| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2009 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Find stationary points |
| Difficulty | Challenging +1.2 This question requires finding stationary points by differentiation (quotient rule in part (a), product rule in part (b)), then analyzing the resulting equations. Part (a) involves showing a quadratic has no real roots; part (b) requires factoring and showing a quadratic discriminant is always positive. While systematic, it demands careful algebraic manipulation across multiple steps and proof-style reasoning about 'for all k/m', making it moderately harder than standard C3 differentiation exercises. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07q Product and quotient rules: differentiation1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Attempt use of quotient rule | *M1 | or equiv; allow numerator wrong way round and denominator errors |
| Obtain \(\frac{(kx^2 + 1)2kx - (kx^2 - 1)2kx}{(kx^2 + 1)^2}\) | A1 | or equiv; with absent brackets implied by subsequent correct working |
| Obtain correct simplified numerator \(4kx\) | A1 | |
| Equate numerator of first derivative to zero | M1 | dep *M |
| State \(x = 0\) or refer to \(4kx\) being linear or observe that, with \(k \neq 0\), only one sol'n | A1\5 | AG or equiv; following numerator of form \(k'kx = 0\), any constant \(k'\) |
| Answer | Marks | Guidance |
|---|---|---|
| Attempt use of product rule | *M1 | |
| Obtain \(me^{mx}(x^2 + mx) + e^{mx}(2x + m)\) | A1 | or equiv |
| Equate to zero and either factorise with factor \(e^{mx}\) or divide through by \(e^{mx}\) | M1 | dep *M |
| Obtain \(mx^2 + (m^2 + 2)x + m = 0\) or equiv | A1 | |
| and observe that \(e^{mx}\) cannot be zero | A1 | |
| Attempt use of discriminant | M1 | using correct \(b^2 - 4ac\) with their \(a, b, c\) |
| Simplify to obtain \(m^2 + 4\) | A1 | |
| Observe that this is positive for all \(m\) and hence two roots | A1\7 | or equiv; AG |
## Part (a)
Attempt use of quotient rule | *M1 | or equiv; allow numerator wrong way round and denominator errors
Obtain $\frac{(kx^2 + 1)2kx - (kx^2 - 1)2kx}{(kx^2 + 1)^2}$ | A1 | or equiv; with absent brackets implied by subsequent correct working
Obtain correct simplified numerator $4kx$ | A1 |
Equate numerator of first derivative to zero | M1 | dep *M
State $x = 0$ or refer to $4kx$ being linear or observe that, with $k \neq 0$, only one sol'n | A1\5 | AG or equiv; following numerator of form $k'kx = 0$, any constant $k'$
## Part (b)
Attempt use of product rule | *M1 |
Obtain $me^{mx}(x^2 + mx) + e^{mx}(2x + m)$ | A1 | or equiv
Equate to zero and either factorise with factor $e^{mx}$ or divide through by $e^{mx}$ | M1 | dep *M
Obtain $mx^2 + (m^2 + 2)x + m = 0$ or equiv | A1 |
and observe that $e^{mx}$ cannot be zero | A1 |
Attempt use of discriminant | M1 | using correct $b^2 - 4ac$ with their $a, b, c$
Simplify to obtain $m^2 + 4$ | A1 |
Observe that this is positive for all $m$ and hence two roots | A1\7 | or equiv; AG\begin{enumerate}[label=(\alph*)]
\item Show that, for all non-zero values of the constant $k$, the curve
$$y = \frac{kx^2 - 1}{kx^2 + 1}$$
has exactly one stationary point. [5]
\item Show that, for all non-zero values of the constant $m$, the curve
$$y = e^{mx}(x^2 + mx)$$
has exactly two stationary points. [7]
\end{enumerate}
\hfill \mbox{\textit{OCR C3 2009 Q9 [12]}}