If the domain of the function f(x) = x ^ 2 - 25 (4 - x ^ 2 ) + _ 10...

Given: $f\left(x\right) = \dfrac{\sqrt{{x}^{2} - 25}}{4 - {x}^{2}} + \log \left({x}^{2} + 2x - 15\right)$

$⇒ {x}^{2} - 25 \ge 0$

$⇒ \left(x - 5\right)\left(x + 5\right) \ge 0$

$⇒ x \in ( - \infty , - 5 ] \cup [ 5, \infty ) . . . \left(i\right)$

$4 - {x}^{2} \ne 0$

$⇒ x \ne 2, - 2 . . . \left(ii\right)$

$⇒ {x}^{2} + 2x - 15 > 0$

$⇒ \left(x + 5\right)\left(x - 3\right) > 0$

$⇒ x \in \left(- \infty , - 5\right) \cup \left(3, \infty\right) . . . \left(iii\right)$

Using \left(i\right), \left(ii\right)\text{and}\left(iii\right)$</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/f964e1ce-17fa-4961-a92b-ed2f16aa59e9.png" style="width: 409px; height: 133px;"/></p><p>$⇒ x \in \left(- \infty , - 5\right) \cup [ 5, \infty )$</p><p>So, on comparing with$x \in \left(- \infty , \alpha\right) \cup [ \beta, \infty )$we get,</p><p>$⇒ {\alpha}^{2} + {\beta}^{3} = {\left(- 5\right)}^{2} + {\left(5\right)}^{3}$</p><p>$⇒ {\alpha}^{2} + {\beta}^{3} = 150$