The equation $2^{2 x} + (a - 1) 2^{x + 1} + a = 0$ has roots reciprocal to each, then set of values of 'a' is

a < 0

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Solution

The correct option is D

Let $2^{x} = t$

f(t) = $t^{2}$ + 2(a-1) t + a = 0 , has positive roots reciprocal to each other

then :f(1) $<$ 0 $\Rightarrow$ 3a $<$ 1 $\Rightarrow$ a $<$ $\frac{1}{3}$

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