Question 2 - A-Level Maths

Edexcel Paper 1 2018 June — Question 2 7 marks

Exam Board	Edexcel
Module	Paper 1 (Paper 1)
Year	2018
Session	June
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Chain Rule
Type	Find stationary points and nature
Difficulty	Moderate -0.8 This is a straightforward application of standard differentiation rules (power rule and chain rule for √x) followed by routine stationary point analysis. The algebra is simple, the verification at x=4 is direct substitution, and the second derivative test is a standard A-level technique requiring no problem-solving insight.
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

A curve $C$ has equation

$$y = x ^ { 2 } - 2 x - 24 \sqrt { x } , \quad x > 0$$

Find (i) $\frac { \mathrm { d } y } { \mathrm {~d} x }$ (ii) $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$
Verify that $C$ has a stationary point when $x = 4$
Determine the nature of this stationary point, giving a reason for your answer.

Show mark scheme Show mark scheme source

Question 2(a):

Answer	Marks	Guidance
Working/Answer	Mark	Guidance
(i) $\frac{\mathrm{d}y}{\mathrm{d}x} = 2x - 2 - 12x^{-\frac{1}{2}}$	M1, A1	1.1b, 1.1b
(ii) $\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2 + 6x^{-\frac{3}{2}}$	B1ft	1.1b

Notes (a)(i): M1 for differentiating to form $Ax + B + Cx^{-\frac{1}{2}}$. A1 for correct answer (coefficients may be unsimplified). (a)(ii): B1ft for correct $\frac{\mathrm{d}^2y}{\mathrm{d}x^2}$ from their $\frac{\mathrm{d}y}{\mathrm{d}x}$ which must have a negative or fractional index.

Question 2(b):

Answer	Marks	Guidance
Working/Answer	Mark	Guidance
Substitutes $x=4$ into their $\frac{\mathrm{d}y}{\mathrm{d}x} = 2\times4 - 2 - 12\times4^{-\frac{1}{2}} = \ldots$	M1	1.1b
Shows $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$ and states "hence there is a stationary point" oe	A1	2.1
Notes: M1 requires evidence \(\left.\frac{\mathrm{d}y}{\mathrm{d}x}\right	_{x=4} = \ldots\) A1 requires reason and minimal conclusion. Allow ✓, QED.

Question 2(c):

Answer	Marks	Guidance
Working/Answer	Mark	Guidance
Substitutes $x=4$ into $\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2 + 6\times4^{-\frac{3}{2}} = (2.75)$	M1	1.1b
$\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2.75 > 0$ and states "hence minimum"	A1ft	2.2a

Notes: M1 for substituting $x=4$ and calculating value or implying its sign. Must appear in (c). A1ft requires correct calculation, valid reason and correct conclusion. Follow through on incorrect $\frac{\mathrm{d}^2y}{\mathrm{d}x^2}$ dependent on having negative or fractional index.

# Question 2(a):

| Working/Answer | Mark | Guidance |
|---|---|---|
| (i) $\frac{\mathrm{d}y}{\mathrm{d}x} = 2x - 2 - 12x^{-\frac{1}{2}}$ | M1, A1 | 1.1b, 1.1b |
| (ii) $\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2 + 6x^{-\frac{3}{2}}$ | B1ft | 1.1b |

**Notes (a)(i):** M1 for differentiating to form $Ax + B + Cx^{-\frac{1}{2}}$. A1 for correct answer (coefficients may be unsimplified). **(a)(ii):** B1ft for correct $\frac{\mathrm{d}^2y}{\mathrm{d}x^2}$ from their $\frac{\mathrm{d}y}{\mathrm{d}x}$ which must have a negative or fractional index.

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# Question 2(b):

| Working/Answer | Mark | Guidance |
|---|---|---|
| Substitutes $x=4$ into their $\frac{\mathrm{d}y}{\mathrm{d}x} = 2\times4 - 2 - 12\times4^{-\frac{1}{2}} = \ldots$ | M1 | 1.1b |
| Shows $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$ and states "hence there is a stationary point" oe | A1 | 2.1 |

**Notes:** M1 requires evidence $\left.\frac{\mathrm{d}y}{\mathrm{d}x}\right|_{x=4} = \ldots$ A1 requires reason and minimal conclusion. Allow ✓, QED.

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# Question 2(c):

| Working/Answer | Mark | Guidance |
|---|---|---|
| Substitutes $x=4$ into $\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2 + 6\times4^{-\frac{3}{2}} = (2.75)$ | M1 | 1.1b |
| $\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 2.75 > 0$ and states "hence minimum" | A1ft | 2.2a |

**Notes:** M1 for substituting $x=4$ and calculating value or implying its sign. Must appear in (c). A1ft requires correct calculation, valid reason and correct conclusion. Follow through on incorrect $\frac{\mathrm{d}^2y}{\mathrm{d}x^2}$ dependent on having negative or fractional index.

---

Show LaTeX source

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

$$y = x ^ { 2 } - 2 x - 24 \sqrt { x } , \quad x > 0$$

(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 $x = 4$\\
(c) Determine the nature of this stationary point, giving a reason for your answer.

\hfill \mbox{\textit{Edexcel Paper 1 2018 Q2 [7]}}

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Q1 3 Q2 7 Q3 4 Q4 4 Q5 5 Q6 10 Q7 7 Q8 5 Q9 10 Q10 8 Q11 10 Q12 10 Q13 7 Q14 10

Edexcel Paper 1 2018 June — Question 2 7 marks

Question 2(a):

Notes (a)(i): M1 for differentiating to form \(Ax + B + Cx^{-\frac{1}{2}}\). A1 for correct answer (coefficients may be unsimplified). (a)(ii): B1ft for correct \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\) from their \(\frac{\mathrm{d}y}{\mathrm{d}x}\) which must have a negative or fractional index.

Question 2(b):

Question 2(c):

Notes: M1 for substituting \(x=4\) and calculating value or implying its sign. Must appear in (c). A1ft requires correct calculation, valid reason and correct conclusion. Follow through on incorrect \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\) dependent on having negative or fractional index.

This paper (14 questions)