| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Find tangent to polynomial curve |
| Difficulty | Moderate -0.3 This is a straightforward C1 curve sketching question requiring basic differentiation and algebraic manipulation. Finding roots is immediate from factored form, the tangent calculation uses standard point-gradient formula, and finding stationary points involves routine differentiation of an expanded polynomial. Slightly easier than average due to the given factored form and verification rather than discovery in part (ii). |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Sketch showing curve with turning points at \(A(2,0)\) and \(B(3,0)\) with appropriate shape | B3 | |
| (ii) \(y = (2-x)(9-6x+x^2)\) | M1 | |
| \(y = 18 - 12x + 2x^2 - 9x + 6x^2 - x^3\) | A1 | |
| \(y = 18 - 21x + 8x^2 - x^3\) | ||
| \(\frac{dy}{dx} = -21 + 16x - 3x^2\) | M1 A1 | |
| Gradient \(= -21 + 32 - 12 = -1\) | M1 | |
| \(\therefore y - 0 = -(x-2)\) | M1 | |
| \(x + y = 2\) | A1 | |
| (iii) \(-21 + 16x - 3x^2 = 0\) | M1 | |
| \(-(3x-7)(x-3) = 0\) | M1 | |
| \(x = 3\) (at \(B\)), \(\frac{7}{3}\) | A1 | |
| \(y = (-\frac{1}{3})(\frac{7}{3})^2 = -\frac{49}{27}\) \(\therefore (\frac{7}{3}, -\frac{49}{27})\) | A1 | (13) |
**(i)** Sketch showing curve with turning points at $A(2,0)$ and $B(3,0)$ with appropriate shape | B3 |
**(ii)** $y = (2-x)(9-6x+x^2)$ | M1 |
$y = 18 - 12x + 2x^2 - 9x + 6x^2 - x^3$ | A1 |
$y = 18 - 21x + 8x^2 - x^3$ | |
$\frac{dy}{dx} = -21 + 16x - 3x^2$ | M1 A1 |
Gradient $= -21 + 32 - 12 = -1$ | M1 |
$\therefore y - 0 = -(x-2)$ | M1 |
$x + y = 2$ | A1 |
**(iii)** $-21 + 16x - 3x^2 = 0$ | M1 |
$-(3x-7)(x-3) = 0$ | M1 |
$x = 3$ (at $B$), $\frac{7}{3}$ | A1 |
$y = (-\frac{1}{3})(\frac{7}{3})^2 = -\frac{49}{27}$ $\therefore (\frac{7}{3}, -\frac{49}{27})$ | A1 | **(13)**
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**Total (72)**10. The curve with equation $y = ( 2 - x ) ( 3 - x ) ^ { 2 }$ crosses the $x$-axis at the point $A$ and touches the $x$-axis at the point $B$.\\
(i) Sketch the curve, showing the coordinates of $A$ and $B$.\\
(ii) Show that the tangent to the curve at $A$ has the equation
$$x + y = 2$$
Given that the curve is stationary at the points $B$ and $C$,\\
(iii) find the exact coordinates of $C$.
\hfill \mbox{\textit{OCR C1 Q10 [13]}}