Question

The maximum value of $ a $ such that the second derivative of $ x^4 + ax^3 + \frac{3x^2}{2} + 1 $ is positive for all real $ x $ is

• A. \( 3 \)
• B. \( -3 \)
• C. \( 2 \)
• D. \( -2 \)

Hint:

For ensuring a quadratic function is positive for all \( x \), check that its discriminant is negative. This guarantees no real roots and that the quadratic expression is always positive.

Answer 1

The Correct Option is C

The given function is: \[ f(x) = x^4 + ax^3 + \frac{3x^2}{2} + 1 \] Step 1: Find the first and second derivatives of \( f(x) \). The first derivative is: \[ f'(x) = \frac{d}{dx} \left( x^4 + ax^3 + \frac{3x^2}{2} + 1 \right) \] \[ f'(x) = 4x^3 + 3ax^2 + 3x \] The second derivative is: \[ f''(x) = \frac{d}{dx} \left( 4x^3 + 3ax^2 + 3x \right) \] \[ f''(x) = 12x^2 + 6ax + 3 \] Step 2: For the second derivative to be positive for all real \( x \), the discriminant of the quadratic must be negative. The discriminant \( \Delta \) of a quadratic equation \( Ax^2 + Bx + C = 0 \) is given by: \[ \Delta = B^2 - 4AC \] For \( f''(x) = 12x^2 + 6ax + 3 \), we have: \[ A = 12, \quad B = 6a, \quad C = 3 \] The discriminant is: \[ \Delta = (6a)^2 - 4(12)(3) = 36a^2 - 144 \] Step 3: Set the discriminant \( \Delta \) less than 0 to ensure that the quadratic has no real roots, which means the second derivative is always positive. \[ 36a^2 - 144<0 \] \[ 36a^2<144 \] \[ a^2<4 \] \[ |a|<2 \] Thus, the maximum value of \( a \) is \( a = 2 \).
Final Answer: \[ \boxed{2} \]