CAIE P2 2021 March — Question 6 10 marks

Exam BoardCAIE
ModuleP2 (Pure Mathematics 2)
Year2021
SessionMarch
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFactor & Remainder Theorem
TypeOne factor, one non-zero remainder
DifficultyModerate -0.3 Part (a) is a standard application of factor and remainder theorems requiring solving two simultaneous equations from p(-2)=0 and p(3)=5. Part (b) adds a mild substitution step with exponentials but remains routine. This is slightly easier than average due to the straightforward algebraic manipulation and lack of conceptual complexity.
Spec1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.06g Equations with exponentials: solve a^x = b

6 The polynomial \(\mathrm { p } ( x )\) is defined by $$\mathrm { p } ( x ) = x ^ { 3 } + a x + b$$ where \(a\) and \(b\) are constants. It is given that \(( x + 2 )\) is a factor of \(\mathrm { p } ( x )\) and that the remainder is 5 when \(\mathrm { p } ( x )\) is divided by \(( x - 3 )\).
  1. Find the values of \(a\) and \(b\).
  2. Hence find the exact root of the equation \(\mathrm { p } \left( \mathrm { e } ^ { 2 y } \right) = 0\).

Question 6(a):
AnswerMarks Guidance
AnswerMarks Guidance
Substitute \(x=-2\) and equate to zero*M1
Substitute \(x=3\) and equate to 5*M1
Obtain \(-8-2a+b=0\) and \(27+3a+b=5\) or equivalentsA1
Solve a pair of relevant linear simultaneous equations for \(a\) or \(b\)DM1 Dependent on at least one M mark
Obtain \(a=-6\) and \(b=-4\)A1
Question 6(b):
AnswerMarks Guidance
AnswerMarks Guidance
Attempt division by \(x+2\) at least as far as \(x^2+kx\)M1
Obtain \(x^2-2x-2\)A1
Obtain (at least) the positive root \(\frac{2+\sqrt{12}}{2}\) or exact equivalentA1
Equate \(e^{2y}\) to positive root, apply logarithms and use power lawM1
Obtain \(\frac{1}{2}\ln\left(\frac{2+\sqrt{12}}{2}\right)\) or \(\frac{1}{2}\ln(1+\sqrt{3})\) or exact equivalentA1
## Question 6(a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Substitute $x=-2$ and equate to zero | *M1 | |
| Substitute $x=3$ and equate to 5 | *M1 | |
| Obtain $-8-2a+b=0$ and $27+3a+b=5$ or equivalents | A1 | |
| Solve a pair of relevant linear simultaneous equations for $a$ or $b$ | DM1 | Dependent on at least one M mark |
| Obtain $a=-6$ and $b=-4$ | A1 | |

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## Question 6(b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt division by $x+2$ at least as far as $x^2+kx$ | M1 | |
| Obtain $x^2-2x-2$ | A1 | |
| Obtain (at least) the positive root $\frac{2+\sqrt{12}}{2}$ or exact equivalent | A1 | |
| Equate $e^{2y}$ to positive root, apply logarithms and use power law | M1 | |
| Obtain $\frac{1}{2}\ln\left(\frac{2+\sqrt{12}}{2}\right)$ or $\frac{1}{2}\ln(1+\sqrt{3})$ or exact equivalent | A1 | |
6 The polynomial $\mathrm { p } ( x )$ is defined by

$$\mathrm { p } ( x ) = x ^ { 3 } + a x + b$$

where $a$ and $b$ are constants. It is given that $( x + 2 )$ is a factor of $\mathrm { p } ( x )$ and that the remainder is 5 when $\mathrm { p } ( x )$ is divided by $( x - 3 )$.
\begin{enumerate}[label=(\alph*)]
\item Find the values of $a$ and $b$.
\item Hence find the exact root of the equation $\mathrm { p } \left( \mathrm { e } ^ { 2 y } \right) = 0$.
\end{enumerate}

\hfill \mbox{\textit{CAIE P2 2021 Q6 [10]}}