The simplest form of the expression $$\frac{{{p^2} - p}}{{2{p^3} + {p^2}}}$$   + $$\frac{{{p^2} - 1}}{{{p^2} + 3p}}$$   + $$\frac{{{p^2}}}{{p + 1}}$$   = ?

Solution (By Examveda Team)

$$\frac{{{p^2} - p}}{{2{p^3} + {p^2}}} + \frac{{{p^2} - 1}}{{{p^2} + 3p}} + \frac{{{p^2}}}{{p + 1}}$$
In such type of question assume values of p
$$\eqalign{ & \therefore {\text{Let }}p = 1 \cr & \therefore \frac{{1 - 1}}{{2 + 1}} + \frac{{1 - 1}}{{1 + 3}} + \frac{1}{{1 + 1}} \cr & = 0 + 0 + \frac{1}{2} \cr & = \frac{1}{2} \cr & {\text{Now check option 'B'}} \cr & \frac{1}{{2{p^2}}} = \frac{1}{2} \cr & {\text{Hence the answer is option 'B'}} \cr} $$