Let `a , b in R,(a in 0)`. If the funtion f defined as
`f(x)={{:(,(2x^(2))/(a),0 le x lt 1),(,a,1 le x lt sqrt2),(,(2b^(2)-4b)/(x^(3)),sqrt2lt x lt oo):}` is a continous in `[0,oo)`. Then, (a,b)=

Options

A. `(sqrt2,1-sqrt3)`
B. `(-sqrt2,1-sqrt3)`
C. `(sqrt2,-1+sqrt3)`
D. `(-sqrt2,1+sqrt3)`

Explanation

To determine the values of \( a \) and \( b \) such that the function \( f \) is continuous on the interval \([0, \infty)\), we need to ensure continuity at the points where the definition of the function changes, specifically at \( x = 1 \) and \( x = \sqrt{2} \).

The function \( f(x) \) is defined as:

\[ f(x) = \begin{cases} \frac{2x^2}{a} & \text{for } 0 \leq x < 1 \\ a & \text{for } 1 \leq x < \sqrt{2} \\ \frac{2b^2 - 4b}{x^3} & \text{for } \sqrt{2} < x < \infty \end{cases} \]

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

We need to check the left-hand limit (LHL) and right-hand limit (RHL) at \( x = 1 \):

1. **LHL at \( x = 1 \)**:
\[ \lim_{x \to 1^-} f(x) = \frac{2(1)^2}{a} = \frac{2}{a} \]

2. **RHL at \( x = 1 \)**:
\[ \lim_{x \to 1^+} f(x) = a \]

Setting LHL equal to RHL for continuity:
\[ \frac{2}{a} = a \]

Multiplying both sides by \( a \) (since \( a \neq 0 \)):
\[ 2 = a^2 \implies a = \sqrt{2} \text{ or } a = -\sqrt{2} \]

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

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

1. **LHL at \( x = \sqrt{2} \)**:
\[ \lim_{x \to \sqrt{2}^-} f(x) = a \]

2. **RHL at \( x = \sqrt{2} \)**:
\[ \lim_{x \to \sqrt{2}^+} f(x) = \frac{2b^2 - 4b}{(\sqrt{2})^3} = \frac{2b^2 - 4b}{2\sqrt{2}} = \frac{b^2 - 2b}{\sqrt{2}} \]

Setting LHL equal to RHL for continuity:
\[ a = \frac{b^2 - 2b}{\sqrt{2}} \]

### Step 3: Solve for \( b \)

Now we substitute the values of \( a \) we found earlier into the equation \( a = \frac{b^2 - 2b}{\sqrt{2}} \):

1. **For \( a = \sqrt{2} \)**:
\[ \sqrt{2} = \frac{b^2 - 2b}{\sqrt{2}} \]
Multiplying both sides by \( \sqrt{2} \):
\[ 2 = b^2 - 2b \]
Rearranging gives:
\[ b^2 - 2b - 2 = 0 \]
Using the quadratic formula:
\[ b = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 + 8}}{2} = \frac{2 \pm \sqrt{12}}{2} = 1 \pm \sqrt{3} \]

2. **For \( a = -\sqrt{2} \)**:
\[ -\sqrt{2} = \frac{b^2 - 2b}{\sqrt{2}} \]
Multiplying both sides by \( \sqrt{2} \):
\[ -2 = b^2 - 2b \]
Rearranging gives:
\[ b^2 - 2b + 2 = 0 \]
The discriminant is negative (\( (-2)^2 - 4 \cdot 1 \cdot 2 < 0 \)), indicating no real solutions for \( b \).

### Conclusion

Thus, the only valid pairs \((a, b)\) are:
\[ (a, b) = \left(\sqrt{2}, 1 + \sqrt{3}\right) \text{ and } \left(\sqrt{2}, 1 - \sqrt{3}\right) \]

### Final Answer
The ordered pairs are:
\[ (a, b) = \left(\sqrt{2}, 1 + \sqrt{3}\right) \text{ and } \left(\sqrt{2}, 1 - \sqrt{3}\right) \]