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

4 < a \leq 5

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0 < a < 4

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a > 4

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a < 3

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Solution

The correct option is A $4 < a \leq 5$
$4^{x} - (a - 3) 2^{x} + (a - 4) = 0$
$x \geq 0, L e t y = 2^{x}$
$y^{2} - (a - 3) y + (a - 4) = 0$
The roots of quadratic must lie between $(0, 1]$
$a) (a - 3)^{2} - 4 (a - 4) \geq 0$
$a^{2} + 9 - 6 a - 4 a + 16 \geq 0$
$a^{2} - 10 a + 25 \geq 0$
$a \geq R$

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

$c) f (0) = a - 4 > 0 a \geq 4$

$d) f (1) = 1 - a + 3 + a - 4 \geq 0, ∴ a \in (4, 5]$

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