General solution of `2^(sinx)+2^(cosx)=2^(1 1/sqrt2)` is

Options

A. `x = n pi - (pi)/(4), n in Z`
B. `x = n pi + (pi)/(4), n in Z`
C. ` x= n pi +(-1)^(n)(pi)/(4), n in Z`
D. `x = 2n pi +- (pi)/(4), n in Z`

Answer

B

Explanation

Using A.M `ge` G.M., we have
`(2^("sin"x) + 2^("cos" x))/(2) ge sqrt(2^("sin"x +"cos"x)) "for all" x in R`
`rArr 2^("sin"x) + 2^("cos"x) ge 2^(1+(1)/(2)("sin" x + "cos"x))"for" all" x in R`
`rArr 2^("sin"x) + 2^("cos"x) ge 2^(1-(1)/(sqrt(2))) " " [ because - sqrt(2) le "sin" x + "cos" x le sqrt(2)]`
Thus, `2^("sin"x) + 2^("cos"x) = 2^(1-(1)/(sqrt(2))) " holds when" 2^("sin"x) = 2^("cos"x)` i.e.
`"sin" x = "cos" x rArr "tan" x = 1 rArr x = n pi + (pi)/(4), n in Z`