Coefficient of x in `f(x)=|(x,(1+sinx)^3,cosx),(1,log(1+x),2),(x^2,(1+x)^2,0)|` is

Options

A. `x-(a+b+c)`
B. `9x^2+a+b+c`
C. `a+b+c`
D. `0`

Answer

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

\[ f(x) = \begin{vmatrix} x & (1 + \sin x)^3 & \cos x \\ 1 & \log(1 + x) & 2 \\ x^2 & (1 + x)^2 & 0 \end{vmatrix} \]

we will evaluate the determinant step by step.

### Step 1: Expand the functions
First, we need to expand the functions \( \sin x \), \( \cos x \), and \( \log(1 + x) \) using their Taylor series around \( x = 0 \).

- \( \sin x = x - \frac{x^3}{6} + O(x^5) \)
- \( \cos x = 1 - \frac{x^2}{2} + O(x^4) \)
- \( \log(1 + x) = x - \frac{x^2}{2} + O(x^3) \)

Now, substituting these expansions into the determinant:

\[ (1 + \sin x)^3 = (1 + x - \frac{x^3}{6} + O(x^5))^3 \]

Using the binomial expansion, we only need the linear term:

\[ (1 + x)^3 = 1 + 3x + 3x^2 + x^3 \]

Thus, we can approximate \( (1 + \sin x)^3 \) as \( 1 + 3x + O(x^2) \).

### Step 2: Substitute the expansions into the determinant
Now we substitute these approximations into the determinant:

\[ f(x) = \begin{vmatrix} x & 1 + 3x & 1 - \frac{x^2}{2} \\ 1 & x - \frac{x^2}{2} & 2 \\ x^2 & (1 + x)^2 & 0 \end{vmatrix} \]

### Step 3: Simplify the determinant
Now we can simplify the determinant. We will focus on the terms that will contribute to the coefficient of \( x \).

\[ = \begin{vmatrix} x & 1 + 3x & 1 - \frac{x^2}{2} \\ 1 & x - \frac{x^2}{2} & 2 \\ x^2 & 1 + 2x & 0 \end{vmatrix} \]

### Step 4: Expand the determinant
We can expand the determinant along the first row:

\[ = x \begin{vmatrix} x - \frac{x^2}{2} & 2 \\ 1 + 2x & 0 \end{vmatrix} - (1 + 3x) \begin{vmatrix} 1 & 2 \\ x^2 & 0 \end{vmatrix} + (1 - \frac{x^2}{2}) \begin{vmatrix} 1 & x - \frac{x^2}{2} \\ x^2 & 1 + 2x \end{vmatrix} \]

### Step 5: Calculate the 2x2 determinants
Calculating the first determinant:

\[ \begin{vmatrix} x - \frac{x^2}{2} & 2 \\ 1 + 2x & 0 \end{vmatrix} = 0 - 2(1 + 2x) = -2 - 4x \]

Calculating the second determinant:

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

Calculating the third determinant:

\[ \begin{vmatrix} 1 & x - \frac{x^2}{2} \\ x^2 & 1 + 2x \end{vmatrix} = 1(1 + 2x) - (x - \frac{x^2}{2})x^2 = 1 + 2x - x^3 + \frac{x^4}{2} \]

### Step 6: Combine the results
Now substituting back into the determinant expansion:

\[ f(x) = x(-2 - 4x) - (1 + 3x)(-2x^2) + (1 - \frac{x^2}{2})(1 + 2x - x^3 + \frac{x^4}{2}) \]

We only need the coefficient of \( x \):

1. From \( x(-2 - 4x) \), we get \( -2x \).
2. From \( - (1 + 3x)(-2x^2) \), we get \( 0 \) (since it contributes to \( x^2 \)).
3. From \( (1 - \frac{x^2}{2})(1 + 2x - x^3 + \frac{x^4}{2}) \), we get \( 2x \).

### Final Calculation
Combining these, we find:

\[ -2x + 2x = 0 \]

Thus, the coefficient of \( x \) in \( f(x) \) is \( -2 \).

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