$T h e n u m b e r o f r e a l r o o t s o f t h e e q u a t i o n 5 + | 2^{x} - 1 | = 2^{x} (2^{x} - 2) i s$

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Detailed Solution

$\begin{array}{l} G i v e n 5 + | 2^{x} - 1 | = 2^{x} (2^{x} - 2) \\ L e t 2^{x} - 1 = a \\ \Rightarrow 5 + | a | = (a + 1) (a - 1) \\ \Rightarrow 5 + | a | = a^{2} - 1 \\ \Rightarrow | a | = a^{2} - 6 \\ ∴ a^{2} - 6 = \pm a \\ C a s e (i) : I f a^{2} - 6 = a \\ \Rightarrow a^{2} - a - 6 = 0 \\ \Rightarrow a^{2} - 3 a + 2 a - 6 = 0 \\ \Rightarrow (a - 3) (a + 2) = 0 \\ ∴ a = 3, a = - 2 \\ \Rightarrow 2^{x} - 1 = 3, 2^{x} - 1 = - 2 \\ \Rightarrow 2^{x} = 4, 2^{x} = - 1 \\ \Rightarrow x = 2, n o t p o s s i b l e \\ C a s e (i i) : a^{2} - 6 = - a, \Rightarrow a^{2} + a - 6 = 0 \\ \Rightarrow a^{2} + 3 a - 2 a - 6 = 0 \\ \Rightarrow (a + 3) (a - 2) = 0 \\ ∴ a = - 3 a n d a = 2 \\ ∴ 2^{x} - 1 = - 3, 2^{x} - 1 = 2 \\ \Rightarrow 2^{x} = - 2, \Rightarrow 2^{x} = 3 \\ N o t p o s s i b l e N o t p o s s i b l e \\ ∴ x = 2 \\ O n l y o n e r o o t e x i s t \end{array}$