| Exam Board | WJEC |
|---|---|
| Module | Further Unit 4 (Further Unit 4) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration using inverse trig and hyperbolic functions |
| Type | Derivative of inverse trig function |
| Difficulty | Standard +0.3 Part (a) is a standard implicit differentiation exercise to derive the derivative of inverse sine with a linear substitution—routine for Further Maths students who know the technique. Part (b) requires recognizing that the domain constraint comes from the square root being real and the arcsine domain, which is straightforward once (a) is complete. This is slightly above average difficulty due to being Further Maths content and requiring careful domain analysis, but it follows a well-practiced method without requiring novel insight. |
| Spec | 1.07s Parametric and implicit differentiation4.08g Derivatives: inverse trig and hyperbolic functions |
| Answer | Marks | Guidance |
|---|---|---|
| a) | METHOD 1: | |
| Writing \(y = \sin^{-1}(2x+5)\) | ||
| \(\therefore \sin y = 2x + 5\) | ||
| \(\frac{1}{2}(\sin y - 5) = x\) | ||
| Differentiating: \(\frac{dx}{dy} = \frac{1}{2}\cos y\) | M1, A1 | Rearrange and attempt to differentiate |
| Since \(\sin^2 y + \cos^2 y = 1\) | ||
| \((2x+5)^2 + \cos^2 y = 1\) | B1 | si |
| \(\cos^2 y = 1 - (2x+5)^2\) | ||
| Therefore: \(\frac{dx}{dy} = (\pm)\frac{1}{2}\sqrt{1-(2x+5)^2}\) | A1 | |
| \(\frac{dy}{dx} = (\pm)\frac{2}{\sqrt{1-(2x+5)^2}}\) | A1 | |
| \(\frac{dy}{dx} = \frac{2}{\sqrt{1-(2x+5)^2}}\) AND valid justification, eg. reference to the gradient of the graph. | A1 | |
| METHOD 2: | ||
| Writing \(y = \sin^{-1}(2x+5)\) | ||
| \(\therefore \sin y = 2x + 5\) | ||
| Differentiating: \(\cos y\frac{dy}{dx} = 2\) | (M1) | Attempt to differentiate |
| A1 | ||
| \(\frac{dy}{dx} = \frac{2}{\cos y}\) | ||
| Since \(\sin^2 y + \cos^2 y = 1\) | ||
| \((2x+5)^2 + \cos^2 y = 1\) | (B1) | si |
| \(\cos^2 y = 1 - (2x+5)^2\) | ||
| Therefore: \(\frac{dy}{dx} = (\pm)\frac{2}{\sqrt{1-(2x+5)^2}}\) | (A1) | |
| \(\frac{dy}{dx} = \frac{2}{\sqrt{1-(2x+5)^2}}\) AND valid justification, eg. reference to the gradient of the graph. | (A1) | |
| (5) | ||
| b) | \(1 - (2x+5)^2 = -4x^2 - 20x - 24\) | |
| METHOD 1: | ||
| When \(-4x^2 - 20x - 24 = 0\) | M1, A1 | |
| \(x = -2\) or \(x = -3\) | ||
| Therefore the range of valid values is \(-3 < x < -2\) | A1 | |
| METHOD 2: | ||
| Valid when \(1 - (2x+5)^2 > 0\) | (M1) | |
| \(-4x^2 - 20x - 24 > 0\) | (A1) | |
| \(4x^2 + 20x + 24 < 0\) | ||
| \(4(x+3)(x+2) < 0\) | (A1) | |
| Therefore \(-3 < x < -2\) | ||
| METHOD 3: | ||
| \(\sin y = 2x + 5\) | ||
| Valid when \(-1 \le 2x + 5 \le 1\) | (M1) | |
| \(-3 \le x \le -2\) | (A1) | |
| Therefore (due to fraction), \(-3 < x < -2\) | (A1) | |
| (3) | ||
| [8] |
a) | **METHOD 1:** | | |
| Writing $y = \sin^{-1}(2x+5)$ | | |
| $\therefore \sin y = 2x + 5$ | | |
| $\frac{1}{2}(\sin y - 5) = x$ | | |
| | | |
| Differentiating: $\frac{dx}{dy} = \frac{1}{2}\cos y$ | M1, A1 | Rearrange and attempt to differentiate |
| | | |
| Since $\sin^2 y + \cos^2 y = 1$ | | |
| $(2x+5)^2 + \cos^2 y = 1$ | B1 | si |
| $\cos^2 y = 1 - (2x+5)^2$ | | |
| | | |
| Therefore: $\frac{dx}{dy} = (\pm)\frac{1}{2}\sqrt{1-(2x+5)^2}$ | A1 | |
| | | |
| $\frac{dy}{dx} = (\pm)\frac{2}{\sqrt{1-(2x+5)^2}}$ | A1 | |
| | | |
| $\frac{dy}{dx} = \frac{2}{\sqrt{1-(2x+5)^2}}$ AND valid justification, eg. reference to the gradient of the graph. | A1 | |
| | | |
| **METHOD 2:** | | |
| Writing $y = \sin^{-1}(2x+5)$ | | |
| $\therefore \sin y = 2x + 5$ | | |
| | | |
| Differentiating: $\cos y\frac{dy}{dx} = 2$ | (M1) | Attempt to differentiate |
| | A1 | |
| $\frac{dy}{dx} = \frac{2}{\cos y}$ | | |
| | | |
| Since $\sin^2 y + \cos^2 y = 1$ | | |
| $(2x+5)^2 + \cos^2 y = 1$ | (B1) | si |
| $\cos^2 y = 1 - (2x+5)^2$ | | |
| | | |
| Therefore: $\frac{dy}{dx} = (\pm)\frac{2}{\sqrt{1-(2x+5)^2}}$ | (A1) | |
| | | |
| $\frac{dy}{dx} = \frac{2}{\sqrt{1-(2x+5)^2}}$ AND valid justification, eg. reference to the gradient of the graph. | (A1) | |
| | (5) | |
b) | $1 - (2x+5)^2 = -4x^2 - 20x - 24$ | | |
| | | |
| **METHOD 1:** | | |
| When $-4x^2 - 20x - 24 = 0$ | M1, A1 | |
| $x = -2$ or $x = -3$ | | |
| | | |
| Therefore the range of valid values is $-3 < x < -2$ | A1 | |
| | | |
| **METHOD 2:** | | |
| Valid when $1 - (2x+5)^2 > 0$ | (M1) | |
| $-4x^2 - 20x - 24 > 0$ | (A1) | |
| $4x^2 + 20x + 24 < 0$ | | |
| $4(x+3)(x+2) < 0$ | (A1) | |
| Therefore $-3 < x < -2$ | | |
| | | |
| **METHOD 3:** | | |
| $\sin y = 2x + 5$ | | |
| Valid when $-1 \le 2x + 5 \le 1$ | (M1) | |
| $-3 \le x \le -2$ | (A1) | |
| Therefore (due to fraction), $-3 < x < -2$ | (A1) | |
| | (3) | |
| | [8] | |\begin{enumerate}[label=(\alph*)]
\item By writing $y = \sin^{-1}(2x + 5)$ as $\sin y = 2x + 5$, show that $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2}{\sqrt{1-(2x+5)^2}}$. [5]
\item Deduce the range of values of $x$ for which $\frac{\mathrm{d}}{\mathrm{d}x}\left(\sin^{-1}(2x+5)\right)$ is valid. [3]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 4 2023 Q10 [8]}}