Pre-U Pre-U 9795/1 Specimen — Question 12 14 marks

Exam BoardPre-U
ModulePre-U 9795/1 (Pre-U Further Mathematics Paper 1)
SessionSpecimen
Marks14
TopicHyperbolic functions
TypeSecond derivative relations with hyperbolics
DifficultyChallenging +1.2 This is a structured parametric differentiation question with hyperbolic functions requiring multiple steps: finding dy/dx using the chain rule, then d²y/dx² using the quotient rule, and solving an equation. While it involves Further Maths content (hyperbolics) and careful algebraic manipulation, the question is highly scaffolded with 'show that' parts guiding students through standard techniques. The difficulty is above average due to the topic and multi-step nature, but below typical Further Maths problem-solving questions.
Spec1.07s Parametric and implicit differentiation4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07d Differentiate/integrate: hyperbolic functions
12 The curve \(C\) is defined parametrically by $$x = t + \ln ( \cosh t ) , \quad y = \sinh t$$
  1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { e } ^ { - t } \cosh ^ { 2 } t\).
  2. Hence show that \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \mathrm { e } ^ { - 2 t } \cosh ^ { 2 } t ( 2 \sinh t - \cosh t )\).
  3. Find the exact value of \(t\) at the point on \(C\) where \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 0\).
(i) \(\frac{dx}{dt} = 1 + \frac{\sinh t}{\cosh t} = \frac{\cosh t + \sinh t}{\cosh t}\) (either form) M1A1
\(= \frac{e^t}{\cosh t}\) (may appear later in this part) A1
\(\frac{dy}{dt} = \cosh t\) B1
\(\frac{dy}{dx} = \cosh t \times \frac{\cosh t}{e^t} = e^{-t}\cosh^2 t\) (AG) M1A1 [6]
(ii) \(\frac{d^2y}{dx^2} = \frac{d}{dt}(e^{-t}\cosh^2 t) \times \frac{dt}{dx}\) M1
\(= (-e^{-t}\cosh^2 t + 2e^{-t}\cosh t\sinh t) \times \frac{\cosh t}{e^t}\) M1A1
\(= e^{-2t}\cosh^2 t(2\sinh t - \cosh t)\) (AG) A1 [4]
(iii) \(e^{-2t}\cosh^2 t(2\sinh t - \cosh t) = 0 \Rightarrow \tanh t = \frac{1}{2}\) M1A1
AnswerMarks Guidance
\(t = \frac{1}{2}\ln\left\frac{2+1}{2-1}\right = \frac{1}{2}\ln 3\) M1A1 [4] 
**(i)** $\frac{dx}{dt} = 1 + \frac{\sinh t}{\cosh t} = \frac{\cosh t + \sinh t}{\cosh t}$ (either form) M1A1

$= \frac{e^t}{\cosh t}$ (may appear later in this part) A1

$\frac{dy}{dt} = \cosh t$ B1

$\frac{dy}{dx} = \cosh t \times \frac{\cosh t}{e^t} = e^{-t}\cosh^2 t$ (AG) M1A1 **[6]**

**(ii)** $\frac{d^2y}{dx^2} = \frac{d}{dt}(e^{-t}\cosh^2 t) \times \frac{dt}{dx}$ M1

$= (-e^{-t}\cosh^2 t + 2e^{-t}\cosh t\sinh t) \times \frac{\cosh t}{e^t}$ M1A1

$= e^{-2t}\cosh^2 t(2\sinh t - \cosh t)$ (AG) A1 **[4]**

**(iii)** $e^{-2t}\cosh^2 t(2\sinh t - \cosh t) = 0 \Rightarrow \tanh t = \frac{1}{2}$ M1A1

$t = \frac{1}{2}\ln\left|\frac{2+1}{2-1}\right| = \frac{1}{2}\ln 3$ M1A1 **[4]**
12 The curve $C$ is defined parametrically by

$$x = t + \ln ( \cosh t ) , \quad y = \sinh t$$

(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { e } ^ { - t } \cosh ^ { 2 } t$.\\
(ii) Hence show that $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \mathrm { e } ^ { - 2 t } \cosh ^ { 2 } t ( 2 \sinh t - \cosh t )$.\\
(iii) Find the exact value of $t$ at the point on $C$ where $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 0$.

\hfill \mbox{\textit{Pre-U Pre-U 9795/1  Q12 [14]}}
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