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 these

Options

A. `(3)/(5)f(0)`
B. `2^(8//5)f(0)`
C. `(64)/(5)f(0)`
D. none of these

Answer

C

Explanation

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)`