| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 12 |
| 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 y-intercept, identifying roots (with multiplicity), and finding a tangent using the product rule. While it has multiple parts and requires careful algebraic manipulation (12 marks total), all techniques are routine for C1 with no problem-solving insight needed—slightly easier than average. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| i) [Graph showing positive cubic with max and min, double root at \(x = -2\) and single root at \(x = \frac{3}{2}\)] | B1, B1, B1 [3] | Positive cubic with max and min. Correct \(y\) intercept – graph must be drawn. Double root shown at \(x = -2\) and single root at \(x = \frac{3}{2}\) with no extras – graph must be drawn. For first mark must clearly be a cubic – must not stop at either axis, do not allow straight line sections/tending to extra turning points etc. |
| ii) \(x^2 + 4x + 4\) or \(2x^3 + 5x^2 - 4x - 12\); \(\frac{dy}{dx} = 6x^2 + 10x - 4\) | B1, M1, A1, M1*, M1dep*, A1n, B1, M1, A1 [9] | Obtain one quadratic factor. Multiply their three term quadratic by linear factor to obtain at least 5 term cubic. If simplified, must be correct. Attempt to differentiate (power of at least one term involving \(x\) reduced by one). When \(x = -1\), gradient = \(-8\). When \(x = -1, y = -5\); \(y + 5 = -8(x+1)\); Correct \(y\) value. Correct equation of straight line through \((-1\), their \(y)\), their gradient from differentiation. \(y\) must have been found, do not allow use of gradient of normal instead of tangent. i.e. \(k(8x + y + 13) = 0\). Must have "=0". Note: If \(x = 1\) used instead of \(x = -1\), then max possible from last 5 marks is M1 M1 only |
| Answer | Marks |
|---|---|
| Correct answer in correct form | [9] |
**i)** [Graph showing positive cubic with max and min, double root at $x = -2$ and single root at $x = \frac{3}{2}$] | B1, B1, B1 [3] | Positive cubic with max and min. Correct $y$ intercept – graph must be drawn. Double root shown at $x = -2$ and single root at $x = \frac{3}{2}$ with no extras – graph must be drawn. For first mark must clearly be a cubic – must not stop at either axis, do not allow straight line sections/tending to extra turning points etc.
**ii)** $x^2 + 4x + 4$ or $2x^3 + 5x^2 - 4x - 12$; $\frac{dy}{dx} = 6x^2 + 10x - 4$ | B1, M1, A1, M1*, M1dep*, A1n, B1, M1, A1 [9] | Obtain one quadratic factor. Multiply their three term quadratic by linear factor to obtain at least 5 term cubic. If simplified, must be correct. Attempt to differentiate (power of at least one term involving $x$ reduced by one). When $x = -1$, gradient = $-8$. When $x = -1, y = -5$; $y + 5 = -8(x+1)$; Correct $y$ value. Correct equation of straight line through $(-1$, their $y)$, their gradient from differentiation. $y$ must have been found, do not allow use of gradient of normal instead of tangent. i.e. $k(8x + y + 13) = 0$. Must have "=0". Note: If $x = 1$ used instead of $x = -1$, then max possible from last 5 marks is M1 M1 only
# Question 10 (continued):
Correct answer in correct form | [9] |
**Guidance at end of mark scheme for Question 5(ii):**
For the inequality $3x^2 - 13x - 10 \geq 0$ with roots $-\frac{2}{3}$ and $5$:
Example markings:
- $-\frac{2}{3} \leq x \geq 5$ scores M1A0
- Allow "$x \leq -\frac{2}{3}, x \geq 5$", "$x \leq -\frac{2}{3}$", "$x \leq -\frac{2}{3}$" or "$x \geq 5$" but do not allow "$x \leq -\frac{2}{3}$"A curve has equation $y = (x + 2)^2(2x - 3)$.
\begin{enumerate}[label=(\roman*)]
\item Sketch the curve, giving the coordinates of all points of intersection with the axes. [3]
\item Find an equation of the tangent to the curve at the point where $x = -1$. Give your answer in the form $ax + by + c = 0$. [9]
\end{enumerate}
\hfill \mbox{\textit{OCR C1 2014 Q10 [12]}}