AQA Further AS Paper 1 2024 June — Question 6 4 marks

Exam BoardAQA
ModuleFurther AS Paper 1 (Further AS Paper 1)
Year2024
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHyperbolic functions
TypeSketch graphs of hyperbolic functions
DifficultyModerate -0.8 Part (a) requires sketching a standard hyperbolic function (cosh x) with its y-intercept, which is direct recall of a basic graph. Part (b) involves solving cosh x = 2 using the standard logarithmic formula, a routine application of a known result. Both parts are straightforward textbook exercises requiring minimal problem-solving, though the hyperbolic functions topic itself is Further Maths content.
Spec4.07b Hyperbolic graphs: sketch and properties4.07e Inverse hyperbolic: definitions, domains, ranges

6
  1. On the axes below, sketch the graph of $$y = \cosh x$$ Indicate the value of any intercept of the curve with the axes. \includegraphics[max width=\textwidth, alt={}, center]{47b12ae4-ca3f-472c-9d15-2ef17a2a4d87-05_1114_1121_552_447} 6
  2. Solve the equation $$\cosh x = 2$$ Give your answers to three significant figures. \includegraphics[max width=\textwidth, alt={}, center]{47b12ae4-ca3f-472c-9d15-2ef17a2a4d87-06_2491_1755_173_123}

Question 6:
Part 6(a):
AnswerMarks Guidance
Correct shape, approximately symmetrical about \(y\)-axis, above \(x\)-axisB1 Draws correct shape
\(y\)-intercept at \((0,1)\) with 1 clearly labelled on \(y\)-axisB1 Indicates one \(y\)-axis intercept at \((0,1)\)
Part 6(b):
AnswerMarks Guidance
\(x = \pm\cosh^{-1}2 = \pm1.32\) (3sf)M1 Obtains at least one correct solution; accept \(\ln(2+\sqrt{3})\) or \(\ln(2-\sqrt{3})\) or equivalent \(\ln\) expression; accept AWRT 1.32 or \(-\)1.32; do not accept \(\cosh^{-1}2\)
\(x = \pm1.32\)A1 Obtains AWRT \(\pm1.32\) with no other solutions given
## Question 6:

### Part 6(a):
| Correct shape, approximately symmetrical about $y$-axis, above $x$-axis | B1 | Draws correct shape |
| $y$-intercept at $(0,1)$ with 1 clearly labelled on $y$-axis | B1 | Indicates one $y$-axis intercept at $(0,1)$ |

### Part 6(b):
| $x = \pm\cosh^{-1}2 = \pm1.32$ (3sf) | M1 | Obtains at least one correct solution; accept $\ln(2+\sqrt{3})$ or $\ln(2-\sqrt{3})$ or equivalent $\ln$ expression; accept AWRT 1.32 or $-$1.32; do not accept $\cosh^{-1}2$ |
| $x = \pm1.32$ | A1 | Obtains AWRT $\pm1.32$ with no other solutions given |

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6
\begin{enumerate}[label=(\alph*)]
\item On the axes below, sketch the graph of

$$y = \cosh x$$

Indicate the value of any intercept of the curve with the axes.\\
\includegraphics[max width=\textwidth, alt={}, center]{47b12ae4-ca3f-472c-9d15-2ef17a2a4d87-05_1114_1121_552_447}

6
\item Solve the equation

$$\cosh x = 2$$

Give your answers to three significant figures.\\

\includegraphics[max width=\textwidth, alt={}, center]{47b12ae4-ca3f-472c-9d15-2ef17a2a4d87-06_2491_1755_173_123}
\end{enumerate}

\hfill \mbox{\textit{AQA Further AS Paper 1 2024 Q6 [4]}}