Question

Let f : [–π/3, 2π/3] → [0,4] be a function defined as f(x)

=$\sqrt{3}$sin x – cos x + 2. Then f ^-1(x) is given by

• A. sin^-1$\left( \frac{x - 2}{2} \right) - \frac{\pi}{6}$
• B. sin^-1$\left( \frac{x - 2}{2} \right) + \frac{\pi}{6}$
• C. $\frac{2\pi}{3}$ + cos^-1$\left( \frac{x - 2}{2} \right)$
• D. None of these

Answer 1

Answer: (B)

sin^-1$\left( \frac{x - 2}{2} \right) + \frac{\pi}{6}$

Answer 2

f(x) = $\sqrt{3}$ sin x – cos x + 2 = 2 sin $\left( x - \frac{\pi}{6} \right)$ + 2.

Since f(x) is one-one and onto, f is invertible.

Now fof ^-1(x) = x

⇒ 2 sin $\left( f^{- 1}(x) - \frac{\pi}{6} \right)$ + 2 = x ⇒ sin $\left( f^{- 1}(x) - \frac{\pi}{6} \right)$ = $\frac{x}{2}$ – 1 ⇒ f^–1(x) = sin^–1$\left( \frac{x}{2} - 1 \right) + \frac{\pi}{6}$, because $\left| \frac{x}{2} - 1 \right|$ ≤ 1 for all x ∈ [0, 4]