The domain of f(x)=(√(cos(sinx))) +(1−x)^(−1) + sin^(−1) ((x^2 +1)/(2x)) is.........=.


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Question Number 44369 by Necxx last updated on 27/Sep/18

$T h e d o m a i n o f$ $Missing \left or extra \right$ $i s . . . . . . . . . = .$
Commented by Necxx last updated on 28/Sep/18

$p l e a s e h e l p$
	Answered by tanmay.chaudhury50@gmail.com last updated on 28/Sep/18
	$sin^(−1) (((x^2 +1)/(2x))) is defined when ∣((x^2 +1)/(2x))∣≤1 ★x^2 +1≤2∣x∣ ∣x∣^2 −2∣x∣+1≤0 (∣x∣−1)^2 ≤0 but (∣x∣−1)^2 can not be less than zero so (∣x∣−1)=0 ∣x∣=1 x=±1 so domain of sin^(−1) ((x^2 +1)/(2x)) xε{±1} ★★domain for (1/(x−1)) xεR−{1} ★★★domain for (√(cos(sinx))) from attachex graph of (√(cos(sinx))) the domain xε(−∞,∞) so required ans is xε [−1,1) the value of x from −1 to +1 excluding 1 pls check...$
	${s i n}^{- 1} (\frac{x^{2} + 1}{2 x}) i s d e f i n e d w h e n ∣ \frac{x^{2} + 1}{2 x} ∣⩽ 1$ $★ x^{2} + 1 ⩽ 2 ∣ x ∣$ $∣ x ∣^{2} - 2 ∣ x ∣ + 1 ⩽ 0$ ${(∣ x ∣ - 1)}^{2} ⩽ 0 b u t {(∣ x ∣ - 1)}^{2} c a n n o t b e l e s s t h a n$ $z e r o s o (∣ x ∣ - 1) = 0$ $∣ x ∣= 1 x = \pm 1$ $s o d o m a i n o f {s i n}^{- 1} \frac{x^{2} + 1}{2 x} x ϵ {\pm 1}$ $★ ★ d o m a i n f o r \frac{1}{x - 1} x ϵ R - {1}$ $★ ★ ★ d o m a i n f o r \sqrt{c o s (s i n x)}$ $f r o m a t t a c h e x g r a p h o f \sqrt{c o s (s i n x)}$ $t h e d o m a i n x ϵ (- \infty, \infty)$ $s o r e q u i r e d a n s i s x ϵ [- 1, 1)$ $t h e v a l u e o f x f r o m - 1 t o + 1 e x c l u d i n g 1$ $p l s c h e c k . . .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 28/Sep/18

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