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Sample Suriname Eindexamen VWO Practice Questions
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1Differentiate f(x) = ln(x² + 4) over the real numbers.
A.1/(x² + 4)
B.2x/(x² + 4)
C.2x ln(x² + 4)
D.2/(x² + 4)
Explanation: Use the chain rule d(ln u)/dx = u′/u. With u = x² + 4 and u′ = 2x, the derivative is 2x/(x² + 4); the logarithm's argument is positive for every real x.
2What is the derivative of f(x) = x²e^(3x)?
A.2xe^(3x)
B.6xe^(3x)
C.3x²e^(3x)
D.xe^(3x)(2 + 3x)
Explanation: The product rule gives f′(x) = 2xe^(3x) + x²·3e^(3x). Factoring yields xe^(3x)(2 + 3x), with both factors' derivatives contributing.
3For x ≠ 3, differentiate f(x) = (2x + 1)/(x - 3).
A.-7/(x - 3)²
B.2/(x - 3)²
C.(4x - 5)/(x - 3)²
D.-5/(x - 3)²
Explanation: The quotient-rule numerator is 2(x - 3) - (2x + 1) = -7. Dividing by the squared denominator gives f′(x) = -7/(x - 3)².
4Evaluate the indefinite integral ∫2x(x² + 1)³ dx.
A.(x² + 1)⁴ + C
B.2(x² + 1)⁴/3 + C
C.(x² + 1)⁴/4 + C
D.x²(x² + 1)⁴/4 + C
Explanation: Set u = x² + 1, so du = 2x dx. Integrating u³ gives u⁴/4 + C, hence (x² + 1)⁴/4 + C. Differentiating this expression recovers the original integrand.
5Evaluate ∫cos x dx from x = 0 to x = π/2, with angles in radians.
A.0
B.1
C.π/2
D.-1
Explanation: An antiderivative of cos x is sin x. The definite integral is sin(π/2) - sin 0 = 1 - 0 = 1.
6Which point is the inflection point of f(x) = x³ - 6x² + 9x + 2?
A.(2, 4)
B.(1, 6)
C.(3, 2)
D.(0, 2)
Explanation: The derivatives are f′(x) = 3x² - 12x + 9 and f″(x) = 6x - 12. The second derivative changes from negative to positive at x = 2, confirming a change in concavity. Since f(2) = 4, the inflection point is (2, 4).
7Find the limit as x tends to positive infinity of (5x³ - 2x + 1)/(2x³ + 7x²).
A.Positive infinity
B.0
C.-2/7
D.5/2
Explanation: Divide numerator and denominator by x³. The terms involving 1/x vanish in the limit, leaving 5/2 as the ratio of leading coefficients.
8Calculate the scalar dot product of u = (2, -1, 3) and v = (1, 4, 2).
A.12
B.-4
C.4
D.√14
Explanation: Multiply corresponding components and add the products. This gives u·v = 2×1 + (-1)×4 + 3×2 = 2 - 4 + 6 = 4.
9A plane passes through (1, 2, 4) and has normal vector (2, -3, 1). Which equation describes it?
A.2x - 3y + z = 0
B.2x - 3y + z = 8
C.2x + 3y + z = 12
D.x - 2y + 4z = 0
Explanation: The point-normal form is 2(x - 1) - 3(y - 2) + (z - 4) = 0. Its constants cancel, leaving 2x - 3y + z = 0, whose coefficients form the required normal vector.
10A model gives an investment's value as A(t) = 5,000e^(0.05t) SRD, with t in years. Using e^0.20 = 1.2214, what is its modelled value at t = 4?
A.SRD 6,000
B.SRD 6,077.53
C.SRD 7,500
D.SRD 6,107
Explanation: Substituting t = 4 gives exponent 0.05 × 4 = 0.20. Thus A(4) = 5,000 × 1.2214 = SRD 6,107 under the supplied continuous-growth model.
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