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  • The value of , so that the function <img alt="\mathrm{ f(x)=\frac{(27-2 x)^{1 / 3}-3}{9-3(24

The value of \mathrm{f(0)}, so that the function \mathrm{ f(x)=\frac{(27-2 x)^{1 / 3}-3}{9-3(243+5 x)^{1 / 5}},(x \neq 0)}is continuous, is given by
 

Option: 1

\frac{2}{3}

 


Option: 2

6


Option: 3

2


Option: 4

4


Answers (1)

Since \mathrm{f(x)} s continuous at \mathrm{x=0}, therefore

\mathrm{ f(0)=\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{(27-2 x)^{1 / 3}-3}{9-3(243+5 x)^{1 / 5}} \quad\left(\text { form } \frac{0}{0}\right) }

\mathrm{ =\lim _{x \rightarrow 0} \frac{\frac{1}{3}(27-2 x)^{-2 / 3}(-2)}{-\frac{3}{5}(243+5 x)^{-4 / 5}(5)}=2 }

Hence option 3 is correct.


 

Posted by

Kshitij

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