If x is real, the function $\frac{(\mathrm{x}-\mathrm{a})(\mathrm{x}-\mathrm{b})}{(\mathrm{x}-\mathrm{c})}$ will assume all real values, provided

Let $\mathrm{y}=\frac{(\mathrm{x}-\mathrm{a})(\mathrm{x}-\mathrm{b})}{(\mathrm{x}-\mathrm{c})}$

or $\mathrm{y}(\mathrm{x}-\mathrm{c})={\mathrm{x}}^2-(\mathrm{a}+\mathrm{b})\mathrm{x}+\mathrm{ab}$

or ${\mathrm{x}}^2-(\mathrm{a}+\mathrm{b}+\mathrm{y})\mathrm{x}+\mathrm{ab}+\mathrm{cy}=0$

$∆={(\mathrm{a}+\mathrm{b}+\mathrm{y})}^2-4(\mathrm{ab}+\mathrm{cy})$

$={\mathrm{y}}^2+2\mathrm{y}(\mathrm{a}+\mathrm{b}-2\mathrm{c})+{(\mathrm{a}-\mathrm{b})}^2$

Since x is real and y assumes all real values, we must have $∆\ge 0$ for all real values of y. The sign of a quadratic in y is same as of first term provided its discriminant ${\mathrm{B}}^2-4\mathrm{AC}<0$

This will be so if $4{(\mathrm{a}+\mathrm{b}-2\mathrm{c})}^2-4{(\mathrm{a}-\mathrm{b})}^2<0$

or $4(\mathrm{a}+\mathrm{b}-2\mathrm{c}+\mathrm{a}-\mathrm{b})(\mathrm{a}+\mathrm{b}-2\mathrm{c}-\mathrm{a}+\mathrm{b})<0$

or $16(\mathrm{a}-\mathrm{c})(\mathrm{b}-\mathrm{c})<0 \mathrm{or} 16(\mathrm{c}-\mathrm{a})(\mathrm{c}-\mathrm{b})=-\mathrm{ve}$

$\therefore \mathrm{c} \mathrm{lies} \mathrm{between} \mathrm{a} \mathrm{and} \mathrm{b} \mathrm{i}.\mathrm{e}., \mathrm{a}<\mathrm{c}<\mathrm{b} ....\left(\mathrm{i}\right)$

$\mathrm{where} \mathrm{a}<\mathrm{b}, \mathrm{but} \mathrm{if} \mathrm{b}<\mathrm{a}$ then the above condition will be 

$\mathrm{b}<\mathrm{c}<\mathrm{a} \mathrm{or} \mathrm{a}>\mathrm{c}>\mathrm{b} ...\left(\mathrm{ii}\right)$

Hence from (i) and (ii) we observe that Option 4 is correct answer.