Question

The interval of x in which the inequality $4^x\le {3.2}^{x+\sqrt{x}}+4^{1+\sqrt{x}}$ is fulfilled is given by set A . The number of integral values in A is

Answer 1

$4^x\le {3.2}^{x+\sqrt{x}}+4^{1+\sqrt{x}}$

$2^{2x}\le {3.2}^{x+\sqrt{x}}+4.2^{2\sqrt{x}}$
$1\le {3.2}^{-x+\sqrt{x}}+4.2^{2(\sqrt{x}-x)}$

Put $2^{-x+\sqrt{x}}=t$

$4t^2+3t-1\ge 0$
$\Rightarrow (4t-1)(t+1)\ge 0$

$\Rightarrow t\in (-\infty ,-1)\cup (\frac{1}{4},\infty )$

But $2^{\sqrt{x}-x}>0$
$\Rightarrow t\in (\frac{1}{4},\infty )$
$\Rightarrow \frac{1}{4}\le 2^{\sqrt{x}-x}$
$\Rightarrow -2\le \sqrt{x}-x$

Let $\sqrt{x}=y$
$\Rightarrow -2\le y-y^2$

$\Rightarrow y^2-y-2\le 0$
$\Rightarrow (y-2)(y+1)\le 0$

$\Rightarrow y\in [-1,2]$
$\Rightarrow \sqrt{x}\in [-1,2]$

But $x\ge 0$
$\Rightarrow x\in [0,4]$

Hence, number of integral solutions =5