Question

If x satisfies the inequality ${\left(\frac{1}{3}\right)}^{\sqrt{x+2}}<3^{-x}$ then the set of all real values of $x$ is

• A. $[-2,0]$
• B. $[0,2)$
• C. $[-2,2)$
• D. $(-2,2)$

Answer 1

The correct option is C $[-2,2)$

Take log of base 3 on both sides.

$\Rightarrow {\log }_3{\left(\frac{1}{3}\right)}^{\sqrt{x+2}}<{\log }_3{3}^{-x}$

$\Rightarrow \sqrt{x+2}{\log }_3\left(\frac{1}{3}\right)<(-x).{\log }_{3}3$

$\Rightarrow \sqrt{x+2}{\log }_3\left(3^{-1}\right)<(-x).{\log }_{3}3$

$\Rightarrow -\sqrt{x+2}<-x\Rightarrow \sqrt{x+2}>x$

Squaring both sides

$(x+2)>x^2\Rightarrow x^2-x-2<0$

$\Rightarrow (x-2)(x+1)<0$

$\Rightarrow xϵ(-1,2)$ $\dots \left(1\right)$

Also $x+2\ge 0$ $[\because x+2>x^2\ge 0]$

$x\ge -2$ $\dots \left(2\right)$

From $\left(1\right)and\left(2\right)$

$-2\le x<2$

$\therefore xϵ[-2,2)$