Question

If both the roots of $(2a-4)9^x-(2a-3)3^x+1=0$ are non-negative, then

• A. 0 < a < 2
• B. 2 < a <
• C. a <
• D. a > 3

Answer 1

The correct option is B

2 < a <

Putting $3^x$ = y, we have

(2a - 4)$y^2$ - (2a - 3)y + 1 = 0

This equation must have real solution

$\Rightarrow$ $(2a-3{)}^3$ - 4 (2a - 4) $\ge$ 0

$\Rightarrow$ $4a^2$ - 20a + 25 $\ge$ 0

$\Rightarrow$ $(2a-5{)}^2\ge 0$. This is true.

y = 1 satisfies the equation

Since $3^x$ is positive and $3^x\ge 3^{\circ },y\ge 1$

Product of the roots = 1 × y > 1 ×

$\Rightarrow \frac{1}{2a-4}$ > 1

$\Rightarrow$ 2a - 4 < 1 $\Rightarrow$ a < $\frac{5}{2}$

Sum of the roots $=\frac{2a-3}{2a-4}$ > 1

$\Rightarrow \frac{(2a-3)-(2a-4)}{2a-4}$ > 0

$\Rightarrow \frac{1}{2a-4}$ > 0

$\Rightarrow$ a > 2 $\Rightarrow$ 2 < a < $\frac{5}{2}$