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JEE Mains · Maths · STD 11 - 12. limits

Let \(f: R-\{0\} \rightarrow R\) be a function such that \(f(x)-6 f\left(\frac{1}{x}\right)=\frac{35}{3 x}-\frac{5}{2}\).
If the \(\lim _{x \rightarrow 0}\left(\frac{1}{\alpha x}+f(x)\right)=\beta ; \alpha, \beta \in R\), then \(\alpha+2 \beta\) is equal to

  1. A \(5\)
  2. B \(3\)
  3. C \(4\)
  4. D \(6\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(4\)

Step-by-step Solution

Detailed explanation

\begin{aligned} & f(x)-6 f\left(\frac{1}{x}\right)=\frac{35}{3 x}-\frac{5}{2}...(1) \\ & 6\left(f\left(\frac{1}{x}\right)-6 f(x)=\frac{35 x}{3}-\frac{5}{2}\right) \\ & 6 f\left(\frac{1}{x}\right)-36 f(x)=\frac{210 x}{3}-\frac{30}{2}...(2) \\ & (1)+(2) \\ & -35…

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