OCR Further Additional Pure 2021 November — Question 5 8 marks

Exam BoardOCR
ModuleFurther Additional Pure (Further Additional Pure)
Year2021
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVector Product and Surfaces
TypeFinding stationary points on surfaces
DifficultyChallenging +1.8 This is a Further Maths question requiring implicit differentiation of a 3D surface and analysis of stationary points. Part (a) involves careful algebraic manipulation of partial derivatives, while part (b) requires recognizing that the equation from (a) cannot equal zero (since RHS is always positive). The multi-step reasoning and conceptual understanding of stationary points in 3D elevate this above standard calculus, though the techniques themselves are systematic once the approach is identified.
Spec8.05d Partial differentiation: first and second order, mixed derivatives
5 The surface \(S\) has equation \(x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = x y z - 1\).
  1. Show that \(( 2 z - x y ) \left( x \frac { \partial z } { \partial x } + y \frac { \partial z } { \partial y } \right) = 2 \left( 1 + z ^ { 2 } \right)\).
  2. Deduce that \(S\) has no stationary point.
Question 5:
AnswerMarks Guidance
5(a) Differentiating x 2 + y 2 + z 2 = x y z − 1 partially w.r.t. either x or y
 z  z  z  z
x): 2 x + 2 z = x y + y z  y): 2 y + 2 z = x y + x z
 x  x  y  y
( )   z  z    z  z 
x. + y.  2 x 2 + y 2 + 2 z x + y = x y x + y + 2 x y z
 
 x  y  x  y
Now x2 + y2 + z2 = xyz−1  xyz−x2 − y2 = z2 +1
 z z ( )
giving (2z−xy)x + y =21+z2
 
AnswerMarks
 x yM1
A1
A1
M1
M1
A1
AnswerMarks
[6]1.1
1.1
1.1
3.1a
1.1
AnswerMarks
2.2aAttempt with LHS correct
First correct
Correct or FT x  y (by symmetry)
  z  z  ( )
 ( 2 z − x y ) x + y = 2 x y z − x 2 − y 2

 x  y
AG legitimately obtained
AnswerMarks
(b) z  z
If both and are zero …
 x  y
AnswerMarks
… then LHS = 0 while RHS  2 (i.e. > 0) ()M1
A1
AnswerMarks
[2]2.1
2.4Considering conditions for stationary points
Contradiction justified
Question 5:
5 | (a) | Differentiating x 2 + y 2 + z 2 = x y z − 1 partially w.r.t. either x or y
 z  z  z  z
x): 2 x + 2 z = x y + y z  y): 2 y + 2 z = x y + x z
 x  x  y  y

( )   z  z    z  z 
x. + y.  2 x 2 + y 2 + 2 z x + y = x y x + y + 2 x y z
 
 x  y  x  y
Now x2 + y2 + z2 = xyz−1  xyz−x2 − y2 = z2 +1
 z z ( )
giving (2z−xy)x + y =21+z2
 
 x y | M1
A1
A1
M1
M1
A1
[6] | 1.1
1.1
1.1
3.1a
1.1
2.2a | Attempt with LHS correct
First correct
Correct or FT x  y (by symmetry)
  z  z  ( )
 ( 2 z − x y ) x + y = 2 x y z − x 2 − y 2

 x  y
AG legitimately obtained
(b) |  z  z
If both and are zero …
 x  y
… then LHS = 0 while RHS  2 (i.e. > 0) () | M1
A1
[2] | 2.1
2.4 | Considering conditions for stationary points
Contradiction justified
5 The surface $S$ has equation $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = x y z - 1$.
\begin{enumerate}[label=(\alph*)]
\item Show that $( 2 z - x y ) \left( x \frac { \partial z } { \partial x } + y \frac { \partial z } { \partial y } \right) = 2 \left( 1 + z ^ { 2 } \right)$.
\item Deduce that $S$ has no stationary point.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Additional Pure 2021 Q5 [8]}}
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