Question

The values of $a$ for which the equation $4^x-2^{x}a-a+3=0$ has at least one solution are given by

• A. $a\in [2,\infty )$
• B. $a\in (2,\infty )$
• C. $a\in (0,2)$
• D. $a\in R$

Answer 1

The correct option is A $a\in [2,\infty )$

GIven $4^x-2^{x}a-a+3=0$

Let $2^x=y$
So, the equation becomes $y^2-ay+3-a=0$
To get a real root $b^2-4ac\ge 0$

$a^2-\mathrm{4.1.}(3-a)\ge$
$a^2+4a-12\ge 0$

$a^2+6a-2a-12\ge 0$

$(a+6)(a-2)\ge 0$
$a>2$ or $a<-6$
As $2^x$ is always positive $a>2$ .