JEE MAIN - Mathematics (2025 - 24th January Evening Shift - No. 6)
For some $a, b,$ let $f(x)=\left|\begin{array}{ccc}\mathrm{a}+\frac{\sin x}{x} & 1 & \mathrm{~b} \\ \mathrm{a} & 1+\frac{\sin x}{x} & \mathrm{~b} \\ \mathrm{a} & 1 & \mathrm{~b}+\frac{\sin x}{x}\end{array}\right|, x \neq 0, \lim \limits_{x \rightarrow 0} f(x)=\lambda+\mu \mathrm{a}+\nu \mathrm{b}.$ Then $(\lambda+\mu+v)^2$ is equal to :
25
16
9
36
Explanation
$$\begin{aligned}
& \lim _{x \rightarrow 0} f(x)=\left|\begin{array}{ccc}
a+1 & 1 & b \\
a & 1+1 & b \\
a & 1 & b+1
\end{array}\right| \\
& =(a+1)(2(b+1)-b)+1(a b-a(b+1))+b a \\
& =(a+1)(b+2)-a+a b \\
& =b+a+2=\lambda+\mu a+v b \\
& \lambda=2, \mu=1, v=1 \Rightarrow(\lambda+\mu+v)^2=16
\end{aligned}$$
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