The function`f(x)={{:(,(x^(2))/(a),0 le x lt 1),(,a,1le x lt sqrt2),(,(2b^(2)-4b)/(x^(2)),sqrt2 le x lt oo):}` is a continuous for `0 le x lt oo`. Then which of the following statements is correct?

Options

A. The number of all possible ordered pairs (a,b) is 3
B. The number of all possible ordered pairs (a,b) is 4
C. The product of all possible pairs ,b is -1
D. The product of all possible values of b is 1

Explanation

To determine the correct statement regarding the function

\[ f(x) = \begin{cases} \frac{x^2}{a} & \text{for } 0 \leq x < 1 \\ a & \text{for } 1 \leq x < \sqrt{2} \\ \frac{2b^2 - 4b}{x^2} & \text{for } \sqrt{2} \leq x < \infty \end{cases} \]

we need to ensure that the function is continuous at the points where the definition of the function changes, specifically at \(x = 1\) and \(x = \sqrt{2}\).

### Step 1: Check continuity at \(x = 1\)

To ensure continuity at \(x = 1\), we need to check that:

\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) \]

Calculating the left-hand limit:

\[ \lim_{x \to 1^-} f(x) = \frac{1^2}{a} = \frac{1}{a} \]

Calculating the right-hand limit:

\[ \lim_{x \to 1^+} f(x) = a \]

Setting these equal for continuity:

\[ \frac{1}{a} = a \]

Multiplying both sides by \(a\) (assuming \(a \neq 0\)) gives:

\[ 1 = a^2 \implies a = \pm 1 \]

### Step 2: Check continuity at \(x = \sqrt{2}\)

Next, we check continuity at \(x = \sqrt{2}\):

\[ \lim_{x \to \sqrt{2}^-} f(x) = a \]

Calculating the right-hand limit:

\[ \lim_{x \to \sqrt{2}^+} f(x) = \frac{2b^2 - 4b}{(\sqrt{2})^2} = \frac{2b^2 - 4b}{2} = b^2 - 2b \]

Setting these equal for continuity:

\[ a = b^2 - 2b \]

### Step 3: Substitute values of \(a\)

Now we substitute the possible values of \(a\) into the equation \(a = b^2 - 2b\):

1. If \(a = 1\):

\[ 1 = b^2 - 2b \implies b^2 - 2b - 1 = 0 \]

Using the quadratic formula:

\[ b = \frac{2 \pm \sqrt{4 + 4}}{2} = 1 \pm \sqrt{2} \]

2. If \(a = -1\):

\[ -1 = b^2 - 2b \implies b^2 - 2b + 1 = 0 \implies (b - 1)^2 = 0 \implies b = 1 \]

### Step 4: Possible pairs of \((a, b)\)

The possible pairs \((a, b)\) are:

1. \((1, 1 + \sqrt{2})\)
2. \((1, 1 - \sqrt{2})\)
3. \((-1, 1)\)

### Step 5: Product of all possible values of \(b\)

The possible values of \(b\) are \(1 + \sqrt{2}\), \(1 - \sqrt{2}\), and \(1\). The product is:

\[ (1 + \sqrt{2})(1 - \sqrt{2})(1) = (1^2 - (\sqrt{2})^2) = 1 - 2 = -1 \]

### Conclusion

Thus, the correct statement regarding the function is that the product of all possible values of \(b\) is \(-1\).