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JEE Main 202524 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
24 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Functions

JEE Main 2025 · 24 January · Morning Shift · Mathematics · Functions

Let f: R-\ 0\ arrow R be a function such that f(x)-6...

Let f:R−{0}→Rf: R-\{0\} \rightarrow R be a function such that f(x)−6f(1x)=353x−52f(x)-6 f\left(\frac{1}{x}\right)=\frac{35}{3 x}-\frac{5}{2}. If the lim⁡x→0(1αx+f(x))=β;α,β∈R\lim _{x \rightarrow 0}\left(\frac{1}{\alpha x}+f(x)\right)=\beta ; \alpha, \beta \in R, then α+2β\alpha+2 \beta is equal to

Options

  1. A55
  2. B33
  3. C44
  4. D66

Correct answer

C. 44

Step-by-step solution

f(x)−6f(1x)=353x−52...(1)6(f(1x)−6f(x)=35x3−52)6f(1x)−36f(x)=210x3−302...(2)(1)+(2)−35f(x)=353[1x+6x]−52(1+6)−f(x)=13(1x+6x)−12f(x)=−13x−2x+12lim⁡x→0[1αx−13x−2x+12]=β⇒α=3β+2β=3+2×12=4\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 f(x)=\frac{35}{3}\left[\frac{1}{x}+6 x\right]-\frac{5}{2}(1+6) \\ & -f(x)=\frac{1}{3}\left(\frac{1}{x}+6 x\right)-\frac{1}{2} \\ & f(x)=-\frac{1}{3 x}-2 x+\frac{1}{2} \\ & \lim _{x \rightarrow 0}\left[\frac{1}{\alpha x}-\frac{1}{3 x}-2 x+\frac{1}{2}\right]=\beta \\ & \Rightarrow \alpha=3 \\ & \beta+2 \beta=3+2 \times \frac{1}{2}=4 \end{aligned}
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