Question

If the function $f\left(x\right)=\frac{{(128a+ax)}^{1/8}-2}{{(32+bx)}^{1/5}-2}$ is continuous at $x=0$, then the value of $\frac{a}{b}$ is

• A. $\frac{3}{5}f\left(0\right)$
• B. $2^{8/5}f\left(0\right)$
• C. $\frac{64}{5}f\left(0\right)$
• D. None of these

Answer 1

Answer: (C)

$\frac{64}{5}f\left(0\right)$

If $f$ is continuous at $x=0,$ then

$f\left(0\right)=\underset{x\to 0}{lim}\frac{{(128a+ax)}^{\frac{1}{8}}-2}{{(32+bx)}^{\frac{1}{5}}-2}$

As $x\to 0,$ the denominator $\to 0.$

Thus, for the limit to exist the numerator must also$\to 0.$

Thus, we have ${\left(128a\right)}^{\frac{1}{8}}=2$

gives $a=2$

Now, we have,

$f\left(0\right)=\underset{x\to 0}{lim}\frac{{(128a+ax)}^{\frac{1}{8}}-2}{{(32+bx)}^{\frac{1}{5}}-2}\left(\frac{0}{0}\right)$

$=\underset{x\to 0}{lim}\frac{\frac{2}{8}{(256+2x)}^{\frac{-7}{8}}}{\frac{b}{5}{(32+bx)}^{\frac{-4}{5}}}=\frac{5}{4b}.\frac{2^{-7}}{2^{-4}}=\frac{5}{32b}$

gives $b=\frac{5}{32f\left(0\right)}$

Hence, we have, $\frac{a}{b}=\frac{64}{5}f\left(0\right)$