Question 5 - A-Level Maths

Edexcel Paper 2 2021 October — Question 5 7 marks

Exam Board	Edexcel
Module	Paper 2 (Paper 2)
Year	2021
Session	October
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Tangents, normals and gradients
Type	Determine nature of stationary points
Difficulty	Moderate -0.8 This is a straightforward differentiation question requiring routine application of the power rule for first and second derivatives, followed by standard stationary point analysis. All steps are mechanical with no problem-solving insight needed—simpler than the average A-level question which typically requires combining multiple techniques or some conceptual understanding.
Spec	1.07d Second derivatives: d^2y/dx^2 notation 1.07i Differentiate x^n: for rational n and sums 1.07n Stationary points: find maxima, minima using derivatives 1.07p Points of inflection: using second derivative

The curve $C$ has equation

$$y = 5 x ^ { 4 } - 24 x ^ { 3 } + 42 x ^ { 2 } - 32 x + 11 \quad x \in \mathbb { R }$$

Find
1. $\frac { \mathrm { d } y } { \mathrm {~d} x }$
2. $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$
1. Verify that $C$ has a stationary point at $x = 1$
2. Show that this stationary point is a point of inflection, giving reasons for your answer.

Show mark scheme Show mark scheme source

Question 5:

Part (a)(i):

Answer	Marks	Guidance
$\frac{dy}{dx} = 20x^3 - 72x^2 + 84x - 32$	M1, A1	M1: $x^n \to x^{n-1}$ for at least one power of $x$

Part (a)(ii):

Answer	Marks	Guidance
$\frac{d^2y}{dx^2} = 60x^2 - 144x + 84$	A1ft	Achieves correct $\frac{d^2y}{dx^2}$ for their $\frac{dy}{dx}$

Part (b)(i):

Answer	Marks	Guidance
$x = 1 \Rightarrow \frac{dy}{dx} = 20 - 72 + 84 - 32$	M1	Substitutes $x=1$ into their $\frac{dy}{dx}$
$\frac{dy}{dx} = 0$ so there is a stationary point at $x = 1$	A1	Obtains $\frac{dy}{dx} = 0$ following correct derivative and makes a minimal conclusion
Alternative: $20x^3 - 72x^2 + 84x - 32 = 4(x-1)^2(5x-8) = 0 \Rightarrow x = \ldots$	M1
When $x=1$, $\frac{dy}{dx} = 0$ so there is a stationary point	A1

Part (b)(ii):

Answer	Marks	Guidance
$\left(\frac{d^2y}{dx^2}\right)_{x=0.8} = \ldots$, $\left(\frac{d^2y}{dx^2}\right)_{x=1.2} = \ldots$	M1	Considers value of second derivative either side of $x=1$. Note: no marks for just evaluating at $x=1$
$\left(\frac{d^2y}{dx^2}\right)_{x=0.8} > 0$, $\left(\frac{d^2y}{dx^2}\right)_{x=1.2} < 0$, hence point of inflection	A1	Fully correct work with correct $\frac{d^2y}{dx^2}$, sign change shown, reasoned conclusion. Interval must be where $x < 1.4$
Alternative 1: $\left(\frac{d^2y}{dx^2}\right)_{x=1} = 0$ (inconclusive), find $\frac{d^3y}{dx^3} = 120x - 144 \Rightarrow \left(\frac{d^3y}{dx^3}\right)_{x=1} = \ldots$	M1	Shows second derivative at $x=1$ is zero then finds third derivative at $x=1$
$\left(\frac{d^2y}{dx^2}\right)_{x=1} = 0$ and $\left(\frac{d^3y}{dx^3}\right)_{x=1} \neq 0$, hence point of inflection	A1	Sufficient reason is "$\neq 0$". For reference $\left(\frac{d^3y}{dx^3}\right)_{x=1} = -24$
Alternative 2: $\left(\frac{dy}{dx}\right)_{x=0.8} = \ldots$, $\left(\frac{dy}{dx}\right)_{x=1.2} = \ldots$	M1	Considers value of first derivative either side of $x=1$
$\left(\frac{dy}{dx}\right)_{x=0.8} < 0$, $\left(\frac{dy}{dx}\right)_{x=1.2} < 0$, hence point of inflection	A1	Sufficient reason is "same sign"/"both negative". Interval must be where $x < 1.4$

## Question 5:

### Part (a)(i):
$\frac{dy}{dx} = 20x^3 - 72x^2 + 84x - 32$ | M1, A1 | M1: $x^n \to x^{n-1}$ for at least one power of $x$

### Part (a)(ii):
$\frac{d^2y}{dx^2} = 60x^2 - 144x + 84$ | A1ft | Achieves correct $\frac{d^2y}{dx^2}$ for their $\frac{dy}{dx}$

### Part (b)(i):
$x = 1 \Rightarrow \frac{dy}{dx} = 20 - 72 + 84 - 32$ | M1 | Substitutes $x=1$ into their $\frac{dy}{dx}$

$\frac{dy}{dx} = 0$ so there is a stationary point at $x = 1$ | A1 | Obtains $\frac{dy}{dx} = 0$ following correct derivative and makes a minimal conclusion

**Alternative:** $20x^3 - 72x^2 + 84x - 32 = 4(x-1)^2(5x-8) = 0 \Rightarrow x = \ldots$ | M1 | |

When $x=1$, $\frac{dy}{dx} = 0$ so there is a stationary point | A1 | |

### Part (b)(ii):
$\left(\frac{d^2y}{dx^2}\right)_{x=0.8} = \ldots$, $\left(\frac{d^2y}{dx^2}\right)_{x=1.2} = \ldots$ | M1 | Considers value of second derivative either side of $x=1$. Note: no marks for just evaluating at $x=1$

$\left(\frac{d^2y}{dx^2}\right)_{x=0.8} > 0$, $\left(\frac{d^2y}{dx^2}\right)_{x=1.2} < 0$, hence point of inflection | A1 | Fully correct work with correct $\frac{d^2y}{dx^2}$, sign change shown, reasoned conclusion. Interval must be where $x < 1.4$

**Alternative 1:** $\left(\frac{d^2y}{dx^2}\right)_{x=1} = 0$ (inconclusive), find $\frac{d^3y}{dx^3} = 120x - 144 \Rightarrow \left(\frac{d^3y}{dx^3}\right)_{x=1} = \ldots$ | M1 | Shows second derivative at $x=1$ is zero then finds third derivative at $x=1$

$\left(\frac{d^2y}{dx^2}\right)_{x=1} = 0$ and $\left(\frac{d^3y}{dx^3}\right)_{x=1} \neq 0$, hence point of inflection | A1 | Sufficient reason is "$\neq 0$". For reference $\left(\frac{d^3y}{dx^3}\right)_{x=1} = -24$

**Alternative 2:** $\left(\frac{dy}{dx}\right)_{x=0.8} = \ldots$, $\left(\frac{dy}{dx}\right)_{x=1.2} = \ldots$ | M1 | Considers value of first derivative either side of $x=1$

$\left(\frac{dy}{dx}\right)_{x=0.8} < 0$, $\left(\frac{dy}{dx}\right)_{x=1.2} < 0$, hence point of inflection | A1 | Sufficient reason is "same sign"/"both negative". Interval must be where $x < 1.4$

Show LaTeX source

\begin{enumerate}
  \item The curve $C$ has equation
\end{enumerate}

$$y = 5 x ^ { 4 } - 24 x ^ { 3 } + 42 x ^ { 2 } - 32 x + 11 \quad x \in \mathbb { R }$$

(a) Find\\
(i) $\frac { \mathrm { d } y } { \mathrm {~d} x }$\\
(ii) $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$\\
(b) (i) Verify that $C$ has a stationary point at $x = 1$\\
(ii) Show that this stationary point is a point of inflection, giving reasons for your answer.

\hfill \mbox{\textit{Edexcel Paper 2 2021 Q5 [7]}}

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Q1 4 Q2 5 Q3 3 Q4 3 Q5 7 Q6 5 Q7 9 Q8 9 Q9 3 Q10 6 Q11 10 Q12 7 Q13 6 Q14 12 Q15 11

Answer	Marks	Guidance
\(\left(\frac{d^2y}{dx^2}\right)_{x=0.8} = \ldots\), \(\left(\frac{d^2y}{dx^2}\right)_{x=1.2} = \ldots\)	M1	Considers value of second derivative either side of \(x=1\). Note: no marks for just evaluating at \(x=1\)
\(\left(\frac{d^2y}{dx^2}\right)_{x=0.8} > 0\), \(\left(\frac{d^2y}{dx^2}\right)_{x=1.2} < 0\), hence point of inflection	A1	Fully correct work with correct \(\frac{d^2y}{dx^2}\), sign change shown, reasoned conclusion. Interval must be where \(x < 1.4\)
Alternative 1: \(\left(\frac{d^2y}{dx^2}\right)_{x=1} = 0\) (inconclusive), find \(\frac{d^3y}{dx^3} = 120x - 144 \Rightarrow \left(\frac{d^3y}{dx^3}\right)_{x=1} = \ldots\)	M1	Shows second derivative at \(x=1\) is zero then finds third derivative at \(x=1\)
\(\left(\frac{d^2y}{dx^2}\right)_{x=1} = 0\) and \(\left(\frac{d^3y}{dx^3}\right)_{x=1} \neq 0\), hence point of inflection	A1	Sufficient reason is "\(\neq 0\)". For reference \(\left(\frac{d^3y}{dx^3}\right)_{x=1} = -24\)
Alternative 2: \(\left(\frac{dy}{dx}\right)_{x=0.8} = \ldots\), \(\left(\frac{dy}{dx}\right)_{x=1.2} = \ldots\)	M1	Considers value of first derivative either side of \(x=1\)
\(\left(\frac{dy}{dx}\right)_{x=0.8} < 0\), \(\left(\frac{dy}{dx}\right)_{x=1.2} < 0\), hence point of inflection	A1	Sufficient reason is "same sign"/"both negative". Interval must be where \(x < 1.4\)

Answer	Marks	Guidance
\(x = 1 \Rightarrow \frac{dy}{dx} = 20 - 72 + 84 - 32\)	M1	Substitutes \(x=1\) into their \(\frac{dy}{dx}\)
\(\frac{dy}{dx} = 0\) so there is a stationary point at \(x = 1\)	A1	Obtains \(\frac{dy}{dx} = 0\) following correct derivative and makes a minimal conclusion
Alternative: \(20x^3 - 72x^2 + 84x - 32 = 4(x-1)^2(5x-8) = 0 \Rightarrow x = \ldots\)	M1
When \(x=1\), \(\frac{dy}{dx} = 0\) so there is a stationary point	A1