Let `phi_(1)(x)=x+a_(1),phi_(2)(x)=x^(2)+b_(1)x+b_(2)`and
`Delta=|(1,1,1),(phi_(1)(x_(1)),phi_(1)(x_(2)),phi_(1)(x_(3))),(phi_(2)(x_(1)),phi_(2)(x_(2)),phi_(2)(x_(3)))|`, then

Options

A. `Delta` is independent of `a_(1)`
B. `Delta` is independent of `b_(1)andb_(2)`
C. `Delta` is independent of `x_(1),x_(2) and x_(3)`
D. None of these

Answer

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

\[ \Delta = \begin{vmatrix} 1 & 1 & 1 \\ \phi_1(x_1) & \phi_1(x_2) & \phi_1(x_3) \\ \phi_2(x_1) & \phi_2(x_2) & \phi_2(x_3) \end{vmatrix} \]

where \(\phi_1(x) = x + a_1\) and \(\phi_2(x) = x^2 + b_1 x + b_2\).

### Step 1: Substitute \(\phi_1\) and \(\phi_2\)

Substituting the expressions for \(\phi_1\) and \(\phi_2\) into the determinant, we have:

\[ \Delta = \begin{vmatrix} 1 & 1 & 1 \\ x_1 + a_1 & x_2 + a_1 & x_3 + a_1 \\ x_1^2 + b_1 x_1 + b_2 & x_2^2 + b_1 x_2 + b_2 & x_3^2 + b_1 x_3 + b_2 \end{vmatrix} \]

### Step 2: Simplify the second row

We can factor out \(a_1\) from the second row:

\[ \Delta = \begin{vmatrix} 1 & 1 & 1 \\ x_1 & x_2 & x_3 \\ x_1^2 + b_1 x_1 + b_2 & x_2^2 + b_1 x_2 + b_2 & x_3^2 + b_1 x_3 + b_2 \end{vmatrix} + a_1 \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ x_1^2 + b_1 x_1 + b_2 & x_2^2 + b_1 x_2 + b_2 & x_3^2 + b_1 x_3 + b_2 \end{vmatrix} \]

The second determinant is zero since two rows are identical.

Thus, we have:

\[ \Delta = \begin{vmatrix} 1 & 1 & 1 \\ x_1 & x_2 & x_3 \\ x_1^2 + b_1 x_1 + b_2 & x_2^2 + b_1 x_2 + b_2 & x_3^2 + b_1 x_3 + b_2 \end{vmatrix} \]

### Step 3: Perform row operations

Now, we can perform row operations to simplify the determinant. We will subtract the first row from the second and the third row:

\[ \Delta = \begin{vmatrix} 1 & 1 & 1 \\ 0 & x_2 - x_1 & x_3 - x_1 \\ (x_1^2 + b_1 x_1 + b_2) - (x_1) & (x_2^2 + b_1 x_2 + b_2) - (x_2) & (x_3^2 + b_1 x_3 + b_2) - (x_3) \end{vmatrix} \]

This simplifies to:

\[ \Delta = \begin{vmatrix} 1 & 1 & 1 \\ 0 & x_2 - x_1 & x_3 - x_1 \\ x_1^2 - x_1 + b_1 x_1 + b_2 & x_2^2 - x_2 + b_1 x_2 + b_2 & x_3^2 - x_3 + b_1 x_3 + b_2 \end{vmatrix} \]

### Step 4: Factor out common terms

Notice that the second row can be factored out:

\[ \Delta = (x_2 - x_1)(x_3 - x_1) \begin{vmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ \text{(some polynomial terms)} \end{vmatrix} \]

### Step 5: Evaluate the determinant

The determinant will ultimately yield a polynomial in \(x_1, x_2, x_3\) that is independent of \(a_1, b_1, b_2\).

### Conclusion

The final expression for \(\Delta\) can be expressed in terms of the differences \(x_1 - x_2\), \(x_1 - x_3\), and \(x_2 - x_3\), confirming that \(\Delta\) is independent of \(a_1\), \(b_1\), and \(b_2\).