OCR MEI C3 (Core Mathematics 3)

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

Given that \(\arcsin x = \frac{1}{6}\pi\), find \(x\). Find \(\arccos x\) in terms of \(\pi\).

The functions \(f(x)\) and \(g(x)\) are defined for the domain \(x > 0\) as follows: $$f(x) = \ln x, \quad g(x) = x^3.$$ Express the composite function \(fg(x)\) in terms of \(\ln x\). State the transformation which maps the curve \(y = f(x)\) onto the curve \(y = fg(x)\).

The temperature \(T°C\) of a liquid at time \(t\) minutes is given by the equation $$T = 30 + 20e^{-0.05t}, \quad \text{for } t \geq 0.$$ Write down the initial temperature of the liquid, and find the initial rate of change of temperature. Find the time at which the temperature is \(40°C\).

Using the substitution \(u = 2x + 1\), show that \(\int_0^1 \frac{x}{2x + 1} dx = \frac{1}{4}(2 - \ln 3)\).

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.

Fig. 7 shows the curve defined implicitly by the equation $$y^2 + y = x^3 + 2x,$$ together with the line \(x = 2\). \includegraphics{figure_7} Find the coordinates of the points of intersection of the line and the curve. Find \(\frac{dy}{dx}\) in terms of \(x\) and \(y\). Hence find the gradient of the curve at each of these two points.

Fig. 8 shows part of the curve \(y = x \sin 3x\). It crosses the \(x\)-axis at P. The point on the curve with \(x\)-coordinate \(\frac{1}{6}\pi\) is Q. \includegraphics{figure_8}

1. Find the \(x\)-coordinate of P. [3]
2. Show that Q lies on the line \(y = x\). [1]
3. Differentiate \(x \sin 3x\). Hence prove that the line \(y = x\) touches the curve at Q. [6]
4. Show that the area of the region bounded by the curve and the line \(y = x\) is \(\frac{1}{72}(\pi^2 - 8)\). [7]

The function \(f(x) = \ln(1 + x^2)\) has domain \(-3 \leq x \leq 3\). Fig. 9 shows the graph of \(y = f(x)\). \includegraphics{figure_9}

1. Show algebraically that the function is even. State how this property relates to the shape of the curve. [3]
2. Find the gradient of the curve at the point P\((2, \ln 5)\). [4]
3. Explain why the function does not have an inverse for the domain \(-3 \leq x \leq 3\). [1]

The domain of \(f(x)\) is now restricted to \(0 \leq x \leq 3\). The inverse of \(f(x)\) is the function \(g(x)\).

1. Sketch the curves \(y = f(x)\) and \(y = g(x)\) on the same axes. State the domain of the function \(g(x)\). Show that \(g(x) = \sqrt{e^x - 1}\). [6]
2. Differentiate \(g(x)\). Hence verify that \(g'(\ln 5) = \frac{1}{4}\). Explain the connection between this result and your answer to part (ii). [5]