Let f:R–{0}→R be a function such that f(x)–6f(1x)=353x–52. If the limx→0(1αx+f(x))=β; α,β∈R, than α+2β is equal to [2025]
(1)
We have, f(x)–6f(1x)=353x–52 ... (i)
Apply x→1x, then
f(1x)–6f(x)=35x3–52 ... (ii)
Using (i) and (ii), we get f(x)=–2x–13x+12
Now, β=limx→0(1αx+f(x))
=limx→0(1αx–2x–13x+12)
=[limx→01x(1α–13)]+12 ⇒ α=3, β=12
Hence, α+2β=3+2×12=4