Question #6ed7e

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Answer 1 · 2017-01-12T04:50:28

Given $f (x) = \frac{1}{log x} and g (x) = \frac{1}{8} + x$ $f(x) = 1/logx and g(x) = 1/8 +x$

We have

$(f \times g) (x) = (\frac{1}{log x}) \times (\frac{1}{8} + x)$ $(fxxg)(x) = (1/logx)xx(1/8+x)$

$\Rightarrow (f \times g) (x) = \frac{\frac{1}{8} + x}{log x}$ $=>(fxxg)(x) = (1/8+x)/logx$

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

$log x \neq 0 \Rightarrow x \neq 1$ $logx!=0=>x!=1$
and $x > 0 and x \in R$ $x>0 and x in R$

So option (c) is valid here.