Question
If $$y = a\ln x + b{x^2} + x$$ has its extremum values at $$x = - 1$$ and $$x = 2,$$ then
A.
$$a = 2,b = - 1$$
B.
$$a = 2,b = - \frac{1}{2}$$
C.
$$a = - 2,b = \frac{1}{2}$$
D.
none of these
Answer :
$$a = 2,b = - \frac{1}{2}$$
Solution :
$$\eqalign{
& y = a\ln x + b{x^2} + x \cr
& {\text{has}}\,{\text{its}}\,{\text{extremum}}\,{\text{values}}\,{\text{at}}\,x = - 1\,{\text{and}}\,2 \cr
& \therefore \frac{{dy}}{{dx}} = 0\,{\text{at}}\,x = - 1\,{\text{and}}\,2 \cr
& \Rightarrow \frac{a}{x} + 2bx + 1 = 0\,{\text{or}}\,2b{x^2} + x + a = 0 \cr
& {\text{has}}\, - 1\,{\text{and}}\,2\,{\text{as}}\,{\text{its}}\,{\text{roots}}{\text{.}} \cr
& \therefore 2b - 1 + a = 0\,......\left( 1 \right) \cr
& 8b + 2 + a = 0\,......\left( 2 \right) \cr
& {\text{Solving}}\,\left( 1 \right)\,{\text{and}}\,\left( 2 \right){\text{we}}\,{\text{get}}\,a = 2,b = - \frac{1}{2} \cr} $$