| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2011 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Find tangent at given point (polynomial/algebraic) |
| Difficulty | Moderate -0.8 This is a straightforward differentiation question requiring the quotient rule (or chain rule), substitution to find the gradient at a point, and using y - y₁ = m(x - x₁) for the tangent equation. Part (ii) is a simple arctan calculation. All steps are routine A-level techniques with no problem-solving insight required, making it easier than average but not trivial since it requires correct application of multiple standard procedures. |
| Spec | 1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Eqn of tangent \(y - 2 = -3(x - 2)\) | B1 B1 M1 A1 [4] | Correct without ×3. For ×3. Correct line eqn. co (for normal M0A0) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\to \theta = \pm108.4°\) (or \(\pm71.6°\)) | M1 A1∨ [2] | Correct link with (\(\pm\) his gradient) co (accept acute or obtuse) or \(-71.6°\) or radians |
$$y = \frac{4}{3x - 4}$$
**(i)** $\frac{dy}{dx} = -4(3x - 4)^{-2} \times 3$
If $x = 2, m = -3$
Eqn of tangent $y - 2 = -3(x - 2)$ | B1 B1 M1 A1 [4] | Correct without ×3. For ×3. Correct line eqn. co (for normal M0A0)
**(ii)** $\tan\theta = \pm(-3)$
$\to \theta = \pm108.4°$ (or $\pm71.6°$) | M1 A1∨ [2] | Correct link with ($\pm$ his gradient) co (accept acute or obtuse) or $-71.6°$ or radians
**or** scalar product, $\tan\theta =$ y-step $÷$ x-step or use of $\tan(A - B)$ M1A1 for eachA curve has equation $y = \frac{4}{3x - 4}$ and $P(2, 2)$ is a point on the curve.
\begin{enumerate}[label=(\roman*)]
\item Find the equation of the tangent to the curve at $P$. [4]
\item Find the angle that this tangent makes with the $x$-axis. [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2011 Q4 [6]}}