The function that is not differentiable at $\mathrm{x}=1$ is

${\mathrm{f}}_1^{\mathrm{\prime }}\left(1\right)=\underset{\mathrm{x}\to 1}{\lim }\frac{\left|\mathrm{x}\right|-1}{\mathrm{x}-1}=1\Rightarrow {\mathrm{f}}_1$ is differentiable at 1.

${\mathrm{f}}_2^{\mathrm{\prime }}\left(1^{-}\right)=\underset{\mathrm{x}\to 1}{\lim }\frac{[1+\sin (\mathrm{x}-1)-1]}{\mathrm{x}-1}=1, {\mathrm{f}}_2^{\mathrm{\prime }}\left(1^{+}\right)=\underset{\mathrm{x}\to 1^{+}}{\lim }\frac{[\mathrm{x}-1]}{\mathrm{x}-1}=1$

$\Rightarrow {\mathrm{f}}_2$ is differentiable at 1.

${\mathrm{f}}_3^{\mathrm{\prime }}\left(1^{-}\right)=\underset{\mathrm{x}\to 1^{-}}{\lim }\frac{\left({\mathrm{x}}^2+7\mathrm{x}-7\right)-1}{\mathrm{x}-1}=\underset{\mathrm{x}\to 1^{-}}{\lim }\frac{(\mathrm{x}+8)(\mathrm{x}-1)}{\mathrm{x}-1}=9$

${\mathrm{f}}_3^{\mathrm{\prime }}\left(1^{+}\right)=\underset{\mathrm{x}\to 1^{+}}{\lim }\frac{(3\mathrm{x}-1)/2-1}{\mathrm{x}-1}=\underset{\mathrm{x}\to 1^{+}}{\lim } \frac{3\mathrm{x}-3}{2\left(\mathrm{x}-1\right)}. \therefore {\mathrm{f}}_3$is not differentiable at 1.