Question

The set of values of ${}^{\prime }{x}^{\prime }$ satisfying the equation $(x^2.2^{x+1}+2^{|x-3|+2})=(x^2.2^{|x-3|+4}+2^{x-1})$ are

• A. $x\in (3,\infty )$
• B. $x\in \{-\frac{1}{2},\frac{1}{2}\}$
• C. $x\in [3,\infty )$
• D. $x\in (3,\infty ]\cup \{-\frac{1}{2},\frac{1}{2}\}$

Answer 1

Answer: (C), (D)

$x\in (3,\infty ]\cup \{-\frac{1}{2},\frac{1}{2}\}$
$x\in [3,\infty )$

The critical point
$x-3=0$
$\therefore$ $x=3$
Consider following cases:
$x<3$
$x^2.2^{x+1}+2^{-(x-3)+2}=x^2{2}^{-(x-3)+4}+2^{x-1}$
${\Rightarrow x}^2.2^{x+1}+2^{-x+5}=x^2{2}^{-x+7}+2^{x-1}{\Rightarrow x}^2.2^{x+1}+2^{-x+5}=x^2{2}^{-x+7}+2^{x-1}\Rightarrow x^2(2^{x+1}-2^{-x+7})=2^{x-1}-2^{-x+5}\Rightarrow x^2.2^{-x+7}(2^{2x-6}-1)=2^{-x+5}(2^{2x-6}-1)\Rightarrow 2^{-x+5}(2^2{x}^2-1)(2^{2x-6}-1)=0\therefore 2^{-x+5}\ne 0$
$x^2=\frac{1}{4},2^{2x-6}=1\therefore x=\pm \frac{1}{2},2x-6=0\therefore x=\pm \frac{1}{2},x\ne 3(\because x<3)x\ge 3x^2.2^{x+1}+2^{x-3+2}=x^2{2}^{x-3+4}+2^{x-1}\Rightarrow x^2.2^{x+1}+2^{x-1}=x^2{2}^{x+1}+2^{x-1}$
which is true for all $x\ge 3$.
$\therefore x^2.2^{x+1}+2^{x-1}=x^2{2}^{x+1}+2^{x-1}x\in [3,\infty )$
Combining both cases, the solution of the given equation is
$x\in [3,\infty )\cup \{-\frac{1}{2},\frac{1}{2}\}$.