The domain of f(x)=cos ^{-1}left(frac{x-3}{2} | vedclass

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(B) The given function is $f(x) = \cos^{-1}\left(\frac{x-3}{2}\right) - \log_{10}(4-x)$.For the logarithmic function $\log_{10}(4-x)$ to be defined,th

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$[1,4)$

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(B) The given function is $f(x) = \cos^{-1}\left(\frac{x-3}{2}\right) - \log_{10}(4-x)$.For the logarithmic function $\log_{10}(4-x)$ to be defined,the argument must be strictly positive:$4 - x > 0 \implies x For the inverse trigonometric function $\cos^{-1}\left(\frac{x-3}{2}\right)$ to be defined,the argument must lie in the interval $[-1, 1]$:$-1 \leq \frac{x-3}{2} \leq 1$.Multiplying by $2$,we get $-2 \leq x - 3 \leq 2$.Adding $3$ to all parts,we get $1 \leq x \leq 5$.The domain of $f(x)$ is the intersection of the two conditions:$x Therefore,the domain is $x \in [1, 4)$.

The domain of $f(x)=\cos ^{-1}\left(\frac{x-3}{2}\right)-\log _{10}(4-x)$ is

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