Edexcel FP1 AS (Further Pure 1 AS) Specimen

Question 1
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  1. (a) Use the substitution \(\mathrm { t } = \tan \left( \frac { \mathrm { x } } { 2 } \right)\) to show that
$$\sec x - \tan x \equiv \frac { 1 - t } { 1 + t } \quad x \neq ( 2 n + 1 ) \frac { \pi } { 2 } , n \in \mathbb { Z }$$ (b) Use the substitution \(\mathrm { t } = \tan \left( \frac { \mathrm { x } } { 2 } \right)\) and the answer to part (a) to prove that $$\frac { 1 - \sin x } { 1 + \sin x } \equiv ( \sec x - \tan x ) ^ { 2 } \quad x \neq ( 2 n + 1 ) \frac { \pi } { 2 } , n \in \mathbb { Z }$$ \section*{Q uestion 1 continued}
Question 2
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  1. The value, V hundred pounds, of a particular stock thours after the opening of trading on a given day is modelled by the differential equation
$$\frac { d V } { d t } = \frac { V ^ { 2 } - t } { t ^ { 2 } + t V } \quad 0 < t < 8.5$$ A trader purchases \(\pounds 300\) of the stock one hour after the opening of trading.
Use two iterations of the approximation formula \(\left( \frac { \mathrm { dy } } { \mathrm { dx } } \right) _ { 0 } \approx \frac { \mathrm { y } _ { 1 } - \mathrm { y } _ { 0 } } { \mathrm {~h} }\) to estimate, to the nearest \(\pounds\), the value of the trader's stock half an hour after it was purchased.
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Question 3
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  1. Use algebra to find the set of values of x for which
$$\frac { 1 } { x } < \frac { x } { x + 2 }$$
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Question 4
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4. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{ff1fc9b0-6514-44e0-a2a3-46aa6411ce10-08_538_807_251_630} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} Figure 1 shows a sketch of a solid sculpture made of glass and concrete. The sculpture is modelled as a parallelepiped. The sculpture is made up of a concrete solid in the shape of a tetrahedron, shown shaded in Figure 1, whose vertices are \(\mathrm { O } ( 0,0,0 ) , \mathrm { A } ( 2,0,0 ) , \mathrm { B } ( 0,3,1 )\) and \(\mathrm { C } ( 1,1,2 )\), where the units are in metres. The rest of the solid parallelepiped is made of glass which is glued to the concrete tetrahedron.
  1. Find the surface area of the glued face of the tetrahedron.
  2. Find the volume of glass contained in this parallelepiped.
  3. Give a reason why the volume of concrete predicted by this model may not be an accurate value for the volume of concrete that was used to make the sculpture. \section*{Q uestion 4 continued}
Question 5
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5. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{ff1fc9b0-6514-44e0-a2a3-46aa6411ce10-10_965_853_212_621} \captionsetup{labelformat=empty} \caption{Figure 2}
\end{figure} Diagram not drawn to scale $$\left[ Y \text { ou may quote without proof that for the general parabola } y ^ { 2 } = 4 a x , \frac { d y } { d x } = \frac { 2 a } { y } \right]$$ The parabola C has equation \(\mathrm { y } ^ { 2 } = 16 \mathrm { x }\)
  1. Deduce that the point \(\mathrm { P } \left( 4 \mathrm { p } ^ { 2 } , 8 \mathrm { p } \right)\) is a general point on C . The line I is the tangent to C at the point P .
  2. Show that an equation for I is $$p y = x + 4 p ^ { 2 }$$ The finite region R , shown shaded in Figure 2, is bounded by the line I , the x -axis and the parabola C.
    The line \(I\) intersects the directrix of \(C\) at the point \(B\), where the \(y\) coordinate of \(B\) is \(\frac { 10 } { 3 }\) Given that \(\mathrm { p } > 0\)
  3. show that the area of R is 36 \section*{Q uestion 5 continued}