The function `f(x) = {{:(x^(2)//a, 0 le x lt 1),(a, 1 le x lt sqrt(2)),((2b^(2) - 4b)//x^(2), sqrt(2) le x lt infty):}` is continuous for `0 le x lt infty`, then the most suitable values of a and b are

Options

A. a=1, b=-1
B. `a=-1, b= 1+sqrt(2)`
C. `a=-1, b=1`
D. none of these

Answer

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

### Step 1: Continuity at \( x = 1 \)

The function is defined as:
- \( f(x) = \frac{x^2}{a} \) for \( 0 \leq x < 1 \)
- \( f(x) = a \) for \( 1 \leq x < \sqrt{2} \)

To ensure continuity at \( x = 1 \), we need:
\[ f(1) = \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} \frac{x^2}{a} \]
Calculating the limit:
\[ \lim_{x \to 1^-} \frac{x^2}{a} = \frac{1^2}{a} = \frac{1}{a} \]
Thus, we require:
\[ f(1) = a = \frac{1}{a} \]
Multiplying both sides by \( a \) (assuming \( a \neq 0 \)):
\[ a^2 = 1 \]
This gives us:
\[ a = 1 \quad \text{or} \quad a = -1 \]

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

Next, we check continuity at \( x = \sqrt{2} \):
- For \( x \) approaching \( \sqrt{2} \) from the left, we have:
\[ f(\sqrt{2}) = a \]
- For \( x \) approaching \( \sqrt{2} \) from the right, we have:
\[ f(x) = \frac{2b^2 - 4b}{x^2} \quad \text{for } x \geq \sqrt{2} \]
Calculating the 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: Finding suitable values for \( a \) and \( b \)

Now we have two equations:
1. \( a^2 = 1 \) gives \( a = 1 \) or \( a = -1 \)
2. \( a = b^2 - 2b \)

#### Case 1: \( a = 1 \)
Substituting \( a = 1 \) into the second equation:
\[ 1 = b^2 - 2b \]
Rearranging gives:
\[ b^2 - 2b - 1 = 0 \]
Using the quadratic formula:
\[ b = \frac{2 \pm \sqrt{4 + 4}}{2} = 1 \pm \sqrt{2} \]

#### Case 2: \( a = -1 \)
Substituting \( a = -1 \) into the second equation:
\[ -1 = b^2 - 2b \]
Rearranging gives:
\[ b^2 - 2b + 1 = 0 \implies (b - 1)^2 = 0 \implies b = 1 \]

### Conclusion

The most suitable values of \( a \) and \( b \) are:
1. \( a = 1 \) and \( b = 1 + \sqrt{2} \) or \( b = 1 - \sqrt{2} \)
2. \( a = -1 \) and \( b = 1 \)

### Final Answer
The most suitable values of \( a \) and \( b \) are:
- \( a = 1 \), \( b = 1 + \sqrt{2} \) or \( b = 1 - \sqrt{2} \)
- \( a = -1 \), \( b = 1 \)