Let f:ℝ−{0}→ℝ be a function such that f(x)−6f1x=353x−52. If the limx→0 1αx+f(x)=β; α,β∈ℝ, then α+2β is equal to :

F(x)−6f(1/x)=353x−52 ....(1)Replace x→1xF(1/x)−6(x)=35x3−52 ....(2)Using (1) and (2)f(x)=−2x−13x+12B=limx→0 1αx+f(x)=limx→0 1αx−2x−13x+12α=3, B=1/2 So, α+2B=3+1=4

Let f:ℝ−{0}→ℝ be a function such that f(x)−6f1x=353x−52. If the limx→0 1αx+f(x)=β; α,β∈ℝ, then α+2β is equal to :

F(x)−6f(1/x)=353x−52 ....(1)Replace x→1xF(1/x)−6(x)=35x3−52 ....(2)Using (1) and (2)f(x)=−2x−13x+12B=limx→0 1αx+f(x)=limx→0 1αx−2x−13x+12α=3, B=1/2 So, α+2B=3+1=4

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Let $f : ℝ - {0} \to ℝ$ be a function such that $f (x) - 6 f (\frac{1}{x}) = \frac{35}{3 x} - \frac{5}{2} . If the lim_{x \to 0} (\frac{1}{αx} + f (x)) = β;$ $α, β \in ℝ, then α + 2 β$ is equal to :

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Detailed Solution

$F (x) - 6 f (1 / x) = \frac{35}{3 x} - \frac{5}{2} . . . . (1)$

Replace $x \to \frac{1}{x}$
$F (1 / x) - 6 (x) = \frac{35 x}{3} - \frac{5}{2} . . . . (2)$
Using (1) and (2)
$\begin{array}{l} f (x) = - 2 x - \frac{1}{3 x} + \frac{1}{2} \\ B = lim_{x \to 0} (\frac{1}{αx} + f (x)) \\ = lim_{x \to 0} (\frac{1}{αx} - 2 x - \frac{1}{3 x} + \frac{1}{2}) \\ α = 3, B = 1 / 2 \\ So, α + 2 B = 3 + 1 = 4 \end{array}$