| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2018 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Inequalities |
| Type | Curve above or below axis |
| Difficulty | Moderate -0.5 Part (i) requires understanding that a quadratic lies above the x-axis when its discriminant is negative (b²-4ac < 0), a standard technique. Part (ii) involves setting the quadratic equal to the line and using discriminant = 0 for tangency, also routine. Both are textbook applications with no novel insight required, making this slightly easier than average. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Either \((x-3)^2 + k - 9 > 0,\ k-9 > 0\) | Either completing the square and using '*their* \(k-9\)' \(> \) or \(\geqslant 0\) OR | |
| or \(2x-6=0 \rightarrow (3,\ k-9),\ k-9>0\) | M1 | Differentiating and setting to 0, using '*their* \(x=3\)' to find \(y\) and using '*their* \(k-9\)' \(>\) or \(\geqslant 0\) OR |
| or \(b^2 < 4ac\) oe \(\rightarrow 36 < 4k\) | Use of discriminant \(<\) or \(\leqslant 0\). Beware use of \(>\) and incorrect algebra. | |
| \(\rightarrow k > 9\) Note: not \(\geqslant\) | A1 | T&I leading to (or no working) correct answer 2/2 otherwise 0/2. |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(x^2 - 6x + k = 7 - 2x \rightarrow x^2 - 4x + k - 7 = 0\) | *M1 | Equates and collects terms |
| Use of \(b^2 - 4ac = 0\): \((16 - 4(k-7) = 0)\) | DM1 | Correct use of discriminant = 0, involving \(k\) from a 3 term quadratic |
| Answer | Marks | Guidance |
|---|---|---|
| \(2x - 6 = -2 \rightarrow x = 2\) \((y = 3)\) | *M1 | Equates their \(\frac{dy}{dx}\) to \(\pm 2\), finds a value for \(x\) |
| \((their\ 3)\) or \(7 - 2(their\ 2) = (their\ 2)^2 - 6(their\ 2) + k\) | DM1 | Substitutes their value(s) into the appropriate equation |
| \(\rightarrow k = 11\) | A1 |
**Question 2(i):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| **Either** $(x-3)^2 + k - 9 > 0,\ k-9 > 0$ | | Either completing the square and using '*their* $k-9$' $> $ or $\geqslant 0$ OR |
| **or** $2x-6=0 \rightarrow (3,\ k-9),\ k-9>0$ | **M1** | Differentiating and setting to 0, using '*their* $x=3$' to find $y$ and using '*their* $k-9$' $>$ or $\geqslant 0$ OR |
| **or** $b^2 < 4ac$ oe $\rightarrow 36 < 4k$ | | Use of discriminant $<$ or $\leqslant 0$. Beware use of $>$ and incorrect algebra. |
| $\rightarrow k > 9$ Note: not $\geqslant$ | **A1** | T&I leading to (or no working) correct answer 2/2 otherwise 0/2. |
| | **2** | |
## Question 2(ii):
**EITHER method:**
$x^2 - 6x + k = 7 - 2x \rightarrow x^2 - 4x + k - 7 = 0$ | *M1 | Equates and collects terms
Use of $b^2 - 4ac = 0$: $(16 - 4(k-7) = 0)$ | DM1 | Correct use of discriminant = 0, involving $k$ from a 3 term quadratic
**OR method:**
$2x - 6 = -2 \rightarrow x = 2$ $(y = 3)$ | *M1 | Equates their $\frac{dy}{dx}$ to $\pm 2$, finds a value for $x$
$(their\ 3)$ or $7 - 2(their\ 2) = (their\ 2)^2 - 6(their\ 2) + k$ | DM1 | Substitutes their value(s) into the appropriate equation
$\rightarrow k = 11$ | A1 |
---2 The equation of a curve is $y = x ^ { 2 } - 6 x + k$, where $k$ is a constant.\\
(i) Find the set of values of $k$ for which the whole of the curve lies above the $x$-axis.\\
(ii) Find the value of $k$ for which the line $y + 2 x = 7$ is a tangent to the curve.\\
\hfill \mbox{\textit{CAIE P1 2018 Q2 [5]}}