Edexcel C1 2018 June — Question 2 7 marks

Exam BoardEdexcel
ModuleC1 (Core Mathematics 1)
Year2018
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTangents, normals and gradients
TypeFind stationary points
DifficultyEasy -1.3 This is a straightforward C1 question testing basic integration and differentiation of simple power functions (rewriting √x as x^(1/2)), followed by solving a linear equation. All steps are routine applications of standard rules with no problem-solving or conceptual challenges required.
Spec1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.08b Integrate x^n: where n != -1 and sums

  1. Given
$$y = 3 \sqrt { x } - 6 x + 4 , \quad x > 0$$
  1. find \(\int y \mathrm {~d} x\), simplifying each term.
    1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\)
    2. Hence find the value of \(x\) such that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\)

Question 2:
Part (a)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(x^n \rightarrow x^{n+1}\), i.e. \(x^{1.5}\) or \(x^2\) or \(x\) seen (not for "\(+ c\)")M1 For \(x^n \rightarrow x^{n+1}\)
Two out of three terms correct un-simplified or simplified (ignore \(+c\))A1
\(2x^{1.5} - 3x^2 + 4x + c\)A1 All correct and simplified on one line including "\(+ c\)". Allow \(\sqrt{x^3}\) for \(x^{1.5}\) but not \(x^1\) for \(x\).
Ignore any spurious integral signs.
Part (b)(i)
*(Mark (b)(i) and (ii) together and must be differentiating the original function not their answer to part (a))*
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(x^n \rightarrow x^{n-1}\), i.e. \(x^{0.5} \rightarrow x^{-0.5}\) or \(6x \rightarrow 6\)M1 For \(x^n \rightarrow x^{n-1}\)
\(\frac{3}{2}x^{-0.5} - 6\)A1 For \(\frac{3}{2}x^{-0.5} - 6\) or equivalent. May be un-simplified. Allow \(\frac{\frac{3}{2}}{\sqrt{x}} - 6\)
Part (b)(ii)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\frac{3}{2}x^{-0.5} - 6 = 0 \Rightarrow x^n = \ldots\)M1 Sets their \(\frac{dy}{dx} = 0\) (may be implied by their working) and reaches \(x^n = C\) (including \(n=1\)) with correct processing allowing sign errors only. May be implied by e.g. \(\sqrt{x} = \frac{1}{4}\) or \(\frac{1}{\sqrt{x}} = 4\)
\(x = \frac{1}{16}\) csoA1 Allow equivalent fractions e.g. \(\frac{9}{144}\) or \(0.0625\). If other solutions are given (e.g. likely to be \(x = 0\) or \(x = -\frac{1}{16}\)) then this mark should be withheld.
# Question 2:

## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x^n \rightarrow x^{n+1}$, i.e. $x^{1.5}$ or $x^2$ or $x$ seen (not for "$+ c$") | M1 | For $x^n \rightarrow x^{n+1}$ |
| Two out of three terms correct un-simplified or simplified (ignore $+c$) | A1 | |
| $2x^{1.5} - 3x^2 + 4x + c$ | A1 | All correct and simplified on one line including "$+ c$". Allow $\sqrt{x^3}$ for $x^{1.5}$ but not $x^1$ for $x$. |

**Ignore any spurious integral signs.**

## Part (b)(i)
*(Mark (b)(i) and (ii) together and must be differentiating the original function not their answer to part (a))*

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x^n \rightarrow x^{n-1}$, i.e. $x^{0.5} \rightarrow x^{-0.5}$ or $6x \rightarrow 6$ | M1 | For $x^n \rightarrow x^{n-1}$ |
| $\frac{3}{2}x^{-0.5} - 6$ | A1 | For $\frac{3}{2}x^{-0.5} - 6$ or equivalent. May be un-simplified. Allow $\frac{\frac{3}{2}}{\sqrt{x}} - 6$ |

## Part (b)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{3}{2}x^{-0.5} - 6 = 0 \Rightarrow x^n = \ldots$ | M1 | Sets their $\frac{dy}{dx} = 0$ (may be implied by their working) and reaches $x^n = C$ (including $n=1$) **with correct processing allowing sign errors only**. May be implied by e.g. $\sqrt{x} = \frac{1}{4}$ or $\frac{1}{\sqrt{x}} = 4$ |
| $x = \frac{1}{16}$ cso | A1 | Allow equivalent fractions e.g. $\frac{9}{144}$ or $0.0625$. If other solutions are given (e.g. likely to be $x = 0$ or $x = -\frac{1}{16}$) then this mark should be withheld. |

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\begin{enumerate}
  \item Given
\end{enumerate}

$$y = 3 \sqrt { x } - 6 x + 4 , \quad x > 0$$

(a) find $\int y \mathrm {~d} x$, simplifying each term.\\
(b) (i) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$\\
(ii) Hence find the value of $x$ such that $\frac { \mathrm { d } y } { \mathrm {~d} x } = 0$\\

\hfill \mbox{\textit{Edexcel C1 2018 Q2 [7]}}