If the equation (2^ 1 ⁻¹ x )^ 2 - (a+ 1 2 ) (2^ 1 ⁻¹ x )^ -a^2=0 has only one real solution then subsets of values of 'a' are

Step-by-step solution

aligned & (2^ ⁻¹ x )^2- (a+ 1 2 ) 2^ ⁻¹ x -a^2=0 & let 2^ ⁻¹ x =t t 2 aligned Now equation t^2- (a+ 1 2 ) t-a^2=0 has one root 2 or greater than 2 and other root less than 2 aligned & f(2) 0 & 4- (a+ 1 2 ) 2-a^2 0 & 4-2 a-1-a^2 0 & a^2+2 a-3 0 & (a+3)(a-1) 0 a -3 or a 1 aligned