JEE Main 2025 — Determinants Question with 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 $\lim x\to 0$, $\begin{aligned} & f(x)=\left|\begin{array}{ccc} a+1 & 1 & b \\ a & 1+1 & b \\ a & 1 & b+1 \end{array}\right|=\lambda+\mu a+v b \\ & R_1 \rightarrow R_1-R_2 \\ & R_2 \rightarrow R_2-R_3 \\ & \left|\begin{array}{ccc} 1 & -1 & 0 \\ 0 & 1 & -1 \\ a & 1 & b+1 \end{array}\right|=\lambda+\mu a+v b \\ & \mathrm{C}_2 \rightarrow \mathrm{C}_1-\mathrm{C}_2 \\ & \left|\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & -1 \\ a & a+1 & b+1 \end{array}\right|=\lambda+\mu a+v b \\ & a+b+2=\lambda+\mu a+v b \\ & \lambda=2, \mu=1, \quad v=1 \\ & (\lambda+\mu+v)=(2+1+1)^2=16 \end{aligned}$