For some aa, bb, let f(x)=∣a+sin⁡xx1ba1+sin⁡xxba1b+sin⁡xx∣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|, x≠0x \neq 0, lim⁡x→0f(x)=λ+μa+νb\lim_{x \to 0} f(x) = \lambda + \mu a + \nu b. Then (λ+μ+ν)2(\lambda + \mu + \nu)^2 is equal to:

Answer: 16

Step-by-step solution

Step 1: Simplify the determinant using row operations

To simplify the determinant, we can perform row operations. Subtracting the first row from the second row (R2→R2−R1R_2 \to R_2 - R_1) and the first row from the third row (R3→R3−R1R_3 \to R_3 - R_1) will introduce zeros, making the determinant easier to calculate. Let k=sin⁡xxk = \frac{\sin x}{x}.

Step 2: Perform row operations

After applying the row operations R2→R2−R1R_2 \to R_2 - R_1 and R3→R3−R1R_3 \to R_3 - R_1, the determinant becomes simpler. Now we can expand the determinant along the first row or first column.

Step 3: Expand the determinant

Expanding the determinant along the first row, we multiply each element by its corresponding cofactor. This gives us a polynomial in terms of aa, bb, and kk.

Step 4: Simplify the expression for f(x)f(x)

Combine the terms to simplify the expression for f(x)f(x). We can factor out k2k^2 from the terms.

Step 5: Evaluate the limit as x→0x \to 0

We know that lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1. Substitute this limit into the simplified expression for f(x)f(x).

Step 6: Calculate the limit and identify coefficients

Comparing this result with the given form λ+μa+νb\lambda + \mu a + \nu b, we can identify the coefficients: λ=2\lambda = 2, μ=1\mu = 1, and ν=1\nu = 1.

Step 7: Calculate (λ+μ+ν)2(\lambda + \mu + \nu)^2

Finally, substitute the values of λ\lambda, μ\mu, and ν\nu into the expression (λ+μ+ν)2(\lambda + \mu + \nu)^2 to find the final answer.

More from this chapter

All Determinants solutions →

Solve your own maths question free →