Edexcel C3 (Core Mathematics 3)

\begin{figure}[h] \end{figure} Figure 2 shows part of the curve \(C\) with equation \(y = \mathrm { f } ( x )\), where $$\mathrm { f } ( x ) = 0.5 \mathrm { e } ^ { x } - x ^ { 2 }$$ The curve \(C\) cuts the \(y\)-axis at \(A\) and there is a minimum at the point \(B\).

1. Find an equation of the tangent to \(C\) at \(A\). The \(x\)-coordinate of \(B\) is approximately 2.15 . A more exact estimate is to be made of this coordinate using iterations \(x _ { n + 1 } = \ln \mathrm { g } \left( x _ { n } \right)\).
2. Show that a possible form for \(\mathrm { g } ( x )\) is \(\mathrm { g } ( x ) = 4 x\).
3. Using \(x _ { n + 1 } = \ln 4 x _ { n }\), with \(x _ { 0 } = 2.15\), calculate \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\). Give the value of \(x _ { 3 }\) to 4 decimal places.

3. (a) Sketch the graph of \(y = | 2 x + a | , a > 0\), showing the coordinates of the points where the graph meets the coordinate axes.
(b) On the same axes, sketch the graph of \(y = \frac { 1 } { x }\).
(c) Explain how your graphs show that there is only one solution of the equation $$x | 2 x + a | - 1 = 0$$ (d) Find, using algebra, the value of \(x\) for which \(x | 2 x + 1 | - 1 = 0\).

4. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of the curve with equation \(y = \mathrm { f } ( x ) , - 1 \leq x \leq 3\). The curve touches the \(x\)-axis at the origin \(O\), crosses the \(x\)-axis at the point \(A ( 2,0 )\) and has a maximum at the point \(B \left( \frac { 4 } { 3 } , 1 \right)\). In separate diagrams, show a sketch of the curve with equation

1. \(y = \mathrm { f } ( x + 1 )\),
2. \(y = | \mathrm { f } ( x ) |\),
3. \(y = \mathrm { f } ( | x | )\),
   marking on each sketch the coordinates of points at which the curve
   1. has a turning point,
   2. meets the \(x\)-axis.

5. (i) Given that \(\sin x = \frac { 3 } { 5 }\), use an appropriate double angle formula to find the exact value of \(\sec 2 x\).
(ii) Prove that $$\cot 2 x + \operatorname { cosec } 2 x \equiv \cot x , \quad \left( x \neq \frac { n \pi } { 2 } , n \in Z \right)$$

1. The function f is defined by \(\mathrm { f } : x \rightarrow \frac { 3 x - 1 } { x - 3 } , x \in j , x \neq 3\).
   1. Prove that \(\mathrm { f } ^ { - 1 } ( x ) = \mathrm { f } ( x )\) for all \(x \in j , x \neq 3\).
   2. Hence find, in terms of \(k , \operatorname { ff } ( k )\), where \(x \neq 3\).
   \begin{figure}[h]
   \captionsetup{labelformat=empty} \caption{Figure 3} \includegraphics[alt={},max width=\textwidth]{909b52e5-2f16-4eab-b691-9d8fcf9bcfd9-5_864_1205_605_242}
   \end{figure} Figure 3 shows a sketch of the one-one function g , defined over the domain \(- 2 \leq x \leq 2\).
2. Find the value of \(\mathrm { fg } ( - 2 )\).
3. Sketch the graph of the inverse function \(\mathrm { g } ^ { - 1 }\) and state its domain. The function h is defined by \(\mathrm { h } : x \mapsto 2 \mathrm {~g} ( x - 1 )\).
4. Sketch the graph of the function h and state its range.

7. (i) (a) Express \(( 12 \cos \theta - 5 \sin \theta )\) in the form \(R \cos ( \theta + \alpha )\), where \(R > 0\) and \(0 < \alpha < 90 ^ { \circ }\).
(b) Hence solve the equation $$12 \cos \theta - 5 \sin \theta = 4$$ for \(0 < \theta < 90 ^ { \circ }\), giving your answer to 1 decimal place.
(ii) Solve $$8 \cot \theta - 3 \tan \theta = 2$$ for \(0 < \theta < 90 ^ { \circ }\), giving your answer to 1 decimal place.

8. The curve \(C\) has equation \(y = \mathrm { f } ( x )\), where $$\mathrm { f } ( x ) = 3 \ln x + \frac { 1 } { x } , \quad x > 0$$ The point \(P\) is a stationary point on \(C\).

1. Calculate the \(x\)-coordinate of \(P\).
2. Show that the \(y\)-coordinate of \(P\) may be expressed in the form \(k - k \ln k\), where \(k\) is a constant to be found. The point \(Q\) on \(C\) has \(x\)-coordinate 1 .
3. Find an equation for the normal to \(C\) at \(Q\). The normal to \(C\) at \(Q\) meets \(C\) again at the point \(R\).
4. Show that the \(x\)-coordinate of \(R\)
   1. satisfies the equation \(6 \ln x + x + \frac { 2 } { x } - 3 = 0\),
   2. lies between 0.13 and 0.14 .