| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2020 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find equation of tangent |
| Difficulty | Standard +0.3 This is a straightforward application of the product rule combined with the standard derivative of arctan, followed by routine tangent line calculation. The product rule setup is clear, and finding where the tangent meets the y-axis requires only basic coordinate geometry. Slightly above average difficulty due to the inverse trig function, but still a standard textbook exercise. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use the product rule | M1 | |
| State or imply derivative of \(\tan^{-1}(\frac{1}{2}x)\) is of the form \(k/(4+x^2)\), where \(k=2\) or \(4\) | M1 | or equivalent |
| Obtain correct derivative in any form, e.g. \(\tan^{-1}\!\left(\frac{1}{2}x\right) + \frac{2x}{x^2+4}\) | A1 | or equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State or imply \(y\)-coordinate is \(\frac{1}{2}\pi\) | B1 | |
| Carry out a complete method for finding \(p\), e.g. by obtaining the equation of the tangent and setting \(x=0\), or by equating the gradient at \(x=2\) to \(\dfrac{\frac{1}{2}\pi - p}{2}\) | M1 | |
| Obtain answer \(p = -1\) | A1 |
## Question 4(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use the product rule | M1 | |
| State or imply derivative of $\tan^{-1}(\frac{1}{2}x)$ is of the form $k/(4+x^2)$, where $k=2$ or $4$ | M1 | or equivalent |
| Obtain correct derivative in any form, e.g. $\tan^{-1}\!\left(\frac{1}{2}x\right) + \frac{2x}{x^2+4}$ | A1 | or equivalent |
## Question 4(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply $y$-coordinate is $\frac{1}{2}\pi$ | B1 | |
| Carry out a complete method for finding $p$, e.g. by obtaining the equation of the tangent and setting $x=0$, or by equating the gradient at $x=2$ to $\dfrac{\frac{1}{2}\pi - p}{2}$ | M1 | |
| Obtain answer $p = -1$ | A1 | |
---4 The equation of a curve is $y = x \tan ^ { - 1 } \left( \frac { 1 } { 2 } x \right)$.
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
\item The tangent to the curve at the point where $x = 2$ meets the $y$-axis at the point with coordinates $( 0 , p )$.
Find $p$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2020 Q4 [6]}}