Question

The curve \( y = ax^3 + bx^2 + cx + 5 \) touches the x-axis at point \( P(-2, 0) \). Then the value of \( c \) is:

• A. \( 4a + 5 \)
• B. \( 4a - 5 \)
• C. \( 5 - 4a \)
• D. \( 0 \)

Hint:

Tangency Conditions}
A curve touches the x-axis at \( x = a \) means: \( f(a) = 0 \), \( f'(a) = 0 \)
Use both the function and its derivative to form equations
Solve system of equations carefully

Answer 1

The Correct Option is A

Since the curve touches the x-axis at \( (-2, 0) \), the point satisfies both: \[ y = 0 \quad \text{and} \quad \frac{dy}{dx} = 0 \text{ at } x = -2 \] Given: \[ y = ax^3 + bx^2 + cx + 5 \] Substitute \( x = -2 \) and \( y = 0 \): \[ a(-8) + b(4) - 2c + 5 = 0 \Rightarrow -8a + 4b - 2c + 5 = 0 \quad \text{(1)} \] Differentiate: \[ \frac{dy}{dx} = 3ax^2 + 2bx + c \Rightarrow \frac{dy}{dx} \text{ at } x = -2 \Rightarrow 12a - 4b + c = 0 \quad \text{(2)} \] Solve equations (1) and (2): From (2): \( c = -12a + 4b \) Substitute into (1): \[ -8a + 4b - 2(-12a + 4b) + 5 = 0 \Rightarrow -8a + 4b + 24a - 8b + 5 = 0 \Rightarrow 16a - 4b + 5 = 0 \Rightarrow 4b = 16a + 5 \] Substitute back into (2): \[ c = -12a + 4b = -12a + (16a + 5) = 4a + 5 \]