Question

Let P(x) be a quadratic polynomial such that \( \begin{vmatrix} P(0) & P(1) \\ P(0) & P(2) \end{vmatrix} = 0 \). Let P(0) = 2 and P(1) + P(2) + P(3) = 14. Then P(4) equals

• A. -14
• B. -6
• C. 30
• D. 16

Hint:

For a quadratic polynomial \(P(x)\), the condition \(P(x_1) = P(x_2)\) implies that the axis of symmetry is at \(x = \frac{x_1+x_2}{2}\). In this case, \(P(1)=P(2)\) means the axis of symmetry is at \(x=1.5\). The vertex x-coordinate is also given by \(-b/2a\), so \(-b/2a = 1.5\), which gives \(b=-3a\), a shortcut to one of the steps.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem requires us to find the specific form of a quadratic polynomial using given conditions, one of which involves a determinant. Once the polynomial is determined, we can evaluate it at the desired point.
Step 2: Key Formula or Approach:
1. Let the quadratic polynomial be \( P(x) = ax^2 + bx + c \). 2. Use the given conditions to form a system of equations to solve for a, b, and c. 3. The determinant condition \( \begin{vmatrix} A & B
C & D \end{vmatrix} = AD - BC \).
Step 3: Detailed Explanation:
First, let's use the condition \( P(0) = 2 \). For \( P(x) = ax^2 + bx + c \): \[ P(0) = a(0)^2 + b(0) + c = c \] So, we have \( c = 2 \). The polynomial is \( P(x) = ax^2 + bx + 2 \). Next, let's evaluate the determinant condition: \[ \begin{vmatrix} P(0) & P(1)
P(0) & P(2) \end{vmatrix} = P(0)P(2) - P(0)P(1) = 0 \] \[ P(0) [P(2) - P(1)] = 0 \] Since we are given \( P(0) = 2 \neq 0 \), it must be that: \[ P(2) - P(1) = 0 \implies P(1) = P(2) \] Now, let's use this with our polynomial form: \[ P(1) = a(1)^2 + b(1) + 2 = a+b+2 \] \[ P(2) = a(2)^2 + b(2) + 2 = 4a+2b+2 \] Setting them equal: \[ a+b+2 = 4a+2b+2 \] \[ 3a + b = 0 \implies b = -3a \] So, the polynomial can be written as \( P(x) = ax^2 - 3ax + 2 \). Now, use the final condition: \( P(1) + P(2) + P(3) = 14 \). We already know \( P(1) = P(2) \). \[ P(1) = a(1)^2 - 3a(1) + 2 = a - 3a + 2 = -2a + 2 \] \[ P(3) = a(3)^2 - 3a(3) + 2 = 9a - 9a + 2 = 2 \] Substituting these into the sum equation: \[ P(1) + P(2) + P(3) = (-2a+2) + (-2a+2) + 2 = 14 \] \[ -4a + 6 = 14 \] \[ -4a = 8 \] \[ a = -2 \] With \( a = -2 \), we can find b: \( b = -3a = -3(-2) = 6 \). The polynomial is \( P(x) = -2x^2 + 6x + 2 \). Finally, we need to find \( P(4) \): \[ P(4) = -2(4)^2 + 6(4) + 2 = -2(16) + 24 + 2 = -32 + 26 = -6 \] Step 4: Final Answer:
The value of P(4) is -6.