| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/1 (Pre-U Mathematics Paper 1) |
| Session | Specimen |
| Marks | 5 |
| Topic | Tangents, normals and gradients |
| Type | Show/verify a given line is a tangent |
| Difficulty | Moderate -0.8 This is a straightforward tangent line question requiring only standard differentiation (chain rule), gradient evaluation at a point, and point-slope form. All steps are routine with no problem-solving insight needed, making it easier than average but not trivial due to the logarithmic manipulation required. |
| Spec | 1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
$\frac{dy}{dx} = \frac{2x}{x^2+3}$ M1, A1
Obtain $\frac{1}{2}$ as gradient B1
Substitute $x = 1$, $y = \ln 4$, and their $m = \frac{1}{2}$ into equation of straight line M1
and obtain $(y - \ln 4) = \frac{1}{2}(x-1)$ or $\ln 4 = \frac{1}{2} + c$
Use of power law of logs to obtain $2y - x = \ln(16) - 1$ AG **cao** A1
**Total: 5**3 Show that the equation of the tangent to the curve $y = \ln \left( x ^ { 2 } + 3 \right)$ at the point $( 1 , \ln 4 )$ is
$$2 y - x = \ln ( 16 ) - 1$$
\hfill \mbox{\textit{Pre-U Pre-U 9794/1 Q3 [5]}}