National Postgraduate Entrance Exam (考研) — 0828 Agricultural Engineering · China · Ministry of Education / NEEA and universities
Original practice questions written to the published syllabus · not past papersThe algebra, calculus, differential equations and numerical methods used in every numerical question of the degree. GATE AG gives it a fixed block of marks every year.
In the exams: GATE AG Section 1, about 12–13 marks a year (2019–26); ICAR AIEEA-PG Unit I; NCEES FE 8–12 questions; allied subject in PRC ABE
China national postgraduate entrance exam (0828 Agricultural Engineering, mathematics paper option)
Where the national mathematics paper is taken, it goes deeper into proofs and theory: convergence of series, multivariable extrema with constraints, line and surface integrals with the Green, Stokes and Gauss theorems, and linear algebra including diagonalisation.
How our questions for National Postgraduate Entrance Exam spread across the units of Engineering Mathematics. Difficulty: 0 easy, 10 medium, 12 hard.
| Eigenvalues, eigenvectors and the Cayley-Hamilton theorem | 3 | |
| Maxima and minima of functions of several variables | 2 | |
| Sequences, series and convergence | 2 | |
| Taylor, Maclaurin and Fourier series | 2 | |
| Line, surface and volume integrals: Green, Stokes and Gauss theorems | 2 | |
| Higher-order linear ordinary differential equations | 2 | |
| Partial differential equations: Laplace, heat and wave equations | 2 | |
| Limits, continuity and partial derivatives | 1 | |
| Euler's theorem on homogeneous functions | 1 | |
| Vector calculus: gradient, divergence and curl | 1 | |
| Laplace transforms | 1 | |
| Numerical solution of equations: bisection and Newton-Raphson | 1 | |
| Numerical integration: trapezoidal and Simpson's rules | 1 | |
| Numerical solution of ODEs: Euler and Runge-Kutta methods | 1 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
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.
Further reading: Kreyszig, Advanced Engineering Mathematics
What is lim((x, y)→(0, 0)) xy/(x² + y²)?
B. The limit does not exist
Along y = mx the expression is m/(1 + m²), which depends on m: 0 along the x-axis, 1/2 along y = x. A limit must be the same along every path, so it does not exist. Answering 0 checks only the axes; 1/2 checks only the line y = x.
Further reading: Kreyszig, Advanced Engineering Mathematics
For f(x, y) = x³ + y³ − 3xy, which statement about its stationary points is correct?
C. (0, 0) is a saddle point and (1, 1) is a local minimum with f = −1
fₓ = 3x² − 3y = 0 and f_y = 3y² − 3x = 0 give (0, 0) and (1, 1). r = 6x, s = −3, t = 6y. At (0, 0): rt − s² = −9 < 0, a saddle. At (1, 1): rt − s² = 36 − 9 = 27 > 0 with r = 6 > 0, a minimum, f = 1 + 1 − 3 = −1. Calling (0, 0) an extremum ignores the sign of rt − s².
Further reading: Kreyszig, Advanced Engineering Mathematics
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.
Further reading: Kreyszig, Advanced Engineering Mathematics
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.
Further reading: Kreyszig, Advanced Engineering Mathematics
A rod of length L = 0.5 m with both ends held at 0 °C has initial temperature u(x, 0) = 50 sin(πx/L) °C and diffusivity α = 1 × 10⁻⁴ m²/s. From the heat equation u_t = α u_xx, what is the mid-point temperature after 600 s?
C. 4.68 °C
The separated solution is u = 50 sin(πx/L) e^(−α(π/L)²t). At the mid-point sin(π/2) = 1, and α π² t/L² = 1e-4 × 9.8696 × 600/0.25 = 2.3687. u = 50e^(−2.3687) = 4.68 °C. Using π instead of π² gives 23.52 °C and L instead of L² gives 15.30 °C; 50 °C ignores the decay.
Further reading: Kreyszig, Advanced Engineering Mathematics
For the wave equation u_tt = c² u_xx on an infinite string with initial displacement f(x) and zero initial velocity, d'Alembert's solution is
D. u = [f(x + ct) + f(x − ct)]/2
The initial shape splits into two half-amplitude copies travelling left and right. Without the ½ the solution at t = 0 would be 2f(x), which breaks the initial condition.
Further reading: Kreyszig, Advanced Engineering Mathematics
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.
Further reading: Chapra and Canale, Numerical Methods for Engineers
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