Question 9 - A-Level Maths

OCR C2 — Question 9 11 marks

Exam Board	OCR
Module	C2 (Core Mathematics 2)
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Factor & Remainder Theorem
Type	Trigonometric substitution equations
Difficulty	Standard +0.3 This is a structured multi-part question that guides students through standard Factor Theorem application, polynomial factorization, and then a routine trigonometric substitution. Parts (i)-(iii) are textbook exercises requiring only direct application of learned techniques. Part (iv) adds a small extension by substituting sin θ for x, but the substitution is obvious and solving sin θ = constant values is standard C2 material. Slightly above average difficulty due to the multi-step nature and the trigonometric extension, but no novel insight required.
Spec	1.02j Manipulate polynomials: expanding, factorising, division, factor theorem 1.05o Trigonometric equations: solve in given intervals

9. $f ( x ) = 2 x ^ { 3 } - 5 x ^ { 2 } + x + 2$.

Show that $( x - 2 )$ is a factor of $\mathrm { f } ( x )$.
Fully factorise $\mathrm { f } ( x )$.
Solve the equation $\mathrm { f } ( x ) = 0$.
Find, in terms of $\pi$, the values of $\theta$ in the interval $0 \leq \theta \leq 2 \pi$ for which $$2 \sin ^ { 3 } \theta - 5 \sin ^ { 2 } \theta + \sin \theta + 2 = 0$$

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Question 9:

Part (i):

Answer	Marks
$f(2) = 16 - 20 + 2 + 2 = 0 \therefore (x-2)$ is a factor	M1 A1

Part (ii):

\[x - 2 \overline{)\, 2x^3 - 5x^2 + x + 2}\]

\[2x^2 - x - 1\]

Answer	Marks
M1 A1

$f(x) = (x-2)(2x^2 - x - 1)$

Answer	Marks
$f(x) = (x-2)(2x+1)(x-1)$	M1 A1

Part (iii):

Answer	Marks
$x = -\frac{1}{2}, 1, 2$	B1

Part (iv):

$\sin\theta = 2$ (no solutions), $-\frac{1}{2}$ or $1$

Answer	Marks	Guidance
$\theta = \pi + \frac{\pi}{6},\, 2\pi - \frac{\pi}{6}$ or $\frac{\pi}{2}$	M1 B1
$\theta = \frac{\pi}{2},\, \frac{7\pi}{6},\, \frac{11\pi}{6}$	A2	(11)

Total: (72)

# Question 9:

## Part (i):
$f(2) = 16 - 20 + 2 + 2 = 0 \therefore (x-2)$ is a factor | M1 A1 |

## Part (ii):
$$x - 2 \overline{)\, 2x^3 - 5x^2 + x + 2}$$
$$2x^2 - x - 1$$
| M1 A1 |
$f(x) = (x-2)(2x^2 - x - 1)$
$f(x) = (x-2)(2x+1)(x-1)$ | M1 A1 |

## Part (iii):
$x = -\frac{1}{2}, 1, 2$ | B1 |

## Part (iv):
$\sin\theta = 2$ (no solutions), $-\frac{1}{2}$ or $1$
$\theta = \pi + \frac{\pi}{6},\, 2\pi - \frac{\pi}{6}$ or $\frac{\pi}{2}$ | M1 B1 |
$\theta = \frac{\pi}{2},\, \frac{7\pi}{6},\, \frac{11\pi}{6}$ | A2 | **(11)**

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**Total: (72)**

Show LaTeX source

9. $f ( x ) = 2 x ^ { 3 } - 5 x ^ { 2 } + x + 2$.\\
(i) Show that $( x - 2 )$ is a factor of $\mathrm { f } ( x )$.\\
(ii) Fully factorise $\mathrm { f } ( x )$.\\
(iii) Solve the equation $\mathrm { f } ( x ) = 0$.\\
(iv) Find, in terms of $\pi$, the values of $\theta$ in the interval $0 \leq \theta \leq 2 \pi$ for which

$$2 \sin ^ { 3 } \theta - 5 \sin ^ { 2 } \theta + \sin \theta + 2 = 0$$

\hfill \mbox{\textit{OCR C2  Q9 [11]}}

This paper (9 questions)

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Q1 6 Q2 6 Q3 7 Q4 7 Q5 7 Q6 9 Q7 9 Q8 10 Q9 11

OCR C2 — Question 9 11 marks

Question 9:

Part (i):

Part (ii):

\[x - 2 \overline{)\, 2x^3 - 5x^2 + x + 2}\]

\[2x^2 - x - 1\]

\(f(x) = (x-2)(2x^2 - x - 1)\)

Part (iii):

Part (iv):

\(\sin\theta = 2\) (no solutions), \(-\frac{1}{2}\) or \(1\)

Total: (72)

This paper (9 questions)

Answer	Marks	Guidance
\(\theta = \pi + \frac{\pi}{6},\, 2\pi - \frac{\pi}{6}\) or \(\frac{\pi}{2}\)	M1 B1
\(\theta = \frac{\pi}{2},\, \frac{7\pi}{6},\, \frac{11\pi}{6}\)	A2	(11)