Question

If $2^{2x+2}-a\cdot 2^{x+2}+5-4a\ge 0$ has atleast one real solution, Then a $ϵ$

• A. $(-\infty ,1]$
• B. $(-3,1]$
• C. $[1,\infty )$
• D. $(\frac{-8}{7},-1]$

Answer 1

The correct option is A

$(-\infty ,1]$

$2^{2x+2}-a\cdot 2^{x+2}+5-4a\ge 0Lett=2^x\therefore 4t^2-4at+5-4a\ge 04t^2+5\ge 4a+4at\frac{4t^2+5}{t+1}\ge 4a\Rightarrow 4a\le \frac{4t^2+5}{t+1}$
$\forall t\text{such that}4a\le \frac{4t^2+5}{t+1}\Rightarrow 4a\le min\frac{4t^2+5}{t+1}$
Let $f\left(t\right)=\frac{4t^2+5}{t+1}\Rightarrow f^{\prime }\left(t\right)=\frac{(t+1)\left(8t\right)-(4t^2+5)}{(t+1{)}^2}$
$f^{\prime }\left(t\right)=\frac{8t^2+8t-4t^2-5}{(t+1{)}^2}\therefore f^{\prime }\left(t\right)=0\Rightarrow 4t^2+8t-5=0\therefore t=\frac{1}{2},\frac{-5}{2}(\text{rejected as}t>0)\therefore f\left(\frac{1}{2}\right)=\frac{4\times \frac{1}{4}+5}{\frac{1}{2}+1}=4\text{Also as}t\to 0\Rightarrow f\left(t\right)\to 5\therefore 4a\le 4=1a\le 1$