CAIE Further Paper 2 2020 November — Question 2 6 marks

Exam BoardCAIE
ModuleFurther Paper 2 (Further Paper 2)
Year2020
SessionNovember
Marks6
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TopicHyperbolic functions
TypeSurface area of revolution with hyperbolics
DifficultyChallenging +1.2 This is a standard surface area of revolution problem using the formula S = 2π∫y√(1+(dy/dx)²)dx with hyperbolics. While it requires knowing that d/dx(cosh x) = sinh x and the identity cosh²x - sinh²x = 1 (which simplifies the integrand to sinh x), the integration is straightforward once set up. It's a direct application of a formula with hyperbolic functions—more challenging than routine A-level due to the Further Maths content, but still a textbook exercise without novel problem-solving.
Spec4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.08d Volumes of revolution: about x and y axes
2 A curve has equation \(\mathrm { y } = \cosh \mathrm { x }\), for \(0 \leqslant x \leqslant \frac { 1 } { 2 }\).
Find, in terms of \(\pi\) and e, the area of the surface generated when the curve is rotated through \(2 \pi\) radians about the \(x\)-axis.
Question 2:
AnswerMarks Guidance
\(\sqrt{1+\left(\frac{dy}{dx}\right)^2} = \sqrt{1+\sinh^2 x} = \cosh x\)M1 A1 Applies \(\cosh^2 x = 1 + \sinh^2 x\)
\(2\pi\int_0^{\frac{1}{2}}\cosh^2 x\, dx\)M1 Correct formula, correct limits
\(= \pi\int_0^{\frac{1}{2}}\cosh 2x + 1\, dx\)M1 Applies \(2\cosh^2 x = \cosh 2x + 1\) or expands \(2(e^x + e^{-x})^2\)
\(= \pi\left[\frac{1}{2}\sinh 2x + x\right]_0^{\frac{1}{2}}\)A1 Correct integration
\(= \frac{1}{4}\pi(e - e^{-1} + 2)\)A1 
## Question 2:

| $\sqrt{1+\left(\frac{dy}{dx}\right)^2} = \sqrt{1+\sinh^2 x} = \cosh x$ | M1 A1 | Applies $\cosh^2 x = 1 + \sinh^2 x$ |
|---|---|---|
| $2\pi\int_0^{\frac{1}{2}}\cosh^2 x\, dx$ | M1 | Correct formula, correct limits |
| $= \pi\int_0^{\frac{1}{2}}\cosh 2x + 1\, dx$ | M1 | Applies $2\cosh^2 x = \cosh 2x + 1$ or expands $2(e^x + e^{-x})^2$ |
| $= \pi\left[\frac{1}{2}\sinh 2x + x\right]_0^{\frac{1}{2}}$ | A1 | Correct integration |
| $= \frac{1}{4}\pi(e - e^{-1} + 2)$ | A1 | |

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2 A curve has equation $\mathrm { y } = \cosh \mathrm { x }$, for $0 \leqslant x \leqslant \frac { 1 } { 2 }$.\\
Find, in terms of $\pi$ and e, the area of the surface generated when the curve is rotated through $2 \pi$ radians about the $x$-axis.\\

\hfill \mbox{\textit{CAIE Further Paper 2 2020 Q2 [6]}}
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Q1 5 Q2 6 Q3 4 Q4 8 Q5 8 Q6 11 Q7 7 Q8 10 Q9 16