The coefficient of x in
f (x) = `|{:(x , 1 + sin x , cos x ),(1 , log (1+x),2),(x^(2) , 1 + x^(2),0):}|, - 1 lt x le 1 `, is

Options

A. 1
B. `-2`
C. `-1`
D. 0

Answer

To find the coefficient of \( x \) in the function

\[ f(x) = \begin{vmatrix} x & 1 + \sin x & \cos x \\ 1 & \log(1+x) & 2 \\ x^2 & 1 + x^2 & 0 \end{vmatrix} \]

we will evaluate the determinant step by step.

### Step 1: Set up the determinant

We start with the determinant:

\[ D = \begin{vmatrix} x & 1 + \sin x & \cos x \\ 1 & \log(1+x) & 2 \\ x^2 & 1 + x^2 & 0 \end{vmatrix} \]

### Step 2: Expand the determinant

Using the formula for the determinant of a \( 3 \times 3 \) matrix, we expand \( D \):

\[ D = x \begin{vmatrix} \log(1+x) & 2 \\ 1 + x^2 & 0 \end{vmatrix} - (1 + \sin x) \begin{vmatrix} 1 & 2 \\ x^2 & 0 \end{vmatrix} + \cos x \begin{vmatrix} 1 & \log(1+x) \\ x^2 & 1 + x^2 \end{vmatrix} \]

### Step 3: Calculate the 2x2 determinants

1. Calculate the first determinant:

\[ \begin{vmatrix} \log(1+x) & 2 \\ 1 + x^2 & 0 \end{vmatrix} = 0 \cdot \log(1+x) - 2(1 + x^2) = -2(1 + x^2) \]

2. Calculate the second determinant:

\[ \begin{vmatrix} 1 & 2 \\ x^2 & 0 \end{vmatrix} = 1 \cdot 0 - 2x^2 = -2x^2 \]

3. Calculate the third determinant:

\[ \begin{vmatrix} 1 & \log(1+x) \\ x^2 & 1 + x^2 \end{vmatrix} = 1(1 + x^2) - x^2 \log(1+x) = 1 + x^2 - x^2 \log(1+x) \]

### Step 4: Substitute back into the determinant

Substituting these back into the expression for \( D \):

\[ D = x(-2(1 + x^2)) - (1 + \sin x)(-2x^2) + \cos x(1 + x^2 - x^2 \log(1+x)) \]

### Step 5: Simplify the expression

Now we simplify \( D \):

\[ D = -2x - 2x^3 + 2x^2 + 2x^2 \sin x + \cos x(1 + x^2 - x^2 \log(1+x)) \]

### Step 6: Find the coefficient of \( x \)

To find the coefficient of \( x \), we look at the terms in \( D \):

- From \( -2x \), the coefficient is \( -2 \).
- The other terms do not contribute to the coefficient of \( x \).

Thus, the coefficient of \( x \) in \( f(x) \) is:

\[ \text{Coefficient of } x = -2 \]

### Final Answer

The coefficient of \( x \) in \( f(x) \) is \( -2 \).

---