Question 9 - A-Level Maths

OCR C2 2007 June — Question 9 14 marks

Exam Board	OCR
Module	C2 (Core Mathematics 2)
Year	2007
Session	June
Marks	14
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Factor & Remainder Theorem
Type	Show equation reduces to polynomial
Difficulty	Standard +0.3 This is a straightforward multi-part question combining routine factor theorem application (substituting x=-1) with standard logarithm manipulation using basic log laws. Part (ii)(a) requires systematic application of log rules to reach the given polynomial form, which is mechanical rather than insightful. The question guides students through each step explicitly, making it slightly easier than average for C2.
Spec	1.02j Manipulate polynomials: expanding, factorising, division, factor theorem 1.06d Natural logarithm: ln(x) function and properties 1.06f Laws of logarithms: addition, subtraction, power rules

9 The polynomial $f ( x )$ is given by $$f ( x ) = x ^ { 3 } + 6 x ^ { 2 } + x - 4 .$$

(a) Show that ( $\mathrm { x } + 1$ ) is a factor of $\mathrm { f } ( \mathrm { x } )$.
(b) Hence find the exact roots of the equation $f ( x ) = 0$.
(a) Show that the equation $$2 \log _ { 2 } ( x + 3 ) + \log _ { 2 } x - \log _ { 2 } ( 4 x + 2 ) = 1$$ can be written in the form $f ( x ) = 0$.
(b) Explain why the equation $$2 \log _ { 2 } ( x + 3 ) + \log _ { 2 } x - \log _ { 2 } ( 4 x + 2 ) = 1$$ has only one real root and state the exact value of this root.

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Answer	Marks	Guidance
(i) (a) $f(-1) = -1 + 6 - 1 - 4 = 0$	B1 1	Confirm $f(-1) = 0$, through any method.

(b) $x = -1$

Answer	Marks	Guidance
$f(x) = (x+1)(x^2 + 5x - 4)$	B1, M1, A1	State $x = -1$ at any point. Attempt complete division by $(x+1)$, or equiv. Obtain $x^2 + 5x + k$.

$-5 \pm \sqrt{25 + 16}$

Answer	Marks	Guidance
$x = \frac{2}{1}(-5 \pm \sqrt{41})$	M1, A1 6	Obtain completely correct quotient. Attempt use of quadratic formula, or equiv, find roots. Obtain $\frac{1}{2}(-5 \pm \sqrt{41})$.
(ii) (a) $\log_2(x+3)^2 + \log_x - \log_2(4x+2) = 1$	B1, M1	State or imply that $2\log(x+3) = \log(x+3)^2$. Add or subtract two, or more, of their algebraic logs correctly.
$\log_2 \frac{(x+3)^2 x}{4x+2} = 1$	A1	Obtain correct equation (or any equivalent, with single term on each side).
$\frac{(x+3)^2 x}{4x+2} = 2$	B1	Use $\log_2 a = 1 \Rightarrow a = 2$ at any point.

$(x^2 + 6x + 9)x = 8x + 4$

Answer	Marks	Guidance
$x^3 + 6x^2 + x - 4 = 0$	A1 5	Confirm given equation correctly.

(b) $x > 0$, otherwise $\log_2 x$ is undefined

Answer	Marks	Guidance
$x = \frac{1}{2}(-5 + \sqrt{41})$	B1, B1∨dep 2	State or imply that $\log x$ only defined for $x > 0$. State $x = \frac{1}{2}(-5 + \sqrt{41})$ (or $x = 0.7$) only, following their single positive root in (i)(b).

(i) (a) $f(-1) = -1 + 6 - 1 - 4 = 0$ | B1 1 | Confirm $f(-1) = 0$, through any method.

(b) $x = -1$

$f(x) = (x+1)(x^2 + 5x - 4)$ | B1, M1, A1 | State $x = -1$ at any point. Attempt complete division by $(x+1)$, or equiv. Obtain $x^2 + 5x + k$.

$-5 \pm \sqrt{25 + 16}$

$x = \frac{2}{1}(-5 \pm \sqrt{41})$ | M1, A1 6 | Obtain completely correct quotient. Attempt use of quadratic formula, or equiv, find roots. Obtain $\frac{1}{2}(-5 \pm \sqrt{41})$.

(ii) (a) $\log_2(x+3)^2 + \log_x - \log_2(4x+2) = 1$ | B1, M1 | State or imply that $2\log(x+3) = \log(x+3)^2$. Add or subtract two, or more, of their algebraic logs correctly.

$\log_2 \frac{(x+3)^2 x}{4x+2} = 1$ | A1 | Obtain correct equation (or any equivalent, with single term on each side).

$\frac{(x+3)^2 x}{4x+2} = 2$ | B1 | Use $\log_2 a = 1 \Rightarrow a = 2$ at any point.

$(x^2 + 6x + 9)x = 8x + 4$

$x^3 + 6x^2 + x - 4 = 0$ | A1 5 | Confirm given equation correctly.

(b) $x > 0$, otherwise $\log_2 x$ is undefined

$x = \frac{1}{2}(-5 + \sqrt{41})$ | B1*, B1∨dep* 2 | State or imply that $\log x$ only defined for $x > 0$. State $x = \frac{1}{2}(-5 + \sqrt{41})$ (or $x = 0.7$) only, following their single positive root in (i)(b).

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9 The polynomial $f ( x )$ is given by

$$f ( x ) = x ^ { 3 } + 6 x ^ { 2 } + x - 4 .$$
\begin{enumerate}[label=(\roman*)]
\item (a) Show that ( $\mathrm { x } + 1$ ) is a factor of $\mathrm { f } ( \mathrm { x } )$.\\
(b) Hence find the exact roots of the equation $f ( x ) = 0$.
\item (a) Show that the equation

$$2 \log _ { 2 } ( x + 3 ) + \log _ { 2 } x - \log _ { 2 } ( 4 x + 2 ) = 1$$

can be written in the form $f ( x ) = 0$.\\
(b) Explain why the equation

$$2 \log _ { 2 } ( x + 3 ) + \log _ { 2 } x - \log _ { 2 } ( 4 x + 2 ) = 1$$

has only one real root and state the exact value of this root.
\end{enumerate}

\hfill \mbox{\textit{OCR C2 2007 Q9 [14]}}

This paper (9 questions)

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Q1 5 Q2 5 Q3 5 Q4 6 Q5 7 Q6 8 Q7 11 Q8 11 Q9 14

OCR C2 2007 June — Question 9 14 marks

(b) \(x = -1\)

\(-5 \pm \sqrt{25 + 16}\)

\((x^2 + 6x + 9)x = 8x + 4\)

(b) \(x > 0\), otherwise \(\log_2 x\) is undefined

This paper (9 questions)

Answer	Marks	Guidance
\(x = \frac{2}{1}(-5 \pm \sqrt{41})\)	M1, A1 6	Obtain completely correct quotient. Attempt use of quadratic formula, or equiv, find roots. Obtain \(\frac{1}{2}(-5 \pm \sqrt{41})\).
(ii) (a) \(\log_2(x+3)^2 + \log_x - \log_2(4x+2) = 1\)	B1, M1	State or imply that \(2\log(x+3) = \log(x+3)^2\). Add or subtract two, or more, of their algebraic logs correctly.
\(\log_2 \frac{(x+3)^2 x}{4x+2} = 1\)	A1	Obtain correct equation (or any equivalent, with single term on each side).
\(\frac{(x+3)^2 x}{4x+2} = 2\)	B1	Use \(\log_2 a = 1 \Rightarrow a = 2\) at any point.