First slide

Functions (XII)

Question

$T h e d o m a i n o f d e f i n i t i o n o f t h e f u n c t i o n f (x) = \sqrt{\sin^{- 1} (2 x) + \frac{π}{6}}$ $f o r r e a l - v a l u e d x i s$

Moderate

A

$[- \frac{1}{4}, \frac{1}{2}]$
B

$[- \frac{1}{2}, \frac{1}{2}]$
C

$(- \frac{1}{2}, \frac{1}{9})$
D

$[- \frac{1}{4}, \frac{1}{4}]$

Solution

$For f (x) = \sqrt{\sin^{- 1} (2 x) + \frac{π}{6}} to be defined and real,$
$\sin^{- 1} 2 x + π / 6 \geq 0$
$or \sin^{- 1} 2 x \geq - \frac{π}{6} - - - - - (1)$
$But we know that - π / 2 \leq \sin^{- 1} 2 x \leq π / 2$ …. (2)
Combining (l) and (2),
$\begin{array}{l} - π / 6 \leq \sin^{- 1} 2 x \leq π / 2 \\ or \sin (- π / 6) \leq 2 x \leq \sin (π / 2) \\ or - 1 / 2 \leq 2 x \leq 1 \\ or - 1 / 4 \leq x \leq 1 / 2 \\ ∴ D o m a i n o f t h e f u n c t i o n = [- \frac{1}{4}, \frac{1}{2}] \end{array}$

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