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100+ Free Concours INP-HB CPGE Practice Questions

Prepare for the Concours d'Entrée aux Classes Préparatoires aux Grandes Écoles (CPGE) — INP-HB Yamoussoukro exam with instant access — no signup required.

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Key Facts: Concours INP-HB CPGE Exam

19,000 FCFA

Official one-track application fee via TrésorPay/TrésorMoney

INP-HB Yamoussoukro

EPGE INP-HB

Premier West African preparatory center for top Grandes Écoles

INP-HB

Yamoussoukro

Main campus location of INP-HB in Côte d'Ivoire

INP-HB Official

Free 100-question English MCQ study bank for Concours INP-HB CPGE. Official exam: French competitive written papers in advanced maths, physics, chemistry, and culture; selective ranking admission. Not an official-format simulation.

Sample Concours INP-HB CPGE Practice Questions

Try these sample questions to test your Concours INP-HB CPGE exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1What is the limit of the function f(x) = (ln(cos x) + x^2 / 2) / x^4 as x approaches 0?
A.-1/12
B.1/12
C.-1/24
D.0
Explanation: Using the Maclaurin expansion of cos(x) up to order 4: cos(x) = 1 - x^2 / 2 + x^4 / 24 + o(x^4). Let u = -x^2 / 2 + x^4 / 24 + o(x^4). Then ln(cos x) = ln(1 + u) = u - u^2 / 2 + o(u^2) = (-x^2 / 2 + x^4 / 24) - 1/2 (-x^2 / 2)^2 + o(x^4) = -x^2 / 2 + x^4 / 24 - x^4 / 8 + o(x^4) = -x^2 / 2 - x^4 / 12 + o(x^4). Therefore, ln(cos x) + x^2 / 2 = -x^4 / 12 + o(x^4), and dividing by x^4 gives the limit -1/12.
2Evaluate the definite integral I = ∫_0^1 (ln(1 + x) / (1 + x^2)) dx.
A.(π / 8) ln 2
B.(π / 4) ln 2
C.(π / 8) ln(1/2)
D.ln 2 / 4
Explanation: Substitute x = tan θ, so dx = (1 + tan^2 θ) dθ = dθ / cos^2 θ. The bounds become 0 to π/4. Then I = ∫_0^(π/4) ln(1 + tan θ) dθ. Using the symmetry substitution θ = π/4 - u: I = ∫_0^(π/4) ln(1 + tan(π/4 - u)) du = ∫_0^(π/4) ln(1 + (1 - tan u)/(1 + tan u)) du = ∫_0^(π/4) ln(2 / (1 + tan u)) du = ∫_0^(π/4) (ln 2 - ln(1 + tan u)) du = (π/4) ln 2 - I. Thus 2I = (π/4) ln 2, yielding I = (π / 8) ln 2.
3Find the general solution to the second-order linear differential equation y'' - 4y' + 4y = e^(2x).
A.y(x) = (C_1 + C_2 x + (1/2)x^2) e^(2x)
B.y(x) = (C_1 + C_2 x + x^2) e^(2x)
C.y(x) = (C_1 e^(2x) + C_2 e^(-2x)) + (1/2) x^2 e^(2x)
D.y(x) = (C_1 + C_2 x) e^(2x) + e^(2x)
Explanation: The characteristic equation is r^2 - 4r + 4 = (r - 2)^2 = 0, giving a double root r = 2. The homogeneous solution is y_h(x) = (C_1 + C_2 x) e^(2x). Since 2 is a root of multiplicity 2 of the characteristic polynomial, the particular solution takes the form y_p(x) = A x^2 e^(2x). Differentiating: y_p' = A(2x + 2x^2)e^(2x), y_p'' = A(2 + 8x + 4x^2)e^(2x). Substituting into y'' - 4y' + 4y: A[(2 + 8x + 4x^2) - 4(2x + 2x^2) + 4(x^2)]e^(2x) = 2A e^(2x) = e^(2x), so A = 1/2. Thus y(x) = (C_1 + C_2 x + (1/2)x^2) e^(2x).
4Consider the real matrix A = [[2, 1, 0], [0, 2, 0], [0, 0, 3]]. Which of the following statements is TRUE regarding matrix A?
A.A is not diagonalizable because the geometric multiplicity of eigenvalue 2 is strictly less than its algebraic multiplicity.
B.A is diagonalizable because it has three distinct eigenvalues.
C.A is diagonalizable because its minimal polynomial has simple roots.
D.A is invertible and therefore must be diagonalizable.
Explanation: The characteristic polynomial of A is χ_A(λ) = (λ - 2)^2 (λ - 3). The eigenvalue λ = 2 has algebraic multiplicity 2. The eigenspace E_2 is the kernel of A - 2I = [[0, 1, 0], [0, 0, 0], [0, 0, 1]]. Any vector v = (x, y, z)^T in E_2 satisfies y = 0 and z = 0, so E_2 = span{(1, 0, 0)^T}, which has dimension 1. Because the geometric multiplicity (dim E_2 = 1) is strictly less than the algebraic multiplicity (2), A is not diagonalizable over R or C.
5Let (u_n) be the sequence defined by u_0 = 0 and u_(n+1) = sqrt(6 + u_n) for all n >= 0. What is the limit of (u_n) as n -> infinity?
A.3
B.sqrt(6)
C.2
D.The sequence diverges to +infinity
Explanation: Consider f(x) = sqrt(6 + x) on [0, +infinity). f is continuous and strictly increasing. The fixed-point equation L = sqrt(6 + L) gives L^2 - L - 6 = 0, so (L - 3)(L + 2) = 0. Since L >= 0, the unique non-negative fixed point is L = 3. By mathematical induction, 0 <= u_n < u_(n+1) < 3 for all n >= 0: u_0 = 0 < u_1 = sqrt(6) < 3, and if 0 <= u_n < 3, then sqrt(6) <= u_(n+1) = sqrt(6 + u_n) < sqrt(6 + 3) = 3. Since (u_n) is strictly increasing and bounded above by 3, it converges to the fixed point L = 3.
6In the complex plane, let A, B, and C be three distinct points with respective affixes z_A, z_B, and z_C. If (z_C - z_A) / (z_B - z_A) = e^(i π / 3), what is the geometric nature of triangle ABC?
A.Direct equilateral triangle
B.Right isosceles triangle at A
C.Indirect equilateral triangle
D.Scalene acute triangle
Explanation: The modulus |(z_C - z_A) / (z_B - z_A)| = |e^(i π / 3)| = 1 implies AC / AB = 1, so AB = AC. The argument arg((z_C - z_A) / (z_B - z_A)) = π / 3 (mod 2π) represents the directed angle (vec(AB), vec(AC)) = +π/3 (+60°). An isosceles triangle with a vertex angle of π/3 is an equilateral triangle, and because the angle is +π/3, it is directly oriented (counter-clockwise).
7What is the determinant of the 3x3 Vandermonde matrix V(a, b, c) = [[1, a, a^2], [1, b, b^2], [1, c, c^2]]?
A.(b - a)(c - a)(c - b)
B.(a - b)(b - c)(c - a)
C.(a + b)(b + c)(c + a)
D.a b c (a - b)(b - c)
Explanation: Subtracting row 1 from rows 2 and 3 yields: det V = [[1, a, a^2], [0, b - a, b^2 - a^2], [0, c - a, c^2 - a^2]]. Expanding along the first column: det V = (b - a)(c^2 - a^2) - (c - a)(b^2 - a^2) = (b - a)(c - a)[(c + a) - (b + a)] = (b - a)(c - a)(c - b).
8Evaluate the integral J = ∫_0^(π/2) x cos(x) dx using integration by parts.
A.π/2 - 1
B.π/2 + 1
C.π - 1
D.1 - π/2
Explanation: Let u = x and v' = cos(x). Then u' = 1 and v = sin(x). Applying integration by parts: ∫_0^(π/2) x cos(x) dx = [x sin(x)]_0^(π/2) - ∫_0^(π/2) sin(x) dx = ((π/2) sin(π/2) - 0) - [-cos(x)]_0^(π/2) = (π/2)(1) - (-cos(π/2) - (-cos(0))) = π/2 - (0 + 1) = π/2 - 1.
9An industrial quality control test in Yamoussoukro detects defective components with a sensitivity of 98% (P(T+|D) = 0.98) and a specificity of 95% (P(T-|D^c) = 0.95). In a production lot, 2% of components are defective (P(D) = 0.02). If a randomly selected component tests positive (T+), what is the probability that it is truly defective?
A.2/7 (~28.6%)
B.98/100 (98%)
C.1/2 (50%)
D.14/19 (~73.7%)
Explanation: By Bayes' theorem: P(D|T+) = (P(T+|D) * P(D)) / P(T+). The total probability of a positive test is P(T+) = P(T+|D)P(D) + P(T+|D^c)P(D^c). Given P(T+|D^c) = 1 - 0.95 = 0.05 and P(D^c) = 0.98: P(T+) = (0.98)(0.02) + (0.05)(0.98) = 0.0196 + 0.0490 = 0.0686. Thus P(D|T+) = 0.0196 / 0.0686 = 196 / 686 = 2 / 7 ≈ 0.2857 (28.6%).
10Let f: E -> F be a linear map between two finite-dimensional real vector spaces. According to the Rank-Nullity Theorem (Théorème du rang), which equality holds?
A.dim(E) = dim(ker(f)) + dim(im(f))
B.dim(F) = dim(ker(f)) + dim(im(f))
C.dim(E) = dim(ker(f)) - dim(im(f))
D.dim(im(f)) = dim(E) + dim(F)
Explanation: The Rank-Nullity Theorem states that for any linear map f defined on a finite-dimensional vector space E, the dimension of the domain E equals the dimension of the kernel (nullity) plus the dimension of the image (rank): dim(E) = dim(ker(f)) + dim(im(f)) = dim(ker(f)) + rg(f).

About the Concours INP-HB CPGE Exam

The Concours d'Entrée aux Classes Préparatoires aux Grandes Écoles (CPGE) of the Institut National Polytechnique Félix Houphouët-Boigny (INP-HB) in Yamoussoukro is West Africa's premier engineering and management preparatory entrance competition. Organized by the École Préparatoire aux Grandes Écoles (EPGE), this two-year intensive program prepares top bachelors from Côte d'Ivoire and the sub-region for national and international engineering and business Grandes Écoles (such as École Polytechnique de Paris, Mines, Centrale, ENSAE, HEC, and INP-HB's own engineering schools ESMG, ESTP, ESI, ESCAE, ESA). This practice bank offers 100 English-language MCQs testing high-level mathematical proofs, physics mechanisms, chemical calculations, and general culture.

Assessment

Two-phase competitive selection: Phase 1: Academic transcript eligibility screening (bulletins de 2nde, 1ère, Tle and Bac results); Phase 2: written competitive papers for the candidates declared admissible. The EPGE preparatory tracks and the BAC series each admits are BCPST (C, D), ECG (A1, B, C, D), PCSI leading to PSI or PC (C, D, E), MPSI leading to MP (C, E), MP2I leading to MPI or MP (C, E), and N2SA (A1, A2, C, D).

Time Limit

Two-phase competition run by the Bureau Central des Concours: online pre-registration and physical file submission in July, admissibility results in early August, and the written admission papers in August.

Passing Score

Rank-ordered admission within available intake quota (concours direct). Final placement depends strictly on cumulative written score and ranking.

Exam Fee

19,000 FCFA for one track (15,000 FCFA registration + 3,000 FCFA dossier + 1,000 FCFA photo via TrésorPay/TrésorMoney; 34,000 FCFA for two tracks) (Institut National Polytechnique Félix Houphouët-Boigny (INP-HB) — École Préparatoire aux Grandes Écoles (EPGE))

Concours INP-HB CPGE Exam Content Outline

Study emphasis ~35%

Advanced Mathematics

Functions, limits, Taylor approximations, integration methods, differential equations, sequences, complex geometry, matrices, systems of equations, combinatorics, and probability.

Study emphasis ~25%

Physics & Mechanics

Rigid body mechanics, Newton's laws, energy methods, harmonic oscillators, resonance, electromagnetic induction, RLC circuits, and wave optics.

Study emphasis ~15%

Chemistry & Thermodynamics

Chemical kinetics, reaction rate laws, pH equilibria, solubility products, redox titrations, and organic functional groups.

Study emphasis ~15%

General Culture & Epistemology

History of scientific ideas, philosophical reasoning, African developmental challenges, and ethical dimensions of technology.

Study emphasis ~10%

Scientific English

Technical reading comprehension, academic syntax, vocabulary in STEM contexts, and logical discourse analysis.

How to Pass the Concours INP-HB CPGE Exam

What You Need to Know

  • Passing score: Rank-ordered admission within available intake quota (concours direct). Final placement depends strictly on cumulative written score and ranking.
  • Assessment: Two-phase competitive selection: Phase 1: Academic transcript eligibility screening (bulletins de 2nde, 1ère, Tle and Bac results); Phase 2: written competitive papers for the candidates declared admissible. The EPGE preparatory tracks and the BAC series each admits are BCPST (C, D), ECG (A1, B, C, D), PCSI leading to PSI or PC (C, D, E), MPSI leading to MP (C, E), MP2I leading to MPI or MP (C, E), and N2SA (A1, A2, C, D).
  • Time limit: Two-phase competition run by the Bureau Central des Concours: online pre-registration and physical file submission in July, admissibility results in early August, and the written admission papers in August.
  • Exam fee: 19,000 FCFA for one track (15,000 FCFA registration + 3,000 FCFA dossier + 1,000 FCFA photo via TrésorPay/TrésorMoney; 34,000 FCFA for two tracks)

Keys to Passing

  • Work through all 100 available questions
  • Review every answer and explanation
  • Track weak areas and revisit them
  • Use our AI tutor for tough concepts

Concours INP-HB CPGE Study Tips from Top Performers

1Work beyond standard Terminale exercises to master complex multi-step mathematical problems and formal derivations.
2Review mechanics thoroughly: free-body diagrams, conservation of mechanical energy, and central force orbits.
3Understand chemical equilibrium: calculate reaction quotients Qr, equilibrium constants K, and pH of polyprotic acid mixtures.
4Develop broad general culture on major scientific milestones, energy transitions, and African industrial development.

Frequently Asked Questions

Is the INP-HB CPGE entrance concours multiple-choice?

No. The official competition consists of rigorous open-ended written problem sets and essays in French. This bank is an English MCQ study adaptation.

Which streams are available in CPGE at INP-HB Yamoussoukro?

The EPGE at INP-HB offers MPSI/MP (Maths-Physics), PCSI/PSI (Physics-Engineering), BCPST (Biology-Chemistry-Physics-Earth Sciences), and ECG (Economics and Commercial General track).

What is the fee for sitting the INP-HB CPGE concours in Côte d'Ivoire?

The official cost is 19,000 FCFA for one competition track (comprising 15,000 FCFA registration, 3,000 FCFA dossier fee, and 1,000 FCFA photo fee), paid via TrésorPay/TrésorMoney.