| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2021 |
| Session | October |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Inverse function graphs and properties |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question on inverse trig functions requiring standard techniques: sketching arcsin (routine), implicit differentiation to find dy/dx (direct application of chain rule), and finding a tangent equation (substitution into point-slope form). All parts are textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Correct shape and position: curve in quadrants 1 and 3, non-zero gradient at origin, gradient \(\to \infty\) at both ends | B1 | Ignore values labelled on axes; mark fullest attempt if multiple attempts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = 2\sin y \Rightarrow \frac{dx}{dy} = 2\cos y\); attempts to use both \(\frac{dy}{dx} = 1 \div \frac{dx}{dy}\) and \(\cos y = \sqrt{1-\sin^2 y} = \sqrt{1 - \frac{x^2}{4}}\) | M1 | Allow \(\frac{dx}{dy} = \pm k\cos y\); allow working leading to \(\frac{\cdots}{\cdots\sqrt{1-\cdots x^2}}\) |
| \(\left(\frac{dy}{dx} =\right) \dfrac{1}{2\sqrt{1-\frac{x^2}{4}}}\) | A1 | Correct unsimplified expression as function of \(x\) |
| \(\left(\frac{dy}{dx} =\right) \dfrac{1}{\sqrt{4-x^2}}\) | A1cso | Fully simplified |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = \arcsin\!\left(\frac{x}{2}\right) \Rightarrow \frac{dy}{dx} = \frac{1}{2} \times \dfrac{1}{\sqrt{1-\left(\frac{x}{2}\right)^2}}\) | M1A1 | \(k\) can equal 1 |
| \(\left(\frac{dy}{dx} =\right) \dfrac{1}{\sqrt{4-x^2}}\) | A1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitutes \(x = \sqrt{2}\) into \(\frac{dy}{dx} = \dfrac{1}{\sqrt{4-x^2}}\), giving \(\dfrac{\sqrt{2}}{2}\) | M1 | — |
| Finds equation of tangent at \(P\): \(y - \frac{\pi}{4} = \frac{\sqrt{2}}{2}\left(x - \sqrt{2}\right)\) | dM1 | Dependent on previous M1 |
| \(y = \dfrac{\sqrt{2}}{2}x - 1 + \dfrac{\pi}{4}\) | A1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts \(\cos 2x = \pm1 \pm 2\sin^2 x\) to write \(y = 4x - 12\sin^2 x\) in terms of \(\cos 2x\) | M1 | — |
| Differentiates \(\cos 2x \to k\sin 2x\), proceeds to \(\frac{dy}{dx} = A + B\sin 2x\) | dM1 | — |
| \(\frac{dy}{dx} = 4 - 12\sin 2x\) or \(4 + (-12)\sin 2x\) | A1 | Condone recovery of slips; states correct \(A\) and \(B\) |
| Maximum gradient \(= 16\) | A1 | Previous A1 must have been scored |
## Question 8:
**Part (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct shape and position: curve in quadrants 1 and 3, non-zero gradient at origin, gradient $\to \infty$ at both ends | B1 | Ignore values labelled on axes; mark fullest attempt if multiple attempts |
**Part (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = 2\sin y \Rightarrow \frac{dx}{dy} = 2\cos y$; attempts to use both $\frac{dy}{dx} = 1 \div \frac{dx}{dy}$ and $\cos y = \sqrt{1-\sin^2 y} = \sqrt{1 - \frac{x^2}{4}}$ | M1 | Allow $\frac{dx}{dy} = \pm k\cos y$; allow working leading to $\frac{\cdots}{\cdots\sqrt{1-\cdots x^2}}$ |
| $\left(\frac{dy}{dx} =\right) \dfrac{1}{2\sqrt{1-\frac{x^2}{4}}}$ | A1 | Correct unsimplified expression as function of $x$ |
| $\left(\frac{dy}{dx} =\right) \dfrac{1}{\sqrt{4-x^2}}$ | A1cso | Fully simplified |
**Alt (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \arcsin\!\left(\frac{x}{2}\right) \Rightarrow \frac{dy}{dx} = \frac{1}{2} \times \dfrac{1}{\sqrt{1-\left(\frac{x}{2}\right)^2}}$ | M1A1 | $k$ can equal 1 |
| $\left(\frac{dy}{dx} =\right) \dfrac{1}{\sqrt{4-x^2}}$ | A1 | — |
**Part (c):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitutes $x = \sqrt{2}$ into $\frac{dy}{dx} = \dfrac{1}{\sqrt{4-x^2}}$, giving $\dfrac{\sqrt{2}}{2}$ | M1 | — |
| Finds equation of tangent at $P$: $y - \frac{\pi}{4} = \frac{\sqrt{2}}{2}\left(x - \sqrt{2}\right)$ | dM1 | Dependent on previous M1 |
| $y = \dfrac{\sqrt{2}}{2}x - 1 + \dfrac{\pi}{4}$ | A1 | — |
---
## Question (ii) — Gradient problem:
**Way Two:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts $\cos 2x = \pm1 \pm 2\sin^2 x$ to write $y = 4x - 12\sin^2 x$ in terms of $\cos 2x$ | M1 | — |
| Differentiates $\cos 2x \to k\sin 2x$, proceeds to $\frac{dy}{dx} = A + B\sin 2x$ | dM1 | — |
| $\frac{dy}{dx} = 4 - 12\sin 2x$ or $4 + (-12)\sin 2x$ | A1 | Condone recovery of slips; states correct $A$ and $B$ |
| Maximum gradient $= 16$ | A1 | Previous A1 must have been scored |8. A curve $C$ has equation $y = \mathrm { f } ( x )$, where
$$f ( x ) = \arcsin \left( \frac { 1 } { 2 } x \right) \quad - 2 \leqslant x \leqslant 2 \quad - \frac { \pi } { 2 } \leqslant y \leqslant \frac { \pi } { 2 }$$
\begin{enumerate}[label=(\alph*)]
\item Sketch $C$.
\item Given $x = 2 \sin y$, show that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { \sqrt { A - x ^ { 2 } } }$$
where $A$ is a constant to be found.
The point $P$ lies on $C$ and has $y$ coordinate $\frac { \pi } { 4 }$
\item Find the equation of the tangent to $C$ at $P$. Write your answer in the form $y = m x + c$, where $m$ and $c$ are constants to be found.\\
(3)
\end{enumerate}
\hfill \mbox{\textit{Edexcel P3 2021 Q8 [7]}}