Let f : mathbb R -\ 0\ rightarrow mathbb R be a function such that f(x)-6 f (1/x )=35/3 x-5/2 . If the lim x rightarrow 0 (1/α x+f(x) )=β ; α, β in mathbb R , then α+2 β is equal to

Mathematics · Limits, Continuity and Differentiability

JEE Main 2025 — 24 January, Morning Shift — Question 3

Let $\mathrm{f}: \mathbb{R}-\{0\} \rightarrow \mathbb{R}$ be a function such that $f(x)-6 f\left(\frac{1}{x}\right)=\frac{35}{3 x}-\frac{5}{2}$ . If the

$\lim _{x \rightarrow 0}\left(\frac{1}{\alpha x}+f(x)\right)=\beta ;$ $\alpha, \beta \in \mathbb{R}$ , then $\alpha+2 \beta$ is equal to

Option A:
3
Option B:
5
Option C:
4
Correct
Option D:
6

Answer: C

Step-by-step solution

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

Replace $\mathrm{x} \rightarrow \frac{1}{\mathrm{x}}$

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

Using (1) & (2)

$f(x)=-2 x-\frac{1}{3 x}+\frac{1}{2}$

$\beta=\lim _{x \rightarrow 0}\left(\frac{1}{\alpha x}+f(x)\right)$

$=\lim _{x \rightarrow 0}\left(\frac{1}{\alpha x}-2 x-\frac{1}{3 x}+\frac{1}{2}\right)$

$\alpha=3, \beta=1 / 2$

So, $\alpha+2\beta=3+1=4$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 24 January, Morning Shift
Subject: Mathematics
Chapter: Limits, Continuity and Differentiability
Topic: Evaluation of Limit of Functions

lim x arrow 0 cosec x (sqrt2 cos ^2 x+3 cos x-sqrtcos ^2 x+sin x+4 ) is

Answer: C

Step-by-step solution

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