Let `f(x)=|[1,1,1] , [3-x,5-3x^2,3x^3-1] , [2x^2-1,3x^5-1,7x^8-1]|` then the equation of `f(x)=0` has

Options

A. f(x) = 0 has at least two real roots
B. f'(x) =0 has at least one real root.
C. f(x) is many-one function
D. none of these

Explanation

To solve the problem, we need to analyze the function given by the determinant:

\[ f(x) = \begin{vmatrix} 1 & 1 & 1 \\ 3 - x & 5 - 3x^2 & 3x^3 - 1 \\ 2x^2 - 1 & 3x^5 - 1 & 7x^8 - 1 \end{vmatrix} \]

We want to find the values of \(x\) for which \(f(x) = 0\).

### Step 1: Evaluate \(f(0)\)

Substituting \(x = 0\):

\[ f(0) = \begin{vmatrix} 1 & 1 & 1 \\ 3 & 5 & -1 \\ -1 & -1 & -1 \end{vmatrix} \]

Calculating the determinant:

\[ = 1 \cdot \begin{vmatrix} 5 & -1 \\ -1 & -1 \end{vmatrix} - 1 \cdot \begin{vmatrix} 3 & -1 \\ -1 & -1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 3 & 5 \\ -1 & -1 \end{vmatrix} \]

Calculating each of the smaller determinants:

1. \(\begin{vmatrix} 5 & -1 \\ -1 & -1 \end{vmatrix} = 5 \cdot (-1) - (-1) \cdot (-1) = -5 - 1 = -6\)

2. \(\begin{vmatrix} 3 & -1 \\ -1 & -1 \end{vmatrix} = 3 \cdot (-1) - (-1) \cdot (-1) = -3 - 1 = -4\)

3. \(\begin{vmatrix} 3 & 5 \\ -1 & -1 \end{vmatrix} = 3 \cdot (-1) - 5 \cdot (-1) = -3 + 5 = 2\)

Now substituting back into the determinant:

\[ f(0) = 1 \cdot (-6) - 1 \cdot (-4) + 1 \cdot 2 = -6 + 4 + 2 = 0 \]

### Step 2: Evaluate \(f(1)\)

Substituting \(x = 1\):

\[ f(1) = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 2 & 2 \\ 1 & 2 & 6 \end{vmatrix} \]

Calculating the determinant:

\[ = 1 \cdot \begin{vmatrix} 2 & 2 \\ 2 & 6 \end{vmatrix} - 1 \cdot \begin{vmatrix} 2 & 2 \\ 1 & 6 \end{vmatrix} + 1 \cdot \begin{vmatrix} 2 & 2 \\ 1 & 2 \end{vmatrix} \]

Calculating each of the smaller determinants:

1. \(\begin{vmatrix} 2 & 2 \\ 2 & 6 \end{vmatrix} = 2 \cdot 6 - 2 \cdot 2 = 12 - 4 = 8\)

2. \(\begin{vmatrix} 2 & 2 \\ 1 & 6 \end{vmatrix} = 2 \cdot 6 - 2 \cdot 1 = 12 - 2 = 10\)

3. \(\begin{vmatrix} 2 & 2 \\ 1 & 2 \end{vmatrix} = 2 \cdot 2 - 2 \cdot 1 = 4 - 2 = 2\)

Now substituting back into the determinant:

\[ f(1) = 1 \cdot 8 - 1 \cdot 10 + 1 \cdot 2 = 8 - 10 + 2 = 0 \]

### Conclusion

We found that \(f(0) = 0\) and \(f(1) = 0\). Therefore, the function \(f(x) = 0\) has at least two roots, which are \(x = 0\) and \(x = 1\).

### Final Answer

The equation \(f(x) = 0\) has at least two real roots.