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Free Practice Questions for CAPESA ISE — Option Économie
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Sample CAPESA ISE — Option Économie Practice Questions
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1What is the limit of the sequence u_n = (1 + 2/n)^n as n tends to +infinity?
A.e^2
B.2e
C.e
D.1
Explanation: Taking logarithms gives ln(u_n) = n ln(1 + 2/n). Since ln(1 + t) = t + o(t) near zero, ln(u_n) tends to 2. Continuity of the exponential function then gives u_n tending to e^2.
2Let f: [a, b] -> R be a continuous function such that f(a) * f(b) < 0. Which fundamental theorem guarantees the existence of at least one c in (a, b) such that f(c) = 0?
A.Mean Value Theorem (Théorème des accroissements finis)
B.Rolle's Theorem
C.Intermediate Value Theorem (Théorème des valeurs intermédiaires)
D.Extreme Value Theorem (Théorème des bornes atteintes)
Explanation: The Intermediate Value Theorem states that if f is continuous on [a, b], it attains every value between f(a) and f(b). Since f(a) and f(b) have opposite signs, 0 lies strictly between them, guaranteeing some c in (a, b) where f(c) = 0.
3For the multivariable function f(x, y) = 2x^2 y + 3xy^3 - 5x + 7, what is the gradient vector grad f(1, 1)?
A.(7, 11)
B.(2, 5)
C.(4, 9)
D.(2, 11)
Explanation: The partial derivatives are df/dx = 4xy + 3y^3 - 5 and df/dy = 2x^2 + 9xy^2. Evaluating at (1, 1) gives 2 and 11 respectively. Thus grad f(1, 1) = (2, 11).
4What is the Hessian matrix H_f(1, 1) of the function f(x, y) = x^3 - 3xy + y^3?
A.[[3, -3], [-3, 3]]
B.[[6, -3], [-3, 6]]
C.[[6, 3], [3, 6]]
D.[[0, -3], [-3, 0]]
Explanation: The first partial derivatives are 3x^2 - 3y and -3x + 3y^2. Differentiating again gives f_xx = 6x, f_yy = 6y and f_xy = f_yx = -3. Evaluating at (1, 1) yields [[6, -3], [-3, 6]].
5Maximize f(x, y) = xy subject to x + 2y = 20 and x, y > 0. What is the maximum value?
A.25
B.40
C.50
D.100
Explanation: Substitute x = 20 - 2y, where 0 < y < 10, to obtain f = 20y - 2y^2 = 50 - 2(y - 5)^2. This expression is at most 50, with equality when y = 5 and x = 10. The completed square establishes a global maximum, rather than merely a stationary point.
6For f(x, y) = x^2 - y^2 + 4xy, what is the nature of the critical point at (0, 0)?
A.Strict local minimum
B.Strict local maximum
C.Non-isolated local minimum
D.Saddle point (point-selle)
Explanation: The first partial derivatives vanish at the origin, and the Hessian determinant is 2(-2) - 4^2 = -20. Its negative sign identifies a saddle point. Directly, f(x, 0) = x^2 is positive off the origin while f(0, y) = -y^2 is negative, so both signs occur arbitrarily nearby.
7A model variable satisfies y' + 2y = 10 and y(0) = 8. Which function meets both the differential equation and the initial condition?
A.8e^{-2x}
B.5 + 3e^{-2x}
C.5 + 3e^{2x}
D.3 + 5e^{-2x}
Explanation: The constant particular solution is 5, and the homogeneous solution is C e^{-2x}. The initial condition requires 5 + C = 8, so C = 3. Substitution gives y' + 2y = -6e^{-2x} + 10 + 6e^{-2x} = 10.
8For y'' - 6y' + 9y = 0, which expression gives the general real solution?
A.C_1 e^{3x} + C_2 e^{-3x}
B.C_1 e^{3x} + C_2 e^{9x}
C.(C_1 + C_2 x)e^{3x}
D.C_1 cos(3x) + C_2 sin(3x)
Explanation: The characteristic polynomial is r^2 - 6r + 9 = (r - 3)^2. A repeated root requires the independent solutions e^{3x} and x e^{3x}. Their linear combinations give (C_1 + C_2 x)e^{3x}.
9Let u_0 = 5 and u_{n+1} = (u_n + 6)/4. Define v_n = u_n - 2. Which statement correctly gives the sum of the series sum_{n=0}^infinity v_n?
A.The sum is 4 because v_n = 3(1/4)^n
B.The sum is 2 because u_n tends to 2
C.The sum is 20/3 because the initial term is 5
D.The series diverges because u_n tends to 2
Explanation: The fixed point of the recurrence is 2, so v_{n+1} = v_n/4. Since v_0 = 3, induction gives v_n = 3(1/4)^n. The centered series is geometric and sums to 3/(1 - 1/4) = 4, even though the uncentered series of u_n would diverge.
10Consider I = integral from 0 to 1 of x ln(x) dx, interpreted as an improper integral at zero. Which conclusion is correct?
A.I diverges because ln(x) tends to minus infinity
B.I = -1/2
C.I = -1/4
D.I = 1/4
Explanation: Integration by parts on [epsilon, 1] gives the antiderivative (x^2/2)ln(x) - x^2/4. As epsilon tends to zero, epsilon^2 ln(epsilon) and epsilon^2 both tend to zero. The integral converges and equals -1/4, consistent with a nonpositive integrand on (0, 1].
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