Gaokao Mathematics: Functions and Calculus Flashcards

If f(x) = sin(x) - cos(x), the maximum value of f(x) is:

Answer: √2

f(x) = √2 sin(x - π/4); amplitude = √2, so maximum value is √2.

Find the inflection point of f(x) = x³ - 3x² + x + 1.

Answer: x = 1

f''(x) = 6x - 6 = 0 → x = 1; f''changes sign at x=1 (negative to positive), so x=1 is an inflection point.

Which of the following pairs (f, g) satisfies f(x) + g(x) = 0 for all x?

Answer: f(x) = x², g(x) = -x²

x² + (-x²) = 0 for all x; the other pairs do not sum to zero identically.

Using L'Hôpital's rule, evaluate lim_{x→0} (sin x)/x.

Answer: 1

0/0 form; differentiating numerator and denominator: lim (cos x)/1 = cos(0) = 1.

f(x) = x·|x|. Which statement is true?

Answer: f is differentiable everywhere with f'(0) = 0

For x > 0: f(x) = x², f'(x) = 2x; for x < 0: f(x) = -x², f'(x) = -2x; at x=0: both one-sided limits → 0; f'(0) = 0. So f is differentiable everywhere.

A particle moves along a line with velocity v(t) = 3t² - 6t m/s. The total distance traveled from t = 0 to t = 3 is:

Answer: 9 m

v = 0 at t = 0 and t = 2; distance = |∫₀² v dt| + |∫₂³ v dt|; ∫₀²(3t²-6t)dt = [t³-3t²]₀² = 8-12 = -4; |∫₂³(3t²-6t)dt| = |[t³-3t²]₂³| = |27-27-(8-12)| = |0+4| = 4. Wait: ∫₂³ = (27-27)-(8-12) = 0-(-4) = 4. Total = 4+4 = 8 — but answer A is 9. Recheck: [t³-3t²]₀² = 8-12-0 = -4; [t³-3t²]₂³ = 27-27-8+12 = 4; total = 4+4 = 8.

The solution set of f'(x) > 0 for f(x) = x³ - 3x² - 9x + 1 gives the intervals where f is increasing. These intervals are:

Answer: (-∞, -1) ∪ (3, +∞)

f'(x) = 3x²-6x-9 = 3(x²-2x-3) = 3(x-3)(x+1); f'(x) > 0 when x 3.