| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2016 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discriminant and conditions for roots |
| Type | Roots given, find equation constants |
| Difficulty | Moderate -0.8 This is a straightforward two-part question testing standard techniques: (a) finding tangent conditions using discriminant (b=0 for mx = 2x²-4x+8) and (b) using factor form to find coefficients and completing the square for vertex. Both parts are routine textbook exercises requiring only direct application of well-practiced methods with no problem-solving insight needed. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = 2x^2 - 4x + 8\); equates with \(y = mx\) and selects \(a, b, c\) | M1 M1 | Equate + solution or use of \(dy/dx\) |
| Uses \(b^2 = 4ac\); \(\rightarrow m = 4\) or \(-12\) | A1 | Use of discriminant for both |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(x) = x^2 + ax + b\); equation of form \((x-1)(x-9)\) | M1 | Any valid method; allow \((x+1)(x+9)\) for M1 |
| \(\rightarrow a = -10, b = 9\) (or using 2 simultaneous equations) | A1 | Must be stated |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Calculus or \(x = \frac{1}{2}(1+9)\) by symmetry | M1 | Any valid method |
| \(\rightarrow (5, -16)\) | A1 |
## Question 6(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = 2x^2 - 4x + 8$; equates with $y = mx$ and selects $a, b, c$ | **M1 M1** | Equate + solution or use of $dy/dx$ |
| Uses $b^2 = 4ac$; $\rightarrow m = 4$ or $-12$ | **A1** | Use of discriminant for both |
## Question 6(b)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(x) = x^2 + ax + b$; equation of form $(x-1)(x-9)$ | **M1** | Any valid method; allow $(x+1)(x+9)$ for M1 |
| $\rightarrow a = -10, b = 9$ (or using 2 simultaneous equations) | **A1** | Must be stated |
## Question 6(b)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Calculus or $x = \frac{1}{2}(1+9)$ by symmetry | **M1** | Any valid method |
| $\rightarrow (5, -16)$ | **A1** | |6
\begin{enumerate}[label=(\alph*)]
\item Find the values of the constant $m$ for which the line $y = m x$ is a tangent to the curve $y = 2 x ^ { 2 } - 4 x + 8$.
\item The function f is defined for $x \in \mathbb { R }$ by $\mathrm { f } ( x ) = x ^ { 2 } + a x + b$, where $a$ and $b$ are constants. The solutions of the equation $\mathrm { f } ( x ) = 0$ are $x = 1$ and $x = 9$. Find
\begin{enumerate}[label=(\roman*)]
\item the values of $a$ and $b$,
\item the coordinates of the vertex of the curve $y = \mathrm { f } ( x )$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2016 Q6 [7]}}