Q124 / 200 · Manipal (MET) 2011 · Mathematics

The value of $f(0)$, so that the function $f(x)=\frac{2-(256-7 x)^{1 / 8}}{(5 x+3 x)^{1 / 5}-2}, x \neq 0$ is continuous everywhere is given by

  • A
    -1
  • B
    1
  • C
    $2^6$
  • Correct answer
    D
    None of these

Solution

For $f(x)$ to be continuous everywhere, we must have $\begin{aligned} f(0) & =\lim _{x \rightarrow 0} f(x) \\ & =\lim _{x \rightarrow 0} \frac{2-(256-7 x)^{1 / 8}}{(5 x+32)^{1 / 5}-2} \\ & =\lim _{x \rightarrow 0} \frac{\frac{7}{8}(256-7 x)^{-7 / 8}}{(5 x+32)^{-4 / 5}} \\ & =\frac{7}{8} \times \frac{2^{-7}}{2^{-4}}=\frac{7}{64}\end{aligned}$

Hence, the correct answer is option: 'D'.

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