If the inequality `x^(2)-2px+q ge0` does not hold for `q=p^(2)-1`, then range of x is given by

Options

A. `x lt p -1`
B. `x gt p+1`
C. `p-1 lt x lt p + 1`
D. none of these

Answer

To solve the inequality \( x^2 - 2px + q \geq 0 \) under the condition that it does not hold for \( q = p^2 - 1 \), we will follow these steps:

### Step 1: Substitute the value of \( q \)
We start by substituting \( q = p^2 - 1 \) into the inequality:

\[ x^2 - 2px + (p^2 - 1) \geq 0 \]

### Step 2: Rewrite the inequality
This simplifies to:

\[ x^2 - 2px + p^2 - 1 \geq 0 \]

### Step 3: Factor the quadratic expression
We can factor the quadratic expression:

\[ (x - p)^2 - 1 \geq 0 \]

This can be rewritten as:

\[ (x - p - 1)(x - p + 1) \geq 0 \]

### Step 4: Analyze the inequality
The inequality \( (x - (p - 1))(x - (p + 1)) \geq 0 \) holds when \( x \) is outside the interval defined by the roots \( p - 1 \) and \( p + 1 \).

### Step 5: Determine the intervals
The critical points are \( x = p - 1 \) and \( x = p + 1 \). The intervals to consider are:

1. \( (-\infty, p - 1) \)
2. \( (p - 1, p + 1) \)
3. \( (p + 1, \infty) \)

### Step 6: Test the intervals
We need to check the sign of the expression in each interval:

- For \( x < p - 1 \), both factors \( (x - (p - 1)) \) and \( (x - (p + 1)) \) are negative, so the product is positive.
- For \( p - 1 < x < p + 1 \), \( (x - (p - 1)) \) is positive and \( (x - (p + 1)) \) is negative, so the product is negative.
- For \( x > p + 1 \), both factors are positive, so the product is positive.

### Step 7: Write the final solution
The inequality \( (x - (p - 1))(x - (p + 1)) \geq 0 \) holds for:

\[ x \in (-\infty, p - 1] \cup [p + 1, \infty) \]

However, since the original inequality does not hold (i.e., is not defined) for \( q = p^2 - 1 \), we exclude the points where the expression equals zero. Thus, the final range of \( x \) is:

\[ x \in (-\infty, p - 1) \cup (p + 1, \infty) \]