1. Tardigrade
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  4. f(a) = log ( (2 + a/2 -a ) ) , 0 < a < 2 then (1/2) f ( (8a/4 +a2) ) =

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Q. $f(a) = \log \left( \frac{2 + a}{2 -a } \right) , 0 < a < 2$ then $\frac{1}{2} f \left( \frac{8a}{4 +a^2} \right) = $

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Solution:

$\frac{1}{2} f \left( \frac{8a}{4 +a^2} \right) = \frac{1}{2} \log \Bigg( \frac{2 + \frac{8a}{4+ a^2}}{2 - \frac{8a}{4 + a^2}} \Bigg)$
$ = \frac{1}{2} \times 2 \, \log \left( \frac{2 + a}{2 -a} \right) = f(a)$