First slide

Functions (XII)

Question

$Let f : [- \frac{π}{3}, \frac{2 π}{3}] \to [0, 4]$ be a function defined as $f (x) = \sqrt{3} \sin x - \cos x + 2 .$ Then $f^{- 1} (x)$ is given by

Moderate

A

$\sin^{- 1} (\frac{x - 2}{2}) - \frac{π}{6}$
B

$\sin^{- 1} (\frac{x - 2}{2}) + \frac{π}{6}$
C

$\frac{2 π}{3} + \cos^{- 1} (\frac{x - 2}{2})$
D

none of these

Solution

$y = f (x) = \sqrt{3} \sin x - \cos x + 2 = 2 \sin (x - \frac{π}{6}) + 2$ …. (1)
Since f(x) is one-one and onto, f is invertible.
$\begin{array}{l} From (1), \sin (x - \frac{π}{6}) = \frac{y - 2}{2} \\ or x = \sin^{- 1} \frac{y - 2}{2} + \frac{π}{6} \\ or f^{- 1} (x) = \sin^{- 1} (\frac{x - 2}{2}) + \frac{π}{6} \end{array}$

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