The function f(x)=`{{:(,x^(2)//a,0lexlt1),(,a,1lexltsqrt2),(,(2b^(2)-4b)/(x^(2)),sqrt2le xltoo):}` and if it is continous at x=1, `sqrt2 , then ` a and b` is equal to

Options

A. `-2`
B. `-4`
C. `-6`
D. `-8`

Explanation

To solve the problem, we need to ensure that the function \( f(x) \) is continuous at the points \( x = 1 \) and \( x = \sqrt{2} \). The function is defined piecewise as follows:

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

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

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

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

Calculating \( \lim_{x \to 1^-} f(x) \):

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

Calculating \( f(1) \):

\[ f(1) = a \]

Setting these equal for continuity:

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

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

\[ 1 = a^2 \]

Thus, we have:

\[ a = \pm 1 \]

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

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

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

Calculating \( \lim_{x \to \sqrt{2}^-} f(x) \):

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

Calculating \( f(\sqrt{2}) \):

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

Setting these equal for continuity:

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

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

Now we have two cases to consider based on the values of \( a \).

#### Case 1: \( a = 1 \)

Substituting \( a = 1 \) into the equation \( 1 = b^2 - 2b \):

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

Using the quadratic formula:

\[ b = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2} \]

#### Case 2: \( a = -1 \)

Substituting \( a = -1 \) into the equation \( -1 = b^2 - 2b \):

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

This simplifies to:

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

### Final Values

Thus, we have two sets of solutions:

1. For \( a = 1 \), \( b = 1 + \sqrt{2} \) or \( b = 1 - \sqrt{2} \).
2. For \( a = -1 \), \( b = 1 \).

### Summary of Results

The values of \( a \) and \( b \) are:

- \( a = 1 \), \( b = 1 + \sqrt{2} \) or \( b = 1 - \sqrt{2} \)
- \( a = -1 \), \( b = 1 \)