Given function

$f(x)=\ln|x|+bx^2+ax, \qquad x\ne 0$

It is given that $f$ has extreme values at $x=-1$ and $x=2$.

Use first derivative condition

For $x\ne 0$, $\frac{d}{dx}(\ln|x|)=\frac{1}{x}$ so $f'(x)=\frac{1}{x}+2bx+a$

Since $x=-1$ and $x=2$ are points of extremum, $f'(-1)=0, \qquad f'(2)=0$

Thus, $f'(-1)=-1-2b+a=0$ $f'(2)=\frac12+4b+a=0$

Solve for $a$ and $b$

From the two equations: $a-2b=1$ $a+4b=-\frac12$

Subtracting, $6b=-\frac32 \implies b=-\frac14$

Then, $a-2\left(-\frac14\right)=1$ $a+\frac12=1$ $a=\frac12$

So Statement-2 is true.

Check nature of extrema using second derivative

$f''(x)=-\frac{1}{x^2}+2b$

Since $b=-\frac14$, $f''(x)=-\frac{1}{x^2}-\frac12$

Now evaluate: $f''(-1)=-1-\frac12=-\frac32<0$ $f''(2)=-\frac14-\frac12=-\frac34<0$

Therefore, $f$ has a local maximum at both $x=-1$ and $x=2$.

Hence Statement-1 is true.

Check whether Statement-2 explains Statement-1

Yes. Once we find $a=\frac12, \quad b=-\frac14,$ the second derivative becomes negative at both $x=-1$ and $x=2$, which proves that both extrema are local maxima.

Therefore Statement-2 is the correct explanation of Statement-1.

Final option

The correct choice is: $\boxed{\text{B}}$