Key Concepts and Formulas

1. Limits of Trigonometric Functions: The fundamental limit $\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$ is crucial for evaluating the entries of the determinant as $x \rightarrow 0$.
2. Properties of Limits with Determinants: For a matrix $A(x)$ whose entries are functions of $x$, if the limits of all entries exist as $x \rightarrow c$, then the limit of the determinant is the determinant of the limits: $\lim_{x \rightarrow c} \det(A(x)) = \det(\lim_{x \rightarrow c} A(x))$.
3. Determinant Calculation: The ability to compute the determinant of a $3 \times 3$ matrix, often simplified using elementary row or column operations before cofactor expansion, is essential.
4. Comparing Coefficients: To find unknown coefficients in an identity, we equate the coefficients of corresponding terms on both sides of the equation.

Step-by-Step Solution

Step 1: Evaluate the Limit of Each Entry in the Determinant

The given function is defined as a determinant: $f(x)=\left|\begin{array}{ccc} \mathrm{a}+\frac{\sin x}{x} & 1 & \mathrm{~b} \\ \mathrm{a} & 1+\frac{\sin x}{x} & \mathrm{~b} \\ \mathrm{a} & 1 & \mathrm{~b}+\frac{\sin x}{x} \end{array}\right|$ We need to find $\lim_{x \rightarrow 0} f(x)$. According to the property of limits with determinants, we can first evaluate the limit of each entry as $x \rightarrow 0$ and then compute the determinant of the resulting matrix.

The key limit to apply is $\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$. Let $S = \frac{\sin x}{x}$. As $x \rightarrow 0$, $S \rightarrow 1$.

Applying this to each entry in the determinant:

• $\lim_{x \rightarrow 0} \left(a + \frac{\sin x}{x}\right) = a + 1$
• $\lim_{x \rightarrow 0} \left(1 + \frac{\sin x}{x}\right) = 1 + 1 = 2$
• $\lim_{x \rightarrow 0} \left(b + \frac{\sin x}{x}\right) = b + 1$
• Constant entries ($a, 1, b$) remain unchanged.

Substituting these limits into the determinant, we get: $\lim_{x \rightarrow 0} f(x) = \left|\begin{array}{ccc} a+1 & 1 & b \\ a & 2 & b \\ a & 1 & b+1 \end{array}\right|$

Step 2: Calculate the Determinant

Now, we calculate the determinant of the resulting matrix, let's call it $\Delta$: $\Delta = \left|\begin{array}{ccc} a+1 & 1 & b \\ a & 2 & b \\ a & 1 & b+1 \end{array}\right|$ To simplify the calculation, we can use elementary row operations to create zeros in a row or column. Let's perform the operation $R_1 \rightarrow R_1 - R_2$. This operation does not change the value of the determinant. $\Delta = \left|\begin{array}{ccc} (a+1)-a & 1-2 & b-b \\ a & 2 & b \\ a & 1 & b+1 \end{array}\right|$ $\Delta = \left|\begin{array}{ccc} 1 & -1 & 0 \\ a & 2 & b \\ a & 1 & b+1 \end{array}\right|$ Now, we expand the determinant along the first row ($R_1$) because it contains a zero, which simplifies the calculation. The formula for cofactor expansion along $R_1$ is $p(tz-uw) - q(sz-uv) + r(sw-tv)$ for a matrix $\left|\begin{array}{ccc} p & q & r \\ s & t & u \\ v & w & z \end{array}\right|$. $\Delta = 1 \cdot \left|\begin{array}{cc} 2 & b \\ 1 & b+1 \end{array}\right| - (-1) \cdot \left|\begin{array}{cc} a & b \\ a & b+1 \end{array}\right| + 0 \cdot \left|\begin{array}{cc} a & 2 \\ a & 1 \end{array}\right|$ Calculate the $2 \times 2$ sub-determinants:

• $\left|\begin{array}{cc} 2 & b \\ 1 & b+1 \end{array}\right| = 2(b+1) - b(1) = 2b + 2 - b = b + 2$
• $\left|\begin{array}{cc} a & b \\ a & b+1 \end{array}\right| = a(b+1) - b(a) = ab + a - ab = a$

Substitute these values back into the expansion: $\Delta = 1 \cdot (b+2) + 1 \cdot (a) + 0$ $\Delta = b + 2 + a$ Rearranging the terms, we get: $\Delta = a + b + 2$ So, $\lim_{x \rightarrow 0} f(x) = a+b+2$.

Step 3: Compare Coefficients to Find $\lambda, \mu, \nu$

We are given that $\lim_{x \rightarrow 0} f(x) = \lambda+\mu a+\nu b$. We found that $\lim_{x \rightarrow 0} f(x) = a+b+2$.

Equating the two expressions: $a+b+2 = \lambda+\mu a+\nu b$ By comparing the coefficients of $a$, $b$, and the constant term on both sides:

• Comparing coefficients of $a$: $1 = \mu \Rightarrow \mu = 1$
• Comparing coefficients of $b$: $1 = \nu \Rightarrow \nu = 1$
• Comparing constant terms: $2 = \lambda \Rightarrow \lambda = 2$

Thus, we have $\lambda=2$, $\mu=1$, and $\nu=1$.

Step 4: Calculate the Final Expression

The problem asks for the value of $(\lambda+\mu+\nu)^2$. Substitute the values we found for $\lambda, \mu,$ and $\nu$: $(\lambda+\mu+\nu)^2 = (2 + 1 + 1)^2$ $= (4)^2$ $= 16$

Common Mistakes & Tips

• Forgetting Limit Properties: Remember that the limit of a determinant can be found by taking the determinant of the limits of its entries, but ensure the individual limits exist.
• Algebraic Errors in Determinants: Determinant calculations can be prone to arithmetic and algebraic errors. Double-check each step, especially signs during cofactor expansion and simplification.
• Inefficient Determinant Calculation: Always look for opportunities to simplify the determinant using row or column operations. Creating zeros in a row or column before expansion significantly reduces the amount of calculation. For example, operations like $R_i \rightarrow R_i - R_j$ or $C_i \rightarrow C_i - C_j$ are very powerful.

Summary

This problem effectively combines concepts from limits and determinants. The first step involves correctly evaluating the limit of each entry in the determinant using the standard limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$. Then, the resulting $3 \times 3$ determinant is calculated, which can be simplified efficiently using row operations. Finally, the coefficients of the resulting linear expression in $a$ and $b$ are compared with the given form $\lambda+\mu a+\nu b$ to determine the values of $\lambda, \mu,$ and $\nu$, leading to the computation of the required expression $(\lambda+\mu+\nu)^2$. The final value is 16.

The final answer is $\boxed{\text{16}}$ which corresponds to option (B).