Application of Derivatives — Questions & Answers

Sample questions from Application of Derivatives

1. Q1. The absolute minimum value of f(x) = 3x^4 - 8x^3 + 12x^2 - 48x + 25 on [0, 3] is:

   • A.25
   • B.-16
   • C.-39✓
   • D.16
   Solution

   f'(x) = 12(x - 2)(x^2 + 2), so the only critical point is x = 2. f(0) = 25, f(2) = 48 - 64 + 48 - 96 + 25 = -39, f(3) = 243 - 216 + 108 - 144 + 25 = 16. The least value is -39.

2. Q2. The function f(x) = sin(x) - cos(x) is strictly increasing on:

   • A.(3π/4, 7π/4)
   • B.(π/4, 5π/4)
   • C.(0, π/2) only
   • D.(0, 3π/4)✓
   Solution

   f'(x) = cos x + sin x = 0 at x = 3π/4, 7π/4 in (0, 2π). f' > 0 on (0, 3π/4), so f increases there.

3. Q3. At a point of local minimum, the first derivative f'(x):

   • A.Changes from positive to negative
   • B.Changes from negative to positive✓
   • C.Remains zero everywhere
   • D.Is always positive
   Solution

   At a local minimum, the function changes from decreasing to increasing, so f'(x) changes sign from negative to positive.

4. Q4. The slope of the tangent to the curve y=f(x) at point (a, f(a)) is given by:

   • A.f(a)
   • B.f'(a)✓
   • C.f''(a)
   • D.1/f'(a)
   Solution

   The derivative f'(a) evaluated at x=a gives the slope of the tangent line to the curve at that point.

5. Q5. The maximum value of f(x) = x * sqrt(4 - x^2) on [-2, 2] is:

   • A.1
   • B.sqrt(2)
   • C.4
   • D.2✓
   Solution

   f'(x) = (4 - 2x^2)/sqrt(4 - x^2) = 0 gives x = sqrt(2). f(sqrt(2)) = sqrt(2)*sqrt(2) = 2, larger than f(±2) = 0.

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