If f(x) `{:(=x^(2)/a", if " 0 le x lt 1 ),(= a", if " 1 le x lt sqrt(2) ),(=(2b^(2)-4b)/(x^(2))", if " sqrt(2) le x ) :}`
is continuous in `(0,oo)` , then the most suitable values of a and b ( in that order ) are

Options

A. `1,-1`
B. `-1,1 + sqrt(2)`
C. `-1,1`
D. `-1,-1`

Answer

To determine the values of \( a \) and \( b \) such that the function \( f(x) \) 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 \( x = \sqrt{2} \).

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

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

1. **Left-hand limit as \( x \to 1^- \)**:
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} \frac{x^2}{a} = \frac{1^2}{a} = \frac{1}{a} \]

2. **Right-hand limit as \( x \to 1^+ \)**:
\[ \lim_{x \to 1^+} f(x) = a \]

Setting the left-hand limit equal to the right-hand limit for continuity:
\[ \frac{1}{a} = a \]

Multiplying both sides by \( a \) (assuming \( a \neq 0 \)):
\[ 1 = a^2 \implies a = 1 \text{ or } a = -1 \]

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

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

1. **Left-hand limit as \( x \to \sqrt{2}^- \)**:
\[ \lim_{x \to \sqrt{2}^-} f(x) = a \]

2. **Right-hand limit as \( x \to \sqrt{2}^+ \)**:
\[ \lim_{x \to \sqrt{2}^+} f(x) = \lim_{x \to \sqrt{2}^+} \frac{2b^2 - 4b}{x^2} = \frac{2b^2 - 4b}{(\sqrt{2})^2} = \frac{2b^2 - 4b}{2} = b^2 - 2b \]

Setting the left-hand limit equal to the right-hand limit for continuity:
\[ a = b^2 - 2b \]

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

Now we have two cases for \( a \):

1. **Case 1: \( a = 1 \)**
\[ 1 = b^2 - 2b \implies b^2 - 2b - 1 = 0 \]
Using the quadratic formula:
\[ b = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2} \]

2. **Case 2: \( a = -1 \)**
\[ -1 = b^2 - 2b \implies b^2 - 2b + 1 = 0 \implies (b - 1)^2 = 0 \implies b = 1 \]

### Conclusion

The suitable values of \( (a, b) \) are:
- From Case 1: \( (1, 1 + \sqrt{2}) \) or \( (1, 1 - \sqrt{2}) \)
- From Case 2: \( (-1, 1) \)

Thus, the most suitable values of \( a \) and \( b \) in order are:
\[ \text{Answer: } (1, 1 + \sqrt{2}) \text{ or } (1, 1 - \sqrt{2}) \text{ or } (-1, 1) \]