If `Delta (x)=|{:(x,1+x^(2),x^(3)),(log(1+x^(2)),e^(x),sinx),(cosx,tanx,sin^(2)x):}|` then

Options

A. `Delta`(x) is divisible by x
B. `Delta` (x)=0
C. `Delta`'(x)=0
D. None of these

Answer

To solve the determinant \( \Delta(x) = \begin{vmatrix} x & 1+x^2 & x^3 \\ \log(1+x^2) & e^x & \sin x \\ \cos x & \tan x & \sin^2 x \end{vmatrix} \), we will follow these steps:

### Step 1: Write the determinant
We start with the determinant as given:

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

### Step 2: Expand the logarithm and exponential functions
We can use the Taylor series expansions for \( \log(1+x^2) \) and \( e^x \):

- The expansion for \( \log(1+x^2) \) around \( x = 0 \) is:
\[ \log(1+x^2) = x^2 - \frac{x^4}{2} + O(x^6) \]

- The expansion for \( e^x \) around \( x = 0 \) is:
\[ e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + O(x^4) \]

### Step 3: Substitute the expansions into the determinant
Now substitute these expansions into the second row of the determinant:

\[ \Delta(x) = \begin{vmatrix} x & 1+x^2 & x^3 \\ x^2 - \frac{x^4}{2} + O(x^6) & 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + O(x^4) & \sin x \\ \cos x & \tan x & \sin^2 x \end{vmatrix} \]

### Step 4: Simplify the determinant
Notice that as \( x \to 0 \), both \( \sin x \) and \( \tan x \) can be approximated by their Taylor series:

- \( \sin x \approx x - \frac{x^3}{6} + O(x^5) \)
- \( \tan x \approx x + \frac{x^3}{3} + O(x^5) \)

Substituting these approximations into the determinant gives:

\[ \Delta(x) = \begin{vmatrix} x & 1+x^2 & x^3 \\ x^2 - \frac{x^4}{2} + O(x^6) & 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + O(x^4) & x - \frac{x^3}{6} + O(x^5) \\ \cos x & x + \frac{x^3}{3} + O(x^5) & x^2 + O(x^4) \end{vmatrix} \]

### Step 5: Factor out \( x \)
From the first column, we can factor out \( x \):

\[ \Delta(x) = x \begin{vmatrix} 1 & 1+x^2 & x^2 \\ \frac{x^2 - \frac{x^4}{2}}{x} & 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + O(x^4) & \frac{x - \frac{x^3}{6}}{x} \\ \frac{\cos x}{x} & 1 + \frac{x^2}{3} + O(x^4) & 1 + O(x^2) \end{vmatrix} \]

### Step 6: Conclude divisibility
Since we have factored out \( x \), we can conclude that \( \Delta(x) \) is divisible by \( x \). Therefore, \( \Delta(0) = 0 \).

### Final Result
Thus, we find that \( \Delta(x) \) is divisible by \( x \).