Question

The set of all values of ${}^{\prime }{x}^{\prime }$ satisfying the inequation $3^{72}{\left(\frac{1}{3}\right)}^x{\left(\frac{1}{3}\right)}^{\sqrt{x}}>1$ are

• A. $x\in (0,16)$
• B. $x\in [0,64)$
• C. $x\in (0,64]$
• D. None of these

Answer 1

Answer: (B)

$x\in [0,64)$

Given Inequation,

$3^{72}{\left(\frac{1}{3}\right)}^x{\left(\frac{1}{3}\right)}^{\sqrt{x}}$

The given inequality is valid only when $x\ge 0$ ......$\left(1\right)$
The given inequality can be written in the form,

$3^{72-x-\sqrt{x}}>1\Rightarrow 72-x-\sqrt{x}>0(\because 3>1)\Rightarrow x+\sqrt{x}-72<0\Rightarrow (\sqrt{x}+9)(\sqrt{x}-8)<0$

But, $\sqrt{x}+9>0$ for all $x\ge 0$

$\therefore \sqrt{x}-8<0\Rightarrow \sqrt{x}<8⟹x<64$ .

$\therefore 0\le x<64$

Thus, the interval $x\in [0,64)$ is the set of all solutions of the given
inequality.