Edexcel C1 2005 January — Question 10 12 marks

Exam BoardEdexcel
ModuleC1 (Core Mathematics 1)
Year2005
SessionJanuary
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCompleting the square and sketching
TypeComplete the square
DifficultyEasy -1.2 This is a straightforward C1 completing the square question with standard follow-up parts. Part (a) is routine algebraic manipulation, part (b) requires basic sketching skills, and part (c) involves solving a simple quadratic equation. All techniques are standard textbook exercises requiring minimal problem-solving insight.
Spec1.02b Surds: manipulation and rationalising denominators1.02e Complete the square: quadratic polynomials and turning points1.02n Sketch curves: simple equations including polynomials

10. Given that $$\mathrm { f } ( x ) = x ^ { 2 } - 6 x + 18 , \quad x \geqslant 0 ,$$
  1. express \(\mathrm { f } ( x )\) in the form \(( x - a ) ^ { 2 } + b\), where \(a\) and \(b\) are integers. The curve \(C\) with equation \(y = \mathrm { f } ( x ) , x \geqslant 0\), meets the \(y\)-axis at \(P\) and has a minimum point at \(Q\).
  2. In the space provided on page 19, sketch the graph of \(C\), showing the coordinates of \(P\) and \(Q\). The line \(y = 41\) meets \(C\) at the point \(R\).
  3. Find the \(x\)-coordinate of \(R\), giving your answer in the form \(p + q \sqrt { } 2\), where \(p\) and \(q\) are integers.

AnswerMarks Guidance
(a) \(x^2 - 6x + 18 = (x-3)^2 + 9\)B1, M1 A1 (3 marks)
(b) "U"-shaped parabolaM1
Vertex in correct quadrantA1ft
\(P\): \((0, 18)\) (or 18 on \(y\)-axis)B1
\(Q\): \((3, 9)\)B1ft (4 marks)
(c) \(x^2 - 6x + 18 = 41\) or \((x-3)^2 + 9 = 41\)M1
Attempt to solve 3 term quadratic \(x = \ldots\)M1
\(x = \frac{6 \pm \sqrt{36 - (4 \times -23)}}{2}\) (or equiv.)A1
\(\sqrt{128} = \sqrt{64 \times 2}\) (or equiv. surd manipulation)M1
\(3 + 4\sqrt{2}\) (Ignore other value)A1 (5 marks)
Total: 12 marks
Notes:
(a) M1 requires \((x \pm a)^2 \pm b \pm 18\), \(a \neq 0\), \(b \neq 0\). Answer only: full marks.
**(a)** $x^2 - 6x + 18 = (x-3)^2 + 9$ | B1, M1 A1 | (3 marks)

**(b)** "U"-shaped parabola | M1 |

Vertex in correct quadrant | A1ft |

$P$: $(0, 18)$ (or 18 on $y$-axis) | B1 |

$Q$: $(3, 9)$ | B1ft | (4 marks)

**(c)** $x^2 - 6x + 18 = 41$ or $(x-3)^2 + 9 = 41$ | M1 |

Attempt to solve 3 term quadratic $x = \ldots$ | M1 |

$x = \frac{6 \pm \sqrt{36 - (4 \times -23)}}{2}$ (or equiv.) | A1 |

$\sqrt{128} = \sqrt{64 \times 2}$ (or equiv. surd manipulation) | M1 |

$3 + 4\sqrt{2}$ (Ignore other value) | A1 | (5 marks)

**Total: 12 marks**

**Notes:**

(a) M1 requires $(x \pm a)^2 \pm b \pm 18$, $a \neq 0$, $b \neq 0$. Answer only: full marks.
10. Given that

$$\mathrm { f } ( x ) = x ^ { 2 } - 6 x + 18 , \quad x \geqslant 0 ,$$
\begin{enumerate}[label=(\alph*)]
\item express $\mathrm { f } ( x )$ in the form $( x - a ) ^ { 2 } + b$, where $a$ and $b$ are integers.

The curve $C$ with equation $y = \mathrm { f } ( x ) , x \geqslant 0$, meets the $y$-axis at $P$ and has a minimum point at $Q$.
\item In the space provided on page 19, sketch the graph of $C$, showing the coordinates of $P$ and $Q$.

The line $y = 41$ meets $C$ at the point $R$.
\item Find the $x$-coordinate of $R$, giving your answer in the form $p + q \sqrt { } 2$, where $p$ and $q$ are integers.
\end{enumerate}

\hfill \mbox{\textit{Edexcel C1 2005 Q10 [12]}}