2 + 3\ln x}\). Find \(\frac{dy}{dx}\). Hence find the exact coordinates of the stationary point of the curve. [7]