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If $f:A\rightarrow B$ is defined as $f(x)=\sqrt 3 sinx+cosx+4$ and $f$ is invertible then find A.

(A) $\large[-\frac{2\pi}{3},\frac{\pi}{3}]$ (B) $\large[\frac{\pi}{6},\frac{5\pi}{6}]$ (C) $\large[-\frac{\pi}{2},\frac{\pi}{2}]$ (D) $\large[-\frac{\pi}{3},\frac{\pi}{3}]$
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Toolbox:
  • $sinAcosB+cosAsinB=sin(A+B)$
Here A is domain of $f(x)$
$f(x)=\sqrt 3sinx+cosx +4$
$=2(\large\frac{\sqrt 3}{2}$ $sinx+\large\frac{1}{2}$ $cosx)+4$
$=2(cos\large\frac{\pi}{6}$ $sinx+sin\large\frac{\pi}{6}$ $cosx)+4$
$=2sin(x+\large\frac{\pi}{6}$$)+4$
Given that $f$ is bijection.
$\Rightarrow\:\large-\frac{\pi}{2}\leq (x+\frac{\pi}{6}) \leq \frac{\pi}{2}$
$\Rightarrow\:\large-\frac{\pi}{2}-\frac{\pi}{6}\leq x \leq \frac{\pi}{2}-\frac{\pi}{6}$
$\Rightarrow\:\large-\frac{2\pi}{3} \leq x\leq \frac{\pi}{3}$
$\Rightarrow\:Domain=A=[\large-\frac{2\pi}{3},\frac{\pi}{3}]$
answered May 28, 2013 by rvidyagovindarajan_1
 

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