CAIE P1 2015 June — Question 10 10 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2015
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVolumes of Revolution
TypeMulti-part: volume and tangent/normal
DifficultyStandard +0.3 Part (i) is a standard volume of revolution calculation requiring substitution and integration of a rational function - routine for this topic. Part (ii) requires finding where a line is normal to the curve by using the perpendicular gradient condition and solving simultaneously, which involves more steps but follows a well-practiced procedure. Overall slightly easier than average due to straightforward algebraic manipulation.
Spec1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations4.08d Volumes of revolution: about x and y axes
The equation of a curve is \(y = \frac{A}{2x - 1}\).
  1. Find, showing all necessary working, the volume obtained when the region bounded by the curve, the \(x\)-axis and the lines \(x = 1\) and \(x = 2\) is rotated through \(360°\) about the \(x\)-axis. [4]
  2. Given that the line \(2y = x + c\) is a normal to the curve, find the possible values of the constant \(c\). [6]
Question 10:
(i)
(ii) ---
10
(i)
AnswerMarks
(ii)4
y =
2x−1.
∫ 16 dx= −16
÷ 2
2
(2x−1) 2x−1
 −8 
Vol = π with limits 1 and 2
2x−1
16π
→
3
m = 1m of tangent = −2
2
dy −4
= ×2
dx (2x−1) 2
dy
Equating their to −2
dx
→ x= 3 or − 1
2 2
(y = 2 or – 2)
5 7
→ c = or −
AnswerMarks
2 2B1
B1
M1
A1
[4]
M1
B1
B1
DM1
A1
A1
AnswerMarks
[6]Correct without the ÷2
For the ÷2 even if first B1 is lost
Use of limits in a changed
expression.
co
Use of m 1 m 2 = −1
Correct without the ×2
For the ×2 even if first B1 is lost
co
co
11
(i)
(ii)
(iii)
(iv)
AnswerMarks
(v)f : xa2x2 −6x+5
2x2 −6x+5− p=0 has no real roots
Uses b2 −4ac→36−8(5− p)
Sets to 0 → p < 1
2
( )
2x2 −6x+5=2 x− 3 2 + 1
2 2
Range of g 1 < g(x) < 13
2
h : xa2x2 −6x+5 for k < x < 4
Smallest k = 3
2
( )
h(x) = 2 x− 3 2+ 1
2 2
Order of operations ±1, ÷2, √, ± 3
2 2
( )
→ Inverse = 3 + x −1
AnswerMarks
2 2 4M1
DM1
A1
[3]
3 × B1
[3]
B1 B1
[2]
B1
[1]
M1
DM1
A1
AnswerMarks
[3]Sets to 0 with p on LHS.
Uses discriminant.
co – must be “<”, not “<”.
co
on (ii) co from sub of x = 4
on (ii)
Using comp square form to try and
get x as subject or y if transposed.
Order must be correct
co (without ±)
Question 10:
--- 10
(i)
(ii) ---
10
(i)
(ii) | 4
y =
2x−1.
∫ 16 dx= −16
÷ 2
2
(2x−1) 2x−1
 −8 
Vol = π with limits 1 and 2
2x−1
16π
→
3
m = 1m of tangent = −2
2
dy −4
= ×2
dx (2x−1) 2
dy
Equating their to −2
dx
→ x= 3 or − 1
2 2
(y = 2 or – 2)
5 7
→ c = or −
2 2 | B1
B1
M1
A1
[4]
M1
B1
B1
DM1
A1
A1
[6] | Correct without the ÷2
For the ÷2 even if first B1 is lost
Use of limits in a changed
expression.
co
Use of m 1 m 2 = −1
Correct without the ×2
For the ×2 even if first B1 is lost
co
co
11
(i)
(ii)
(iii)
(iv)
(v) | f : xa2x2 −6x+5
2x2 −6x+5− p=0 has no real roots
Uses b2 −4ac→36−8(5− p)
Sets to 0 → p < 1
2
( )
2x2 −6x+5=2 x− 3 2 + 1
2 2
Range of g 1 < g(x) < 13
2
h : xa2x2 −6x+5 for k < x < 4
Smallest k = 3
2
( )
h(x) = 2 x− 3 2+ 1
2 2
Order of operations ±1, ÷2, √, ± 3
2 2
( )
→ Inverse = 3 + x −1
2 2 4 | M1
DM1
A1
[3]
3 × B1
[3]
B1 B1
[2]
B1
[1]
M1
DM1
A1
[3] | Sets to 0 with p on LHS.
Uses discriminant.
co – must be “<”, not “<”.
co
on (ii) co from sub of x = 4
on (ii)
Using comp square form to try and
get x as subject or y if transposed.
Order must be correct
co (without ±)
The equation of a curve is $y = \frac{A}{2x - 1}$.

\begin{enumerate}[label=(\roman*)]
\item Find, showing all necessary working, the volume obtained when the region bounded by the curve, the $x$-axis and the lines $x = 1$ and $x = 2$ is rotated through $360°$ about the $x$-axis. [4]
\item Given that the line $2y = x + c$ is a normal to the curve, find the possible values of the constant $c$. [6]
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2015 Q10 [10]}}
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