| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Tangent meets curve/axis — further geometry |
| Difficulty | Standard +0.3 This is a standard C3 exponential function question requiring routine techniques: finding range (recognizing e^x > 0), finding intercepts (substituting x=0 and y=0), differentiating e^(3x+1) using chain rule, and finding tangent equations. Part (iv) involves solving simultaneous equations of two tangents, which adds mild algebraic complexity but remains straightforward. All techniques are core C3 material with no novel insight required, making it slightly easier than average. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.07m Tangents and normals: gradient and equations |
| Answer | Marks |
|---|---|
| \(f(x) > -2\) | B1 |
| Answer | Marks |
|---|---|
| \(x = 0, y = e - 2 \therefore P(0, e-2)\) | B1 |
| \(y = 0, 0 = e^{3x+1} - 2\) | |
| \(3x + 1 = \ln 2\) | M1 |
| \(x = \frac{1}{3}(\ln 2 - 1) \therefore Q(\frac{1}{3}(\ln 2 - 1), 0)\) | A1 |
| Answer | Marks |
|---|---|
| \(f'(x) = 3e^{3x+1}\) | M1 |
| at \(P\), grad \(= 3e\) | A1 |
| \(\therefore y - (e-2) = 3e(x - 0)\) | M1 |
| \(y = 3ex + e - 2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| at \(Q\), grad \(= 6\) | B1 | |
| tangent at \(Q: y - 0 = 6(x - \frac{1}{3}(\ln 2 - 1))\) | M1 | |
| \(y = 6x - 2\ln 2 + 2\) | ||
| intersect: \(3ex + e - 2 = 6x - 2\ln 2 + 2\) | ||
| \(x(3e - 6) = 4 - e - 2\ln 2\) | M1 | |
| \(x = \frac{4 - e - 2\ln 2}{3e - 6} = -0.0485 \text{ (3sf)}\) | A1 | (12) |
## (i)
$f(x) > -2$ | B1 |
## (ii)
$x = 0, y = e - 2 \therefore P(0, e-2)$ | B1 |
$y = 0, 0 = e^{3x+1} - 2$ | |
$3x + 1 = \ln 2$ | M1 |
$x = \frac{1}{3}(\ln 2 - 1) \therefore Q(\frac{1}{3}(\ln 2 - 1), 0)$ | A1 |
## (iii)
$f'(x) = 3e^{3x+1}$ | M1 |
at $P$, grad $= 3e$ | A1 |
$\therefore y - (e-2) = 3e(x - 0)$ | M1 |
$y = 3ex + e - 2$ | A1 |
## (iv)
at $Q$, grad $= 6$ | B1 |
tangent at $Q: y - 0 = 6(x - \frac{1}{3}(\ln 2 - 1))$ | M1 |
$y = 6x - 2\ln 2 + 2$ | |
intersect: $3ex + e - 2 = 6x - 2\ln 2 + 2$ | |
$x(3e - 6) = 4 - e - 2\ln 2$ | M1 |
$x = \frac{4 - e - 2\ln 2}{3e - 6} = -0.0485 \text{ (3sf)}$ | A1 | (12)
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**Total (72)**$$\text{f}(x) = e^{3x + 1} - 2, \quad x \in \mathbb{R}.$$
\begin{enumerate}[label=(\roman*)]
\item State the range of f. [1]
\end{enumerate}
The curve $y = \text{f}(x)$ meets the $y$-axis at the point $P$ and the $x$-axis at the point $Q$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the exact coordinates of $P$ and $Q$. [3]
\item Show that the tangent to the curve at $P$ has the equation
$$y = 3ex + e - 2.$$ [4]
\item Find to 3 significant figures the $x$-coordinate of the point where the tangent to the curve at $P$ meets the tangent to the curve at $Q$. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR C3 Q9 [12]}}