sry everyone i wasn't able to reply as i had gone abroad.

 

 

Let 5x + 3 = t

x = (t -3)/5

=> f(x) = f((t-3)/5) 

also f(x) = f(t)  --------------- given

hence f(t) = f((t-3)/5)

=> f(x) = f((x-3)/5) 

now if we put (x-3)/5 = k then f(k) =f((k -3)/5)

therefore we can write

  f(x) = f((x-3)/5) = f[{(x-3/5) - 3}/5] = f[(x-3 + 3*5)/5^2] = f({(x - 3 + 3*5)/5^2) - 3}/5)

= f[(x - 3 + 3*5 + 3*5^2)/5^3

repeating this n times we get

f(x) = f[(x-(3 + 3*5 +3*5² + 3*5³ + 3*5⁴ ......3*5^n-1)/5ⁿ)]

= f(x - 3/4(5ⁿ- 1)/5ⁿ)  -------------- (using sum of g.p)

= f[{(4x + 3)/4*5ⁿ} - 3/4]

 

as function is independent of n i.e for any value of n f(x) remains f(x) hence

by putting limits on l.h.s & r.h.s

lim n tending to infinity f(x) remains f(x)

& on r.h.s if n tends to infinity then f[{(4x + 3)/4*5ⁿ} - 3/4] becomes f(-3/4)

therefore 

i hope u all understood the solution

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