Find the number of integral vaues of 'a' for which the range of function `f (x) = (x ^(2) -ax +1)/(x ^(2) -3x+2) `is `(-oo,oo),`

Answer

To find the number of integral values of 'a' for which the range of the function

\[ f(x) = \frac{x^2 - ax + 1}{x^2 - 3x + 2} \]

is \((-∞, ∞)\), we will follow these steps:

### Step 1: Set up the equation
We start by letting \(y = f(x)\):

\[ y = \frac{x^2 - ax + 1}{x^2 - 3x + 2} \]

### Step 2: Cross-multiply
Cross-multiplying gives us:

\[ y(x^2 - 3x + 2) = x^2 - ax + 1 \]

Rearranging this, we get:

\[ yx^2 - 3yx + 2y = x^2 - ax + 1 \]

### Step 3: Rearrange to form a quadratic equation
Rearranging the equation leads to:

\[ (y - 1)x^2 + (a - 3y)x + (2y - 1) = 0 \]

### Step 4: Determine conditions for real roots
For \(f(x)\) to have a range of \((-∞, ∞)\), the quadratic in \(x\) must have real roots for all values of \(y\). This requires the discriminant \(D\) to be non-negative:

\[ D = (a - 3y)^2 - 4(y - 1)(2y - 1) \geq 0 \]

### Step 5: Expand the discriminant
Expanding the discriminant gives:

\[ D = (a - 3y)^2 - 4(2y^2 - 3y + 1) \]

This simplifies to:

\[ D = (a - 3y)^2 - 8y^2 + 12y - 4 \]

### Step 6: Combine like terms
Combining like terms results in:

\[ D = a^2 - 6ay + 9y^2 - 8y^2 + 12y - 4 = a^2 - 6ay + y^2 + 12y - 4 \]

### Step 7: Formulate as a quadratic in \(y\)
This can be rewritten as:

\[ (1 - 8)y^2 + (-6a + 12)y + (a^2 - 4) \geq 0 \]

### Step 8: Conditions for \(y\)
For this quadratic in \(y\) to be non-negative for all \(y\), the coefficient of \(y^2\) must be non-positive (i.e., \(1 - 8 \leq 0\)), and the discriminant must be less than or equal to zero:

1. \(1 - 8 \leq 0\) is always true.
2. The discriminant must satisfy:

\[ (-6a + 12)^2 - 4(1)(a^2 - 4) \leq 0 \]

### Step 9: Solve the discriminant inequality
Expanding this gives:

\[ 36a^2 - 144a + 144 - 4a^2 + 16 \leq 0 \]

This simplifies to:

\[ 32a^2 - 144a + 160 \leq 0 \]

### Step 10: Factor the quadratic
Dividing through by 16 gives:

\[ 2a^2 - 9a + 10 \leq 0 \]

### Step 11: Find the roots
Using the quadratic formula:

\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{9 \pm \sqrt{(-9)^2 - 4 \cdot 2 \cdot 10}}{2 \cdot 2} \]

Calculating the discriminant:

\[ 81 - 80 = 1 \]

Thus, the roots are:

\[ a = \frac{9 \pm 1}{4} = \{2.5, 2\} \]

### Step 12: Determine the interval
The quadratic opens upwards, so the solution to \(2a^2 - 9a + 10 \leq 0\) is:

\[ 2 \leq a \leq 2.5 \]

### Step 13: Find integral values
The only integral value of \(a\) in this range is:

\[ a = 2 \]

### Conclusion
Thus, the number of integral values of \(a\) for which the range of the function is \((-∞, ∞)\) is **1**.

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