Question

If all the real solutions of the equation $4^x-(a-3)2^x+(a-4)=0$ are non positive, then

• A. 4 < a £ 5
• B. 0 < a< 4
• C. a> 4
• D. a< 3

Answer 1

The correct option is A 4 < a £ 5

$4^x-(a-3)2^x+(a-4)=0$
$x\ge 0,Lety=2^x$
$y^2-(a-3)y+(a-4)=0$
The roots of quadratic must lie between $(0,1]$
$a\left)\right(a-3{)}^2-4(a-4)\ge 0$
$a^2+9-6a-4a+16\ge 0$
$a^2-10a+25\ge 0$
$a\ge R$

$b)0\le \frac{a-3}{2}\le 1$
$0<a-3\le 2$
$0<a\le 5$

$c\left)f\right(0)=a-4>0a\ge 4$

$d\left)f\right(1)=1-a+3+a-4\ge 0,\therefore a\in (4,5]$