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Question Number 60095 by Kunal12588 last updated on 17/May/19

Commented by Kunal12588 last updated on 17/May/19

$f i n d D o m a i n$
	Answered by tanmay last updated on 17/May/19

	${l o g}_{10} s i n (x - 3) + \sqrt{16 - x^{2}}$ $g (x) = 16 - x^{2}$ $g (x) = (4 + x) (4 - x)$ $g (x) = 0 [[w h e n x = - 4 a n d x = 4$ $g (x) < 0 w h e n x > 4$ $g (x) < 0 w h e n x < - 4$ $g (x) > 0 w h e n 4 > x > - 4$ $s o f o r \sqrt{16 - x^{2}} 4 ⩾ x ⩾ - 4 \to x \in [- 4, 4]$ ${l o g}_{10} s i n (x - 3)$ $s i n (x - 3) > 0$ $s o w h e n (x - 3) > 0 s i n (x - 3) > 0$ $s o {l o g}_{10} s i n (x - 3) f e s i b l e w h e n x > 3 x \in (3, \infty)$ $s o c o m b i n i n g$ $x \in [- 4, 4] \cap (3, \infty) \to x \in (3, 4]$
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