Let f(x)=|x|+1 and g(x)=[x]-1 , where [.] is the greatest integer...

and $$\begin{aligned} f(x) & =|x|+1 \\ g(x) & =[x]-1 \\ h(x) & =\frac{f(x)}{g(x)}=\frac{|x|+1}{[x]-1}\end{aligned}$$ (1) For, $x < 0$$\begin{aligned} & f(x)=-x+1 \\ & \therefore \quad f(x)=-1 \end{aligned}$$f(x)$ is differentiable $\forall x < 0\Rightarrow$ Statement (1) is correct. (2) At $x=0.0001g(x)$ is continuous everywhere except for integer value as [ $x$ ] is continuous $\forall x \in \mathrm{R} except integers. So,$g(x)$is continuous at$x=0.0001$. Statement (2) is correct.$\begin{aligned} &\text{(3)} \quad g(x)=[x]-1 \\ & {[x] \Rightarrow \text{ Always } a b \text{ integer }} \\ & \qquad \begin{aligned} g^{\prime}(x) & =\frac{d}{d x}[x]-0 \\ g^{\prime}(x) & =0-0=0 \end{aligned}\end{aligned}\Rightarrow Statement (3) is incorrect. Hence, (1) and (2) only correct.