For some $a,b$, let $f(x)=\begin{vmatrix}a+\frac{\sin x}x&1&b\a&1+\frac{\sin x}x&b\a&1&b+\frac{\sin x}x\end{vmatrix}$, $x
eq0$, $\lim_{x o0}f(x)=\lambda+\mu a+
u b$. Then $(\lambda+\mu+
u)^2$ is equal to :

Question

For some $a,b$, let $f(x)=\begin{vmatrix}a+\frac{\sin x}x&1&b\a&1+\frac{\sin x}x&b\a&1&b+\frac{\sin x}x\end{vmatrix}$, $x
eq0$, $\lim_{x	o0}f(x)=\lambda+\mu a+
u b$. Then $(\lambda+\mu+
u)^2$ is equal to :

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] Write the matrix as $sI+M$ with $s=\frac{\sin x}x$ and $M$ having every row equal to $(a,1,b)$ (rank 1). Then $\det=s^3+s^2\operatorname{tr}M=s^3+s^2(a+1+b)$. As $x	o0$, $s	o1$: limit $=2+a+b$. So $\lambda=2$, $\mu=
u=1$ and $(\lambda+\mu+
u)^2=16$. Answer (1).

Q. For some $a,b$, let $f(x)=\begin{vmatrix}a+\frac{\sin x}x&1&b\\a&1+\frac{\sin x}x&b\\a&1&b+\frac{\sin x}x\end{vmatrix}$, $x\neq0$, $\lim_{x\to0}f(x)=\lambda+\mu a+\nu b$. Then $(\lambda+\mu+\nu)^2$ is equal to :

Explanation

Reviewed by Dr. Abubacker

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