Let a, b $ \in $R be such that the function f given by $f\left( x \right) = \ln \left| x \right| + b{x^2} + ax,x \ne 0$ has extreme values at x = -1 and x = 2.
Statement – 1: f has a local minima at x = -1 and at x = 2
Statement – 2: a = $\dfrac{1}{2}$, b = $\dfrac{{ - 1}}{4}$
$\left( a \right)$ Statement 1 is false, statement 2 is true.
$\left( b \right)$ Statement 1 is true, statement 2 is true. Statement 2 is a correct explanation for statement 1.
$\left( c \right)$ Statement 1 is true, statement 2 is true. Statement 2 is not a correct explanation for statement 1.
$\left( d \right)$ Statement 1 is true, statement 2 is false.

Hint: In this particular question to find the maxima and minima differentiate the given function w.r.t x and equate to zero and solve for given extreme values of x, then again differentiate the given function and calculate its value on previous calculated x value if we got positive than it is a minima and if we got negative than it is a maxima so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given function
$f\left( x \right) = \ln \left| x \right| + b{x^2} + ax,x \ne 0$
Has extreme values at x = -1 and x = 2.
Now differentiate the given equation w.r.t x we have,
$ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = \dfrac{d}{{dx}}\left( {\ln \left| x \right| + b{x^2} + ax} \right)$
Now as we know that $\dfrac{d}{{dx}}\ln x = \dfrac{1}{x},\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$ so we have,
$ \Rightarrow f'\left( x \right) = \dfrac{1}{x} + 2bx + a = 0$
Now the extreme values of the given function is, x = -1, x = 2.
$ \Rightarrow f'\left( { - 1} \right) = \dfrac{1}{{ - 1}} - 2b + a = - 1 - 2b + a = 0$
$ \Rightarrow a - 2b = 1$................. (1)
$ \Rightarrow f'\left( 2 \right) = \dfrac{1}{2} + 2b\left( 2 \right) + a = a + 4b + \dfrac{1}{2} = 0$
$ \Rightarrow a + 4b = - \dfrac{1}{2}$................. (2)
Now subtract equation (1) from equation (2) we have,
$ \Rightarrow a + 4b - a + 2b = - \dfrac{1}{2} - 1$
$ \Rightarrow 6b = - \dfrac{3}{2}$
$ \Rightarrow b = - \dfrac{1}{4}$
Now from equation (1) we have,
$ \Rightarrow a - 2\left( {\dfrac{{ - 1}}{4}} \right) = 1$
$ \Rightarrow a = 1 - \dfrac{1}{2} = \dfrac{1}{2}$
So the f’(x) becomes
$ \Rightarrow f'\left( x \right) = \dfrac{1}{x} - 2\left( {\dfrac{1}{4}} \right)x + \dfrac{1}{2} = \dfrac{1}{x} - \dfrac{1}{2}x + \dfrac{1}{2}$
Now again differentiate this equation we have,
$ \Rightarrow \dfrac{d}{{dx}}f'\left( x \right) = f''\left( x \right) = \dfrac{d}{{dx}}\left( {\dfrac{1}{x} - \dfrac{1}{2}x + \dfrac{1}{2}} \right)$
$ \Rightarrow f''\left( x \right) = \dfrac{{ - 1}}{{{x^2}}} - \dfrac{1}{2}$
$ \Rightarrow f''\left( { - 1} \right) = \dfrac{{ - 1}}{{{{\left( { - 1} \right)}^2}}} - \dfrac{1}{2} = - 1 - \dfrac{1}{2} = \dfrac{{ - 3}}{2}$ = negative so it is a maxima.
And
$ \Rightarrow f''\left( 2 \right) = \dfrac{{ - 1}}{{{2^2}}} - \dfrac{1}{2} = - \dfrac{1}{4} - \dfrac{1}{2} = \dfrac{{ - 3}}{4}$ = negative so it is also a maxima.

Hence the given function is maximum at x = -1, and at x = 2.
So this is the required answer.
Hence Statement 1 is true, statement 2 is true. Statement 2 is a correct explanation for statement 1.
Hence option (b) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the basic differentiation properties such as so differentiate the given function according to these properties as above, after that we get two equations in terms of a and b solve them as above and substitute the values in the f’(x) and again differentiate, so if the value of double differentiation is negative on given extremities then the function is maximum at given extremities otherwise not.