Pre-U Pre-U 9795/1 2019 Specimen — Question 4 3 marks

Exam BoardPre-U
ModulePre-U 9795/1 (Pre-U Further Mathematics Paper 1)
Year2019
SessionSpecimen
Marks3
TopicHyperbolic functions
TypeSecond derivative relations with hyperbolics
DifficultyChallenging +1.2 This is a multi-step further maths question requiring differentiation of hyperbolic functions, manipulation to express the derivative in terms of y, and then recognizing the reverse substitution for integration. While it requires several techniques and the connection between parts, the individual steps are fairly standard for further maths students familiar with hyperbolic identities (cosh²x - sinh²x = 1) and the chain rule. The integration requires insight to recognize the substitution but follows naturally from part (a).
Spec1.08h Integration by substitution4.07d Differentiate/integrate: hyperbolic functions
4
  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. Hence or otherwise find \(\int \frac { 2 t } { \sqrt { 1 + t ^ { 4 } } } \mathrm {~d} t\).
(a) \(y = (\sinh x)^{\frac{1}{2}} \Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{2}(\sinh x)^{-\frac{1}{2}}\cdot\cosh x\) OR \(y^2 = \sinh x \Rightarrow 2y\frac{\mathrm{d}y}{\mathrm{d}x} = \cosh x\) M1A1
\(= \frac{\sqrt{1+y^4}}{2y}\) A1
Total: 3
(b) \(\int\frac{2y}{\sqrt{1+y^4}}\,\mathrm{d}y = \int 1\,\mathrm{d}x\)
By separating variables in (a)'s answer M1
\(\Rightarrow x = \int\frac{2y}{\sqrt{1+y^4}}\,\mathrm{d}y\) A1
But \(x = \sinh^{-1}y^2\) so \(\int\frac{2t}{\sqrt{1+t^4}}\,\mathrm{d}t = \sinh^{-1}(t^2) + C\)
condone missing "\(+ C\)" A1
OR ALTERNATE SOLUTION 1
Set \(t^2 = \sinh\theta\), \(2t\,\mathrm{d}t = \cosh\theta\,\mathrm{d}\theta\) M1 for full substitution M1
\(\int\frac{2t}{\sqrt{1+t^4}}\,\mathrm{d}t = \int\frac{\cosh\theta}{\sqrt{1+\sinh^2\theta}}\,\mathrm{d}\theta\) A1
\(\int 1\,\mathrm{d}\theta = \theta = \sinh^{-1}(t^2) + C\) A1
OR ALTERNATE SOLUTION 2
Set \(t^2 = \tan\theta\), \(2t\,\mathrm{d}t = \sec^2\theta\,\mathrm{d}\theta\) M1 for full substitution M1
\(\int\frac{2t}{\sqrt{1+t^4}}\,\mathrm{d}t = \int\frac{\sec^2\theta}{\sqrt{1+\tan^2\theta}}\,\mathrm{d}\theta = \int\sec\theta\,\mathrm{d}\theta\) A1
AnswerMarks Guidance
\(= \ln\sec\theta + \tan\theta + C = \ln\left
Total: 3
**(a)** $y = (\sinh x)^{\frac{1}{2}} \Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{2}(\sinh x)^{-\frac{1}{2}}\cdot\cosh x$ **OR** $y^2 = \sinh x \Rightarrow 2y\frac{\mathrm{d}y}{\mathrm{d}x} = \cosh x$ **M1A1**

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

**Total: 3**

**(b)** $\int\frac{2y}{\sqrt{1+y^4}}\,\mathrm{d}y = \int 1\,\mathrm{d}x$

By separating variables in **(a)**'s answer **M1**

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

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

condone missing "$+ C$" **A1**

**OR ALTERNATE SOLUTION 1**

Set $t^2 = \sinh\theta$, $2t\,\mathrm{d}t = \cosh\theta\,\mathrm{d}\theta$ M1 for full substitution **M1**

$\int\frac{2t}{\sqrt{1+t^4}}\,\mathrm{d}t = \int\frac{\cosh\theta}{\sqrt{1+\sinh^2\theta}}\,\mathrm{d}\theta$ **A1**

$\int 1\,\mathrm{d}\theta = \theta = \sinh^{-1}(t^2) + C$ **A1**

**OR ALTERNATE SOLUTION 2**

Set $t^2 = \tan\theta$, $2t\,\mathrm{d}t = \sec^2\theta\,\mathrm{d}\theta$ M1 for full substitution **M1**

$\int\frac{2t}{\sqrt{1+t^4}}\,\mathrm{d}t = \int\frac{\sec^2\theta}{\sqrt{1+\tan^2\theta}}\,\mathrm{d}\theta = \int\sec\theta\,\mathrm{d}\theta$ **A1**

$= \ln|\sec\theta + \tan\theta| + C = \ln\left|t^2 + \sqrt{1+t^4}\right| + C$ **A1**

**Total: 3**
4
\begin{enumerate}[label=(\alph*)]
\item Given that $y = \sqrt { \sinh x }$ for $x \geqslant 0$, express $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $y$ only.
\item Hence or otherwise find $\int \frac { 2 t } { \sqrt { 1 + t ^ { 4 } } } \mathrm {~d} t$.
\end{enumerate}

\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2019 Q4 [3]}}
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