CAIE P1 2016 November — Question 11 12 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2016
SessionNovember
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVolumes of Revolution
TypeMulti-part: volume and stationary points
DifficultyStandard +0.3 Part (i) requires standard differentiation of a sum involving negative power and linear function, solving dy/dx=0, and using second derivative test—routine calculus. Part (ii) is a straightforward volume of revolution calculation using the standard formula with polynomial expansion. Both parts follow textbook procedures with no novel insight required, making this slightly easier than average.
Spec1.07n Stationary points: find maxima, minima using derivatives1.07p Points of inflection: using second derivative4.08d Volumes of revolution: about x and y axes

A curve has equation \(y = (kx - 3)^{-1} + (kx - 3)\), where \(k\) is a non-zero constant.
  1. Find the \(x\)-coordinates of the stationary points in terms of \(k\), and determine the nature of each stationary point. Justify your answers. [7]
  1. \includegraphics{figure_3} The diagram shows part of the curve for the case when \(k = 1\). Showing all necessary working, find the volume obtained when the region between the curve, the \(x\)-axis, the \(y\)-axis and the line \(x = 2\), shown shaded in the diagram, is rotated through \(360°\) about the \(x\)-axis. [5]

Question 11:

(ii) ---
11 (i)
AnswerMarks
(ii)dy
=−k(kx−3)−2
Attempt diffn. and equate to 0 +k =0
dx
(kx−3)2 =1 or k3x2 −6k2x+8k (=0)
2 4
x= or
k k
d2y
=2k2(kx−3)−3
dx2
2 d2y ( −2k2)
When x= , = <0 MAX All previous
k dx2
4 d2y ( 2k2)
When x= , = >0 MIN working correct
k dx2
2
V =(π)∫(x−3)−1 +(x−3) dx
 
=(π)∫[(x−3)−2 +(x−3)2
+2]dx
 (x−3)3 
=(π)−(x−3)−1 + (+2x) Condone missing 2x
 3 
 
 1 1 
=(π) 1− +4− −9+0
  
 3 3 
AnswerMarks
=40π/3 oe or 41.9*M1
DM1
*A1*A1
B1
DB1
DB1
*M1
A1
A1
DM1
AnswerMarks
A1[7]
[5](kx−3)−2
Must contain + other
term(s)
Simplify to a quadratic
Legitimately obtained
Ak2(kx−3)−3
Ft must contain
where A>0
Convincing alt. methods (values
either side) must show which
values used & cannot use
x=3/k
Attempt to expand y² and then
integrate
Or
 x3 
−(x−3)−1 + −3x2 +9x+2x
3
 
Apply limits 0→2
2 missing → 28π/3 scores
M1A0A1M1A0
Question 11:
--- 11 (i)
(ii) ---
11 (i)
(ii) | dy
=−k(kx−3)−2
Attempt diffn. and equate to 0 +k =0
dx
(kx−3)2 =1 or k3x2 −6k2x+8k (=0)
2 4
x= or
k k
d2y
=2k2(kx−3)−3
dx2
2 d2y ( −2k2)
When x= , = <0 MAX All previous
k dx2
4 d2y ( 2k2)
When x= , = >0 MIN working correct
k dx2
2
V =(π)∫(x−3)−1 +(x−3) dx
 
=(π)∫[(x−3)−2 +(x−3)2
+2]dx
 (x−3)3 
=(π)−(x−3)−1 + (+2x) Condone missing 2x
 3 
 
 1 1 
=(π) 1− +4− −9+0
  
 3 3 
=40π/3 oe or 41.9 | *M1
DM1
*A1*A1
B1
DB1
DB1
*M1
A1
A1
DM1
A1 | [7]
[5] | (kx−3)−2
Must contain + other
term(s)
Simplify to a quadratic
Legitimately obtained
Ak2(kx−3)−3
Ft must contain
where A>0
Convincing alt. methods (values
either side) must show which
values used & cannot use
x=3/k
Attempt to expand y² and then
integrate
Or
 x3 
−(x−3)−1 + −3x2 +9x+2x
3
 
Apply limits 0→2
2 missing → 28π/3 scores
M1A0A1M1A0
A curve has equation $y = (kx - 3)^{-1} + (kx - 3)$, where $k$ is a non-zero constant.

\begin{enumerate}[label=(\roman*)]
\item Find the $x$-coordinates of the stationary points in terms of $k$, and determine the nature of each stationary point. Justify your answers. [7]
\end{enumerate}

\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item \includegraphics{figure_3}

The diagram shows part of the curve for the case when $k = 1$. Showing all necessary working, find the volume obtained when the region between the curve, the $x$-axis, the $y$-axis and the line $x = 2$, shown shaded in the diagram, is rotated through $360°$ about the $x$-axis. [5]
\end{enumerate}

\hfill \mbox{\textit{CAIE P1 2016 Q11 [12]}}