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Find the domain of $f(x)=\sqrt{cos(sinx)}+sin^{-1}\bigg(\large\frac{1+x^2}{2x}\bigg)$

(A) $R$ the set of Real Numbers.

(B) $(-1,1)$

(C) $\{-1,1\}$

(D) $[-1,1]$

1 Answer

Given: $f(x)=\sqrt{cos(sinx)}+sin^{-1}\bigg(\large\frac{1+x^2}{2x}\bigg)$
For domain $ cos(sinx)\geq 0$ and $ -1\leq \large\frac{1+x^2}{2x}$$\leq 1$
Since $-1\leq sinx\leq 1$, $cos(sinx) \geq 0\:\:\forall x$
and $-2x\leq {1+x^2}\leq 2x$
$\Rightarrow\:(1+x^2)\leq |2x|$
$\Rightarrow\:1+|x|^2\leq 2|x|$
$\Rightarrow\:|x|^2-2|x|+1\leq 0$
$\Rightarrow\:(|x|-1)^2\leq 0$
$\Rightarrow\:|x|-1=0\:\:\:or\:\:\:|x|=1$
$\Rightarrow\:x=\{-1,1\}$
answered May 30, 2013 by rvidyagovindarajan_1
 

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