OCR MEI C2 — Question 2 13 marks

Exam BoardOCR MEI
ModuleC2 (Core Mathematics 2)
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStationary points and optimisation
TypeSecond derivative test justification
DifficultyStandard +0.3 This is a standard multi-part calculus question covering routine techniques: finding stationary points using first/second derivatives, verifying a gradient calculation, finding a point with given gradient, and computing a triangle area. All steps are straightforward applications of C2 methods with no novel problem-solving required, making it slightly easier than average.
Spec1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.07p Points of inflection: using second derivative1.08e Area between curve and x-axis: using definite integrals

2 A cubic curve has equation \(y = x ^ { 3 } - 3 x ^ { 2 } + 1\).
  1. Use calculus to find the coordinates of the turning points on this curve. Determine the nature of these turning points.
  2. Show that the tangent to the curve at the point where \(x = - 1\) has gradient 9 . Find the coordinates of the other point, P , on the curve at which the tangent has gradient 9 and find the equation of the normal to the curve at P . Show that the area of the triangle bounded by the normal at P and the \(x\) - and \(y\)-axes is 8 square units.

Question 2
Part i
B1: condone one error
Part ii
M1: use of \(y' = 0\)
AnswerMarks
A2: \((0, 1)\) or \((2, -3)\)A1 for one correct or \(x = 0, 2\)
T1: sign of \(y''\) used to test or \(y'\) either sideDep't on M1 or \(y\) either side or clear cubic sketch
B1: \(y' = 3x^2 - 6x\)
M1: \(y'(-1) = 3 + 6 = 9\)
A1: \(3x^2 - 6x = 9\)
AnswerMarks
B1: \(x = 3\)SC B1 for \((0,1)\) from their \(y'\)
B1: At P \(y = 1\)
M1: implies the M1
B1: grad normal \(= -\frac{1}{9}\) cao
AnswerMarks
A1: \(y - 1 = -\frac{1}{9}(x - 3)\)ft their \((3, 1)\) and their grad, not 9
B1: intercepts 12 and \(\frac{4}{3}\) or use of \(\int_0^3 \left(12 - \frac{4}{9}x\right) dx\) (their normal)
M1: (area calculation)
AnswerMarks
A1: \(\frac{1}{2} \times 12 \times \frac{4}{3}\) caoft their normal (linear)
Total: 13 marks
# Question 2

## Part i
B1: condone one error

## Part ii
M1: use of $y' = 0$
A2: $(0, 1)$ or $(2, -3)$ | A1 for one correct or $x = 0, 2$
T1: sign of $y''$ used to test or $y'$ either side | Dep't on M1 or $y$ either side or clear cubic sketch
B1: $y' = 3x^2 - 6x$
M1: $y'(-1) = 3 + 6 = 9$
A1: $3x^2 - 6x = 9$
B1: $x = 3$ | SC B1 for $(0,1)$ from their $y'$
B1: At P $y = 1$
M1: implies the M1
B1: grad normal $= -\frac{1}{9}$ cao
A1: $y - 1 = -\frac{1}{9}(x - 3)$ | ft their $(3, 1)$ and their grad, not 9
B1: intercepts 12 and $\frac{4}{3}$ or use of $\int_0^3 \left(12 - \frac{4}{9}x\right) dx$ (their normal)
M1: (area calculation)
A1: $\frac{1}{2} \times 12 \times \frac{4}{3}$ cao | ft their normal (linear)

**Total: 13 marks**
2 A cubic curve has equation $y = x ^ { 3 } - 3 x ^ { 2 } + 1$.\\
(i) Use calculus to find the coordinates of the turning points on this curve. Determine the nature of these turning points.\\
(ii) Show that the tangent to the curve at the point where $x = - 1$ has gradient 9 .

Find the coordinates of the other point, P , on the curve at which the tangent has gradient 9 and find the equation of the normal to the curve at P .

Show that the area of the triangle bounded by the normal at P and the $x$ - and $y$-axes is 8 square units.

\hfill \mbox{\textit{OCR MEI C2  Q2 [13]}}