If `a, b gt 0 and Delta (x)= |(x,a,a),(b,x,a),(b,b,x)|`, then

Options

A. `Delta(x)` is increasing on `(-sqrt(ab), sqrt(ab))`
B. `Delta (x)` is decreasing on `(sqrt(ab), oo)`
C. `Delta(x)` has a local manimum at `x = sqrt(ab)`
D. none of these

Answer

To solve the problem, we need to evaluate the determinant \( \Delta(x) = \begin{vmatrix} x & a & a \\ b & x & a \\ b & b & x \end{vmatrix} \) and analyze its behavior.

### Step 1: Calculate the determinant

We will expand the determinant using the first row:

\[ \Delta(x) = x \begin{vmatrix} x & a \\ b & x \end{vmatrix} - a \begin{vmatrix} b & a \\ b & x \end{vmatrix} + a \begin{vmatrix} b & x \\ b & b \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} x & a \\ b & x \end{vmatrix} = x^2 - ab \)
2. \( \begin{vmatrix} b & a \\ b & x \end{vmatrix} = bx - ab \)
3. \( \begin{vmatrix} b & x \\ b & b \end{vmatrix} = b^2 - bx \)

Substituting these back into the expression for \( \Delta(x) \):

\[ \Delta(x) = x(x^2 - ab) - a(bx - ab) + a(b^2 - bx) \]

Simplifying this:

\[ \Delta(x) = x^3 - abx - abx + a^2b + ab^2 - abx \]

Combining like terms:

\[ \Delta(x) = x^3 - 3abx + a^2b + ab^2 \]

### Step 2: Find the first derivative

Next, we find the first derivative \( \Delta'(x) \):

\[ \Delta'(x) = \frac{d}{dx}(x^3 - 3abx + a^2b + ab^2) = 3x^2 - 3ab \]

### Step 3: Set the first derivative to zero

To find critical points, we set \( \Delta'(x) = 0 \):

\[ 3x^2 - 3ab = 0 \implies x^2 = ab \implies x = \pm \sqrt{ab} \]

### Step 4: Analyze the sign of the first derivative

We will analyze the sign of \( \Delta'(x) \):

- For \( x < -\sqrt{ab} \): \( \Delta'(x) > 0 \) (increasing)
- For \( -\sqrt{ab} < x < \sqrt{ab} \): \( \Delta'(x) < 0 \) (decreasing)
- For \( x > \sqrt{ab} \): \( \Delta'(x) > 0 \) (increasing)

### Step 5: Find the second derivative

Now, we find the second derivative \( \Delta''(x) \):

\[ \Delta''(x) = \frac{d}{dx}(3x^2 - 3ab) = 6x \]

### Step 6: Determine local maxima and minima

Evaluate the second derivative at the critical points:

1. At \( x = \sqrt{ab} \):
\[ \Delta''(\sqrt{ab}) = 6\sqrt{ab} > 0 \quad \text{(local minimum)} \]

2. At \( x = -\sqrt{ab} \):
\[ \Delta''(-\sqrt{ab}) = -6\sqrt{ab} < 0 \quad \text{(local maximum)} \]

### Conclusion

Thus, we conclude that:

- \( \Delta(x) \) has a local maximum at \( x = -\sqrt{ab} \)
- \( \Delta(x) \) has a local minimum at \( x = \sqrt{ab} \)