If the quadratic polynomial `P (x)=(p-3)x ^(2) -2px+3p-6` ranges from `[0,oo)` for every `x in R,` then the value of p can be:

Options

A. 3
B. 4
C. 6
D. 7

Answer

To solve the problem, we need to analyze the quadratic polynomial \( P(x) = (p-3)x^2 - 2px + (3p-6) \) and determine the values of \( p \) such that the polynomial ranges from \( [0, \infty) \) for every \( x \in \mathbb{R} \).

### Step 1: Identify the conditions for the polynomial to range from \( [0, \infty) \)

For the quadratic polynomial \( P(x) \) to be non-negative for all \( x \), the following conditions must be satisfied:
1. The coefficient of \( x^2 \) must be positive.
2. The discriminant must be non-positive (i.e., the quadratic has no real roots or a double root).

### Step 2: Check the coefficient of \( x^2 \)

The coefficient of \( x^2 \) is \( p - 3 \). For this to be positive:
\[ p - 3 > 0 \implies p > 3 \]

### Step 3: Calculate the discriminant

The discriminant \( D \) of the quadratic \( P(x) \) is given by:
\[ D = b^2 - 4ac \]
where \( a = p - 3 \), \( b = -2p \), and \( c = 3p - 6 \).

Substituting these values into the discriminant formula:
\[ D = (-2p)^2 - 4(p - 3)(3p - 6) \]
\[ D = 4p^2 - 4[(p - 3)(3p - 6)] \]

### Step 4: Expand and simplify the discriminant

Expanding \( (p - 3)(3p - 6) \):
\[ (p - 3)(3p - 6) = 3p^2 - 6p - 9p + 18 = 3p^2 - 15p + 18 \]
Now substituting back into the discriminant:
\[ D = 4p^2 - 4(3p^2 - 15p + 18) \]
\[ D = 4p^2 - 12p^2 + 60p - 72 \]
\[ D = -8p^2 + 60p - 72 \]

### Step 5: Set the discriminant less than or equal to zero

To ensure the quadratic does not have real roots:
\[ -8p^2 + 60p - 72 \leq 0 \]

### Step 6: Solve the quadratic inequality

First, we can find the roots of the equation:
\[ -8p^2 + 60p - 72 = 0 \]
Dividing the entire equation by -4:
\[ 2p^2 - 15p + 18 = 0 \]

Using the quadratic formula \( p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ p = \frac{15 \pm \sqrt{(-15)^2 - 4 \cdot 2 \cdot 18}}{2 \cdot 2} \]
\[ p = \frac{15 \pm \sqrt{225 - 144}}{4} \]
\[ p = \frac{15 \pm \sqrt{81}}{4} \]
\[ p = \frac{15 \pm 9}{4} \]

Calculating the roots:
1. \( p = \frac{24}{4} = 6 \)
2. \( p = \frac{6}{4} = \frac{3}{2} \)

### Step 7: Determine the valid range for \( p \)

The roots of the quadratic \( 2p^2 - 15p + 18 = 0 \) are \( p = \frac{3}{2} \) and \( p = 6 \). The quadratic opens upwards (since the coefficient of \( p^2 \) is positive), so it is less than or equal to zero between its roots:
\[ \frac{3}{2} \leq p \leq 6 \]

### Step 8: Combine conditions

We also have the condition \( p > 3 \). Therefore, combining the two conditions:
\[ 3 < p \leq 6 \]

### Conclusion

The only integer value of \( p \) that satisfies this condition is:
\[ p = 6 \]

Thus, the value of \( p \) can be \( 6 \).