Question

If the function \( f(x) = x^3 + ax^2 + bx + 40 \) satisfies the conditions of Rolle's theorem on the interval \( [-5, 4] \) and \( -5, 4 \) are two roots of the equation \( f(x) = 0 \), then one of the values of \( c \) as stated in that theorem is:

• A. 3
• B. \( \frac{1 + \sqrt{67}}{3} \)
• C. \( \frac{1 + \sqrt{65}}{3} \)
• D. \( -2 \) \bigskip

Hint:

Rolle's Theorem helps in finding the point \( c \) where the derivative of the function equals zero. Always use the system of equations derived from the function's roots and the differentiability condition to solve for \( a \) and \( b \).

Answer 1

The Correct Option is B

We are given that \( f(x) = x^3 + ax^2 + bx + 40 \) satisfies the conditions of Rolle’s Theorem. Rolle’s Theorem states that if a function is continuous on the closed interval \([a, b]\), differentiable on the open interval \((a, b)\), and \( f(a) = f(b) \), then there exists a point \( c \in (a, b) \) such that \( f'(c) = 0 \). Given: - \( f(x) = x^3 + ax^2 + bx + 40 \) - The roots of the equation \( f(x) = 0 \) are \( x = -5 \) and \( x = 4 \). Step 1: Use the fact that \( f(-5) = 0 \) and \( f(4) = 0 \). Since \( f(-5) = 0 \), substitute \( x = -5 \) in the equation \( f(x) \): \[ f(-5) = (-5)^3 + a(-5)^2 + b(-5) + 40 = 0. \] \[ -125 + 25a - 5b + 40 = 0. \] \[ 25a - 5b - 85 = 0 \quad \text{(Equation 1)}. \] Similarly, substitute \( x = 4 \) into \( f(x) \): \[ f(4) = (4)^3 + a(4)^2 + b(4) + 40 = 0. \] \[ 64 + 16a + 4b + 40 = 0. \] \[ 16a + 4b + 104 = 0 \quad \text{(Equation 2)}. \] Step 2: Solve the system of equations. From Equation 1: \[ 25a - 5b = 85. \] From Equation 2: \[ 16a + 4b = -104. \] Multiply Equation 1 by 4 and Equation 2 by 5 to eliminate \( b \): \[ 100a - 20b = 340 \quad \text{(Equation 3)}. \] \[ 80a + 20b = -520 \quad \text{(Equation 4)}. \] Add Equation 3 and Equation 4: \[ 100a - 20b + 80a + 20b = 340 - 520. \] \[ 180a = -180. \] \[ a = -1. \] Substitute \( a = -1 \) into Equation 1: \[ 25(-1) - 5b = 85, \] \[ -25 - 5b = 85, \] \[ -5b = 110, \] \[ b = -22. \] Step 3: Find the value of \( c \). Now, differentiate \( f(x) \): \[ f'(x) = 3x^2 + 2ax + b. \] Substitute \( a = -1 \) and \( b = -22 \): \[ f'(x) = 3x^2 - 2x - 22. \] Since \( f'(c) = 0 \), solve for \( c \): \[ 3c^2 - 2c - 22 = 0. \] Use the quadratic formula: \[ c = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(3)(-22)}}{2(3)}. \] \[ c = \frac{2 \pm \sqrt{4 + 264}}{6}, \] \[ c = \frac{2 \pm \sqrt{268}}{6}, \] \[ c = \frac{2 \pm \sqrt{4 \times 67}}{6}, \] \[ c = \frac{2 \pm 2\sqrt{67}}{6}, \] \[ c = \frac{1 \pm \sqrt{67}}{3}. \] Thus, one of the values of \( c \) is \( \frac{1 + \sqrt{67}}{3} \). \bigskip