Road to A

I and II

I, III and IV

IV only

All of the above

$$\left(-\infty,\infty\right)$$

$$\left[1,\frac{3}{2}\right]$$

$$\left(1,\frac{3}{2}\right)$$

$$\left(-\infty,1\right)\cup\left(\frac{3}{2},\infty\right)$$

$$\left[\frac{4}{5},1\right)\cup\left(1,2\right]$$

$$\left(\frac{4}{5},1\right)\cup\left(1,2\right)$$

$$\left(-\infty,\infty\right)$$

$$\left(-\infty,\frac{4}{5}\right)\cup\left(1,2\right)$$

$$x=-1$$

$$x=-1,\ x=-\frac{3}{2}$$

$$x=1,\ x=-\frac{3}{2}$$

$$x=1$$

$$\left(-\infty,-4\right)\cup\left(4,\infty\right)$$

$$\left(-\infty,4\right)$$

$$\left(-2,2\right)$$

$$\left[-2,2\right]$$

$$g\left(x\right)=\frac{2\left(x+2\right)}{x+1}$$

$$g\left(x\right)=\frac{2\left(x-2\right)}{x+1}$$

$$g\left(x\right)=\frac{2\left(x+2\right)}{x-1}$$

$$g\left(x\right)=\frac{2\left(x-2\right)}{x-1}$$

$$\frac{\ln4}{2}$$

$$-\ln2$$

$$\ln2$$

$$\frac{\ln2}{2}$$

-2

0

-3

Undefined

I and II

II and III

I, II and III

IV only

$$\lim_{x\rightarrow0}\frac{\sqrt[]{x-1+h}-\sqrt[]{x-1}}{h}$$

$$\lim_{h\rightarrow0}\frac{\sqrt[]{x-1+h}-\sqrt[]{x-1}}{h}$$

$$\lim_{x\rightarrow0}\frac{\sqrt[]{x-1-h}-\sqrt[]{x-1}}{h}$$

$$\lim_{h\rightarrow0}\frac{\sqrt[]{x-1-h}-\sqrt[]{x-1}}{h}$$