Composite transformation sketch

5 The graph with equation \(y = \mathrm { f } ( x )\) is transformed to the graph with equation \(y = \mathrm { g } ( x )\) by a stretch in the \(x\)-direction with factor 0.5 , followed by a translation of \(\binom { 0 } { 1 }\).

1. The diagram below shows the graph of \(y = \mathrm { f } ( x )\). On the diagram sketch the graph of \(y = \mathrm { g } ( x )\).
   \includegraphics[max width=\textwidth, alt={}, center]{5d26c357-ea9f-47d9-8eca-2152901cf2f1-07_613_1527_623_342}
2. Find an expression for \(\mathrm { g } ( x )\) in terms of \(\mathrm { f } ( x )\).

9. In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable. \begin{figure}[h] \end{figure} Figure 5 shows a sketch of the curve with equation \(y = \mathrm { f } ( x )\) where $$\mathrm { f } ( x ) = \sqrt { x } \quad x > 0$$ The point \(P ( 9,3 )\) lies on the curve and is shown in Figure 5.
On the next page there is a copy of Figure 5 called Diagram 1.

1. On Diagram 1, sketch and clearly label the graphs of $$y = \mathrm { f } ( 2 x ) \text { and } y = \mathrm { f } ( x ) + 3$$ Show on each graph the coordinates of the point to which \(P\) is transformed. The graph of \(y = \mathrm { f } ( 2 x )\) meets the graph of \(y = \mathrm { f } ( x ) + 3\) at the point \(Q\).
2. Show that the \(x\) coordinate of \(Q\) is the solution of $$\sqrt { x } = 3 ( \sqrt { 2 } + 1 )$$
3. Hence find, in simplest form, the coordinates of \(Q\).
   
   \includegraphics[max width=\textwidth, alt={}]{f1e1d4f5-dd27-4839-a6f3-f6906666302c-27_599_720_274_660}
   \section*{Diagram 1} Turn over for a copy of Diagram 1 if you need to redraw your graphs. Only use this copy if you need to redraw your graphs.
   \includegraphics[max width=\textwidth, alt={}, center]{f1e1d4f5-dd27-4839-a6f3-f6906666302c-29_600_718_1991_660} Copy of Diagram 1

5. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of the curve with equation \(y = \mathrm { f } ( x )\) where $$\mathrm { f } ( x ) = \frac { x } { x - 2 } , \quad x \neq 2$$ The curve passes through the origin and has two asymptotes, with equations \(y = 1\) and \(x = 2\), as shown in Figure 1.

1. In the space below, sketch the curve with equation \(y = \mathrm { f } ( x - 1 )\) and state the equations of the asymptotes of this curve.
2. Find the coordinates of the points where the curve with equation \(y = \mathrm { f } ( x - 1 )\) crosses the coordinate axes.

5. \begin{figure}[h] \end{figure} Figure 1 shows the sketch of a curve with equation \(y = \mathrm { f } ( x ) , x \in \mathbb { R }\). The curve crosses the \(y\)-axis at \(( 0,4 )\) and crosses the \(x\)-axis at \(( 5,0 )\). The curve has a single turning point, a maximum, at (2, 7). The line with equation \(y = 1\) is the only asymptote to the curve.

1. State the coordinates of the turning point on the curve with equation \(y = \mathrm { f } ( x - 2 )\).
2. State the solution of the equation f( \(2 x\) ) \(= 0\)
3. State the equation of the asymptote to the curve with equation \(y = \mathrm { f } ( - x )\). Given that the line with equation \(y = k\), where \(k\) is a constant, meets the curve \(y = \mathrm { f } ( x )\) at only one point,
4. state the set of possible values for \(k\).

11. \begin{figure}[h] \end{figure} Figure 4 shows a sketch of part of the curve with equation \(y = \mathrm { f } ( x ) , \quad x \in \mathbb { R }\)
The curve meets the coordinate axes at the points \(A ( 0 , - 3 )\) and \(B \left( - \frac { 1 } { 3 } \ln 4,0 \right)\) and the curve has an asymptote with equation \(y = - 4\) In separate diagrams, sketch the graph with equation

1. \(y = | f ( x ) |\)
2. \(y = 2 \mathrm { f } ( x ) + 6\) On each sketch, give the exact coordinates of the points where the curve crosses or meets the coordinate axes and the equation of any asymptote. Given that $$\begin{array} { l l } \mathrm { f } ( x ) = \mathrm { e } ^ { - 3 x } - 4 , & x \in \mathbb { R }
   \mathrm {~g} ( x ) = \ln \left( \frac { 1 } { x + 2 } \right) , & x > - 2 \end{array}$$
3. state the range of f,
4. find \(\mathrm { f } ^ { - 1 } ( x )\),
5. express \(f g ( x )\) as a polynomial in \(x\).

5. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of part of the curve with equation \(y = \mathrm { f } ( x ) , x \in \mathbb { R }\).
The curve meets the coordinate axes at the points \(A ( 0,1 - k )\) and \(B \left( \frac { 1 } { 2 } \ln k , 0 \right)\), where \(k\) is a constant and \(k > 1\), as shown in Figure 2. On separate diagrams, sketch the curve with equation

1. \(y = | f ( x ) |\),
2. \(y = \mathrm { f } ^ { - 1 } ( x )\). Show on each sketch the coordinates, in terms of \(k\), of each point at which the curve meets or cuts the axes. Given that \(\mathrm { f } ( x ) = \mathrm { e } ^ { 2 x } - k\),
3. state the range of f ,
4. find \(\mathrm { f } ^ { - 1 } ( x )\),
5. write down the domain of \(\mathrm { f } ^ { - 1 }\).

6. \begin{figure}[h] \end{figure} Figure 3 shows a sketch of the graph of \(y = \mathrm { f } ( x )\) where $$\mathrm { f } ( x ) = \left\{ \begin{array} { r r } 5 - 2 x , & x \leqslant 4
\mathrm { e } ^ { 2 x - 8 } - 4 , & x > 4 \end{array} \right.$$

1. State the range of \(\mathrm { f } ( x )\).
2. Determine the exact value of ff(0).
3. Solve \(\mathrm { f } ( x ) = 21\) Give each answer as an exact answer.
4. Explain why the function f does not have an inverse.

7
\includegraphics[max width=\textwidth, alt={}, center]{d858728a-3371-4755-880c-54f96c5e5156-3_465_748_1133_717} The diagram shows the curve with equation \(y = \cos ^ { - 1 } x\).

1. Sketch the curve with equation \(y = 3 \cos ^ { - 1 } ( x - 1 )\), showing the coordinates of the points where the curve meets the axes.
2. By drawing an appropriate straight line on your sketch in part (i), show that the equation \(3 \cos ^ { - 1 } ( x - 1 ) = x\) has exactly one root.
3. Show by calculation that the root of the equation \(3 \cos ^ { - 1 } ( x - 1 ) = x\) lies between 1.8 and 1.9 .
4. The sequence defined by $$x _ { 1 } = 2 , \quad x _ { n + 1 } = 1 + \cos \left( \frac { 1 } { 3 } x _ { n } \right)$$ converges to a number \(\alpha\). Find the value of \(\alpha\) correct to 2 decimal places and explain why \(\alpha\) is the root of the equation \(3 \cos ^ { - 1 } ( x - 1 ) = x\).

4 Fig. 4 shows the curve \(y = f ( x )\), where \(f ( x ) = \sqrt { 1 - 9 x ^ { 2 } } , - a \leqslant x \leqslant a\). \begin{figure}[h] \end{figure}

1. Find the value of \(a\).
2. Write down the range of \(\mathrm { f } ( x )\).
3. Sketch the curve \(y = \mathrm { f } \left( \frac { 1 } { 3 } x \right) - 1\).

6. \begin{figure}[h] \end{figure} Figure 1 shows the curve with equation \(y = \mathrm { f } ( x )\). The curve crosses the axes at \(( p , 0 )\) and \(( 0 , q )\) and the lines \(x = 1\) and \(y = 2\) are asymptotes of the curve.

1. Showing the coordinates of any points of intersection with the axes and the equations of any asymptotes, sketch on separate diagrams the graphs of
   1. \(y = | \mathrm { f } ( x ) |\),
   2. \(y = 2 \mathrm { f } ( x + 1 )\). Given also that $$\mathrm { f } ( x ) \equiv \frac { 2 x - 1 } { x - 1 } , \quad x \in \mathbb { R } , \quad x \neq 1$$
2. find the values of \(p\) and \(q\),
3. find an expression for \(\mathrm { f } ^ { - 1 } ( x )\).