Question

The possible value(s) of $a$ for which $e^{2x}-(a-2)e^x+a<0$ holds true for atleast one $x\in (0,\infty )$, are:

• A. $7$
• B. $8$
• C. $9$
• D. $10$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $8$
C $9$
D $10$

$e^{2x}-(a-2)e^x+a<0$
let $y=e^x$
$=>y^2-(a-2)y+a<0$
$y^2-ay+2y+a<0$
$a>\frac{y^2+2y}{y-1}$
$a>y+3+\frac{3}{y-1}$
$a>y-1+\frac{3}{y-1}+4$ $-\left(1\right)$
Now for $x>0$, $y>1$
$=>y-1>0$
$=>y-1$ and $\frac{3}{y-1}$ are positive real numbers.
Using Arthematic mean-Geomatric mean inequality we can write
$\frac{y-1+\frac{3}{y-1}}{2}\ge \sqrt{y-1.\frac{3}{y-1}}$
or, $y-1+\frac{3}{y-1}\ge 2\sqrt{3}$
Putting this result in equation (1):
$a>2\sqrt{3}+4$
Now as $2\sqrt{3}$ is nearly equal to $3.464$, we can say that:
$a>7.464$
From given options $a$ can take values $8,9,10$