Correct answer: A. k > 0
First, rewrite the equation so that it is equal to 0: k * 9^x - (2k+1) * 3^x + k - 1 = 0.
This equation can be viewed as a quadratic equation in terms of 3^x. Let u = 3^x. Since x must be a real number, u must be a positive real number (u > 0). Substituting u into the equation gives:
k * u^2 - (2k+1) * u + (k-1) = 0.
For this quadratic equation to have at least one real solution for x, it must have at least one positive real root for u.
First, for the roots for u to be real, the discriminant (D) must be non-negative.
D = b^2 - 4ac = (-(2k+1))^2 - 4(k)(k-1) = (4k^2 + 4k + 1) - (4k^2 - 4k) = 8k + 1.
The condition D ≥ 0 implies 8k + 1 ≥ 0, or k ≥ -1/8.
Also, note that if k=0, the equation is not quadratic. It becomes -(1)u - 1 = 0, so u = -1. This is not a positive root, so k cannot be 0.
Now, we need to determine when at least one root u is positive, given k ≥ -1/8. Let's analyze the sum of the roots (S) and the product of the roots (P) of the quadratic in u:
S = -b/a = (2k+1)/k
P = c/a = (k-1)/k
There are two main scenarios for having at least one positive root:
1. The two roots have opposite signs (one positive, one negative). This occurs when their product P is negative.
P < 0 => (k-1)/k < 0. This is true when 0 < k < 1. This entire interval satisfies the condition k ≥ -1/8. Thus, if 0 < k < 1, there is one positive root for u.
2. Both roots are positive. This occurs when their product P is positive and their sum S is positive.
P > 0 => (k-1)/k > 0 => k < 0 or k > 1.
S > 0 => (2k+1)/k > 0 => k < -1/2 or k > 0.
For both P > 0 and S > 0, the intersection of these conditions is k < -1/2 or k > 1. We must also satisfy the discriminant condition k ≥ -1/8. Applying this, the condition k < -1/2 is eliminated, leaving only k > 1. Thus, if k > 1, there are two positive roots for u.
A boundary case is when one root is 0 and the other is positive. This occurs when P = 0 and S > 0. P = 0 when k=1. If k=1, S = (2(1)+1)/1 = 3, which is positive. The roots are u=0 and u=3. The root u=3 is positive, so k=1 is a valid value.
Combining the conditions from all scenarios: (0 < k < 1) or (k > 1) or (k = 1). This simplifies to k > 0.
Choice A is correct. Choice B is incorrect because it only ensures the roots for u are real, not that at least one is positive. For instance, if k = -1/8, the only root is u=-3, which is not positive. Choice C is incorrect because it only accounts for the case where both roots are positive, ignoring the case where one is positive and one is negative. Choice D is incorrect because it represents the conditions for S>0 and P>0 without considering the constraint from the discriminant, which would eliminate k < -1/2.