Edexcel PURE 2024 October — Question 8

Exam BoardEdexcel
ModulePURE
Year2024
SessionOctober
PaperDownload PDF ↗
TopicAreas Between Curves
TypeTwo Curves Intersection Area
DifficultyStandard +0.3 This is a standard A-level integration question requiring finding intersection points by solving a quartic equation (which factors nicely), then computing the area between curves using definite integration. The algebra is straightforward and the integration involves only basic polynomial and reciprocal terms. Slightly easier than average due to the clean factorization and routine techniques.
Spec1.08d Evaluate definite integrals: between limits1.08f Area between two curves: using integration
  1. In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{e7412e14-6a5a-4545-8d6b-4bceb141cc15-20_762_851_376_607} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} Figure 1 shows a sketch of part of the curve \(C _ { 1 }\) with equation $$y = x ^ { 2 } + 3 \quad x > 0$$ and part of the curve \(C _ { 2 }\) with equation $$y = 13 - \frac { 9 } { x ^ { 2 } } \quad x > 0$$ The curves \(C _ { 1 }\) and \(C _ { 2 }\) intersect at the points \(P\) and \(Q\) as shown in Figure 1 .
  1. Use algebra to find the \(x\) coordinate of \(P\) and the \(x\) coordinate of \(Q\). The finite region \(R\), shown shaded in Figure 1, is bounded by \(C _ { 1 }\) and \(C _ { 2 }\)
  2. Use algebraic integration to find the exact area of \(R\).
Question 8:
AnswerMarks
8(a)9
x2 +3=13−  x4 +3x2 =13x2 −9
x2
AnswerMarks
 x 4 − 1 0 x 2 + 9 = 0M1A1
( ) ( )
x 4 − 1 0 x 2 + 9 = 0  x 2 − 1 x 2 − 9 = 0  x 2 = . ..
AnswerMarks
 x = . . .M1
x = 1 , x = 3A1
(4)
Part (a) may also be done without obtaining a quartic:
9 9 9
y = x2 +3, y =13−  x2 +3=13−  x2 −10+ =0
x2 x2 x2

x −
9
x
 
x −
1
x

= 0
x −
9
x
= 0  x 2 = 9  x = 3 or x −
1
x
= 0  x 2 = 1  x = 1
M1: Solves simultaneously and attempts to factorise to x
x
x
x
0
  
−
 
−

=
A1: Correct factorisation
M1: Attempts to solve via x2
A1: x = 1 and x = 3. Both correct and no other values. Condone any confusion with which is P and
which is Q and condone e.g. P = 1, Q = 3
Part (a) may also be done via y e.g.:
y = x 2 + 3 , y = 1 3 −
9
x 2
 y = 1 3 −
y
9
− 3
 y 2 − 1 6 y + 4 8 = 0
y2 −16y+48=0 y=4,12
y = 4  x = 1 , y = 1 2  x = 3
PMT
M1: Solves simultaneously to obtain a 3TQ in y
A1: Correct 3TQ in y
M1: Solves their 3TQ in y and uses at least one value of y to find a value for x
A1: x = 1 and x = 3. Both correct and no other values. Condone any confusion with which is P and
which is Q and condone e.g. P = 1, Q = 3
AnswerMarks
(b)  1 3 − 9 − ( x 2 + 3 )  d x = 1 3 x + 9 − x 3 − 3 x ( + c )
x 2 x 3
or
  1 3 − 9  d x = 1 3 x + 9 ( + c ) ,  ( x 2 + 3 ) d x = x 3 + 3 x ( + c )
AnswerMarks
x 2 x 3M1A1
3
 9 x3 9 33  1
10x+ −  =10 ( 3 )+ − − 10+9−  =...
 x 3  3 3  3
1
or
 9 3 x3  3  1 
13x+ − +3x =39+3−( 13+9 )−9+9− +3=...
    
 x  3   3 
AnswerMarks
1 1dM1
16
=
AnswerMarks
3A1
(4)

Total 8

Question 8:
--- 8(a) ---
8(a) | 9
x2 +3=13−  x4 +3x2 =13x2 −9
x2
 x 4 − 1 0 x 2 + 9 = 0 | M1A1
( ) ( )
x 4 − 1 0 x 2 + 9 = 0  x 2 − 1 x 2 − 9 = 0  x 2 = . ..
 x = . . . | M1
x = 1 , x = 3 | A1
(4)
Part (a) may also be done without obtaining a quartic:
9 9 9
y = x2 +3, y =13−  x2 +3=13−  x2 −10+ =0
x2 x2 x2


x −
9
x
 
x −
1
x

= 0
x −
9
x
= 0  x 2 = 9  x = 3 or x −
1
x
= 0  x 2 = 1  x = 1
M1: Solves simultaneously and attempts to factorise to x
x
x
x
0
  
−
 
−

=
A1: Correct factorisation
M1: Attempts to solve via x2
A1: x = 1 and x = 3. Both correct and no other values. Condone any confusion with which is P and
which is Q and condone e.g. P = 1, Q = 3
Part (a) may also be done via y e.g.:
y = x 2 + 3 , y = 1 3 −
9
x 2
 y = 1 3 −
y
9
− 3
 y 2 − 1 6 y + 4 8 = 0
y2 −16y+48=0 y=4,12
y = 4  x = 1 , y = 1 2  x = 3
PMT
M1: Solves simultaneously to obtain a 3TQ in y
A1: Correct 3TQ in y
M1: Solves their 3TQ in y and uses at least one value of y to find a value for x
A1: x = 1 and x = 3. Both correct and no other values. Condone any confusion with which is P and
which is Q and condone e.g. P = 1, Q = 3
(b) |   1 3 − 9 − ( x 2 + 3 )  d x = 1 3 x + 9 − x 3 − 3 x ( + c )
x 2 x 3
or
  1 3 − 9  d x = 1 3 x + 9 ( + c ) ,  ( x 2 + 3 ) d x = x 3 + 3 x ( + c )
x 2 x 3 | M1A1
3
 9 x3 9 33  1
10x+ −  =10 ( 3 )+ − − 10+9−  =...
 x 3  3 3  3
1
or
 9 3 x3  3  1 
13x+ − +3x =39+3−( 13+9 )−9+9− +3=...
    
 x  3   3 
1 1 | dM1
16
=
3 | A1
(4)
Total 8
\begin{enumerate}
  \item In this question you must show all stages of your working.
\end{enumerate}

Solutions relying entirely on calculator technology are not acceptable.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{e7412e14-6a5a-4545-8d6b-4bceb141cc15-20_762_851_376_607}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}

Figure 1 shows a sketch of part of the curve $C _ { 1 }$ with equation

$$y = x ^ { 2 } + 3 \quad x > 0$$

and part of the curve $C _ { 2 }$ with equation

$$y = 13 - \frac { 9 } { x ^ { 2 } } \quad x > 0$$

The curves $C _ { 1 }$ and $C _ { 2 }$ intersect at the points $P$ and $Q$ as shown in Figure 1 .\\
(a) Use algebra to find the $x$ coordinate of $P$ and the $x$ coordinate of $Q$.

The finite region $R$, shown shaded in Figure 1, is bounded by $C _ { 1 }$ and $C _ { 2 }$\\
(b) Use algebraic integration to find the exact area of $R$.

\hfill \mbox{\textit{Edexcel PURE 2024 Q8}}
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