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JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Sets Relations and Functions

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

, then

α + 2 β

is equal to

(A)3
(B)5
(C)4✓ Correct
(D)6

Answer: (C)

Solution

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

Replacing

x \to \frac{1}{x}

:

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

Multiplying (2) by

6

and adding to (1):

- 35 f (x) = \frac{35}{3 x} + 70 x - \frac{35}{2} \Rightarrow f (x) = - \frac{1}{3 x} - 2 x + \frac{1}{2} .

x \to 0 lim (\frac{1}{α x} + f (x)) = x \to 0 lim (\frac{1}{α x} - \frac{1}{3 x} - 2 x + \frac{1}{2}) .

For this limit to exist (be finite), the

\frac{1}{x}

terms must cancel:

\frac{1}{α} = \frac{1}{3} \Rightarrow α = 3

. Then

β = x \to 0 lim (- 2 x + \frac{1}{2}) = \frac{1}{2} .

α + 2 β = 3 + 2 (\frac{1}{2}) = 4.

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Single correct option

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