OCR Further Additional Pure 2019 June — Question 2 11 marks

Exam BoardOCR
ModuleFurther Additional Pure (Further Additional Pure)
Year2019
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVector Product and Surfaces
TypeClassifying stationary points on surfaces
DifficultyStandard +0.8 This is a Further Maths question on multivariable calculus requiring computation of five partial derivatives, verification of a stationary point, and classification using the second derivative test (Hessian determinant). While systematic, it demands careful differentiation of products involving trig functions and proper application of the discriminant D = f_xx·f_yy - (f_xy)^2, which is beyond standard A-level and involves multiple computational steps with potential for algebraic errors.
Spec8.05d Partial differentiation: first and second order, mixed derivatives8.05e Stationary points: where partial derivatives are zero8.05f Nature of stationary points: classify using Hessian matrix
2 A surface has equation \(z = \mathrm { f } ( x , y )\) where \(\mathrm { f } ( x , y ) = x ^ { 2 } \sin y + 2 y \cos x\).
  1. Determine \(\mathrm { f } _ { x } , \mathrm { f } _ { y } , \mathrm { f } _ { x x } , \mathrm { f } _ { y y } , \mathrm { f } _ { x y }\) and \(\mathrm { f } _ { y x }\).
    1. Verify that \(z\) has a stationary point at \(\left( \frac { 1 } { 2 } \pi , \frac { 1 } { 2 } \pi , \frac { 1 } { 4 } \pi ^ { 2 } \right)\).
    2. Determine the nature of this stationary point.
Question 2:
AnswerMarks Guidance
2(a) f =2xsiny−2ysinx f = x2cosy+2cosx
x y
f = 2sin y−2ycosx f =−x2sin y
xx yy
f =f =2xcosy−2sinx
AnswerMarks
xy yx1.1a
1.1
1.1 2.5
AnswerMarks
1.1B1 B1
B1 B1
B1
AnswerMarks Guidance
[5]Both must be written down, or stated equal
(b)i When x = y = 1π, f = f =0
2 x y
⇒ stationary point
( )2×1+2× ( )
Visible check/calculation that z = 1π 1π ×0 = 1π2
AnswerMarks
2 2 41.1a
2.2a
AnswerMarks
2.1M1
A1
B1
AnswerMarks
[3]Allow statement only
iif f
xx xy
AnswerMarks
H= attempted with numerical values
f f
yx yy
2 −2
AnswerMarks
H= = –1π2 −4 (or –8.935-ish)
−2 − 1 π2 2
4
AnswerMarks Guidance
⇒ Saddle-point sinceH < 0
2.1
AnswerMarks
2.2aM1
A1
B1
AnswerMarks
[3]Considering (the determinant of) the Hessian matrix, or
equivalent
AnswerMarks Guidance
Correctly calculated value ofH
FT conclusion fromH < 0
(a)7x ≡ 6 ≡ 25 ≡ 44 ≡ 63 …
⇒ x ≡ 9 (mod 19)1.1
1.1M1
A1Adding 19s to RHS to reach a multiple of 7
Must be a general result (i.e. not just x = 9)
Alt.
AnswerMarks Guidance
7 – 1 (mod 19) is 11 and multiplying through by thisM1
⇒ x ≡ 9 (mod 19)A1 Must be a general result (i.e. not just x = 9)
[2]No requirement to justify by hcf(7, 19) = 1 e.g.
x ≡ 3 (mod 4) ⇒ x is odd while x ≡ 4 (mod 6) ⇒ x is even
AnswerMarks
Statement that there are no solutions3.1a
2.1M1
A1Consideration of parity (i.e. mod 2) or equivalent
For valid reason
Alt.
AnswerMarks Guidance
h = hcf(4, 6) = 2 and solutions exist provided h(4 – 3) M1
But 2 ∤ 1 hence no solutionsA1 Conclusion with justification
[2]
Question 2:
2 | (a) | f =2xsiny−2ysinx f = x2cosy+2cosx
x y
f = 2sin y−2ycosx f =−x2sin y
xx yy
f =f =2xcosy−2sinx
xy yx | 1.1a
1.1
1.1 2.5
1.1 | B1 B1
B1 B1
B1
[5] | Both must be written down, or stated equal
(b) | i | When x = y = 1π, f = f =0
2 x y
⇒ stationary point
( )2×1+2× ( )
Visible check/calculation that z = 1π 1π ×0 = 1π2
2 2 4 | 1.1a
2.2a
2.1 | M1
A1
B1
[3] | Allow statement only
ii | f f
xx xy
| H | = attempted with numerical values
f f
yx yy
2 −2
| H | = = –1π2 −4 (or –8.935-ish)
−2 − 1 π2 2
4
⇒ Saddle-point since | H | < 0 | 1.1
2.1
2.2a | M1
A1
B1
[3] | Considering (the determinant of) the Hessian matrix, or
equivalent
Correctly calculated value of | H |
FT conclusion from | H | < 0
(a) | 7x ≡ 6 ≡ 25 ≡ 44 ≡ 63 …
⇒ x ≡ 9 (mod 19) | 1.1
1.1 | M1
A1 | Adding 19s to RHS to reach a multiple of 7
Must be a general result (i.e. not just x = 9)
Alt.
7 – 1 (mod 19) is 11 and multiplying through by this | M1
⇒ x ≡ 9 (mod 19) | A1 | Must be a general result (i.e. not just x = 9)
[2] | No requirement to justify by hcf(7, 19) = 1 e.g.
x ≡ 3 (mod 4) ⇒ x is odd while x ≡ 4 (mod 6) ⇒ x is even
Statement that there are no solutions | 3.1a
2.1 | M1
A1 | Consideration of parity (i.e. mod 2) or equivalent
For valid reason
Alt.
h = hcf(4, 6) = 2 and solutions exist provided h | (4 – 3) | M1 | Considering h and “b – a”
But 2 ∤ 1 hence no solutions | A1 | Conclusion with justification
[2]
2 A surface has equation $z = \mathrm { f } ( x , y )$ where $\mathrm { f } ( x , y ) = x ^ { 2 } \sin y + 2 y \cos x$.
\begin{enumerate}[label=(\alph*)]
\item Determine $\mathrm { f } _ { x } , \mathrm { f } _ { y } , \mathrm { f } _ { x x } , \mathrm { f } _ { y y } , \mathrm { f } _ { x y }$ and $\mathrm { f } _ { y x }$.
\item \begin{enumerate}[label=(\roman*)]
\item Verify that $z$ has a stationary point at $\left( \frac { 1 } { 2 } \pi , \frac { 1 } { 2 } \pi , \frac { 1 } { 4 } \pi ^ { 2 } \right)$.
\item Determine the nature of this stationary point.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{OCR Further Additional Pure 2019 Q2 [11]}}
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