For some \( a, b \), let \( f(x) = \left| \begin{matrix} a + \frac{\sin x}{x} & 1 & b \\ a & 1 + \frac{\sin x}{x} & b \\ a & 1 & b + \frac{\sin x}{x} \end{matrix} \right| \), where \( x \neq 0 \), \( \lim_{x \to 0} f(x) = \lambda + \mu a + \nu b \).
Then \( (\lambda + \mu + \nu)^2 \) is equal to:
For some \( a, b \), let \( f(x) = \left| \begin{matrix} a + \frac{\sin x}{x} & 1 & b \\ a & 1 + \frac{\sin x}{x} & b \\ a & 1 & b + \frac{\sin x}{x} \end{matrix} \right| \), where \( x \neq 0 \), \( \lim_{x \to 0} f(x) = \lambda + \mu a + \nu b \).
Then \( (\lambda + \mu + \nu)^2 \) is equal to:
Show Hint
- \( 25 \)
- \( 16 \)
- \( 36 \)
- \( 9 \)
The Correct Option is D
Solution and Explanation
First, compute the determinant of the matrix as \( x \to 0 \) and then take the limit to find the value of \( \lambda + \mu + \nu \). The limit and determinant calculation gives the value 3 for \( \lambda + \mu + \nu \), so squaring this gives 9.
Final Answer: \( (\lambda + \mu + \nu)^2 = 9 \).
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