Question

The equation 2^2x + (a –1) 2^x+1 + a = 0 has roots of opposite signs then exhaustive set of values of 'a' is-

• A. a < 0
• B. a Î (–1, 0)
• C. a Î (–¥, 1/3)
• D. a Î (0, 1/3)

Answer 1

Answer: (C)

a Î (–¥, 1/3)

Answer 2

Let 2^x = t

Ž t² + 2 (a –1) t + a = 0

Now t = 1 should lie between the roots of this equation

Ž 1 + 2(a –1) + a < 0 Ž a < $\frac{1}{3}$