| Exam Board | WJEC |
|---|---|
| Module | Further Unit 4 (Further Unit 4) |
| Year | 2019 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Differentiate inverse hyperbolic functions |
| Difficulty | Standard +0.8 This is a Further Maths question testing differentiation of inverse trigonometric and hyperbolic functions. Part (a) requires recognizing sin⁻¹(cos θ) = π/2 - θ and showing the derivative is constant. Part (b) applies product rule with arctan differentiation. Part (c) combines inverse hyperbolic differentiation with finding a normal equation. While the techniques are standard for Further Maths, the combination of inverse functions and multi-step application places this moderately above average difficulty. |
| Spec | 1.07q Product and quotient rules: differentiation4.08g Derivatives: inverse trig and hyperbolic functions |
| Answer | Marks |
|---|---|
| METHOD 1: | |
| \(\frac{dy}{d\theta} = \frac{1}{\sqrt{1-(\cos\theta)^2}} \times (-\sin\theta)\) | M1A1 |
| \(\frac{dy}{d\theta} = \frac{1}{\sin\theta} \times (-\sin\theta)\) | A1 |
| \(\frac{dy}{d\theta} = -1\) | A1 |
| METHOD 2: | |
| \(\sin y = \cos\theta\) | (M1) |
| \(\cos y \frac{dy}{d\theta} = -\sin\theta\) | (A1) |
| \(\frac{dy}{d\theta} = \frac{-\sin\theta}{\cos y}\) | |
| \(\cos^2 y = 1 - \sin^2 y = \sqrt{1 - \cos^2\theta} = \sin\theta\) | (A1) |
| \(\therefore \cos y = \sqrt{1 - \sin^2 y} = \sin\theta\) | (A1) |
| \(\frac{dy}{d\theta} = \frac{-\sin\theta}{\sin\theta} = -1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Differentiating | M1 | |
| \(\frac{dy}{dx} = x^3 \times \frac{1}{1+(4x)^2} \times 4 + 3x^2 \times \tan^{-1}(4x)\) | A1A1 | A1 for each term |
| When \(x = \frac{\pi}{2}\), \(\frac{dy}{dx} = 10.8(42)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| METHOD 1: | ||
| \(\frac{dy}{dx} = \frac{1}{1-(1-x)^2} \times -1\) | M1A1 | |
| METHOD 2: | ||
| \(\tanh y = 1 - x\) | (M1) | |
| \(\text{sech}^2 y \frac{dy}{dx} = -1\) | (A1) | |
| \(\frac{dy}{dx} = \frac{-1}{\text{sech}^2 y} = \frac{-1}{1-\tanh^2 y} = \frac{-1}{1-(1-x)^2}\) | ||
| THEN: | ||
| When \(x = 1.7\), \(\frac{dy}{dx} = -\frac{100}{51}\) and \(y = -0.8673\) | B1B1 | B1 for each part; FT value of \(\frac{dy}{dx}\) |
| Gradient of normal \(= \frac{51}{100}\) | B1 | |
| Equation: \(y + 0.8673 = \frac{51}{100}(x - 1.7)\) | B1 | cao |
## Part a)
METHOD 1: | |
$\frac{dy}{d\theta} = \frac{1}{\sqrt{1-(\cos\theta)^2}} \times (-\sin\theta)$ | M1A1 |
$\frac{dy}{d\theta} = \frac{1}{\sin\theta} \times (-\sin\theta)$ | A1 |
$\frac{dy}{d\theta} = -1$ | A1 |
METHOD 2: | |
$\sin y = \cos\theta$ | (M1) |
$\cos y \frac{dy}{d\theta} = -\sin\theta$ | (A1) |
$\frac{dy}{d\theta} = \frac{-\sin\theta}{\cos y}$ | |
$\cos^2 y = 1 - \sin^2 y = \sqrt{1 - \cos^2\theta} = \sin\theta$ | (A1) |
$\therefore \cos y = \sqrt{1 - \sin^2 y} = \sin\theta$ | (A1) |
$\frac{dy}{d\theta} = \frac{-\sin\theta}{\sin\theta} = -1$ | |
## Part b)
Differentiating | M1 |
$\frac{dy}{dx} = x^3 \times \frac{1}{1+(4x)^2} \times 4 + 3x^2 \times \tan^{-1}(4x)$ | A1A1 | A1 for each term
When $x = \frac{\pi}{2}$, $\frac{dy}{dx} = 10.8(42)$ | A1 |
## Part c)
METHOD 1: | |
$\frac{dy}{dx} = \frac{1}{1-(1-x)^2} \times -1$ | M1A1 |
METHOD 2: | |
$\tanh y = 1 - x$ | (M1) |
$\text{sech}^2 y \frac{dy}{dx} = -1$ | (A1) |
$\frac{dy}{dx} = \frac{-1}{\text{sech}^2 y} = \frac{-1}{1-\tanh^2 y} = \frac{-1}{1-(1-x)^2}$ | |
THEN: | |
When $x = 1.7$, $\frac{dy}{dx} = -\frac{100}{51}$ and $y = -0.8673$ | B1B1 | B1 for each part; FT value of $\frac{dy}{dx}$
Gradient of normal $= \frac{51}{100}$ | B1 |
Equation: $y + 0.8673 = \frac{51}{100}(x - 1.7)$ | B1 | cao\begin{enumerate}[label=(\alph*)]
\item Given that $y = \sin^{-1}(\cos \theta)$, where $0 \leqslant \theta \leqslant \pi$, show that $\frac{\mathrm{d}y}{\mathrm{d}\theta} = k$, where the value of $k$ is to be determined. [4]
\item Find the value of the gradient of the curve $y = x^3 \tan^{-1} 4x$ when $x = \frac{\pi}{2}$. [4]
\item Find the equation of the normal to the curve $y = \tanh^{-1}(1 - x)$ when $x = 1.7$. [6]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 4 2019 Q9 [14]}}