# Question #6ed7e

Jan 12, 2017

Given $f \left(x\right) = \frac{1}{\log} x \mathmr{and} g \left(x\right) = \frac{1}{8} + x$

We have

$\left(f \times g\right) \left(x\right) = \left(\frac{1}{\log} x\right) \times \left(\frac{1}{8} + x\right)$

$\implies \left(f \times g\right) \left(x\right) = \frac{\frac{1}{8} + x}{\log} x$

$\left(f \times g\right) \left(x\right)$ will be defined when

$\log x \ne 0 \implies x \ne 1$
and $x > 0 \mathmr{and} x \in R$

So option (c) is valid here.