Question
[Maximum mark: 7]
Consider the functions \(f(x)=\sqrt{3}sin x + cos x\) where 0 ≤ x ≤ π and g(x) = 2x where x ∈ R.
(a) Find ( f • g)(x).
(b) Solve the equation ( f • g)(x) = 2 cos 2x where 0 ≤ x ≤ π.
Answer/Explanation
Ans: (a) ( f • g)(x) = f(2x)
\(f(2x)=\sqrt{3}sin2x+cos 2x\)
(b) \(\sqrt{3}sin2x+cos 2x = 2cos 2x\)
\(\sqrt{3}sin2x=cos 2x\)
recognising to use tan or cot
\(tan 2x = \frac{1}{\sqrt{3}}OR cot 2x = \sqrt{3}\) (values may be seen in right triangle)
\(\left ( arctan \left ( \frac{1}{\sqrt{3}} \right )= \right )\frac{\pi }{6}\) (seen anywhere) (accept degrees)
\(2x = \frac{\pi }{6},\frac{7\pi }{6}\)
\(x = \frac{\pi }{12},\frac{7\pi }{12}\)
| Note: Do not award the final A1 if any additional solutions are seen. Award A1A0 for correct answers in degrees. Award A0A0 for correct answers in degrees with additional values. |
Question
Find the least positive value of $x$ for which $\cos \left(\frac{x}{2}+\frac{\pi}{3}\right)=\frac{1}{\sqrt{2}}$.
Answer/Explanation
determines $\frac{\pi}{4}$ (or $45^{\circ}$ ) as the first quadrant (reference) angle attempts to solve $\frac{x}{2}+\frac{\pi}{3}=\frac{\pi}{4}$
Note: Award $\boldsymbol{M 1}$ for attempting to solve $\frac{x}{2}+\frac{\pi}{3}=\frac{\pi}{4}, \frac{7 \pi}{4}(, \ldots)$
$\frac{x}{2}+\frac{\pi}{3}=\frac{\pi}{4} \Rightarrow x<0$ and so $\frac{\pi}{4}$ is rejected
$\begin{aligned} & \frac{x}{2}+\frac{\pi}{3}=2 \pi-\frac{\pi}{4}\left(=\frac{7 \pi}{4}\right) \\ & x=\frac{17 \pi}{6} \quad \text { (must be in radians) }\end{aligned}$