Question

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

• A. $(-\infty ,-1)\cup (7,\infty )$
• B. (-1,5)
• C. (1,7)
• D. (-1,7)

Answer 1

Answer: (A)

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