| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2015 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discriminant and conditions for roots |
| Type | Condition on k, prove inequality |
| Difficulty | Standard +0.3 This is a multi-part question on quadratic functions requiring completion of the square for range (routine), Vieta's formulas for roots (standard), and discriminant manipulation for the no real roots condition (straightforward algebra). All techniques are standard P1 material with no novel insight required, making it slightly 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 unknown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x^2 + 6x - 8 = (x+3)^2 - 17\) | B1 B1 | B1 for \((x+3)^2\), B1 for \(-17\); or B1 for \(x=-3\), B1 \(y=-17\) |
| or \(2x+6=0 \rightarrow x=-3 \rightarrow y=-17\) | B1\(^\checkmark\) [3] | Following through visible method |
| \(\rightarrow\) Range \(f(x) \geqslant -17\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((x-k)(x+2k) = 0\) | M1 | Realises the link between roots and the equation, comparing coefficients of \(x\) |
| \(\equiv x^2 + 5x + b = 0\) | ||
| \(\rightarrow k = 5\) | A1 | |
| \(\rightarrow b = -2k^2 = -50\) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((x+a)^2 + a(x+a) + b = a\) | M1 | Replaces "\(x\)" by "\(x+a\)" in 2 terms |
| Uses \(b^2 - 4ac \rightarrow 9a^2 - 4(2a^2 + b - a)\) | DM1 | Any use of discriminant |
| \(\rightarrow a^2 < 4(b-a)\) | A1 [3] |
## Question 8:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x^2 + 6x - 8 = (x+3)^2 - 17$ | **B1 B1** | B1 for $(x+3)^2$, B1 for $-17$; or B1 for $x=-3$, B1 $y=-17$ |
| or $2x+6=0 \rightarrow x=-3 \rightarrow y=-17$ | **B1$^\checkmark$** [3] | Following through visible method |
| $\rightarrow$ Range $f(x) \geqslant -17$ | | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(x-k)(x+2k) = 0$ | **M1** | Realises the link between roots and the equation, comparing coefficients of $x$ |
| $\equiv x^2 + 5x + b = 0$ | | |
| $\rightarrow k = 5$ | **A1** | |
| $\rightarrow b = -2k^2 = -50$ | **A1** [3] | |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(x+a)^2 + a(x+a) + b = a$ | **M1** | Replaces "$x$" by "$x+a$" in 2 terms |
| Uses $b^2 - 4ac \rightarrow 9a^2 - 4(2a^2 + b - a)$ | **DM1** | Any use of discriminant |
| $\rightarrow a^2 < 4(b-a)$ | **A1** [3] | |
---8 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.\\
(i) In the case where $a = 6$ and $b = - 8$, find the range of f .\\
(ii) In the case where $a = 5$, 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 $k$.\\
(iii) 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 2015 Q8 [9]}}