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Exam-Style Questions.Problems adapted from questions set for previous Mathematics exams. |
1. | IB Analysis and Approaches |
Solve for x where \( -\pi \le x \le \pi \).
$$ \sin{2x} = \cos{x} $$2. | IB Analysis and Approaches |
(a) Show that \(2x+15+\dfrac{40}{x-3}= \dfrac{2x^2+9x-5}{x-3}, \quad x \in \mathbb{R}, x \neq 3\)
(b) Hence or otherwise, solve the equation \( 2\cos{2\theta}+15+\dfrac{40}{\cos{2\theta}-3}=0, \quad \text{ for } 0 \le \theta \le \pi\)
3. | A-Level |
(a) Solve the following trigonometric equation for \(–360° \lt x \lt 360°\):
$$ 5 \sin^2 x + 2\sin x + 3 = 7 \cos^2 x $$giving your answers to the nearest integer.
(b) Hence find the smallest positive solution of the equation
$$ 5 \sin^2(3\theta + 20°) + 2\sin (3\theta + 20°) + 3 = 7 \cos^2 (3\theta + 20°) $$giving your answer to 2 decimal places.
4. | A-Level |
The function \(f\) is defined as \(f(x) = 12x^3 - 5x^2 -11x + 6 \).
(a) Use the Factor Theorem to show that \( (4x-3) \) is a factor of \(f(x)\)
(b) Express \(f(x)\) as a product of linear factors.
(d) Hence solve the equation \( g(\theta ) = 0 \), giving your answers, in radians, in the interval \(0 \le \theta \le 2 \pi \).
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