If f(x) = \ cases 4 + ax - 4 - ax x , & - 1 ≤ x < 0 (3x + 2)/(x -...

Given that $f$\left(x\right)$is continuous in the interval$[ - 1, 1 ]$, so, it will be continuous at$x = 0$.<br /><br />Now$f\left({0}^{+}\right) = \dfrac{- 1}{4}$<br /><br />$f\left({0}^{-}\right) =\lim_{x \to {0}^{-}} \dfrac{2ax}{x ( \sqrt{4 + ax} + \sqrt{4 - ax} )} ($rationalizing)<br /><br />$= \lim_{x \to {0}^{-}} \dfrac{2a}{4} = \dfrac{a}{2}$<br /><br />Now$f\left({0}^{+}\right) =f ( {0}^{-} )$($∵f ( x )$is continuous at point 0)<br /><br />$⇒ \dfrac{a}{2} = \dfrac{- 1}{4}$⇒ a =$\dfrac{- 1}{2}