OCR MEI C3 (Core Mathematics 3)

You are given that \(y^2 = 4x + 7\).

1. Use implicit differentiation to find \(\frac{dy}{dx}\) in terms of \(y\). [2]
2. Make \(x\) the subject of the equation. Find \(\frac{dx}{dy}\) and hence show that in this case \(\frac{dx}{dy} = \frac{1}{\frac{dx}{dy}}\). [3]

1. Expand \((e^x + e^{-x})^2\). [1]
2. Hence find \(\int (e^x + e^{-x})^2 dx\). [3]

1. Sketch the graph of \(y = |3x - 6|\). [2]
2. Solve the equation \(|3x - 6| = x + 4\) and illustrate your answer on your graph. [4]

Find \(\int x \sin 3x dx\). [4]

Make \(x\) the subject of \(t = \ln \sqrt{\frac{5}{(x-3)}}\). [4]

The function f(x) is defined as \(f(x) = \frac{\ln x}{x}\). The graph of the function is shown in Fig. 6. \includegraphics{figure_6}

1. Give the coordinates of the point, P, where the curve crosses the \(x\)-axis. [1]
2. Use calculus to find the coordinates of the stationary point, Q, and show that it is a maximum. [6]

An oil slick is circular with radius \(r\) km and area \(A\) km\(^2\). The radius increases with time at a rate given by \(\frac{dr}{dt} = 0.5\), in kilometres per hour.

1. Show that \(\frac{dA}{dt} = \pi r\). [4]
2. Find the rate of increase of the area of the slick at a time when the radius is 6 km. [2]

Fig. 8 shows the graph of \(y = x\sqrt{1 + x}\). The point P on the curve is on the \(x\)-axis. \includegraphics{figure_8}

1. Write down the coordinates of P. [1]
2. Show that \(\frac{dy}{dx} = \frac{3x + 2}{2\sqrt{1 + x}}\). [4]
3. Hence find the coordinates of the turning point on the curve. What can you say about the gradient of the curve at P? [4]
4. By using a suitable substitution, show that \(\int_0^0 x\sqrt{1 + x} dx = \int_0^1 \left(u^{\frac{3}{2}} - u^{\frac{1}{2}}\right) du\). Evaluate this integral, giving your answer in an exact form. What does this value represent? [7]
5. Use your answer to part (ii) to differentiate \(y = x\sqrt{1 + x} \sin 2x\) with respect to \(x\). (You need not simplify your result.) [2]

The functions f(x) and g(x) are defined by $$f(x) = x^2, \quad g(x) = 2x - 1,$$ for all real values of \(x\).

1. State the ranges of f(x) and g(x). Explain why f(x) has no inverse. [3]
2. Find an expression for the inverse function g\(^{-1}\)(x) in terms of \(x\). Sketch the graphs of \(y = g(x)\) and \(y = g^{-1}(x)\) on the same axes. [4]
3. Find expressions for gf(x) and fg(x). [2]
4. Solve the equation gf(x) = fg(x). Sketch the graphs of \(y = gf(x)\) and \(y = fg(x)\) on the same axes to illustrate your answer. [4]
5. Show that the equation f(x + a) = g\(^{-1}\)(x) has no solution if \(a > \frac{1}{4}\). [5]