| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2011 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Sketch then find derivative/gradient/tangent |
| Difficulty | Moderate -0.3 This is a straightforward C1 curve sketching question requiring standard techniques: expanding to find intercepts, applying product rule for differentiation, finding a tangent equation, and using parallel tangents (equal gradients). All steps are routine with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.07q Product and quotient rules: differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Shape (cubic in this orientation) | B1 | |
| Touching x-axis at −3 | B1 | |
| Crossing at −1 on x-axis | B1 | |
| Intersection at 9 on y-axis | B1 | |
| Notes | (a) Crossing at −3 is B0. Touching at −1 is B0. | |
| (b) \(y = (x + 1)(x^2 + 6x + 9) = x^3 + 7x^2 + 15x + 9\) or equiv. (possibly unsimplified) | B1 | |
| Differentiates their polynomial correctly – may be unsimplified | M1 | |
| \(\frac{dy}{dx} = 3x^2 + 14x + 15\) | A1 cso | (*) |
| Notes | (b) M: This needs to be correct differentiation here. A1: Fully correct simplified answer. | |
| (c) At \(x = -5\): \(\frac{dy}{dx} = 75 - 70 + 15 = 20\) | B1 | |
| At \(x = -5\): \(y = -16\) | B1 | |
| \(y - ("-16") = "20"(x - (-5))\) or \(y = "20x" + c\) with \((-5, "-16")\) used to find c | M1 | |
| \(y = 20x + 84\) | A1 | |
| Notes | (c) M: If the −5 and "−16" are the wrong way round or – omitted the M mark can still be given if a correct formula is seen, (e.g. \(y - y_1 = m(x - x_1)\)) is seen, otherwise M0. m should be numerical and not 0 or infinity and should not have involved negative reciprocal. (d) 1st M: Putting the derivative expression equal to their value for gradient. 2nd M: Attempt to solve quadratic (see notes) This may be implied by correct answer. | |
| (d) Parallel: \(3x^2 + 14x + 15 = "20"\) | M1 | |
| \((3x - 1)(x + 5) = 0\), \(x = ...\) | M1 | |
| \(x = \frac{1}{3}\) | A1 |
(a) Shape (cubic in this orientation) | B1 |
Touching x-axis at −3 | B1 |
Crossing at −1 on x-axis | B1 |
Intersection at 9 on y-axis | B1 |
Notes | | (a) Crossing at −3 is B0. Touching at −1 is B0.
(b) $y = (x + 1)(x^2 + 6x + 9) = x^3 + 7x^2 + 15x + 9$ or equiv. (possibly unsimplified) | B1 |
Differentiates their polynomial correctly – may be unsimplified | M1 |
$\frac{dy}{dx} = 3x^2 + 14x + 15$ | A1 cso | (*)
Notes | | (b) M: This needs to be correct differentiation here. A1: Fully correct simplified answer.
(c) At $x = -5$: $\frac{dy}{dx} = 75 - 70 + 15 = 20$ | B1 |
At $x = -5$: $y = -16$ | B1 |
$y - ("-16") = "20"(x - (-5))$ or $y = "20x" + c$ with $(-5, "-16")$ used to find c | M1 |
$y = 20x + 84$ | A1 |
Notes | | (c) M: If the −5 and "−16" are the wrong way round or – omitted the M mark can still be given if a correct formula is seen, (e.g. $y - y_1 = m(x - x_1)$) is seen, otherwise M0. m should be numerical and not 0 or infinity and should not have involved negative reciprocal. (d) 1st M: Putting the derivative expression equal to their value for gradient. 2nd M: Attempt to solve quadratic (see notes) This may be implied by correct answer.
(d) Parallel: $3x^2 + 14x + 15 = "20"$ | M1 |
$(3x - 1)(x + 5) = 0$, $x = ...$ | M1 |
$x = \frac{1}{3}$ | A1 |
---10. The curve $C$ has equation
$$y = ( x + 1 ) ( x + 3 ) ^ { 2 }$$
\begin{enumerate}[label=(\alph*)]
\item Sketch $C$, showing the coordinates of the points at which $C$ meets the axes.
\item Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = 3 x ^ { 2 } + 14 x + 15$.
The point $A$, with $x$-coordinate - 5 , lies on $C$.
\item Find the equation of the tangent to $C$ at $A$, giving your answer in the form $y = m x + c$, where $m$ and $c$ are constants.
Another point $B$ also lies on $C$. The tangents to $C$ at $A$ and $B$ are parallel.
\item Find the $x$-coordinate of $B$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2011 Q10 [14]}}