Question

If If $f(x)=\left\{\begin{array}{ll}\frac{\sqrt{1+\sqrt{5+x}-a}}{(x-4)} & 0 \leq x<4 \text { is continuous at } x=4 \\ b & x>4\end{array}\right.$ ,

then value of $\frac{1}{a b}$ is equal to

Answer 1

$\because f\left(x\right)$ is continuous at $x=4$
$\therefore \underset{x \rightarrow 4^{-}}{l i m}f\left(x\right)=\underset{x \rightarrow 4^{+}}{l i m}f\left(x\right)=f\left(4\right)$
$\underset{x \rightarrow 4^{-}}{l i m}\frac{\sqrt{1 + \sqrt{5 + x}} - a}{x - 4}=b$
for numerator to be zero $\Rightarrow a=2$
now after rationalization,
$\underset{x \rightarrow 4^{-}}{l i m}\frac{\left(\sqrt{1 + \sqrt{5 + x}} - 2\right) \left(\sqrt{1 + \sqrt{5 + x}} + 2\right)}{\left(x - 4\right) \left(\sqrt{1 + \sqrt{5 + x}} + 2\right)}=b$
$\underset{x \rightarrow 4^{-}}{l i m}\frac{1 + \sqrt{5 + x} - 4}{\left(x - 4\right) 4}=b$
again on rationalization,
$\underset{x \rightarrow 4^{-}}{l i m}\frac{\left(5 + x\right) - 9}{\left(x - 4\right) \times 4 \left(\sqrt{5 + x} + 3\right)}=b$
$\frac{1}{4 \times 6}=b\Rightarrow b=\frac{1}{24}$
$\frac{1}{a b}=\frac{24}{2}=12$