For some $a, b$, let $$f(x)=\left|\begin{array}{ccc} a+\frac{\sin x}{x} & 1 & b \\ a & 1+\frac{\sin x}{x} & b \\ a & 1 & b+\frac{\sin x}{x} \end{array}\right|, \quad x \neq 0,$$ $\lim _{x \rightarrow 0} f(x)=\lambda+\mu a+\nu b$. Then $(\lambda+\mu+\nu)^{2}$ is equal to:

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Limits

For some

a, b

, let

f (x) = a + \frac{s i n x}{x} a a 1 1 + \frac{s i n x}{x} 1 b b b + \frac{s i n x}{x}, x \neq = 0,

lim_{x \to 0} f (x) = λ + μ a + ν b

. Then

(λ + μ + ν)^{2}

is equal to:

Answer: (D)

x \to 0

\frac{sin x}{x} \to 1

, so the limiting determinant becomes

x \to 0 lim f (x) = a + 1 a a 121 b b b + 1 .

Expand along the first row:

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

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

= ab + 2 a + b + 2 - a - ab = a + b + 2.

Comparing with

λ + μ a + ν b

gives

λ = 2, μ = 1, ν = 1

, so

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

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Single correct option

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