if `f(x)=x^2-ax+3 , x` is rational and `f(x)=2-x , x` is irrational is continuous at exactly two points, then the possible values of `a `

Options

A. `(2,oo)`
B. `(-oo , 3)`
C. `(-oo,-1),cup ( 3, oo) `
D. none of these

Answer

To solve the problem, we need to analyze the function \( f(x) \) which is defined differently for rational and irrational numbers.

Given:
- \( f(x) = x^2 - ax + 3 \) for rational \( x \)
- \( f(x) = 2 - x \) for irrational \( x \)

We want to find the values of \( a \) such that \( f(x) \) is continuous at exactly two points.

### Step 1: Set the two expressions equal
To find the points of continuity, we need to set the two expressions equal to each other:

\[ x^2 - ax + 3 = 2 - x \]

### Step 2: Rearrange the equation
Rearranging the equation gives us:

\[ x^2 - ax + x + 3 - 2 = 0 \]

This simplifies to:

\[ x^2 + (1 - a)x + 1 = 0 \]

### Step 3: Determine the discriminant
For the quadratic equation \( x^2 + (1 - a)x + 1 = 0 \) to have exactly two distinct real roots, the discriminant must be greater than zero. The discriminant \( D \) is given by:

\[ D = b^2 - 4ac \]

Here, \( a = 1 \), \( b = 1 - a \), and \( c = 1 \). Thus,

\[ D = (1 - a)^2 - 4 \cdot 1 \cdot 1 \]

This simplifies to:

\[ D = (1 - a)^2 - 4 \]

### Step 4: Set the discriminant greater than zero
To ensure there are exactly two points of continuity, we need:

\[ (1 - a)^2 - 4 > 0 \]

### Step 5: Solve the inequality
This can be factored as:

\[ (1 - a - 2)(1 - a + 2) > 0 \]

Which simplifies to:

\[ (-a - 1)(-a + 3) > 0 \]

### Step 6: Find the critical points
The critical points from the factors are:

\[ a + 1 = 0 \quad \Rightarrow \quad a = -1 \]
\[ -a + 3 = 0 \quad \Rightarrow \quad a = 3 \]

### Step 7: Test intervals
We need to test the intervals determined by the critical points \( a = -1 \) and \( a = 3 \):

1. **Interval 1:** \( (-\infty, -1) \)
- Choose \( a = -2 \): \( (-(-2) - 1)(-(-2) + 3) = (2 - 1)(2 + 3) = 1 \cdot 5 > 0 \) (True)

2. **Interval 2:** \( (-1, 3) \)
- Choose \( a = 0 \): \( (-(0) - 1)(-(0) + 3) = (-1)(3) = -3 < 0 \) (False)

3. **Interval 3:** \( (3, \infty) \)
- Choose \( a = 4 \): \( (-(4) - 1)(-(4) + 3) = (-5)(-1) = 5 > 0 \) (True)

### Step 8: Conclusion
The solution to the inequality is:

\[ a \in (-\infty, -1) \cup (3, \infty) \]

Thus, the possible values of \( a \) are:

\[ \boxed{(-\infty, -1) \cup (3, \infty)} \]