For some a, b , let f(x)=| array ccc a+( x)/(x) & 1 & ~b a & 1+( x)/(x) & ~b a & 1 & ~b+( x)/(x) array |, x ≠ 0, _ x arrow 0 f(x)=λ+μ a+ b . Then (λ+μ+ )^2 is equal to :

Correct answer: A. 16 _ x arrow 0 | array ccc a+( x)/(x) & 1 & b a & 1+( x)/(x) & b a & 1 & b+( x)/(x) array |=λ+μ a+v b At x arrow 0 , aligned & f(x)=| array ccc a+1 & 1 & b a & 1+1 & b a & 1 & b+1 array |=λ+μ a+v b & R_1 arrow R_1-R_2 & R_2 arrow R_2-R_3 & | array ccc 1 & -1 & 0 0 & 1 & -1 a & 1 & b+1 array |=λ+μ a+v b & C_2 arrow C_1-C_2 & | array ccc 1 & 0 & 0 0 & 1 & -1 a & a+1 & b+1 array |=λ+μ a+v b & a+b+2=λ+μ a+v b & λ=2, μ=1, v=1 & (λ+μ+v)=(2+1+1)^2=16 aligned

JEE Main 202524 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 24 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Determinants

JEE Main 2025 · 24 January · Evening Shift · Mathematics · Determinants

For some a, b , let f(x)=| array ccc a+( x)/(x) & 1 & ~b a & 1+(...

For some

\mathrm{a}, \mathrm{b}

, let

f(x)=\left|

\begin{array}{ccc}\mathrm{a}+\frac{\sin x}{x} & 1 & \mathrm{~b} \\ \mathrm{a} & 1+\frac{\sin x}{x} & \mathrm{~b} \\ \mathrm{a} & 1 & \mathrm{~b}+\frac{\sin x}{x}\end{array}

\right|, x \neq 0, \lim _{x \rightarrow 0} f(x)=\lambda+\mu \mathrm{a}+\nu \mathrm{b}

. Then

(\lambda+\mu+\nu)^2

is equal to :

Options

Correct answer

A. 16

Step-by-step solution

\lim _{x \rightarrow 0}\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|=\lambda+\mu a+v b

\lim x \rightarrow 0

\begin{aligned} & f(x)=\left|\begin{array}{ccc} a+1 & 1 & b \\ a & 1+1 & b \\ a & 1 & b+1 \end{array}\right|=\lambda+\mu a+v b \\ & R_1 \rightarrow R_1-R_2 \\ & R_2 \rightarrow R_2-R_3 \\ & \left|\begin{array}{ccc} 1 & -1 & 0 \\ 0 & 1 & -1 \\ a & 1 & b+1 \end{array}\right|=\lambda+\mu a+v b \\ & \mathrm{C}_2 \rightarrow \mathrm{C}_1-\mathrm{C}_2 \\ & \left|\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & -1 \\ a & a+1 & b+1 \end{array}\right|=\lambda+\mu a+v b \\ & a+b+2=\lambda+\mu a+v b \\ & \lambda=2, \mu=1, \quad v=1 \\ & (\lambda+\mu+v)=(2+1+1)^2=16 \end{aligned}

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