periodic functions (arihant)

\large f(x+\lambda )= 1+ \sqrt{2f(x)-f^{2}(x )} → (1.)

substituing : \large x=x+\lambda , in the given equation we have,

\large f(x+2\lambda )= 1+ \sqrt{2f(x+\lambda)-f^{2}(x+\lambda )} →(2)

now squaring equation (1) and substituting:

\large (f(x)-1)^{2} , in place of \large {2f(x+\lambda )-f^{2}(x+\lambda)}

we finally have,

\large f(x+2\lambda) = f(x)

hence period = \large 2\lambda

f(x-1)+f(x+1)=√3f(x)→(1)

substituting. x=x+1 , in eq. (1)

f(x) + f(x+2) = √3f(x+1) → (2)

substituting. x=x+2 , in eq. (1)

f(x+1) + f(x+3) =√3f(x+2) →(3)

adding (1) + (3) and rearranging the terms (with the help of eq. (2) )

f(x-1) + f(x+3) =f(x+1) → (4)

substituting x=x+2 in eq. (4)

f(x+1) + f(x+5) = f(x+3) →(5)

adding (4) + (5)

f(x-1) + f(x+5) =0 →(6)

substituting x= x+1 in (6)

f(x) + f(x+6) =0 → (7)

substituting x= x+6 in eq. (7)

f(x+6) + f (x+12 )=0

(6) - (7)

f(x) = f (x+12)

hence period = 12.