Question

If \(f(x) = \begin{bmatrix} \cos x & x &1 \\ 2 \sin x & x & 2x \\ \sin x & x & x \end{bmatrix}\). Then \(\lim_{x \to 0} \frac{f(x)}{x^2}\) is:

• A. 2
• B. 0
• C. 3
• D. 6

Answer 1

The Correct Option is B

1. Understand the problem:

Given the determinant function \( f(x) = \begin{vmatrix} \cos x & x & 1 \\ 2\sin x & x & 2x \\ \sin x & x & x \end{vmatrix} \), we need to evaluate \( \lim_{x \to 0} \frac{f(x)}{x^2} \).

2. Expand the determinant \( f(x) \):

Expand along the first row:

\[ f(x) = \cos x \begin{vmatrix} x & 2x \\ x & x \end{vmatrix} - x \begin{vmatrix} 2\sin x & 2x \\ \sin x & x \end{vmatrix} + 1 \begin{vmatrix} 2\sin x & x \\ \sin x & x \end{vmatrix} \]

Simplify each minor:

\[ f(x) = \cos x (x^2 - 2x^2) - x (2x\sin x - 2x\sin x) + (2x\sin x - x\sin x) = \cos x (-x^2) - x (0) + x\sin x = -x^2 \cos x + x\sin x \]

3. Compute the limit:

\[ \lim_{x \to 0} \frac{f(x)}{x^2} = \lim_{x \to 0} \frac{-x^2 \cos x + x\sin x}{x^2} = \lim_{x \to 0} \left( -\cos x + \frac{\sin x}{x} \right) = -1 + 1 = 0 \]

Correct Answer: (B) 0

Answer 2

First, let's compute the determinant of the matrix:

\[ f(x) = \cos(x) \cdot (x^2 - 2x^2) - x \cdot (2x\sin(x) - x\sin(x)) + 1 \cdot (2x\sin(x) - x\sin(x)) \] \[ f(x) = \cos(x) \cdot (-x^2) - x \cdot (x\sin(x)) + 1 \cdot (x\sin(x)) \] \[ f(x) = -x^2 \cdot \cos(x) - x^2 \cdot \sin(x) + x \cdot \sin(x) \] \[ f(x) = -x^2 \cdot (\cos(x) + \sin(x)) + x \cdot \sin(x) \]

Now, we need to find the limit of \( \frac{f(x)}{x^2} \) as \( x \) approaches 0.

\[ \lim_{x \to 0} \frac{f(x)}{x^2} = \lim_{x \to 0} \frac{-x^2(\cos(x) + \sin(x)) + x\sin(x)}{x^2} \] \[ = \lim_{x \to 0} \left[ -(\cos(x) + \sin(x)) + \frac{\sin(x)}{x} \right] \]

As \( x \to 0 \):

\[ \cos(x) \to 1, \quad \sin(x) \to 0, \quad \frac{\sin(x)}{x} \to 1 \]

Therefore:

\[ \lim_{x \to 0} \left[ -(\cos(x) + \sin(x)) + \frac{\sin(x)}{x} \right] = -(1 + 0) + 1 = -1 + 1 = 0 \]