The set of values of a for which ` (a - 1) x^(2) - (a + 1) x + a - 1`
`ge 0 ` ture for all ` x ge 2` is

Options

A. `(-infty, 1)`
B. `(1, (7)/(3))`
C. `((7)/(3), infty)`
D. none of these

Explanation

To solve the inequality \( (a - 1)x^2 - (a + 1)x + (a - 1) \geq 0 \) for all \( x \geq 2 \), we will analyze the quadratic expression and determine the conditions on \( a \).

### Step 1: Identify the quadratic expression
The given expression is:
\[ f(x) = (a - 1)x^2 - (a + 1)x + (a - 1) \]
We want to find the values of \( a \) such that \( f(x) \geq 0 \) for all \( x \geq 2 \).

### Step 2: Determine the conditions for the quadratic to be non-negative
A quadratic \( Ax^2 + Bx + C \) is non-negative for all \( x \) if:
1. \( A \geq 0 \)
2. The discriminant \( D = B^2 - 4AC \leq 0 \)

Here, \( A = a - 1 \), \( B = -(a + 1) \), and \( C = a - 1 \).

### Step 3: Set conditions for \( A \)
For \( f(x) \) to be non-negative, we need:
\[ a - 1 \geq 0 \implies a \geq 1 \]

### Step 4: Set conditions for the discriminant
Now, we calculate the discriminant:
\[ D = (-(a + 1))^2 - 4(a - 1)(a - 1) \]
\[ D = (a + 1)^2 - 4(a - 1)^2 \]
Expanding both terms:
\[ D = (a^2 + 2a + 1) - 4(a^2 - 2a + 1) \]
\[ D = a^2 + 2a + 1 - 4a^2 + 8a - 4 \]
\[ D = -3a^2 + 10a - 3 \]
We need \( D \leq 0 \):
\[ -3a^2 + 10a - 3 \leq 0 \]

### Step 5: Solve the quadratic inequality
To solve \( -3a^2 + 10a - 3 \leq 0 \), we first find the roots of the equation:
\[ -3a^2 + 10a - 3 = 0 \]
Using the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ a = \frac{-10 \pm \sqrt{10^2 - 4 \cdot (-3) \cdot (-3)}}{2 \cdot (-3)} \]
\[ a = \frac{-10 \pm \sqrt{100 - 36}}{-6} \]
\[ a = \frac{-10 \pm \sqrt{64}}{-6} \]
\[ a = \frac{-10 \pm 8}{-6} \]
Calculating the two roots:
1. \( a_1 = \frac{-2}{-6} = \frac{1}{3} \)
2. \( a_2 = \frac{-18}{-6} = 3 \)

### Step 6: Determine the intervals
The quadratic \( -3a^2 + 10a - 3 \) opens downwards (since the coefficient of \( a^2 \) is negative), so it is non-positive between its roots:
\[ \frac{1}{3} \leq a \leq 3 \]

### Step 7: Combine conditions
Now, combining \( a \geq 1 \) with \( \frac{1}{3} \leq a \leq 3 \):
\[ 1 \leq a \leq 3 \]

### Final Answer
Thus, the set of values of \( a \) for which the inequality holds is:
\[ \boxed{[1, 3]} \]