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Problem 540
540
A function f is defined by \(f : x \mapsto 3 - 2 \tan\left(\frac{1}{2}x\right)\) for \(0 \leq x < \pi\).
State the range of \(f\). [1]
State the exact value of \(f\left(\frac{2}{3}\pi\right)\). [1]
Sketch the graph of \(y = f(x)\). [2]
Obtain an expression, in terms of \(x\), for \(f^{-1}(x)\). [3]
Solution
(i) The function \(f(x) = 3 - 2 \tan\left(\frac{1}{2}x\right)\) is defined for \(0 \leq x < \pi\). The range of \(\tan\left(\frac{1}{2}x\right)\) is \((-\infty, \infty)\), so \(f(x)\) can take any value less than or equal to 3. Thus, the range is \(f \leq 3\).
(ii) To find \(f\left(\frac{2}{3}\pi\right)\), substitute \(x = \frac{2}{3}\pi\) into the function: \(f\left(\frac{2}{3}\pi\right) = 3 - 2 \tan\left(\frac{1}{2} \times \frac{2}{3}\pi\right) = 3 - 2 \tan\left(\frac{1}{3}\pi\right) = 3 - 2\sqrt{3}\).
(iii) The graph of \(y = f(x)\) starts at \(y = 3\) when \(x = 0\) and decreases as \(x\) increases, with no turning points, tending tangentially towards \(x = \pi\).
(iv) To find \(f^{-1}(x)\), start with \(y = 3 - 2 \tan\left(\frac{x}{2}\right)\). Rearrange to make \(x\) the subject: \(y = 3 - 2 \tan\left(\frac{x}{2}\right) \Rightarrow 2 \tan\left(\frac{x}{2}\right) = 3 - y \Rightarrow \tan\left(\frac{x}{2}\right) = \frac{3-y}{2} \Rightarrow \frac{x}{2} = \tan^{-1}\left(\frac{3-y}{2}\right) \Rightarrow x = 2 \tan^{-1}\left(\frac{3-y}{2}\right)\). Thus, \(f^{-1}(x) = 2 \tan^{-1}\left(\frac{3-x}{2}\right)\).
