Question

If $y=aln\left|x\right|+b{x}^2+x$ 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 1

The correct option is B $a=2,b=-\frac{1}{2}$

Given, $f\left(x\right)=a{\log }_e\mid x\mid +b{x}^2+x$
$f^{\prime }\left(x\right)=\frac{a}{x}+2bx+1$
Given $x=-1,2$ are extremum of $f\left(x\right)$
$\Rightarrow f^{\prime }(-1)=0=f^{\prime }\left(2\right)$
$\Rightarrow -a-2b+1=0.....\left(1\right)$ and $\frac{a}{2}+4b+1=0.....\left(2\right)$
Solving (1) and (2) we get, $a=2,b=-\frac{1}{2}$