If the domain of the function f (x) = _ e ( 2x + 3 4 x ^ 2 + x - 3...

Given,

$f\left(x\right) = \log_{e}\left(\dfrac{2x + 3}{4{x}^{2} + x - 3}\right) + \cos^{- 1}\left(\dfrac{2x - 1}{x + 2}\right)$

Now, using the definition of $log$ function and $\cos^{- 1}$ function we get,

$\dfrac{2x + 3}{4{x}^{2} + x - 3} > 0$ and $- 1 \le \dfrac{2x - 1}{x + 2} \le 1$

$⇒ \dfrac{2x + 3}{( 4x - 3 ) ( x + 1 )} > 0, \dfrac{3x + 1}{x + 2} \ge 0 \text{ and } \dfrac{x - 3}{x + 2} \le 0$

 

Now, solving $\dfrac{2x + 3}{( 4x - 3 ) ( x + 1 )} > 0$ we get,

$⇒ x \in \left(\dfrac{- 3}{2}, - 1\right) \cup \left(\dfrac{3}{4}, \infty\right) . . . . \left(i\right)$

Similarly for $\dfrac{3x + 1}{x + 2} \ge 0$ we get,

$x \in ( - \infty , - 2 ) \cup \left[\dfrac{- 1}{3}, \infty\right) . . . \left(ii\right)$

And for $\dfrac{x - 3}{x + 2} \le 0$ we get,

$x \in ( - 2, 3 ] . . . \left(iii\right)$

Now, taking intersection of \left(i\right) \cap \left(ii\right) \cap \left(iii\right)$we get,</p><p>$( \dfrac{3}{4}, 3 ]$</p><p>So, on comparing with$( \alpha, \beta ]$we get,$\alpha = \dfrac{3}{4} \text{ and } \beta = 3$</p><p>Hence,$5\beta - 4\alpha = 15 - 3 = 12