OCR MEI Paper 3 2022 June — Question 10 5 marks

Exam BoardOCR MEI
ModulePaper 3 (Paper 3)
Year2022
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStationary points and optimisation
TypeClassify nature of stationary points
DifficultyStandard +0.3 This is a straightforward stationary point question requiring standard differentiation of polynomial and trigonometric terms, then sign-checking the derivative at given bounds and using the second derivative test. The algebraic manipulation is routine and the question provides helpful bounds to verify rather than requiring independent solving.
Spec1.07i Differentiate x^n: for rational n and sums1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx1.09a Sign change methods: locate roots

10 In this question you must show detailed reasoning. Fig. C2.2 indicates that the curve \(\mathrm { y } = \frac { 4 \mathrm { x } ( \pi - \mathrm { x } ) } { \pi ^ { 2 } } - \sin \mathrm { x }\) has a stationary point near \(x = 3\).
  • Verify that the \(x\)-coordinate of this stationary point is between 2.6 and 2.7.
  • Show that this stationary point is a maximum turning point.

Question 10:
AnswerMarks Guidance
\(\frac{dy}{dx} = \frac{4}{\pi} - \frac{8x}{\pi^2} - \cos x\)M1 3.1a
\(x = 2.6\), \(\frac{dy}{dx} = 0.0226...\)M1 3.1a
\(x = 2.7\), \(\frac{dy}{dx} = -0.0112...\)A1 1.1
Gradient zero for a value between 2.6 and 2.7E1 2.2a
Gradient positive, zero, negative so maxB1 2.4
[5 marks]
## Question 10:

$\frac{dy}{dx} = \frac{4}{\pi} - \frac{8x}{\pi^2} - \cos x$ | M1 | 3.1a | Attempt to differentiate. Both an $x$ term and a $\cos$ term needed, condone other errors. If $\pi^2$ is treated as a variable and incorrectly differentiated e.g. to $2\pi$ then M0. Similarly with $\pi$.

$x = 2.6$, $\frac{dy}{dx} = 0.0226...$ | M1 | 3.1a | At least one substitution into their expression

$x = 2.7$, $\frac{dy}{dx} = -0.0112...$ | A1 | 1.1 | Both correct (at least 2 d.p., rounded or truncated)

Gradient zero for a value between 2.6 and 2.7 | E1 | 2.2a | Can be implied by 'sign change' or sketch. Dependent on M2 but can be earned following **M2A0**

Gradient positive, zero, negative so max | B1 | 2.4 | Allow $2^{nd}$ derivative used at turning point found BC ($x = 2.67$, $2^{nd}$ deriv $= -0.356$) or 2 relevant values e.g. $-0.295$ and $-0.383$ seen and stated as $< 0$ therefore max

**[5 marks]**

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10 In this question you must show detailed reasoning.
Fig. C2.2 indicates that the curve $\mathrm { y } = \frac { 4 \mathrm { x } ( \pi - \mathrm { x } ) } { \pi ^ { 2 } } - \sin \mathrm { x }$ has a stationary point near $x = 3$.

\begin{itemize}
  \item Verify that the $x$-coordinate of this stationary point is between 2.6 and 2.7.
  \item Show that this stationary point is a maximum turning point.
\end{itemize}

\hfill \mbox{\textit{OCR MEI Paper 3 2022 Q10 [5]}}