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  4. The value of x ∈ R mid[ log (1.6)1-x2 -(0.625)6(1+x)] ∈ R is

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Q. The value of $\left\{x \in R \mid\left[\log (1.6)^{1-x^{2}} -(0.625)^{6(1+x)}\right] \in R \right\}$ is

EAMCETEAMCET 2013

Solution:

$\left. x \in R \mid \log \left[(1.6)^{1-x^{2}}-(0.625)^{6(1+x)}\right] \in R \right\}$
Now, $(1.6)^{1-x^{2}}>(0.625)^{6(1+x)}$
$\Rightarrow (1.6)^{1-x^{2}}>(0.625)^{6(1+x)}$
$=\left(\frac{8}{5}\right)^{1-x^{2}}>\left(\frac{8}{5}\right)^{-6(1+x)}$
$\therefore 1-x^{2}>-6(1+x)$
$\Rightarrow x^{2}-6 x-7<0$
$\Rightarrow (x-7)(x+1)<0$
$\Rightarrow x \in(-\infty,-1) \cup(7, \infty)$
Hence,
$\left. x \in R \left|\log \left[(1.6)^{1-x^{2}}-(0.625)^{6(1+x)}\right]\right| \in R \right\}$
$=(-\infty,-1) \cup(7, \infty)$