| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2016 |
| Session | March |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Find constant then solve inequality or further work |
| Difficulty | Moderate -0.5 This is a straightforward application of the factor theorem requiring students to substitute x = -1/2 to find a, then factorise and solve an inequality. All steps are routine A-level techniques with no novel insight required, making it slightly easier than average but still requiring multiple standard procedures. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Substitute \(x = -\frac{1}{2}\) and equate to zero, or divide by \((2x+1)\) and equate constant remainder to zero | M1 | |
| Obtain \(a = 3\) | A1 | [2] |
| (ii) (a) Commence division by \((2x+1)\) reaching a partial quotient of \(2x^2 + kx\) | M1 | |
| Obtain factorisation \((2x+1)(2x^2 - x + 2)\) | A1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| (b) State or imply critical value \(x = -\frac{1}{2}\) | B1 | |
| Show that \(2x^2 - x + 2\) is always positive, or that the gradient of \(4x^3 + 3x + 2\) is always positive | B1* | |
| Justify final answer \(x > -\frac{1}{2}\) | B1(dep*) | [3] |
(i) Substitute $x = -\frac{1}{2}$ and equate to zero, or divide by $(2x+1)$ and equate constant remainder to zero | M1 |
Obtain $a = 3$ | A1 | [2]
(ii) (a) Commence division by $(2x+1)$ reaching a partial quotient of $2x^2 + kx$ | M1 |
Obtain factorisation $(2x+1)(2x^2 - x + 2)$ | A1 | [2] |
[The M1 is earned if inspection reaches an unknown factor $2x^2 + Bx + C$ and an equation in $B$ and/or $C$, or an unknown factor $Ax^2 + Bx + 2$ and an equation in $A$ and/or $B$.]
(b) State or imply critical value $x = -\frac{1}{2}$ | B1 |
Show that $2x^2 - x + 2$ is always positive, or that the gradient of $4x^3 + 3x + 2$ is always positive | B1* |
Justify final answer $x > -\frac{1}{2}$ | B1(dep*) | [3]4 The polynomial $4 x ^ { 3 } + a x + 2$, where $a$ is a constant, is denoted by $\mathrm { p } ( x )$. It is given that ( $2 x + 1$ ) is a factor of $\mathrm { p } ( x )$.\\
(i) Find the value of $a$.\\
(ii) When $a$ has this value,
\begin{enumerate}[label=(\alph*)]
\item factorise $\mathrm { p } ( x )$,
\item solve the inequality $\mathrm { p } ( x ) > 0$, justifying your answer.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2016 Q4 [7]}}