If the function `f(x)=((128 a+a x)^(1//8)-2)/((32+b x)^(1//5)-2)`is continuous at `x=0`, then the value of `a//b`is`3/5f(0)`b. `2^(8//5)f(0)`c. `(64)/5f(0)`d. none of theseA. `(3)/(5)f(0)`B. `2^(8//5)f(0)`C. `(64)/(5)f(0)`D. none of these

If the function `f(x)=((128 a+a x)^(1//8)-2)/((32+b x)^(1//5)-2)`is co

1.	If the function `f(x)=((128 a+a x)^(1//8)-2)/((32+b x)^(1//5)-2)`is continuous at `x=0`, then the value of `a//b`is`3/5f(0)`b. `2^(8//5)f(0)`c. `(64)/5f(0)`d. none of theseA. `(3)/(5)f(0)`B. `2^(8//5)f(0)`C. `(64)/(5)f(0)`D. none of these
Answer» Correct Answer - C If f is continuous at x = 0, then `f(0)=underset(xrarr0)(lim)((128a+ax)^(1//8)-2)/((32+bx)^(1//5)-2)` As `xrarr0`, the denominator `rarr0`. Thus, for limit to exist the numberator must also `rarr0`. Thus, we have `(128a)^(1//8)=2` or a = 2. Now, we have `f(0)=underset(xrarr0)(lim)((256+2x)^(1//8)-2)/((32+bx)^(1//5)-2)" "((0)/(0))` `rArr" "f(0)=underset(xrarr0)(lim)((2)/(8)(256+2x)^(-7//8))/((b)/(5)(32+bx)^(-4//5))=(5)/(4b).(2^(-7))/(2^(-4))=(5)/(32b)` `rArr" "b=(5)/(32f(0))` ltBrgt Hence, we have `(a)/(b)=(64)/(5)f(0)`