Consider the function $f:\left( { - \infty ,\infty ,} \right) \to $ $\left( { - \infty ,\infty ,} \right)$ defined by $f\left( x \right) = \dfrac{{\left[ {{x^2} - ax + 1} \right]}}{{\left[ {{x^2} + ax + 1} \right]}}$ , $0 < a < 2$ . Which of the following is true.
A. ${\left( {2 + a} \right)^2}f''(1) - {\left( {2 - a} \right)^2}f''( - 1) = 0$
B. ${\left( {2 - a} \right)^2}f''(1) - {\left( {2 + a} \right)^2}f''( - 1) = 0$
C. $f'(1)f'( - 1) = {\left( {2 - a} \right)^2}$
D. $f'(1)f'( - 1) = - {\left( {2 + a} \right)^2}$

Hint: When we look to the options it is clear that we need to find the first derivative and then find the second derivative to get the required solution. We use differentiation properties to get simplified expressions.

Formula Used:
Quotient rule: $\left(\dfrac{f\left(x\right)}{g\left(x\right)}\right)'=\dfrac{g\left(x\right)f'\left(x\right)-g'\left(x\right)f\left(x\right)}{g\left(x\right)^2}$

Complete step by step solution:
$f(x)$ is given in the question $f\left( x \right) = \dfrac{{\left[ {{x^2} - ax + 1} \right]}}{{\left[ {{x^2} + ax + 1} \right]}}$ hence for finding the derivative we will simplify
$f(x)$which will help us to differentiate the function in the easiest way.
$f\left( x \right) = \dfrac{{\left[ {{x^2} - ax + 1} \right]}}{{\left[ {{x^2} + ax + 1} \right]}}$
For simplifying lets add and subtract $ax$ in the numerator as well as denominator.
$f(x) = \dfrac{{\left( {{x^2} + ax + 1} \right) - 2ax}}{{{x^2} + ax + 1}}$
$f(x) = 1 - \dfrac{{2ax}}{{{x^2} + ax + 1}}$
Now differentiating $f(x)$ for the First time with respect to $x$
$\left(\dfrac{f\left(x\right)}{g\left(x\right)}\right)'=\dfrac{g\left(x\right)f'\left(x\right)-g'\left(x\right)f\left(x\right)}{g\left(x\right)^2}$
$f'(x) = - \left[ {\dfrac{{\left( {{x^2} + ax + 1} \right)2a - 2ax(2x + a)}}{{{{\left( {{x^2} + ax + 1} \right)}^2}}}} \right]$
$ = - \left[ {\dfrac{{ - 2a{x^2} + 2a}}{{{{({x^2} + ax + 1)}^2}}}} \right]$
$ = 2a\left[ {\dfrac{{{x^2} - 1}}{{{{({x^2} + ax + 1)}^2}}}} \right]$ ---------- (i)
Now calculating the second derivative of $f(x)$where we will be differentiating $f'(x)$ again with respect to $x$ .
$f'(x) = - \left[ {\dfrac{{\left( {{x^2} + ax + 1} \right)2a - 2ax(2x + a)}}{{{{\left( {{x^2} + ax + 1} \right)}^2}}}} \right]$
Differentiating $f(x)$ using The Quotient Rule.
$f''(x) = 2a\left[ {\dfrac{{{{\left( {{x^2} + ax + 1} \right)}^2}2x - ({x^2} - 1)2\left( {{x^2} + ax + 1} \right)(2x + a)}}{{{{\left( {{x^2} + ax + 1} \right)}^4}}}} \right]$
$f''(x) = 2a\left[ {\dfrac{{2x\left( {{x^2} + ax + 1} \right) - 2({x^2} - 1)(2x + a)}}{{{{\left( {{x^2} + ax + 1} \right)}^3}}}} \right]$
$f''(x) = 2a\left[ {\dfrac{{2x\left( {{x^2} + ax + 1} \right) - 2({x^2} - 1)(2x + a)}}{{{{\left( {{x^2} + ax + 1} \right)}^3}}}} \right]$ -------- (ii)
Now in $f''(x)$ substituting the values.
$f''\left( 1 \right) = \dfrac{{4a\left( {a + 2} \right)}}{{{{\left( {a + 2} \right)}^3}}}= \dfrac{{4a}}{{{{\left( {a + 2} \right)}^2}}}$
$f''\left( 1 \right) = \dfrac{{4a}}{{{{\left( {a + 2} \right)}^2}}}$ ------ (iii)
$f''\left( { - 1} \right) = \dfrac{{ - 4a\left( {2 - a} \right)}}{{{{\left( {2 - a} \right)}^3}}}$
$f''\left( { - 1} \right) = \dfrac{{ - 4a}}{{{{\left( {2 - a} \right)}^2}}}$ ------ (iv)
Now equation (iii) and (iv) can be simplified as –
Equation (iii) will be simplified as –
 $f''\left( 1 \right) = \dfrac{{4a}}{{{{\left( {a + 2} \right)}^2}}}$ ------ (iii)
$4a = f''\left( 1 \right){\left( {a + 2} \right)^2}$
Equation (iv) will be simplified as –
$f''\left( { - 1} \right) = \dfrac{{ - 4a}}{{{{\left( {2 - a} \right)}^2}}}$ ------ (iv)
$4a = - {\left( {2 - a} \right)^2}f''\left( { - 1} \right)$
From both the simplified equation we will equate each other –
$4a = f''\left( 1 \right){\left( {a + 2} \right)^2}$------ (v)
$4a = - {\left( {2 - a} \right)^2}f''\left( { - 1} \right)$------ (vi)
$f''\left( 1 \right){\left( {a + 2} \right)^2} = - {\left( {2 - a} \right)^2}f''\left( { - 1} \right)$
$f''\left( 1 \right){\left( {a + 2} \right)^2} + {\left( {a - 2} \right)^2}f''\left( { - 1} \right) = 0$
$f''\left( 1 \right){\left( {2 + a} \right)^2} + {\left( {2 - a} \right)^2}f''\left( { - 1} \right) = 0$

Option ‘A’ is correct

Note: To solve the given question we have to find $f’’(x)$ and calculate the second order derivative at $ x = 1$ and $x = -1$. Then substitute the values and form an equation to get the required solution.