1. Tardigrade
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  4. Let f(x)=√ log 2((10 x-4/4-x2))-1. Then sum of all integers in domain of f(x) is

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Q. Let $f(x)=\sqrt{\log _2\left(\frac{10 x-4}{4-x^2}\right)-1}$. Then sum of all integers in domain of $f(x)$ is

Relations and Functions - Part 2

Solution:

$ \log _2\left(\frac{10 x-4}{4-x^2}\right) \geq 1 \Rightarrow \frac{10 x-4}{4-x^2}-2 \geq 0 \Rightarrow \frac{10 x-4-8+2 x^2}{4-x^2} \geq 0$
$\Rightarrow \frac{2 x^2+10 x-12}{x^2-4} \leq 0 $
$\therefore x \in[-6,-2) \cup[1,2) $
$\therefore x=\{-6,-5,-4,-3,1\}$
Hence sum $=-17$