| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2020 |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Solving quadratics and applications |
| Type | Quadratic function range and roots analysis |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question on basic quadratic properties. Part (a) asks for the range (likely completing the square), part (b) uses sum/product of roots with a simple constraint, and part (c) is a routine discriminant condition. All parts involve standard techniques with no novel insight required, making this easier than average. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02e Complete the square: quadratic polynomials and turning points1.02f Solve quadratic equations: including in a function of unknown1.02u Functions: definition and vocabulary (domain, range, mapping) |
| Answer | Marks | Guidance |
|---|---|---|
| \(x^2 + 6x - 8 = (x+3)^2 - 17\) OR \(2x + 6 = 0 \rightarrow x = -3 \rightarrow y = -17\) | B1B1 | B1 for \((x+3)^2\), B1 for \(-17\) OR B1 for \(x = -3\), B1 for \(y = -17\) |
| Range \(f(x) \geqslant -17\) | B1FT | FT; following through visible method |
| Answer | Marks | Guidance |
|---|---|---|
| \((x-k)(x+2k) = 0 \equiv x^2 + 5x + b = 0\) | M1 | Realises the link between roots and the equation |
| \(k = 5\) | A1 | Comparing coefficients of \(x\) |
| \(b = -2k^2 = -50\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \((x+a)^2 + a(x+a) + b = a\) | M1* | Replaces '\(x\)' by '\(x + a\)' in 2 terms |
| Uses \(b^2 - 4ac\), \(9a^2 - 4(2a^2 + b - a)\) | DM1 | Any use of discriminant |
| \(a^2 < 4(b-a)\) | A1 | AG |
# Question 11:
## Part 11(a):
| $x^2 + 6x - 8 = (x+3)^2 - 17$ **OR** $2x + 6 = 0 \rightarrow x = -3 \rightarrow y = -17$ | B1B1 | B1 for $(x+3)^2$, B1 for $-17$ **OR** B1 for $x = -3$, B1 for $y = -17$ |
| Range $f(x) \geqslant -17$ | B1FT | FT; following through visible method |
## Part 11(b):
| $(x-k)(x+2k) = 0 \equiv x^2 + 5x + b = 0$ | M1 | Realises the link between roots and the equation |
| $k = 5$ | A1 | Comparing coefficients of $x$ |
| $b = -2k^2 = -50$ | A1 | |
## Part 11(c):
| $(x+a)^2 + a(x+a) + b = a$ | M1* | Replaces '$x$' by '$x + a$' in 2 terms |
| Uses $b^2 - 4ac$, $9a^2 - 4(2a^2 + b - a)$ | DM1 | Any use of discriminant |
| $a^2 < 4(b-a)$ | A1 | AG |
---11 The function f is defined, for $x \in \mathbb { R }$, by $\mathrm { f } : x \mapsto x ^ { 2 } + a x + b$, where $a$ and $b$ are constants.\\
(a) It is given that $a = 6$ and $b = - 8$.
Find the range of f .\\
(b) It is given instead that $a = 5$ and that the roots of the equation $\mathrm { f } ( x ) = 0$ are $k$ and $- 2 k$, where $k$ is a constant.
Find the values of $b$ and $k$.\\
(c) Show that if the equation $\mathrm { f } ( x + a ) = a$ has no real roots then $a ^ { 2 } < 4 ( b - a )$.\\
\hfill \mbox{\textit{CAIE P1 2020 Q11 [9]}}