Edexcel C3 (Core Mathematics 3) 2016 June

1. The functions \(f\) and \(g\) are defined by

$$\begin{aligned} & \mathrm { f } : x \rightarrow 7 x - 1 , \quad x \in \mathbb { R }
& \mathrm {~g} : x \rightarrow \frac { 4 } { x - 2 } , \quad x \neq 2 , x \in \mathbb { R } \end{aligned}$$

1. Solve the equation \(\operatorname { fg } ( x ) = x\)
2. Hence, or otherwise, find the largest value of \(a\) such that \(\mathrm { g } ( a ) = \mathrm { f } ^ { - 1 } ( a )\)
   

2.

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\), writing your answer as a single fraction in its simplest form.
2. Hence find the set of values of \(x\) for which \(\frac { \mathrm { d } y } { \mathrm {~d} x } < 0\)
   2. $$y = \frac { 4 x } { x ^ { 2 } + 5 }$$

1. (a) Express \(2 \cos \theta - \sin \theta\) in the form \(R \cos ( \theta + \alpha )\), where \(R\) and \(\alpha\) are constants, \(R > 0\) and \(0 < \alpha < 90 ^ { \circ }\). Give the exact value of \(R\) and give the value of \(\alpha\) to 2 decimal places.
   (b) Hence solve, for \(0 \leqslant \theta < 360 ^ { \circ }\),

$$\frac { 2 } { 2 \cos \theta - \sin \theta - 1 } = 15$$ Give your answers to one decimal place.
(c) Use your solutions to parts (a) and (b) to deduce the smallest positive value of \(\theta\) for which $$\frac { 2 } { 2 \cos \theta + \sin \theta - 1 } = 15$$ Give your answer to one decimal place.

4. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve with equation \(y = g ( x )\), where $$\mathrm { g } ( x ) = \left| 4 \mathrm { e } ^ { 2 x } - 25 \right| , \quad x \in \mathbb { R }$$ The curve cuts the \(y\)-axis at the point \(A\) and meets the \(x\)-axis at the point \(B\). The curve has an asymptote \(y = k\), where \(k\) is a constant, as shown in Figure 1

1. Find, giving each answer in its simplest form,
   1. the \(y\) coordinate of the point \(A\),
   2. the exact \(x\) coordinate of the point \(B\),
   3. the value of the constant \(k\). The equation \(\mathrm { g } ( x ) = 2 x + 43\) has a positive root at \(x = \alpha\)
2. Show that \(\alpha\) is a solution of \(x = \frac { 1 } { 2 } \ln \left( \frac { 1 } { 2 } x + 17 \right)\) The iteration formula $$x _ { n + 1 } = \frac { 1 } { 2 } \ln \left( \frac { 1 } { 2 } x _ { n } + 17 \right)$$ can be used to find an approximation for \(\alpha\)
3. Taking \(x _ { 0 } = 1.4\) find the values of \(x _ { 1 }\) and \(x _ { 2 }\) Give each answer to 4 decimal places.
4. By choosing a suitable interval, show that \(\alpha = 1.437\) to 3 decimal places. \includegraphics[max width=\textwidth, alt={}, center]{d3ba2776-eedb-48f0-834f-41aa454afba3-07_2258_47_315_37}

5. (i) Find, using calculus, the \(x\) coordinate of the turning point of the curve with equation $$y = \mathrm { e } ^ { 3 x } \cos 4 x , \quad \frac { \pi } { 4 } \leqslant x < \frac { \pi } { 2 }$$ Give your answer to 4 decimal places.
(ii) Given \(x = \sin ^ { 2 } 2 y , \quad 0 < y < \frac { \pi } { 4 }\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) as a function of \(y\). Write your answer in the form $$\frac { \mathrm { d } y } { \mathrm {~d} x } = p \operatorname { cosec } ( q y ) , \quad 0 < y < \frac { \pi } { 4 }$$ where \(p\) and \(q\) are constants to be determined. \includegraphics[max width=\textwidth, alt={}, center]{d3ba2776-eedb-48f0-834f-41aa454afba3-09_2258_47_315_37}

6. $$f ( x ) = \frac { x ^ { 4 } + x ^ { 3 } - 3 x ^ { 2 } + 7 x - 6 } { x ^ { 2 } + x - 6 } , \quad x > 2 , x \in \mathbb { R }$$

1. Given that $$\frac { x ^ { 4 } + x ^ { 3 } - 3 x ^ { 2 } + 7 x - 6 } { x ^ { 2 } + x - 6 } \equiv x ^ { 2 } + A + \frac { B } { x - 2 }$$ find the values of the constants \(A\) and \(B\).
2. Hence or otherwise, using calculus, find an equation of the normal to the curve with equation \(y = \mathrm { f } ( x )\) at the point where \(x = 3\)

7. (a) For \(- \frac { \pi } { 2 } \leqslant y \leqslant \frac { \pi } { 2 }\), sketch the graph of \(y = \mathrm { g } ( x )\) where $$g ( x ) = \arcsin x \quad - 1 \leqslant x \leqslant 1$$ (b) Find the exact value of \(x\) for which $$3 g ( x + 1 ) + \pi = 0$$

1. (a) Prove that

$$2 \cot 2 x + \tan x \equiv \cot x \quad x \neq \frac { n \pi } { 2 } , n \in \mathbb { Z }$$ (b) Hence, or otherwise, solve, for \(- \pi \leqslant x < \pi\), $$6 \cot 2 x + 3 \tan x = \operatorname { cosec } ^ { 2 } x - 2$$ Give your answers to 3 decimal places.
(Solutions based entirely on graphical or numerical methods are not acceptable.) \includegraphics[max width=\textwidth, alt={}, center]{d3ba2776-eedb-48f0-834f-41aa454afba3-14_2258_47_315_37}

9. The amount of an antibiotic in the bloodstream, from a given dose, is modelled by the formula $$x = D \mathrm { e } ^ { - 0.2 t }$$ where \(x\) is the amount of the antibiotic in the bloodstream in milligrams, \(D\) is the dose given in milligrams and \(t\) is the time in hours after the antibiotic has been given. A first dose of 15 mg of the antibiotic is given.

1. Use the model to find the amount of the antibiotic in the bloodstream 4 hours after the dose is given. Give your answer in mg to 3 decimal places. A second dose of 15 mg is given 5 hours after the first dose has been given. Using the same model for the second dose,
2. show that the total amount of the antibiotic in the bloodstream 2 hours after the second dose is given is 13.754 mg to 3 decimal places. No more doses of the antibiotic are given. At time \(T\) hours after the second dose is given, the total amount of the antibiotic in the bloodstream is 7.5 mg .
3. Show that \(T = a \ln \left( b + \frac { b } { \mathrm { e } } \right)\), where \(a\) and \(b\) are integers to be determined.
   

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   \includegraphics[max width=\textwidth, alt={}, center]{d3ba2776-eedb-48f0-834f-41aa454afba3-16_2258_47_315_37}