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

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detailed solution

Correct option is C

f0=Ltx→0 256+ax18−232+bx15−2Use L – Hospital Rule=Ltx→0 18256+ax−78.a1532+bx−45.b∴ab=645f0

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Let f(x)=(256+ax)18−2(32+bx)15−2. If f is continuous at x=0 then the value of ab is

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