Question

If $f\left(x\right)=a\log \left|x\right|+b{x}^2+x$ has is extremum value 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 $a=2,b=-\frac{1}{2}$

We have $f\left(x\right)=a\log \left|x\right|+b{x}^2+x$
On differentiating w.r.t $x$ we get
$f^{\prime }\left(x\right)=\frac{a}{x}+2bx+1$
Since, $f\left(x\right)$ attains its extremum value at $x=-1,2$
$\therefore f^{\prime }(-1)=-\frac{a}{1}-2b+1$
$\Rightarrow -a-2b+1=\mathrm{0...}\left(i\right)$
and $\frac{a}{2}+4b+1=\mathrm{0...}\left(ii\right)$
From Eqs(i) and (ii)
$\Rightarrow a=2,b=-\frac{1}{2}$