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  • If the function <img alt="\mathrm{f x)=\frac{(128 a+a x)^{15}-2}{(32+b x)^{15}-2}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bf%20x%29%3D%5Cfrac%7B%28128%20a&plus;a%20x%2

If the function \mathrm{f x)=\frac{(128 a+a x)^{15}-2}{(32+b x)^{15}-2}} is continuous at \mathrm{A=0}, then the value of \mathrm{\frac{a}{b}} is

Option: 1

\frac{3}{5} f(0)


Option: 2

\frac{3}{5} f(0)


Option: 3

\frac{64}{5} f(0)


Option: 4

none of these


Answers (1)

If \mathrm{f} is continuous at \mathrm{x=0}, then
\mathrm{f(0)=\lim _{x \rightarrow 0} \frac{(128 a+a x)^{1 / 8}-2}{(32+b x)^{1 / 5}-2}}

As \mathrm{x \rightarrow 0}, the denominator \mathrm{\rightarrow 0}.

Thus, for limit to exist the numerator must also \mathrm{\rightarrow 0}.

Thus, we have \mathrm{(128 a)^{1 / 8}=2\: or\: a=2}

Now, we have
\mathrm{ f(0)=\lim _{x \rightarrow 0} \frac{(256+2 x)^{1 / 8}-2}{(32+b x)^{1 / 5}-2} \quad\left(\frac{0}{0}\right)}
\mathrm{\Rightarrow f(0)=\lim _{x \rightarrow 0} \frac{\frac{2}{8}(256+2 x)^{-7 / 8}}{\frac{b}{5}(32+b x)^{-4 / 5}}=\frac{5}{4 b} \cdot \frac{2^{-7}}{2^{-4}}=\frac{5}{32 b}}
\mathrm{\Rightarrow \quad b=\frac{5}{32 f(0)}}
Hence, we have \mathrm{\frac{a}{b}=\frac{64}{5} f(0)}

Posted by

Sumit Saini

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