Edexcel C3 (Core Mathematics 3)

The curve \(C\) has equation \(y = 2e^x + 3x^2 + 2\). The point \(A\) with coordinates \((0, 4)\) lies on \(C\). Find the equation of the tangent to \(C\) at \(A\). [5]

Express \(\frac{x}{(x+1)(x+3)} + \frac{x+12}{x^2-9}\) as a single fraction in its simplest form. [6]

The functions f and g are defined by \(\text{f: } x \mapsto x^2 - 2x + 3, x \in \mathbb{R}, 0 \leq x \leq 4,\) \(\text{g: } x \mapsto \lambda x^2 + 1, \text{ where } \lambda \text{ is a constant, } x \in \mathbb{R}.\)

1. Find the range of f. [3]
2. Given that gf(2) = 16, find the value of \(\lambda\). [3]

1. Sketch, on the same set of axes, the graphs of $$y = 2 - e^{-x} \text{ and } y = \sqrt{x}.$$ [3] [It is not necessary to find the coordinates of any points of intersection with the axes.] Given that f(x) = \(e^{-x} + \sqrt{x} - 2\), \(x \geq 0\),
2. explain how your graphs show that the equation f(x) = 0 has only one solution, [1]
3. show that the solution of f(x) = 0 lies between \(x = 3\) and \(x = 4\). [2]

The iterative formula \(x_{n+1} = (2 - e^{-x_n})^2\) is used to solve the equation f(x) = 0.

1. Taking \(x_0 = 4\), write down the values of \(x_1\), \(x_2\), \(x_3\) and \(x_4\), and hence find an approximation to the solution of f(x) = 0, giving your answer to 3 decimal places. [4]

\includegraphics{figure_1} Figure 1 shows a sketch of the curve with equation \(y = e^{-x} - 1\).

1. Copy Fig. 1 and on the same axes sketch the graph of \(y = \frac{1}{2}|x - 1|\). Show the coordinates of the points where the graph meets the axes. [2]

The \(x\)-coordinate of the point of intersection of the graph is \(\alpha\).

1. Show that \(x = \alpha\) is a root of the equation \(x + 2e^{-x} - 3 = 0\). [3]
2. Show that \(-1 < \alpha < 0\). [2]

The iterative formula \(x_{n+1} = -\ln[\frac{1}{2}(3 - x_n)]\) is used to solve the equation \(x + 2e^{-x} - 3 = 0\).

1. Starting with \(x_0 = -1\), find the values of \(x_1\) and \(x_2\). [2]
2. Show that, to 2 decimal places, \(\alpha = -0.58\). [2]

f(x) = \(x^2 - 2x - 3\), \(x \in \mathbb{R}\), \(x \geq 1\).

1. Find the range of f. [1]
2. Write down the domain and range of \(f^{-1}\). [2]
3. Sketch the graph of \(f^{-1}\), indicating clearly the coordinates of any point at which the graph intersects the coordinate axes. [4]

Given that g(x) = \(|x - 4|\), \(x \in \mathbb{R}\),

1. find an expression for gf(x). [2]
2. Solve gf(x) = 8. [5]

f(x) = \(x + \frac{e^x}{5}\), \(x \in \mathbb{R}\).

1. Find f'(x). [2]

The curve \(C\), with equation \(y = \)f(x), crosses the \(y\)-axis at the point \(A\).

1. Find an equation for the tangent to \(C\) at \(A\). [3]
2. Complete the table, giving the values of \(\sqrt{x + \frac{e^x}{5}}\) to 2 decimal places.

\(x\)	0	0.5	1	1.5	2
\(\sqrt{x + \frac{e^x}{5}}\)	0.45	0.91

[2]

1. Express \(2\cos\theta + 5\sin\theta\) in the form \(R\cos(\theta - \alpha)\), where \(R > 0\) and \(0 < \alpha < \frac{\pi}{2}\). Give the values of \(R\) and \(\alpha\) to 3 significant figures. [3]
2. Find the maximum and minimum values of \(2\cos\theta + 5\sin\theta\) and the smallest possible value of \(\theta\) for which the maximum occurs. [2]

The temperature \(T\) °C, of an unheated building is modelled using the equation $$T = 15 + 2\cos\left(\frac{\pi t}{12}\right) + 5\sin\left(\frac{\pi t}{12}\right), \quad 0 \leq t < 24,$$ where \(t\) hours is the number of hours after 1200.

1. Calculate the maximum temperature predicted by this model and the value of \(t\) when this maximum occurs. [4]
2. Calculate, to the nearest half hour, the times when the temperature is predicted to be 12 °C. [6]