The function \(\mathrm { f } ( x )\) is defined by
$$f ( x ) = \sqrt { 1 + 2 x } \text { for } x \geqslant - \frac { 1 } { 2 }$$
Find an expression for \(\mathrm { f } ^ { - 1 } ( x )\) and state the domain of this inverse function.
Explain why \(\mathrm { g } ( x ) = 1 + x ^ { 2 }\), with domain all real numbers, has no inverse function.
3 In this question you must show detailed reasoning.
The diagram shows the curve with equation \(y = x ^ { 5 }\) and the square \(O A B C\) where the points \(A , B\) and \(C\) have coordinates \(( 1,0 ) , ( 1,1 )\) and \(( 0,1 )\) respectively.
The curve cuts the square into two parts.
\includegraphics[max width=\textwidth, alt={}, center]{60e1e785-c34b-48ef-a63f-13a25fee186e-04_658_780_1318_230}
Show that the relationship between the areas of the two parts of the square is
\(\frac { \text { Area to left of curve } } { \text { Area below curve } } = 5\).
5 In this question you must show detailed reasoning.
Using the substitution \(\mathrm { u } = \mathrm { x } + 1\), find the value of the positive integer \(c\) such that \(\int _ { \mathrm { c } } ^ { \mathrm { c } + 4 } \frac { \mathrm { x } } { ( \mathrm { x } + 1 ) ^ { 2 } } \mathrm { dx } = \ln 3 - \frac { 1 } { 3 }\).
6 In this question you must show detailed reasoning.
Solve the equation \(\tan x - 3 \cot x = 2\) for values of \(x\) in the interval \(0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }\).
8 In this question you must show detailed reasoning.
Express \(\cos x + \sqrt { 3 } \sin x\) in the form \(\mathrm { R } \sin ( \mathrm { x } + \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { 1 } { 2 } \pi\). Give the values of \(R\) and \(\alpha\) in exact form.
Hence solve the equation \(\cos x = \sqrt { 3 } ( 1 - \sin x )\) for values of \(x\) in the interval \(- \pi \leqslant x \leqslant \pi\). Give the roots of this equation in exact form.
9 This question is about the equation \(\mathrm { f } ( x ) = 0\), where \(\mathrm { f } ( x ) = x ^ { 4 } - x - \frac { 1 } { 3 x - 2 }\).
Fig. 9.1 shows the curve \(y = f ( x )\).
Fig. 9.1
\includegraphics[max width=\textwidth, alt={}, center]{60e1e785-c34b-48ef-a63f-13a25fee186e-06_940_929_518_239}
Show, by calculation, that the equation \(\mathrm { f } ( x ) = 0\) has a root between \(x = 1\) and \(x = 2\).
Fig. 9.2 shows part of a spreadsheet being used to find a root of the equation.
\begin{table}[h]
10 The diagram below shows the curve \(y = f ( x )\).
\includegraphics[max width=\textwidth, alt={}, center]{60e1e785-c34b-48ef-a63f-13a25fee186e-07_942_679_1500_242}
Sketch the graph of the gradient function, \(y = f ^ { \prime } ( x )\), on the copy of the diagram in the Printed Answer Booklet.
11 Fig. 11.1 shows the curve with equation \(\mathrm { y } = \mathrm { g } ( \mathrm { x } )\) where \(\mathrm { g } ( x ) = x \sin x + \cos x\) and the curve of the gradient function \(\mathrm { y } = \mathrm { g } ^ { \prime } ( \mathrm { x } )\) for \(- 2 \pi \leqslant x \leqslant 2 \pi\).
\begin{figure}[h]
Show that the \(x\)-coordinates of the points on the curve \(y = g ( x )\) where the gradient is 1 satisfy the equation \(\frac { 1 } { x } - \cos x = 0\).
Fig. 11.2 shows part of the curve with equation \(y = \frac { 1 } { x } - \cos x\).
\begin{figure}[h]
Use the Newton-Raphson method with a suitable starting value to find the smallest positive \(x\)-coordinate of a point on the curve \(y = x \sin x + \cos x\) where the gradient is 1 .
You should write down at least the following.
The iteration you use
The starting value
The solution correct to \(\mathbf { 4 }\) decimal places
Explain why \(x _ { 1 } = 3\) is not a suitable starting value for the Newton-Raphson method in part (b).
13 Substitute appropriate values of \(t _ { 1 }\) and \(t _ { 2 }\) to verify that \(t _ { 1 } t _ { 2 }\) gives the correct value for the \(y\)-coordinate of the point of intersection of the tangents at the points A and B in Fig. \(\mathbf { C 1 . }\)
14 Substitute appropriate values of \(t _ { 1 }\) and \(t _ { 2 }\) to verify that the expression \(t _ { 1 } ^ { 2 } + t _ { 2 } ^ { 2 } + t _ { 1 } t _ { 2 } + \frac { 1 } { 2 }\) gives the correct value for the \(y\)-coordinate of the point of intersection of the normals at the points A and B in Fig. C2.
Show that, for the curve \(y = a x ^ { 2 } + b x + c\), the equation of the tangent at the point with \(x\)-coordinate \(t\) is \(\mathrm { y } = ( 2 \mathrm { at } + \mathrm { b } ) \mathrm { x } - \mathrm { at } ^ { 2 } + \mathrm { c }\).
Hence show that for the curve with equation \(y = a x ^ { 2 } + b x + c\), the tangents at two points, \(P\) and Q , on the curve cross at a point which has \(x\)-coordinate equal to the mean of the \(x\)-coordinates of points P and Q , as given in lines 11 to 14 .
16 Show that the expression \(a \left( \frac { x _ { P } + x _ { Q } } { 2 } \right) ^ { 2 } + b \left( \frac { x _ { P } + x _ { Q } } { 2 } \right) + c - a \left( \frac { x _ { P } - x _ { Q } } { 2 } \right) ^ { 2 }\) is equivalent to \(a x _ { P } x _ { Q } + b \left( \frac { x _ { P } + x _ { Q } } { 2 } \right) + c\), as given in lines 15 and 16 .
17 Show that, for the curve \(y = x ^ { 2 }\), the equation of the normal at the point \(\left( t , t ^ { 2 } \right)\) is \(y = - \frac { x } { 2 t } + t ^ { 2 } + \frac { 1 } { 2 }\), as given in line 27.
18 A student is investigating the intersection points of tangents to the curve \(y = 6 x ^ { 2 } - 7 x + 1\). She uses software to draw tangents at pairs of points with \(x\)-coordinates differing by 5 .
Find the equation of the curve that all the intersection points lie on.