China · Ministry of Education / NEEA and universities · pg · secondary source
Politics 100 + foreign language 100 + two subject papers of 150 each (a national mathematics paper or a specialist foundation paper).
0828 农业工程 is a first-level discipline in the 2022 catalogue (verified on the MOE page). The general format is from a secondary source; which mathematics paper 0828 uses and the specialist papers are set by each university and are unverified.
Official source: https://www.moe.gov.cn/srcsite/A22/moe_833/202209/t20220914_660828.html
Original questions written to this exam's published syllabus and pattern. Not past papers.
1. By the Cayley-Hamilton theorem, the matrix A = [[1, 2], [3, 4]] satisfies A² = ?
D. 5A + 2I
Characteristic polynomial: λ² − (trace)λ + det = λ² − 5λ − 2 (det = 4 − 6 = −2). Cayley-Hamilton says A² − 5A − 2I = 0, so A² = 5A + 2I. Direct check: A² = [[7, 10], [15, 22]] and 5A + 2I = [[7, 10], [15, 22]]. 5A − 2I comes from writing det as +2 instead of −2.
Engineering Mathematics · Eigenvalues, eigenvectors and the Cayley-Hamilton theorem
2. The series Σ n²/2ⁿ (n = 1 to ∞) converges by the ratio test. What is its sum?
D. 6
Ratio test: a(n+1)/a(n) = [(n + 1)²/n²]/2 → 1/2 < 1, so it converges. Using Σ n xⁿ = x/(1 − x)² and Σ n² xⁿ = x(1 + x)/(1 − x)³ with x = 1/2: (1/2)(3/2)/(1/8) = 6. 2 is Σ n/2ⁿ (the first-power sum); 4 comes from dropping the (1 + x) factor.
Engineering Mathematics · Sequences, series and convergence
3. What is a particular integral of y'' + y = sin x?
B. −(x/2)cos x
sin x already solves the homogeneous equation (resonance), so the trial must be multiplied by x. The standard result is PI = −(x/2)cos x; check: y'' + y = sin x. −sin x/2 uses the non-resonant formula 1/(1 − a²) wrongly; (x/2)cos x has the wrong sign.
Engineering Mathematics · Higher-order linear ordinary differential equations
4. Explicit Euler is applied to the stiff equation y' = −50y. For the numerical solution not to grow, the step size must satisfy
A. h < 0.04
Euler gives y(n+1) = (1 − 50h)y(n). Stability needs |1 − 50h| < 1, i.e. 0 < h < 2/50 = 0.04. h < 0.02 is the condition for no oscillation (1 − 50h > 0), stricter than stability; h < 2 forgets to divide by λ. h < 0.4 misplaces the decimal in 2/50.
Engineering Mathematics · Numerical solution of ODEs: Euler and Runge-Kutta methods
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