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 May 22nd, 2019, 02:37 AM #11 Member   Joined: Jul 2017 From: KOLKATA Posts: 50 Thanks: 3 So range of y should be (-0,1] -1/sqrt(2) Right ?
 May 22nd, 2019, 02:54 AM #12 Global Moderator   Joined: Dec 2006 Posts: 20,968 Thanks: 2217 Select choice (c) (as given in the question) as the correct answer.
May 22nd, 2019, 01:45 PM   #13
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Quote:
 Originally Posted by skeeter because $x = 9$ forms a removable discontinuity for $y = \sin[2 \cdot f(x)]$ (see original equation for $f$) ... mea culpa. domain should be $x \in [-7, 9) \cup (9, \infty)$ good catch by skipjack.
I had looked at the question after the discontinuity had been removed.

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