Question

Let $f\left(x\right)=\frac{(256+ax{)}^{1/8}-2}{(32+bx{)}^{1/5}-2}$. If $f$ is continuous at $x=0$, then the value of $a/b$ is:

• A. $\frac{8}{5}f\left(0\right)$
• B. $\frac{32}{5}f\left(0\right)$
• C. $\frac{64}{5}f\left(0\right)$
• D. $\frac{16}{5}f\left(0\right)$

Answer 1

The correct option is C $\frac{64}{5}f\left(0\right)$

$f\left(x\right)=\frac{(256+ax{)}^{1/8}-2}{(32+bx{)}^{1/5}-2}$
Given , $f$ is continuous at $x=0$.
$\underset{x\to 0}{lim}f\left(x\right)=f\left(0\right)$
Now, $\underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}\frac{(256+ax{)}^{1/8}-2}{(32+bx{)}^{1/5}-2}$
It is of the form $\frac{0}{0}$, so applying L-Hospital's rule
$=\underset{x\to 0}{lim}\frac{\frac{1}{8}a(256+ax{)}^{-7/8}}{\frac{1}{5}b(32+bx{)}^{-4/5}}$
$=\frac{5a}{64b}$
Hence,$\frac{5a}{64b}=f\left(0\right)$
$\Rightarrow \frac{a}{b}=\frac{64}{5}f\left(0\right)$