Explanation

(b)
Given,
$\tan^{- 1}{\left( \frac{1}{2x + 1} \right) + \tan^{- 1}\left( \frac{1}{4x + 1} \right)}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \tan^{- 1}\left( \frac{2}{x^{2}} \right)$
$\Rightarrow \ \ \ {\ \ \tan^{- 1}}\left( \frac{\frac{1}{2x + 1} + \frac{1}{4x + 1}}{1 - \frac{1}{2x + 1} \times \frac{1}{4x + 1}} \right)$
$\Rightarrow \ \ \ \ \ \tan^{- 1}\left( \frac{2}{x^{2}} \right)$
$\Rightarrow \ \tan^{- 1}\left( \frac{6x + 2}{8x^{2} + 6x} \right) = \tan^{- 1}\left( \frac{2}{x^{2}} \right)$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{6x + 2}{8x^{2} + 6x} = \frac{2}{x^{2}}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 6x^{3} + 2x^{2} = 16x^{2} + 12x$
$\Rightarrow \ \ \ 6x^{3} - 14x^{2} - 12x = 0$
$\Rightarrow \ \ \ \ 2x\left( 3x^{2} - 7x - 6 \right) = 0$
$\Rightarrow 2x(3x + 2)(x - 3) = 0$
$\Rightarrow \ \ x = 0,\ - \frac{2}{3},\ 3$
But $x = - \frac{2}{3}$ does not satisfy the given relation