OCR MEI C2 — Question 3 5 marks

Exam BoardOCR MEI
ModuleC2 (Core Mathematics 2)
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStationary points and optimisation
TypeSecond derivative test justification
DifficultyModerate -0.8 This is a straightforward application of basic differentiation rules (power rule) to find stationary points and classify them using the second derivative test. The function is simple, requiring only differentiation of x and x^(-1), solving a trivial equation (1 - 1/x² = 0), and verifying the second derivative is positive. This is easier than average as it's a standard textbook exercise with clear steps and no problem-solving insight required.
Spec1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.07p Points of inflection: using second derivative
3 A curve has equation \(y = x + \frac { 1 } { x }\).
Use calculus to show that the curve has a turning point at \(x = 1\).
Show also that this point is a minimum.
Question 3:
AnswerMarks
\(y = x + x^{-1}\) soiB1
\(y' = 1 - \frac{1}{x^2}\)B1
subs \(x = 1\) to get \(y' = 0\)B1
\(y'' = 2x^{-3}\) attemptedM1ft
Stating \(y'' > 0\) so min caoA1
Note: \(1 - x^{-2}\) is acceptable
Or solving \(1 - x^{-2} = 0\) to obtain \(x = 1\)
Or checking \(y'\) before and after \(x = 1\)
AnswerMarks
Valid conclusion
First quadrant sketch scoresB2

Total: 5 marks

Question 3:

$y = x + x^{-1}$ soi | B1

$y' = 1 - \frac{1}{x^2}$ | B1

subs $x = 1$ to get $y' = 0$ | B1

$y'' = 2x^{-3}$ attempted | M1ft

Stating $y'' > 0$ so min cao | A1

Note: $1 - x^{-2}$ is acceptable

Or solving $1 - x^{-2} = 0$ to obtain $x = 1$

Or checking $y'$ before and after $x = 1$

Valid conclusion | 

First quadrant sketch scores | B2

Total: 5 marks
3 A curve has equation $y = x + \frac { 1 } { x }$.\\
Use calculus to show that the curve has a turning point at $x = 1$.\\
Show also that this point is a minimum.

\hfill \mbox{\textit{OCR MEI C2  Q3 [5]}}
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Q1 12 Q2 13 Q3 5 Q4 12 Q5 5