If `f:Rto(1,10)`, where `f(x)=(x^(2)+1-alpha)/(x^(2)+2)` is an onto function, then the value `alpha` is

Options

A. 19
B. `-19`
C. 10
D. none of these

Answer

To solve the problem, we need to determine the value of \( \alpha \) such that the function \( f(x) = \frac{x^2 + 1 - \alpha}{x^2 + 2} \) is onto the interval \( (1, 10) \).

### Step 1: Set up the equation
We start by setting \( f(x) = y \):
\[ y = \frac{x^2 + 1 - \alpha}{x^2 + 2} \]
Cross-multiplying gives:
\[ y(x^2 + 2) = x^2 + 1 - \alpha \]
Rearranging this, we have:
\[ yx^2 + 2y - x^2 - 1 + \alpha = 0 \]
This can be rewritten as:
\[ (y - 1)x^2 + (2y + \alpha - 1) = 0 \]

### Step 2: Determine conditions for real roots
For \( f(x) \) to be onto, the quadratic in \( x^2 \) must have real roots for all \( y \) in the interval \( (1, 10) \). The discriminant \( D \) of the quadratic must be non-negative:
\[ D = b^2 - 4ac \]
Here, \( a = y - 1 \), \( b = 0 \), and \( c = 2y + \alpha - 1 \). Thus, we have:
\[ D = 0^2 - 4(y - 1)(2y + \alpha - 1) \geq 0 \]
This simplifies to:
\[ -4(y - 1)(2y + \alpha - 1) \geq 0 \]
This implies:
\[ (y - 1)(2y + \alpha - 1) \leq 0 \]

### Step 3: Analyze the inequality
The product \( (y - 1)(2y + \alpha - 1) \leq 0 \) indicates that one factor must be non-positive while the other is non-negative.

1. **For \( y - 1 \leq 0 \)**: This occurs when \( y \leq 1 \), which is outside our interval \( (1, 10) \).
2. **For \( y - 1 \geq 0 \)**: This occurs when \( y \geq 1 \).

Thus, we focus on the second factor:
\[ 2y + \alpha - 1 \leq 0 \]
This gives:
\[ \alpha \leq 1 - 2y \]

### Step 4: Find the bounds for \( \alpha \)
Now, we need to analyze this inequality for the maximum value of \( y \) in the interval \( (1, 10) \):
- At \( y = 10 \):
\[ \alpha \leq 1 - 2(10) = 1 - 20 = -19 \]
- At \( y = 1 \):
\[ \alpha \leq 1 - 2(1) = 1 - 2 = -1 \]

### Step 5: Determine the value of \( \alpha \)
For \( f(x) \) to be onto \( (1, 10) \), we need \( \alpha \) to be such that the quadratic has real roots for all \( y \) in the interval. Therefore, we need:
\[ \alpha = -1 \]

### Conclusion
Thus, the value of \( \alpha \) such that \( f(x) \) is an onto function from \( \mathbb{R} \) to \( (1, 10) \) is:
\[ \alpha = -1 \]