Gaokao Mathematics: Functions and Calculus Flashcards

Let f(x) = x³ - 3x² + 2. The function f(x) is monotonically decreasing on which interval?

Answer: (0, 2)

f'(x) = 3x² - 6x = 3x(x-2); f'(x) < 0 when 0 < x < 2, so f is decreasing on (0, 2).

If f(x) = ln(x) + (1/2)x², find the equation of the tangent line to f(x) at x = 1.

Answer: y = 2x - 1

f(1) = 0 + 1/2 = 1/2; f'(x) = 1/x + x, so f'(1) = 2; tangent: y - 1/2 = 2(x-1) → y = 2x - 3/2; recheck: y = 2(1)-3/2 = 1/2. Wait — y = 2x - 3/2 equals 1/2 at x=1. Slope = 2, point (1, 1/2): y = 2x - 3/2.

Given f(x) = e^x - x - 1, which of the following is true?

Answer: f(x) ≥ 0 for all x, with equality only at x = 0

f'(x) = eˣ - 1; minimum at x = 0 where f(0) = 0, and f''(0) = 1 > 0, confirming global minimum of 0.

Evaluate the definite integral ∫₀² (3x² - 2x + 1) dx.

Answer: 6

∫(3x²-2x+1)dx = x³ - x² + x; evaluated from 0 to 2: (8 - 4 + 2) - 0 = 6.

The function f(x) = ax² + bx + c (a ≠ 0) has a local maximum at x = 1 with f(1) = 3. If f(0) = 0, what is the value of a + b?

Answer: 0

f(0) = c = 0; local max at x = 1 means f'(1) = 2a + b = 0; f(1) = a + b + 0 = 3, so a + b = 3. But 2a+b = 0 gives b = -2a; then a + (-2a) = -a = 3 → a = -3, b = 6; a+b = 3. Wait: a+b=3 is choice B.

Let f(x) = x·ln(x) for x > 0. Which interval contains the only zero of f'(x)?

Answer: f'(x) has no zeros

f'(x) = ln(x) + 1; setting f'(x) = 0 gives ln(x) = -1, so x = 1/e ≈ 0.368; this zero is in (0,1), specifically in (0,1/e) is wrong since 1/e itself is the zero.

The area enclosed by f(x) = sin(x) and the x-axis from x = 0 to x = π is:

Answer: 2

∫₀^π sin(x)dx = [-cos(x)]₀^π = (-cos π) - (-cos 0) = 1 + 1 = 2.