The domain of a function f(y) = frac{cos^{-1} | vedclass

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(C) For the function $f(y) = \frac{\cos^{-1}(y-5)}{\sqrt{25-y^2}}$ to be defined,the following conditions must be satisfied: $1$. The argument of $\co

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$[4, 5)$

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(C) For the function $f(y) = \frac{\cos^{-1}(y-5)}{\sqrt{25-y^2}}$ to be defined,the following conditions must be satisfied: $1$. The argument of $\cos^{-1}$ must be in the interval $[-1, 1]$,so $-1 \leq y-5 \leq 1$. Adding $5$ to all sides gives $4 \leq y \leq 6$. $2$. The denominator must be non-zero and the expression inside the square root must be positive,so $25-y^2 > 0$,which implies $y^2 $3$. Taking the intersection of the two conditions: $y \in [4, 6]$ and $y \in (-5, 5)$. $4$. The intersection is $4 \leq y Therefore,the domain is $[4, 5)$.

$A$. $f(x)=\frac{\|x+2\|}{x+2}, x \neq-2$	$1$. $[\frac{1}{3}, 1]$
$B$. $g(x)=\|[x]\|, x \in R$	$2$. $Z$
$C$. $h(x)=\|x-[x]\|, x \in R$	$3$. $W$
$D$. $f(x)=\frac{1}{2-\sin 3x}, x \in R$	$4$. $[0, 1)$
	$5$. $\{-1, 1\}$

The domain of a function $f(y) = \frac{\cos^{-1}(y-5)}{\sqrt{25-y^2}}$ is

Similar Questions

Consider the following lists.
$A$. $f(x)=\frac{|x+2|}{x+2}, x \neq-2$ $1$. $[\frac{1}{3}, 1]$
$B$. $g(x)=|[x]|, x \in R$ $2$. $Z$
$C$. $h(x)=|x-[x]|, x \in R$ $3$. $W$
$D$. $f(x)=\frac{1}{2-\sin 3x}, x \in R$ $4$. $[0, 1)$
$5$. $\{-1, 1\}$

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