CAIE P1 2015 June — Question 10 11 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2015
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAreas by integration
TypeArea between curve and line
DifficultyStandard +0.3 This is a straightforward integration question requiring finding a normal line equation (using chain rule differentiation) and calculating two areas by integration. While it involves multiple steps (differentiation, normal equation, two integrations), each technique is standard P1 material with no novel insights required. The 'show areas are equal' adds mild problem-solving but the path is clear.
Spec1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07m Tangents and normals: gradient and equations1.08e Area between curve and x-axis: using definite integrals1.08h Integration by substitution

\includegraphics{figure_10} The diagram shows part of the curve \(y = \frac{8}{\sqrt{(3x + 4)}}\). The curve intersects the \(y\)-axis at \(A(0, 4)\). The normal to the curve at \(A\) intersects the line \(x = 4\) at the point \(B\).
  1. Find the coordinates of \(B\). [5]
  2. Show, with all necessary working, that the areas of the regions marked \(P\) and \(Q\) are equal. [6]

Question 10:

(i)
(ii) ---
10
(i)
AnswerMarks
(ii)8
y =
3x+4
dy −4
= × 3 aef
dx (3x+4)2 3
3 2
→ m (x=0) = − Perpendicular m (x=0) =
2 3
2
Eqn of normal y−4= (x−0)
3
 20
Meets x = 4 at B 4, 
 3 
8 8 (3x+4))
∫ dx= ÷3
(3x+4) 1
2
32
Limits from 0 to 4 → Area P =
3
Area Q = Trapezium – P
Area of Trapezium =
1 20 64
4+ ×4=
2 3  3
32
→ Areas of P and Q are both
AnswerMarks
3B1
B1
M1
M1
A1
[5]
B1 B1
M1 A1
M1
A1
AnswerMarks
[6]Without the “×3”
For “×3” even if 1st B mark lost.
Use of m 1 m 2 = −1 after attempting
dy
to find (x=0)
dx
Unsimplified line equation
cao
Without “÷3”. For “÷3”
Correct use of correct limits. cao
Correct method for area of
trapezium
All correct.
Question 10:
--- 10
(i)
(ii) ---
10
(i)
(ii) | 8
y =
3x+4
dy −4
= × 3 aef
dx (3x+4)2 3
3 2
→ m (x=0) = − Perpendicular m (x=0) =
2 3
2
Eqn of normal y−4= (x−0)
3
 20
Meets x = 4 at B 4, 
 3 
8 8 (3x+4))
∫ dx= ÷3
(3x+4) 1
2
32
Limits from 0 to 4 → Area P =
3
Area Q = Trapezium – P
Area of Trapezium =
1 20 64
4+ ×4=
2 3  3
32
→ Areas of P and Q are both
3 | B1
B1
M1
M1
A1
[5]
B1 B1
M1 A1
M1
A1
[6] | Without the “×3”
For “×3” even if 1st B mark lost.
Use of m 1 m 2 = −1 after attempting
dy
to find (x=0)
dx
Unsimplified line equation
cao
Without “÷3”. For “÷3”
Correct use of correct limits. cao
Correct method for area of
trapezium
All correct.
\includegraphics{figure_10}

The diagram shows part of the curve $y = \frac{8}{\sqrt{(3x + 4)}}$. The curve intersects the $y$-axis at $A(0, 4)$. The normal to the curve at $A$ intersects the line $x = 4$ at the point $B$.

\begin{enumerate}[label=(\roman*)]
\item Find the coordinates of $B$. [5]
\item Show, with all necessary working, that the areas of the regions marked $P$ and $Q$ are equal. [6]
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2015 Q10 [11]}}