Question

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

(A)  4 < a ≤ 5

(B)  0 < a < 4

(C)  a > 4 

(D)  a < 3

Answer 1

Correct option  (A) 4 < a ≤ 5

4^x - (a - 3)^2x + (a - 4)  = 0,x ≤ 0

Let y = 2^x . Then 

y² - (a - 3)y + (a - 4) = 0 

The roots of quadratic equation must lie between (0, 1].

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