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If $f:R\rightarrow S$ defined by $\:f(x)=\sin x-\sqrt3\; \cos x+1\;$ is onto function then $S$ = ?

$\begin{array}{1 1} [0,1] \\ A\subset B \\B\subset A \\ A=B \\ A\:and\:B\; are\; disjoint\; sets. \end{array}$

Can you answer this question?
 
 

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Toolbox:
  • $-1\leq \sin x \leq 1$
Ans- (C)
$f(x)=\sin x-\sqrt3 \cos x+1$
$=2(\large\frac{1}{2}$$\sin x-\frac{1}{2}\sqrt3 \cos x)+1$
$=2(\sin(x-\frac{\pi}{3}))+1$
We know that $-1 \leq \sin(x-\frac{\pi}{3})\leq 1$
$\Rightarrow\:-2 \leq 2\sin(x-\frac{\pi}{3})\leq 2$
$\Rightarrow\:-1 \leq 2\sin(x-\frac{\pi}{3})+1\leq 3$
$\Rightarrow\: $ range of $ f(x)=S=[-1,3]$
answered May 4, 2013 by rvidyagovindarajan_1
edited May 17, 2014 by rohanmaheshwari0831_1
 

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