If $f(x)=\left|\log _{10} x\right|$, then at $x=1$
(a) $f(x)$ is continuous and $f\left(1^{+}\right)=\log _{10} e$
(b) $f(x)$ is continuous and $f\left(1^{+}\right)=\log _{10} e$
(c) $f(x)$ is continuous and $f\left(1^{-}\right)=\log _{10} e$
(d) $f(x)$ is continuous and $f\left(1^{-}\right)=-\log _{10} e$
(a) $f(x)$ is continuous and $f^{\prime}\left(1^{+}\right)=\log _{10} e$
(d) $f(x)$ is continuous and $f^{\prime}\left(1^{-}\right)=-\log _{10} e$
Given: $f(x)=\left|\log _{10} x\right|=\left|\frac{\log _{e} x}{\log _{e} 10}\right|=\left|\left(\log _{e} x\right) \times\left(\log _{10} e\right)\right|=\left(\log _{10} e\right)\left|\log _{e} x\right|$
$\Rightarrow f^{\prime}\left(1^{+}\right)=\lim _{\mathrm{h} \rightarrow 0} \frac{f(1+h)-f(1)}{h}=\lim _{\mathrm{h} \rightarrow 0} \frac{\left(\log _{10} e\right)\left|\log _{e}(1+h)\right|-\left(\log _{10} e\right)\left|\log _{e} 1\right|}{h}=\left(\log _{10} e\right) \lim _{\mathrm{h} \rightarrow 0} \frac{\left|\log _{e}(1+h)\right|}{h}=\left(\log _{10}(e)\right) \times 1=\left(\log _{10} e\right)$
Also,
$f^{\prime}\left(1^{-}\right)=\lim _{\mathrm{h} \rightarrow 0} \frac{f(1-h)-f(1)}{h}=\lim _{\mathrm{h} \rightarrow 0} \frac{\left(\log _{10} e\right)\left|\log _{e}(1-h)\right|-\left(\log _{10} e\right)\left|\log _{e} 1\right|}{h}=-\left(\log _{10} e\right) \lim _{\mathrm{h} \rightarrow 0} \frac{\left|\log _{e}(1-h)\right|}{-h}=-\left(\log _{10} e\right) \times 1=-\left(\log _{10} e\right)$