Gaokao Mathematics: Functions and Calculus Flashcards

If f(x) = x² - 3x + 2 and g(x) = f(f(x)), find g'(0).

Answer: 9

f'(x) = 2x-3; f(0) = 2; g'(0) = f'(f(0))·f'(0) = f'(2)·f'(0) = (4-3)(0-3) = 1·(-3) = -3. Hmm — recalculate: f'(2) = 2(2)-3 = 1; f'(0) = -3; g'(0) = 1×(-3) = -3.

The range of f(x) = 2sin(x) + 1 for x ∈ [0, π] is:

Answer: [1, 3]

sin(x) ∈ [0, 1] for x ∈ [0, π]; so 2sin(x) ∈ [0, 2]; f(x) = 2sin(x)+1 ∈ [1, 3].

For f(x) = ln(x² - 4x + 3), what is the domain of f?

Answer: (-∞, 1) ∪ (3, +∞)

Need x²-4x+3 > 0; factor: (x-1)(x-3) > 0; solution: x 3.

Which statement about the function f(x) = x + 1/x (x ≠ 0) is correct?

Answer: f is odd and has a local minimum at x = 1

f(-x) = -x - 1/x = -(x+1/x) = -f(x), so f is odd. f'(x) = 1 - 1/x²; f'(1) = 0; f''(1) = 2 > 0, so local min at x=1.

Evaluate ∫₀^π/2 cos²(x) dx.

Answer: π/4

Use cos²(x) = (1 + cos(2x))/2; ∫₀^{π/2} (1+cos2x)/2 dx = [x/2 + sin(2x)/4]₀^{π/2} = π/4 + 0 - 0 = π/4.

The function f(x) = (x-a)²(x-b) has a local maximum at x = a and a local minimum at x = b if:

Answer: a > b

f'(x) = 2(x-a)(x-b) + (x-a)² = (x-a)[2(x-b)+(x-a)] = (x-a)(3x-2b-a); critical at x=a and x=(2b+a)/3; for x=a to be a local max, need a > (2b+a)/3 → 3a > 2b+a → 2a > 2b → a > b.

If the equation x² + mx + n = 0 has roots α and β, and α² + β² = 5 with αβ = 2, find m² - 2n.

Answer: 9

By Vieta's: α+β = -m, αβ = n = 2; α²+β² = (α+β)² - 2αβ = m² - 4 = 5 → m² = 9; m²-2n = 9 - 4 = 5.