| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simultaneous equations |
| Type | Tangency condition for line and curve |
| Difficulty | Standard +0.3 Part (a) requires substituting two x-values into simultaneous equations to find k and a—straightforward algebra. Part (b) involves the tangency condition (discriminant = 0) for a quadratic, which is a standard technique. Both parts are routine applications of well-practiced methods with no novel insight required, making this slightly easier than the average A-level question. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02q Use intersection points: of graphs to solve equations1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(4\times 0^2 - 0 + \frac{1}{2}k^2 = 0 - a\) | M1 | Equating the equations of curve and line and substituting \(x = 0\). Condone slight errors e.g. \(\pm\) sign errors |
| \(4\times\left(\frac{3}{4}\right)^2 - \frac{3}{4}k + \frac{1}{2}k^2 = \frac{3}{4} - a\) | M1 | Equating the equations of curve and line and substituting \(x = \frac{3}{4}\). Condone slight errors e.g. \(\pm\) sign errors |
| \(k = 2,\ a = -2\) | A1 A1 | WWW |
| Alternative: \((x-0)\left(x-\frac{3}{4}\right) = 0\) or \(x(4x-3) = 0 [\Rightarrow 4x^2 - 3x = 0]\) | \*M1 | Use \(0, \frac{3}{4}\) to form a quadratic equation. Do not allow \((x+0)\left(x+\frac{3}{4}\right)=0\) |
| \(4x^2 - kx + \frac{1}{2}k^2 = x - a\) leading to \(4x^2 - (k+1)x + \frac{1}{2}k^2 + a[=0]\) | DM1 | Equating the equations of curve and line and rearranging so that terms are all on same side. Condone slight errors e.g. \(\pm\) sign errors |
| \(k = 2,\ a = -2\) | A1 A1 | WWW |
| Alternative: \(-\frac{b}{a} = \frac{3}{4} + 0\) and \(\frac{c}{a} = 0\times\frac{3}{4}\) | \*M1 | Using sum and product of roots. Condone \(\pm\) sign errors |
| \(\frac{k+1}{4} = \frac{3}{4}\) and \(\frac{\frac{1}{2}k^2 + a}{4} = 0\) | DM1 | Equating the equations of curve and line and equating to \(\frac{3}{4}\) and \(0\) |
| \(k = 2,\ a = -2\) | A1 A1 | WWW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(4x^2 - kx + \frac{1}{2}k^2 = x + \frac{7}{2} \Rightarrow 4x^2 - kx - x + \frac{1}{2}k^2 - \frac{7}{2}[=0]\) | \*M1 | OE. Substitute \(a = -\frac{7}{2}\) and rearrange so that terms are all on same side. Condone \(\pm\) sign errors. Watch for multiples |
| \((k+1)^2 - 4\times 4\left(\frac{1}{2}k^2 - \frac{7}{2}\right)\) | \*DM1 | Use of \(b^2 - 4ac\) with the coefficients from *their* 3-term quadratic. Both coefficients '\(b\)' and '\(c\)' must consist of two components |
| \(\Rightarrow 7k^2 - 2k - 57\) | A1 | OE |
| \((k-3)(7k+19)\) or other valid method | DM1 | Factorising or use of formula or completing the square. Must be evidence of an attempt to solve. Dependent upon both previous method marks |
| \(k = 3,\ k = -\frac{19}{7}\) | A1 | OE e.g. AWRT \(-2.71\). No ISW if inequalities used. SC: If second DM1 not scored, SC B1 available for correct final answers |
| Alternative: \(8x - k = 1\) and \(4x^2 - kx + \frac{1}{2}k^2 = x + \frac{7}{2}\) | \*M1 | Equating gradients and equating line and curve |
| \(4x^2 - (8x-1)x + \frac{1}{2}(8x-1)^2 = x + \frac{7}{2}\) or \(4\left(\frac{k+1}{8}\right)^2 - k\left(\frac{k+1}{8}\right) + \frac{1}{2}k^2 = \frac{k+1}{8} + \frac{7}{2}\) | \*DM1 | Forming an equation in \(x\) or \(k\) only |
| \(28x^2 - 8x - 3\) or \(7k^2 - 2k - 57\) | A1 | OE. A correct 3 term quadratic in \(x\) or \(k\) only |
| \((14x+3)(2x-1)\) or \((k-3)(7k+19)\) or other valid method | DM1 | OE. Factorising or use of formula or completing the square. Must be evidence of attempt to solve. Dependent upon both previous method marks |
| \(k = 3,\ k = -\frac{19}{7}\) | A1 | OE |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $4\times 0^2 - 0 + \frac{1}{2}k^2 = 0 - a$ | **M1** | Equating the equations of curve and line and substituting $x = 0$. Condone slight errors e.g. $\pm$ sign errors |
| $4\times\left(\frac{3}{4}\right)^2 - \frac{3}{4}k + \frac{1}{2}k^2 = \frac{3}{4} - a$ | **M1** | Equating the equations of curve and line and substituting $x = \frac{3}{4}$. Condone slight errors e.g. $\pm$ sign errors |
| $k = 2,\ a = -2$ | **A1 A1** | WWW |
| **Alternative:** $(x-0)\left(x-\frac{3}{4}\right) = 0$ or $x(4x-3) = 0 [\Rightarrow 4x^2 - 3x = 0]$ | **\*M1** | Use $0, \frac{3}{4}$ to form a quadratic equation. Do not allow $(x+0)\left(x+\frac{3}{4}\right)=0$ |
| $4x^2 - kx + \frac{1}{2}k^2 = x - a$ leading to $4x^2 - (k+1)x + \frac{1}{2}k^2 + a[=0]$ | **DM1** | Equating the equations of curve and line and rearranging so that terms are all on same side. Condone slight errors e.g. $\pm$ sign errors |
| $k = 2,\ a = -2$ | **A1 A1** | WWW |
| **Alternative:** $-\frac{b}{a} = \frac{3}{4} + 0$ and $\frac{c}{a} = 0\times\frac{3}{4}$ | **\*M1** | Using sum and product of roots. Condone $\pm$ sign errors |
| $\frac{k+1}{4} = \frac{3}{4}$ and $\frac{\frac{1}{2}k^2 + a}{4} = 0$ | **DM1** | Equating the equations of curve and line and equating to $\frac{3}{4}$ and $0$ |
| $k = 2,\ a = -2$ | **A1 A1** | WWW |
---
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $4x^2 - kx + \frac{1}{2}k^2 = x + \frac{7}{2} \Rightarrow 4x^2 - kx - x + \frac{1}{2}k^2 - \frac{7}{2}[=0]$ | **\*M1** | OE. Substitute $a = -\frac{7}{2}$ and rearrange so that terms are all on same side. Condone $\pm$ sign errors. Watch for multiples |
| $(k+1)^2 - 4\times 4\left(\frac{1}{2}k^2 - \frac{7}{2}\right)$ | **\*DM1** | Use of $b^2 - 4ac$ with the coefficients from *their* 3-term quadratic. Both coefficients '$b$' and '$c$' must consist of two components |
| $\Rightarrow 7k^2 - 2k - 57$ | **A1** | OE |
| $(k-3)(7k+19)$ or other valid method | **DM1** | Factorising or use of formula or completing the square. Must be evidence of an attempt to solve. Dependent upon both previous method marks |
| $k = 3,\ k = -\frac{19}{7}$ | **A1** | OE e.g. AWRT $-2.71$. No ISW if inequalities used. **SC:** If second DM1 not scored, **SC B1** available for correct final answers |
| **Alternative:** $8x - k = 1$ and $4x^2 - kx + \frac{1}{2}k^2 = x + \frac{7}{2}$ | **\*M1** | Equating gradients and equating line and curve |
| $4x^2 - (8x-1)x + \frac{1}{2}(8x-1)^2 = x + \frac{7}{2}$ or $4\left(\frac{k+1}{8}\right)^2 - k\left(\frac{k+1}{8}\right) + \frac{1}{2}k^2 = \frac{k+1}{8} + \frac{7}{2}$ | **\*DM1** | Forming an equation in $x$ or $k$ only |
| $28x^2 - 8x - 3$ or $7k^2 - 2k - 57$ | **A1** | OE. A correct 3 term quadratic in $x$ or $k$ only |
| $(14x+3)(2x-1)$ or $(k-3)(7k+19)$ or other valid method | **DM1** | OE. Factorising or use of formula or completing the square. Must be evidence of attempt to solve. Dependent upon both previous method marks |
| $k = 3,\ k = -\frac{19}{7}$ | **A1** | OE |
---5 The equation of a curve is $y = 4 x ^ { 2 } - k x + \frac { 1 } { 2 } k ^ { 2 }$ and the equation of a line is $y = x - a$, where $k$ and $a$ are constants.
\begin{enumerate}[label=(\alph*)]
\item Given that the curve and the line intersect at the points with $x$-coordinates 0 and $\frac { 3 } { 4 }$, find the values of $k$ and $a$.
\item Given instead that $a = - \frac { 7 } { 2 }$, find the values of $k$ for which the line is a tangent to the curve. [5]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2022 Q5 [9]}}