Pre-U Pre-U 9795/1 2012 June — Question 3 3 marks

Exam BoardPre-U
ModulePre-U 9795/1 (Pre-U Further Mathematics Paper 1)
Year2012
SessionJune
Marks3
TopicHyperbolic functions
TypeSecond derivative relations with hyperbolics
DifficultyStandard +0.8 Part (i) requires differentiating a composite hyperbolic function and algebraic manipulation using the identity cosh²x - sinh²x = 1 to eliminate x. Part (ii) is a non-standard integral requiring the substitution t² = sinh u, which is not immediately obvious. Both parts demand more insight than routine hyperbolic function questions, placing this above average difficulty but not at the extreme end for Further Maths.
Spec4.07d Differentiate/integrate: hyperbolic functions4.08h Integration: inverse trig/hyperbolic substitutions
3
  1. Given that \(y = \sqrt { \sinh x }\) for \(x \geqslant 0\), express \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(y\) only.
  2. Find \(\int \frac { 2 t } { \sqrt { 1 + t ^ { 4 } } } \mathrm {~d} t\).
(i) \(y = (\sinh x)^{\frac{1}{2}} \Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{2}(\sinh x)^{-\frac{1}{2}}\cdot\cosh x\) OR \(y^2 = \sinh x \Rightarrow 2y\frac{\mathrm{d}y}{\mathrm{d}x} = \cosh x\) M1 A1
\(= \frac{\sqrt{1+y^4}}{2y}\) A1
[3]
(ii) \(\int\frac{2y}{\sqrt{1+y^4}}\,\mathrm{d}y = \int 1\,\mathrm{d}x\) By Sep\(^{\mathrm{n}}\). Vars. in (i)'s answer M1
\(\Rightarrow x = \int\frac{2y}{\sqrt{1+y^4}}\,\mathrm{d}y\) A1
But \(x = \sinh^{-1}y^2\) so \(\int\frac{2t}{\sqrt{1+t^4}}\,\mathrm{d}x = \sinh^{-1}(t^2) + C\) condone missing "\(+ C\)" A1
ALT.1 Set \(t^2 = \sinh\theta\), \(2t\,\mathrm{d}t = \cosh\theta\,\mathrm{d}\theta\) M1 Full substn.
\(\int\frac{2t}{\sqrt{1+t^4}}\,\mathrm{d}t = \int\frac{\cosh\theta}{\sqrt{1+\sinh^2\theta}}\,\mathrm{d}\theta\) A1 \(= \int 1\,\mathrm{d}\theta = \theta = \sinh^{-1}(t^2)\) A1
ALT.2 Set \(t^2 = \tan\theta\), \(2t\,\mathrm{d}t = \sec^2\theta\,\mathrm{d}\theta\) M1 Full substn.
\(\int\frac{2t}{\sqrt{1+t^4}}\,\mathrm{d}t = \int\frac{\sec^2\theta}{\sqrt{1+\tan^2\theta}}\,\mathrm{d}\theta\) A1 \(= \int\sec\theta\,\mathrm{d}\theta\)
AnswerMarks Guidance
\(= \ln\sec\theta + \tan\theta = \ln
[3]
**(i)** $y = (\sinh x)^{\frac{1}{2}} \Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{2}(\sinh x)^{-\frac{1}{2}}\cdot\cosh x$ **OR** $y^2 = \sinh x \Rightarrow 2y\frac{\mathrm{d}y}{\mathrm{d}x} = \cosh x$ **M1 A1**

$= \frac{\sqrt{1+y^4}}{2y}$ **A1**

**[3]**

**(ii)** $\int\frac{2y}{\sqrt{1+y^4}}\,\mathrm{d}y = \int 1\,\mathrm{d}x$ By Sep$^{\mathrm{n}}$. Vars. in **(i)**'s answer **M1**

$\Rightarrow x = \int\frac{2y}{\sqrt{1+y^4}}\,\mathrm{d}y$ **A1**

But $x = \sinh^{-1}y^2$ so $\int\frac{2t}{\sqrt{1+t^4}}\,\mathrm{d}x = \sinh^{-1}(t^2) + C$ condone missing "$+ C$" **A1**

**ALT.1** Set $t^2 = \sinh\theta$, $2t\,\mathrm{d}t = \cosh\theta\,\mathrm{d}\theta$ **M1** Full substn.
$\int\frac{2t}{\sqrt{1+t^4}}\,\mathrm{d}t = \int\frac{\cosh\theta}{\sqrt{1+\sinh^2\theta}}\,\mathrm{d}\theta$ **A1** $= \int 1\,\mathrm{d}\theta = \theta = \sinh^{-1}(t^2)$ **A1**

**ALT.2** Set $t^2 = \tan\theta$, $2t\,\mathrm{d}t = \sec^2\theta\,\mathrm{d}\theta$ **M1** Full substn.
$\int\frac{2t}{\sqrt{1+t^4}}\,\mathrm{d}t = \int\frac{\sec^2\theta}{\sqrt{1+\tan^2\theta}}\,\mathrm{d}\theta$ **A1** $= \int\sec\theta\,\mathrm{d}\theta$
$= \ln|\sec\theta + \tan\theta| = \ln|t^2 + \sqrt{1+t^4}|$ **A1**

**[3]**
3 (i) Given that $y = \sqrt { \sinh x }$ for $x \geqslant 0$, express $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $y$ only.\\
(ii) Find $\int \frac { 2 t } { \sqrt { 1 + t ^ { 4 } } } \mathrm {~d} t$.

\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2012 Q3 [3]}}
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