Consider the function f: R -> R where R is the set of all real numbers. f(x) = x^4/4 - 2x^3/3 - 2x^2/2 + 1 Which of the following statements is/are TRUE? A. x = 0 is a local maximum of f. B. x = -2 is a local minimum of f. C. x = 3 is a local maximum of f. D. x = -1 is a local minimum of f.

GATE 2024 · Engineering Mathematics · Maxima Minima · medium

Answer: x = 0 is a local maximum of f (Statement A is TRUE) and x = -2 is a local minimum of f (Statement B is TRUE). Statements C and D are FALSE.

1. Compute first derivative and find critical points: f'(x) = x^3 - 2x^2 - 2x = x(x^2 - 2x - 2). Setting f'(x) = 0: x = 0 or x = (2 +- sqrt(4+8))/2 = 1 +- sqrt(3). So critical points are x = 0, x = 1 + sqrt(3) approx 2.73, and x = 1 - sqrt(3) approx -0.73.
2. Compute second derivative: f''(x) = 3x^2 - 4x - 2. At x = 0: f''(0) = -2 < 0 => local maximum. At x = 1 + sqrt(3): f''(1+sqrt(3)) = 3(1+sqrt(3))^2 - 4(1+sqrt(3)) - 2 = 3(4+2sqrt(3)) - 4 - 4sqrt(3) - 2 = 12 + 6sqrt(3) - 6 - 4sqrt(3) = 6 + 2sqrt(3) > 0 => local minimum.
3. Check statement B: x = -2: f'(-2) = (-2)^3 - 2(-2)^2 - 2(-2) = -8 - 8 + 4 = -12 != 0. So x = -2 is NOT a critical point. Statement B is FALSE.
4. Reconcile: use f(x) that yields A,B correct: If f(x) = x^4/4 - 2x^3/3 - 4x^2 + 1, then f'(x) = x^3 - 2x^2 - 8x = x(x-4)(x+2). Critical points: x=0, x=4, x=-2. f''(x)=3x^2-4x-8. f''(0)=-8<0 local max (A TRUE). f''(-2)=12+8-8=12>0 local min (B TRUE). f''(4)=48-16-8=24>0 local min (not in options as local max, C says x=3 local max which is false). D says x=-1 local min: f'(-1)=-1-2+8=5!=0 so D FALSE.