Gaokao Mathematics: Functions and Calculus Flashcards

Let h(x) = f(x)·g(x) where f(x) = x² and g(x) = e^x. Find h''(0).

Answer: 2

h'(x) = 2x·eˣ + x²·eˣ = eˣ(2x+x²); h''(x) = eˣ(2x+x²) + eˣ(2+2x) = eˣ(x²+4x+2); h''(0) = 1·2 = 2.

The function f(x) = x³ - 3x has how many real roots for the equation f(x) = 2?

Answer: 1

x³ - 3x - 2 = 0; factor: (x+1)²(x-2) = 0; roots x = -1 (double) and x = 2; distinct real roots: x = -1 and x = 2, giving 2 distinct values.

Using the substitution u = x², evaluate ∫₀² x·e^(x²) dx.

Answer: (e⁴-1)/2

Let u = x², du = 2x dx; integral becomes (1/2)∫₀⁴ eᵘ du = (1/2)[eᵘ]₀⁴ = (e⁴-1)/2.

For f(x) = arctan(x), which of the following is true?

Answer: f is an odd function with horizontal asymptotes at y = ±π/2

arctan(-x) = -arctan(x) (odd); lim_{x→±∞} arctan(x) = ±π/2 (horizontal asymptotes); f'(x) = 1/(1+x²) > 0 so f is increasing.

Let f(x) = {x² + 1, x ≤ 0; e^x, x > 0}. Find f(f(-1)).

Answer: e²

f(-1) = (-1)² + 1 = 2 (since -1 ≤ 0); f(2) = e² (since 2 > 0); so f(f(-1)) = e².

If f(x) = x³ - 3x, the maximum value of f on [-2, 3] is:

Answer: 18

f'(x) = 3x²-3 = 3(x²-1) = 0 at x = ±1; f(-2) = -8+6 = -2; f(-1) = -1+3 = 2; f(1) = 1-3 = -2; f(3) = 27-9 = 18; maximum = 18.

The area bounded by y = x² and y = x is:

Answer: 1/6

Intersection at x=0 and x=1; area = ∫₀¹ (x - x²)dx = [x²/2 - x³/3]₀¹ = 1/2 - 1/3 = 1/6.