If $f(x)=a \log |x|+b x^2+x$ has its extremum values at $x=-1$ and $x=2$, then
Answer & explanation
Correct answer: option 2
We have,
$f(x)=a \log |x|+b x^2+x \Rightarrow f^{\prime}(x)=\frac{a}{x}+2 b x+1$
Since f(x) attains its extremum values at x = -1, 2
∴ $f^{\prime}(-1)=0$ and $f^{\prime}(2)=0$
$\Rightarrow -a-2 b+1=0$ and $\frac{a}{2}+4 b+1=0$
$\Rightarrow a=2$ and $b=-1 / 2$