Find the value of parameter for which the inequality 4^x-a2^x-a+3<=0 is satisfied by at least one real value of x.

Thanks!

Dear Palak J,

Ans:  your inequation is 4^x-a2^x-(a-3)<=0

make a hole square and we get,  [2^x-a/2]^2<=(a/2)^2  + a - 2

now the value of a square is >=0

hence the RHS term of the above equation is i.e (a/2)^2 + a - 2 >=0

solving we get,  a € R - (-6<=a<=2)

or a>=2 && a<=-6

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Soumyajit Das IIT Kharagpur