Question

If $3^x+2^{2x} \geq 5^x$, 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 $x - 3 \geq 0$ i.e., $x \geq 3$, then $x$ is not negative
$\therefore $ this possibility is ruled out.
$\therefore $ $x - 3 < 0\,$ i.e., $\, x < 3$
$\therefore $ given equation becomes
$x^2\cdot 2^{x+ 1} + 2^{3-x + 2} = x^22^{3 -x + 4} +2^{x -1}$
$\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.
$\therefore \, x \neq \pm \frac{1}{2}$ and $x \neq 3 $
Hence there is no negative integral solution.