Edexcel Paper 1 Specimen — Question 1 7 marks

Exam BoardEdexcel
ModulePaper 1 (Paper 1)
SessionSpecimen
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTangents, normals and gradients
TypeDetermine nature of stationary points
DifficultyEasy -1.2 This is a straightforward differentiation question requiring only routine application of the power rule twice, identification of a stationary point by substitution, and standard second derivative test. All steps are mechanical with no problem-solving or insight required, making it easier than average.
Spec1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta
  1. The curve \(C\) has equation
$$y = 3 x ^ { 4 } - 8 x ^ { 3 } - 3$$
  1. Find (i) \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) (ii) \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\)
  2. Verify that \(C\) has a stationary point when \(x = 2\)
  3. Determine the nature of this stationary point, giving a reason for your answer.
Question 1
1(a)(i)
AnswerMarks Guidance
Differentiates to a cubic formM1 1.1b
\(\frac{dy}{dx} = 12x^3 - 24x^2\)A1 1.1b
1(a)(ii)
AnswerMarks Guidance
\(\frac{d^2y}{dx^2} = 36x^2 - 48x\)A1ft 1.1b
Guidance: A1ft is achieved for correct \(\frac{d^2y}{dx^2}\) from their \(\frac{dy}{dx} = 12x^3 - 24x^2\)
1(b)
AnswerMarks Guidance
Substitutes \(x = 2\) into their \(\frac{dy}{dx}\)M1 1.1b
\(\frac{dy}{dx} = 0\) and states "hence there is a stationary point"A1 2.1
Guidance: All aspects of the proof must be correct
1(c)
AnswerMarks Guidance
Substitutes \(x = 2\) into their \(\frac{d^2y}{dx^2}\)M1 1.1b
Alternatively: Calculates the gradient of \(C\) either side of \(x = 2\)
AnswerMarks Guidance
\(\frac{d^2y}{dx^2} = 48 > 0\) and states "hence the stationary point is a minimum"A1ft 2.2a
Guidance: A1ft for a correct calculation, a valid reason and a correct conclusion. Follow through on an incorrect \(\frac{d^2y}{dx^2}\)
# Question 1

## 1(a)(i)
Differentiates to a cubic form | M1 | 1.1b

$\frac{dy}{dx} = 12x^3 - 24x^2$ | A1 | 1.1b

## 1(a)(ii)
$\frac{d^2y}{dx^2} = 36x^2 - 48x$ | A1ft | 1.1b

**Guidance:** A1ft is achieved for correct $\frac{d^2y}{dx^2}$ from their $\frac{dy}{dx} = 12x^3 - 24x^2$

## 1(b)
Substitutes $x = 2$ into their $\frac{dy}{dx}$ | M1 | 1.1b

$\frac{dy}{dx} = 0$ and states "hence there is a stationary point" | A1 | 2.1

**Guidance:** All aspects of the proof must be correct

## 1(c)
Substitutes $x = 2$ into their $\frac{d^2y}{dx^2}$ | M1 | 1.1b

Alternatively: Calculates the gradient of $C$ either side of $x = 2$

$\frac{d^2y}{dx^2} = 48 > 0$ and states "hence the stationary point is a minimum" | A1ft | 2.2a

**Guidance:** A1ft for a correct calculation, a valid reason and a correct conclusion. Follow through on an incorrect $\frac{d^2y}{dx^2}$
\begin{enumerate}
  \item The curve $C$ has equation
\end{enumerate}

$$y = 3 x ^ { 4 } - 8 x ^ { 3 } - 3$$

(a) Find (i) $\frac { \mathrm { d } y } { \mathrm {~d} x }$\\
(ii) $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$\\
(b) Verify that $C$ has a stationary point when $x = 2$\\
(c) Determine the nature of this stationary point, giving a reason for your answer.

\hfill \mbox{\textit{Edexcel Paper 1  Q1 [7]}}
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