Question

The least positive value of $a$ for which $4^x-a\cdot 2^x-a+3\le 0$ is satisfied by atleast one real value of $x$ is

Answer 1

Let $f\left(x\right)=4^x-a\cdot 2^x-a+3$ and $x_1,x_2$ be the roots of $f\left(x\right)=0$
Let $t=2^x,t\in (0,\infty )\forall x\in R$
$\Rightarrow g\left(t\right)=t^2-at-a+3$
Let $t_1,t_2$ be the roots of $g\left(t\right)=0$

For $f\left(x\right)\le 0$ to have atleast one real $x$
$g\left(t\right)\le 0$ should have one real $t,$ where $t\in (0,\infty )$

Required cases:
Case $I:$ Both roots are positive


So,
$\left(1\right)D\ge 0\Rightarrow a^2+4a-12\ge 0$
$\Rightarrow (a+6)(a-2)\ge 0$
$\Rightarrow a\in (-\infty ,-6]\cup [2,\infty )$

$\left(2\right)\frac{-B}{2A}>0\Rightarrow \frac{a}{2}>0\Rightarrow a>0$

$\left(3\right)g\left(0\right)>0\Rightarrow -a+3>0\Rightarrow a<3$ $\therefore a\in [2,3)...\left(i\right)$

Case $II:$ One root is positive other root is negative


$0$ lies in between the roots
So, $g\left(0\right)<0\Rightarrow a>3...\left(ii\right)$

Case $III:$ One root is $0$ and other root is greater than $0$


$g\left(0\right)=0\Rightarrow a=3$
$-\frac{B}{2A}>0\Rightarrow \frac{a}{2}>0\Rightarrow a>0$
$\therefore a=3...\left(iii\right)$

From $\left(i\right),\left(ii\right)$ and $\left(iii\right)$
$\therefore a\in [2,\infty )$