Question

If \( (2, a) \) and \( (b, 19) \) are two stationary points of the curve \( y = 2x^3 - 15x^2 + 36x + c \), then \( a + b + c = \dots \)

• A. \( -20 \)
• B. \( 15 \)
• C. \( -12 \)
• D. \( 24 \)

Hint:

For stationary points, differentiate the equation of the curve and solve for where the derivative equals zero. Then substitute back into the original equation.

Answer 1

The Correct Option is B

We are given the curve equation \( y = 2x^3 - 15x^2 + 36x + c \) and that \( (2, a) \) and \( (b, 19) \) are stationary points. Step 1: To find stationary points, take the derivative of the equation: \[ y' = 6x^2 - 30x + 36 \] At stationary points, \( y' = 0 \), so solve for \( x \) when \( y' = 0 \). Step 2: Substitute \( x = 2 \) and \( x = b \) into the derivative equation and solve to find \( a \) and \( b \). Step 3: Use the equation for the curve to solve for \( a \), \( b \), and \( c \). After solving, we find \( a + b + c = 15 \). % Final Answer The value of \( a + b + c \) is \( 15 \).

Answer 2

Step 1: Find the derivative of the curve
Given curve: \( y = 2x^3 - 15x^2 + 36x + c \)
Derivative:
\[ \frac{dy}{dx} = 6x^2 - 30x + 36 \]

Step 2: Stationary points satisfy \(\frac{dy}{dx} = 0\)
\[ 6x^2 - 30x + 36 = 0 \implies x^2 - 5x + 6 = 0 \]
Factorizing:
\[ (x - 2)(x - 3) = 0 \implies x = 2, 3 \]

Step 3: Find corresponding \( y \)-values
For \( x = 2 \):
\[ a = y(2) = 2(2)^3 - 15(2)^2 + 36(2) + c = 16 - 60 + 72 + c = 28 + c \]
For \( x = 3 \):
\[ 19 = y(3) = 2(3)^3 - 15(3)^2 + 36(3) + c = 54 - 135 + 108 + c = 27 + c \]

Step 4: Find \( c \) using the known \( y \)-value at \( x=3 \)
\[ 19 = 27 + c \implies c = 19 - 27 = -8 \]

Step 5: Calculate \( a \) and sum \( a + b + c \)
We know \( b = 3 \) (from \( x=3 \))
\[ a = 28 + c = 28 - 8 = 20 \]
Sum:
\[ a + b + c = 20 + 3 + (-8) = 15 \]