| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2016 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Find tangent given derivative expression |
| Difficulty | Standard +0.3 This is a straightforward multi-part calculus question requiring standard techniques: substituting a point into the derivative to find gradient, writing tangent equation, integrating to find the curve equation, then using a second point to find the constant. All steps are routine A-level procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07m Tangents and normals: gradient and equations1.08a Fundamental theorem of calculus: integration as reverse of differentiation |
A curve is such that $\frac{dy}{dx} = \frac{2}{a}x^{-\frac{1}{2}} + ax^{-\frac{3}{2}}$, where $a$ is a positive constant. The point $A(a^2, 3)$ lies on the curve. Find, in terms of $a$,
\begin{enumerate}[label=(\roman*)]
\item the equation of the tangent to the curve at $A$, simplifying your answer, [3]
\item the equation of the curve. [4]
\end{enumerate}
It is now given that $B(16, 8)$ also lies on the curve.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Find the value of $a$ and, using this value, find the distance $AB$. [5]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2016 Q10 [12]}}