For some a, b, let \(f(x)=\left|\begin{array}{ccc}\mathrm{a}+\frac{\sin x}{x} & 1 & \mathrm{~b} \\ \mathrm{a} &a…
For some a, b, let \(f(x)=\left|\begin{array}{ccc}\mathrm{a}+\frac{\sin x}{x} & 1 & \mathrm{~b} \\ \mathrm{a} & 1+\frac{\sin x}{x} & \mathrm{~b} \\ \text { a } & 1 & \mathrm{~b}+\frac{\sin x}{x}\end{array}\right|, x \neq 0, \lim _{x \rightarrow 0} f(x)=\lambda+\mu \mathrm{a}+\nu \mathrm{b}\). Then \((\lambda+\mu+\nu)^2\) is equal to :
[JEE Main 2025, 24 Jan (Shift 2)]
16
\(f\left(x\right)=\left|\begin{matrix}\mathrm{a}+\frac{\sin x}{x} & 1 & \mathrm{b} \\ \mathrm{a} & 1+\frac{\sin x}{x} & \mathrm{b} \\ \mathrm{a} & 1 & \mathrm{b}+\frac{\sin x}{x}\end{matrix}\right|\\ \lim _{x\to 0}f\left(x\right)=\left|\begin{matrix}a+1 & 1 & b \\ a & 1+1 & b \\ a & 1 & b+1\end{matrix}\right|\\ =\left(a+1\right)\left(2\left(b+1\right)-b\right)+1\left(ab-a\left(b+1\right)\right)-ba\\ =\left(a+1\right)\left(b+2\right)-a-ab\\ =b+a+2=\lambda +\mu a+vb\\ \text{On comparing, we get}\\ \lambda =2,\mu =1,v=1\\ \Rightarrow {\left(\lambda +\mu +v\right)}^{2}=16\)
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