For some $\mathrm{a}, \mathrm{b}$, let $f(x)=\left|\begin{array}{ccc}\mathrm{a}+\frac{\sin x}{x} & 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 :
  1. 16
  2. 25
  3. 9
  4. 36

Solution

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

Asked in: JEE Main 2025 (24 Jan Shift 2)

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