Let be a function such that . If the , than is equal to [2025]

Question

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 $\lim_{x \to 0} (\frac{1}{α x} + f (x)) = β; α, β \in R$ , than $α + 2 β$ is equal to [2025]

Smart Achievers · Accepted Answer

(1)

We have, $f (x) - 6 f (\frac{1}{x}) = \frac{35}{3 x} - \frac{5}{2}$ ... (i)

Apply $x \to \frac{1}{x}$ , then

$\lim_{x \to 0} (\frac{1}{α x} + f (x)) = β; α, β \in R$ ... (ii)

Using (i) and (ii), we get $f (x) = - 2 x - \frac{1}{3 x} + \frac{1}{2}$

Now, $β = \lim_{x \to 0} (\frac{1}{α x} + f (x))$

$= \lim_{x \to 0} (\frac{1}{α x} - 2 x - \frac{1}{3 x} + \frac{1}{2})$

$= [\lim_{x \to 0} \frac{1}{x} (\frac{1}{α} - \frac{1}{3})] + \frac{1}{2} \Rightarrow α = 3, β = \frac{1}{2}$

Hence, $α + 2 β = 3 + 2 \times \frac{1}{2} = 4$

Solution

Q.
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 $\lim_{x \to 0} (\frac{1}{α x} + f (x)) = β; α, β \in R$ , than $α + 2 β$ is equal to [2025]

1 4

2 3

3 5

4 6

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Solution

Q. 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 4

2 3

3 5

4 6

Ans. (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

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Q.
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 $\lim_{x \to 0} (\frac{1}{α x} + f (x)) = β; α, β \in R$ , than $α + 2 β$ is equal to [2025]