JEE Main 2025 Jan 24 Shift 2, Mathematics Q15: Methods to Evaluate Limits

JEE Main2025Jan 24, Shift 2Mathematics

For some $a, b$ , let $f (x) = a + \frac{sin x}{x} 1 b a 1 + \frac{sin x}{x} b a 1 b + \frac{sin x}{x}$ , $x \neq = 0$ , $x \to 0 lim f (x) = λ + μ a + ν b$ . Then $(λ + μ + ν)^{2}$ is equal to:

A
$25$
B
$9$
C
$36$
D
$16$

Solution

Since $x \to 0 lim \frac{sin x}{x} = 1$ , the limit determinant is

$x \to 0 lim f (x) = a + 11 b a 2 b a 1 b + 1$

Expand along row 1:

$= (a + 1) [2 (b + 1) - b] - 1 [a (b + 1) - ab] + b [a - 2 a]$

$= (a + 1) (b + 2) - (a) + b (- a)$

$= (a + 1) (b + 2) - a - ab$

$= ab + 2 a + b + 2 - a - ab$

$= a + b + 2$

Matching with $λ + μ a + ν b$ : $λ = 2, μ = 1, ν = 1$ .

$(λ + μ + ν)^{2} = (2 + 1 + 1)^{2} = 16$

Correct option: (4)

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JEE Main 2025 Jan 24 Shift 2, Mathematics Q15: Methods to Evaluate Limits

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