| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2018 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Inequalities |
| Type | Tangent condition for line-curve |
| Difficulty | Standard +0.3 This is a standard discriminant problem requiring students to set up a quadratic equation, show b²-4ac≥0 for all k (part i), then find when the discriminant equals zero for tangency (part ii). The algebra is straightforward and the method is a common textbook exercise, making it slightly easier than average. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.02f Solve quadratic equations: including in a function of unknown1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| For their 3-term quadratic, a recognisable application of \(b^2 - 4ac\) | M1 | Expect \(2x^2 - x(3+k) + 1 - k^2 (= 0)\) oe for the 3-term quad. |
| \((b^2 - 4ac =) (3+k)^2 - 4(2)(1-k^2)\) oe | A1 | Must be correct. Ignore any RHS |
| \(9k^2 + 6k + 1\) | A1 | Ignore any RHS |
| \((3k+1)^2 \geqslant 0\) Do not allow \(> 0\). Hence curve and line meet. AG | A1 | Allow \((9)\left(k + \frac{1}{3}\right)^2 \geqslant 0\). Conclusion required. |
| ALT: Attempt solution of 3-term quadratic | M1 | |
| Solutions \(x = k+1\), \(\frac{1}{2}(1-k)\) | A1A1 | |
| Which exist for all values of \(k\). Hence curve and line meet. AG | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(k = -\frac{1}{3}\) | B1 | ALT: \(dy/dx = 4x - 3 \Rightarrow 4x - 3 = k\) |
| Sub (one of) their \(k = -\frac{1}{3}\) into either line \(1 \rightarrow 2x^2 - \frac{8}{3}x + \frac{8}{9} (= 0)\) or into the derivative of line \(1 \rightarrow 4x - (3+k)(= 0)\) | M1 | Sub \(k = 4x - 3\) into line \(1 \rightarrow 2x^2 - x(4x) + 1 - (4x-3)^2 (= 0)\) |
| \(x = \frac{2}{3}\) Do not allow unsubstantiated \(\left(\frac{2}{3}, -\frac{1}{9}\right)\) following \(k = -\frac{1}{3}\) | A1 | \(x = \frac{2}{3}\), \(y = -\frac{1}{9}\) (both required) [from \(-18x^2 + 24x - 8 (= 0)\) oe] |
| \(y = -\frac{1}{9}\) Do not allow unsubstantiated \(\left(\frac{2}{3}, -\frac{1}{9}\right)\) following \(k = -\frac{1}{3}\) | A1 | \(k = -\frac{1}{3}\) |
## Question 9(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| For their 3-term quadratic, a recognisable application of $b^2 - 4ac$ | M1 | Expect $2x^2 - x(3+k) + 1 - k^2 (= 0)$ oe for the 3-term quad. |
| $(b^2 - 4ac =) (3+k)^2 - 4(2)(1-k^2)$ oe | A1 | Must be correct. Ignore any RHS |
| $9k^2 + 6k + 1$ | A1 | Ignore any RHS |
| $(3k+1)^2 \geqslant 0$ Do not allow $> 0$. Hence curve and line meet. **AG** | A1 | Allow $(9)\left(k + \frac{1}{3}\right)^2 \geqslant 0$. Conclusion required. |
| ALT: Attempt solution of 3-term quadratic | M1 | |
| Solutions $x = k+1$, $\frac{1}{2}(1-k)$ | A1A1 | |
| Which exist for all values of $k$. Hence curve and line meet. **AG** | A1 | |
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## Question 9(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $k = -\frac{1}{3}$ | B1 | ALT: $dy/dx = 4x - 3 \Rightarrow 4x - 3 = k$ |
| Sub (one of) their $k = -\frac{1}{3}$ into either line $1 \rightarrow 2x^2 - \frac{8}{3}x + \frac{8}{9} (= 0)$ or into the derivative of line $1 \rightarrow 4x - (3+k)(= 0)$ | M1 | Sub $k = 4x - 3$ into line $1 \rightarrow 2x^2 - x(4x) + 1 - (4x-3)^2 (= 0)$ |
| $x = \frac{2}{3}$ Do not allow unsubstantiated $\left(\frac{2}{3}, -\frac{1}{9}\right)$ following $k = -\frac{1}{3}$ | A1 | $x = \frac{2}{3}$, $y = -\frac{1}{9}$ (both required) [from $-18x^2 + 24x - 8 (= 0)$ oe] |
| $y = -\frac{1}{9}$ Do not allow unsubstantiated $\left(\frac{2}{3}, -\frac{1}{9}\right)$ following $k = -\frac{1}{3}$ | A1 | $k = -\frac{1}{3}$ |
---9 A curve has equation $y = 2 x ^ { 2 } - 3 x + 1$ and a line has equation $y = k x + k ^ { 2 }$, where $k$ is a constant.\\
(i) Show that, for all values of $k$, the curve and the line meet.\\
(ii) State the value of $k$ for which the line is a tangent to the curve and find the coordinates of the point where the line touches the curve.\\
\hfill \mbox{\textit{CAIE P1 2018 Q9 [8]}}