If f(x) = \ cases 1 + 5 + x - a (x - 4) & 0 ≤ x then value of...

∵ f$\left(x\right)$ is continuous at $x = 4$
$∴ \lim_{x \to {4}^{-}}f\left(x\right) = \lim_{x \to {4}^{+}}f\left(x\right) = f\left(4\right)$
$\lim_{x \to {4}^{-}}\dfrac{\sqrt{1 + \sqrt{5 + x}} - a}{x - 4} = b$
for numerator to be zero $⇒ a = 2$
now after rationalization,
$\lim_{x \to {4}^{-}}\dfrac{\left(\sqrt{1 + \sqrt{5 + x}} - 2\right)\left(\sqrt{1 + \sqrt{5 + x}} + 2\right)}{\left(x - 4\right)\left(\sqrt{1 + \sqrt{5 + x}} + 2\right)} = b$
$\lim_{x \to {4}^{-}}\dfrac{1 + \sqrt{5 + x} - 4}{\left(x - 4\right)4} = b$
again on rationalization,

$\lim_{x \to {4}^{-}}\dfrac{\left(5 + x\right) - 9}{\left(x - 4\right) \times 4\left(\sqrt{5 + x} + 3\right)} = b$
$\dfrac{1}{4 \times 6} = b ⇒ b = \dfrac{1}{24}$
$\dfrac{1}{ab} = \dfrac{24}{2} = 12$