Question

If $3^x+2^{2x}\ge 5^x$, then the solution set for $x$ is

• A. $(-\infty ,2)$
• B. $[2,\infty )$
• C. $(0,2)$
• D. $\{2\}$

Answer 1

Answer: (A)

We have
$x^2\cdot 2^{x+1}+2^{|x-3|+2}-x^2\cdot 2^{|x-3|+4}+2^{x-1}$
If $x-3\ge 0$ i.e., $x\ge 3$, then $x$ is not negative
$\therefore$ this possibility is ruled out.
$\therefore$ $x-3<0$ i.e., $x<3$
$\therefore$ given equation becomes
$x^2\cdot 2^{x+1}+2^{3-x+2}=x^2{2}^{3-x+4}+2^{x-1}$
$\Rightarrow$ $x^2\cdot 2^{x+1}+2^{5-x}=2^2\cdot 2^{7-x}+2^{x-1}$
$\Rightarrow$ $x^2\cdot 2^{x+1}-x^2\cdot 2^{7-x}+2^{5-x}-2^{x-1}=0$
$\Rightarrow$ $x^2\cdot 2^2(2^{x-1}-2^{5-x})+(2^{5-x}-2^{x-1})=0$
$\Rightarrow$ $(4x^2-1)(2^{x+1}-2^{5-x})=0$
$\Rightarrow$ $(4x^2-1)(2^{2x-6}-1)=0$
$\Rightarrow x^2=\frac{1}{4}$ and $2^{2x-6}=1$
$\Rightarrow 2x-6=0$
$\Rightarrow x=3$
$\therefore x=\pm \frac{1}{2}$ and $x=3$
But $x<3$ and $x$ is a negative integer.
$\therefore x\ne \pm \frac{1}{2}$ and $x\ne 3$
Hence there is no negative integral solution.