CAIE P3 2014 June — Question 5 7 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2014
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFactor & Remainder Theorem
TypeMultiple unknowns with derivative condition
DifficultyStandard +0.8 This question requires understanding of the product rule for differentiation, factor theorem, and solving simultaneous equations. Part (i) is a proof requiring algebraic manipulation of derivatives. Part (ii) combines multiple concepts: applying factor theorem twice (using f(2)=0 and f'(2)=0 from part i), then solving the resulting system. While systematic, it demands more conceptual insight than routine factor theorem questions and involves non-trivial algebraic manipulation across 7 marks total.
Spec1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.07a Derivative as gradient: of tangent to curve

  1. The polynomial \(f(x)\) is of the form \((x - 2)^2g(x)\), where \(g(x)\) is another polynomial. Show that \((x - 2)\) is a factor of \(f'(x)\). [2]
  2. The polynomial \(x^5 + ax^4 + 3x^3 + bx^2 + a\), where \(a\) and \(b\) are constants, has a factor \((x - 2)^2\). Using the factor theorem and the result of part (i), or otherwise, find the values of \(a\) and \(b\). [5]

AnswerMarks
(i) Differentiate \(f(x)\) and obtain \(f'(x) = (x - 2)^2 g'(x) + 2(x - 2)g(x)\)B1
Conclude that \((x - 2)\) is a factor of \(f'(x)\)B1
Total: 2 marks
(ii) Either: Substitute \(x = 2\), equate to zero and state a correct equation, e.g. \(32 + 16a + 24 + 4b + a = 0\)B1
Differentiate polynomial, substitute \(x = 2\) and equate to zero or divide by \((x - 2)\) and equate constant remainder to zeroM1*
Obtain a correct equation, e.g. \(80 + 32a + 36 + 4b = 0\)A1
Or1: Identify given polynomial with \((x - 2)^2(x^3 + Ax^2 + Bx + C)\) and obtain an equation in \(a\) and/or \(b\)M1*
Obtain a correct equation, e.g. \(\frac{1}{4}a - 4(4 + a) + 4 = 3\)A1
Obtain a second correct equation, e.g. \(-\frac{1}{4}a + 4(4 + a) = b\)A1
Or2: Divide given polynomial by \((x - 2)^2\) and obtain an equation in \(a\) and \(b\)M1*
Obtain a correct equation, e.g. \(29 + 8a + b + 0\)A1
Obtain a second correct equation, e.g. \(176 + 47a + 4b = 0\)A1
Solve for \(a\) or for \(b\)M1(dep*)
Obtain \(a = -4\) and \(b = 3\)A1
Total: 5 marks
**(i)** Differentiate $f(x)$ and obtain $f'(x) = (x - 2)^2 g'(x) + 2(x - 2)g(x)$ | B1 |
Conclude that $(x - 2)$ is a factor of $f'(x)$ | B1 |
| Total: 2 marks |

**(ii)** Either: Substitute $x = 2$, equate to zero and state a correct equation, e.g. $32 + 16a + 24 + 4b + a = 0$ | B1 |
Differentiate polynomial, substitute $x = 2$ and equate to zero or divide by $(x - 2)$ and equate constant remainder to zero | M1* |
Obtain a correct equation, e.g. $80 + 32a + 36 + 4b = 0$ | A1 |

Or1: Identify given polynomial with $(x - 2)^2(x^3 + Ax^2 + Bx + C)$ and obtain an equation in $a$ and/or $b$ | M1* |
Obtain a correct equation, e.g. $\frac{1}{4}a - 4(4 + a) + 4 = 3$ | A1 |
Obtain a second correct equation, e.g. $-\frac{1}{4}a + 4(4 + a) = b$ | A1 |

Or2: Divide given polynomial by $(x - 2)^2$ and obtain an equation in $a$ and $b$ | M1* |
Obtain a correct equation, e.g. $29 + 8a + b + 0$ | A1 |
Obtain a second correct equation, e.g. $176 + 47a + 4b = 0$ | A1 |

Solve for $a$ or for $b$ | M1(dep*) |
Obtain $a = -4$ and $b = 3$ | A1 |
| Total: 5 marks |

---
\begin{enumerate}[label=(\roman*)]
\item The polynomial $f(x)$ is of the form $(x - 2)^2g(x)$, where $g(x)$ is another polynomial. Show that $(x - 2)$ is a factor of $f'(x)$. [2]
\item The polynomial $x^5 + ax^4 + 3x^3 + bx^2 + a$, where $a$ and $b$ are constants, has a factor $(x - 2)^2$. Using the factor theorem and the result of part (i), or otherwise, find the values of $a$ and $b$. [5]
\end{enumerate}

\hfill \mbox{\textit{CAIE P3 2014 Q5 [7]}}