For some a, b,let f(x) = left|begin{array}{cc | vedclass

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(D) Given $f(x) = \left|\begin{array}{ccc} a+\frac{\sin x}{x} & 1 & b \ a & 1+\frac{\sin x}{x} & b \ a & 1 & b+\frac{\sin x}{x} \end{array}ight|$.

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$16$

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(D) Given $f(x) = \left|\begin{array}{ccc} a+\frac{\sin x}{x} & 1 & b \ a & 1+\frac{\sin x}{x} & b \ a & 1 & b+\frac{\sin x}{x} \end{array}ight|$.As $x ightarrow 0$,$\frac{\sin x}{x} ightarrow 1$. Let $k = \frac{\sin x}{x} ightarrow 1$.Then $\lim_{x ightarrow 0} f(x) = \left|\begin{array}{ccc} a+1 & 1 & b \ a & 1+1 & b \ a & 1 & b+1 \end{array}ight| = \left|\begin{array}{ccc} a+1 & 1 & b \ a & 2 & b \ a & 1 & b+1 \end{array}ight|$.Applying $C_1 ightarrow C_1 - C_2$ and $C_3 ightarrow C_3 - C_2$ is not necessary,let's expand directly:$= (a+1)[2(b+1) - b] - 1[a(b+1) - ab] + b[a - 2a]$$= (a+1)(b+2) - 1[ab + a - ab] + b(-a)$$= ab + 2a + b + 2 - a - ab$$= a + b + 2$.Comparing with $\lambda + \mu a + 
u b$,we get $\lambda = 2, \mu = 1, 
u = 1$.Therefore,$(\lambda + \mu + 
u)^2 = (2 + 1 + 1)^2 = 4^2 = 16$.

For some $a, b$,let $f(x) = \left|\begin{array}{ccc} a+\frac{\sin x}{x} & 1 & b \\ a & 1+\frac{\sin x}{x} & b \\ a & 1 & b+\frac{\sin x}{x} \end{array}\right|, \quad x \neq 0$. If $\lim_{x \rightarrow 0} f(x) = \lambda + \mu a + \nu b$,then $(\lambda + \mu + \nu)^2$ is equal to:

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