OCR C1 — Question 8 9 marks

Exam BoardOCR
ModuleC1 (Core Mathematics 1)
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStationary points and optimisation
TypeSecond derivative test justification
DifficultyModerate -0.3 This is a straightforward stationary points question requiring differentiation of a power function (including fractional power), solving f'(x)=0, and using the second derivative test. All steps are routine C1 techniques with no problem-solving insight needed, making it slightly easier than average, though the fractional power adds minor computational care.
Spec1.07d Second derivatives: d^2y/dx^2 notation1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx

8. $$f ( x ) = 2 - x + 3 x ^ { \frac { 2 } { 3 } } , \quad x > 0 .$$
  1. Find \(f ^ { \prime } ( x )\) and \(f ^ { \prime \prime } ( x )\).
  2. Find the coordinates of the turning point of the curve \(y = \mathrm { f } ( x )\).
  3. Determine whether the turning point is a maximum or minimum point.

Question 8:
Part (i):
AnswerMarks Guidance
AnswerMark Notes
\(f'(x) = -1 + 2x^{-\frac{1}{3}}\)M1 A1
\(f''(x) = -\frac{2}{3}x^{-\frac{4}{3}}\)A1
Part (ii):
AnswerMarks Guidance
AnswerMark Notes
For TP, \(-1 + 2x^{-\frac{1}{3}} = 0\)M1
\(x^{\frac{1}{3}} = 2\)M1
\(x = 8 \quad \therefore (8,6)\)A2
Part (iii):
AnswerMarks Guidance
AnswerMark Notes
\(f''(8) = -\frac{1}{24}\), \(f''(x) < 0 \therefore\) maximumM1 A1 (9)
# Question 8:

## Part (i):
| Answer | Mark | Notes |
|--------|------|-------|
| $f'(x) = -1 + 2x^{-\frac{1}{3}}$ | M1 A1 | |
| $f''(x) = -\frac{2}{3}x^{-\frac{4}{3}}$ | A1 | |

## Part (ii):
| Answer | Mark | Notes |
|--------|------|-------|
| For TP, $-1 + 2x^{-\frac{1}{3}} = 0$ | M1 | |
| $x^{\frac{1}{3}} = 2$ | M1 | |
| $x = 8 \quad \therefore (8,6)$ | A2 | |

## Part (iii):
| Answer | Mark | Notes |
|--------|------|-------|
| $f''(8) = -\frac{1}{24}$, $f''(x) < 0 \therefore$ maximum | M1 A1 | **(9)** |

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8.

$$f ( x ) = 2 - x + 3 x ^ { \frac { 2 } { 3 } } , \quad x > 0 .$$

(i) Find $f ^ { \prime } ( x )$ and $f ^ { \prime \prime } ( x )$.\\
(ii) Find the coordinates of the turning point of the curve $y = \mathrm { f } ( x )$.\\
(iii) Determine whether the turning point is a maximum or minimum point.\\

\hfill \mbox{\textit{OCR C1  Q8 [9]}}