Elementary Mathematics II (Calculus) — 2025 Past Questions
Lagos State University MAT102: 90 questions.
Topics: Derivatives as limits of rate of change, differentiation techniques, maxima and minima, Integration, definite integrals, reduction formulae, Derivatives and Differentiation Techniques, Applications to areas, volumes (Trapezium and Simpson's rules), Integration, Limits and Continuity, Functions of a real variable, graphs, limits, and continuity, Maxima and Minima, Curve sketching, integration, definite integrals, reduction formulae, Derivatives and Rates of Change, Definite Integrals, Applications to Areas and Volumes, Functions of a Real Variable
- The rate of change of the area of a circle with respect to its radius is
- Diameter
- Perimeter
- Area × radius
- 2 × diameter
- Let F(x) = ∫₁ˣ (t³ - 4t² + t + 2) dt. Find F′(4).
- 4
- 5
- 6
- 7
- Compute ∫₄⁹ (1/√t) dt.
- 1
- 2
- 3
- 4
- Evaluate limₙ→∞ (10n² - 3n + 7)/(2n + 6)².
- 1
- 1.5
- 2
- 2.5
- Compute [f(2 + h) - f(2)]/h, where f(x) = 2x² - x + 3.
- 7 - h
- 7 + h
- 7 - 2h
- 7 + 2h
- Compute f′(1) where f(x) = 2√(3x² - 4x + 5).
- 2
- -2
- 1
- 3
- Find the average rate of change of g(t) = t² - 3t + 4 on the interval [2,4].
- 4
- 6
- 8
- 10
- Suppose g(4) = 4, g′(4) = -2, and f(x) = g(x)/x. Find f′(4).
- -5/8
- -7/8
- -3/4
- -5/4
- The second derivative of (x + 4)(11x + 1)(3x - 2) is
- 198x + 226
- 226x² + 198x
- 226x + 198
- 198x² - 226x
- If y = f(x), then at turning points
- f(x) = 0
- dy/dx = 0
- dy/dx > 0
- dy/dx = 1
- Which of these is not correct about the first principle?
- dy/dx = [f(x+h) - f(x)]/x
- dy/dx = [f(x-h) + f(x)]/h
- dy/dx = lim_{h→0} {[f(x+h) - f(x)]/h}
- dy/dx = lim_{x→0} {[f(x+h) - f(x)]/x}
- What is the limit of (x³ - 1)/(x² - 1) as x → 1?
- ∞
- -∞
- 3/2
- 1
- Evaluate lim_{x→π/2} (sin x)/(cos x).
- Not possible
- π/2
- 0
- ∞
- If y = 112, then dy/dx is
- 112
- 112y
- 112x
- 211
- If f(x,y) = 1 + x²/y + y³/x, calculate f(20,14).
- 5837/35
- 20378/35
- 697/7
- 10009/5
- Given g(x) = 2 + 3x - 8x² and h(x) = (x - 1)², then g(h(x)) is
- Very difficult
- 3 + 26x - 45x² - 8x⁴
- -3 + 26x - 45x² + 32x³ - 8x⁴
- 3 - 26x - 45x² - 32x³ - 8x⁴
- At minimum points of y = f(x), which of the following is correct?
- d²y/dx² = 1
- d²y/dx² < 0
- d²y/dx² = 0
- d²y/dx² > 0
- The equation y = x³(x - 4) has ______ number of inflexion.
- 1
- 2
- 3
- 4
- If y = θⁿ, where n ∈ ℤ⁺, which of the following is not correct?
- θ²y″ - n²y′ + ny = 0
- θy′ = ny
- θ²y″ = n(n-1)y
- y + y′ = nθⁿ(n⁻¹ + θ⁻¹)
- Differentiate sin²(x² + 1).
- 4x cos²(x² + 1)
- 2 cos(x² + 1)
- 4x sin(x² + 1) cos(x² + 1)
- 2x sin²(x² + 1) cos(x² + 1) + 2x
- The first derivative of cosᵐ ψ with respect to ψ is
- -m cosᵐ⁻¹ ψ sin ψ
- m cosᵐ⁻¹ ψ sinᵐ ψ
- -m sinᵐ⁻¹ ψ cosᵐ ψ
- m sinᵐ ψ
- Find d/dx [arcsin((1 - x²)/(1 + x²))].
- Not possible
- 2(1 + x²)
- (1 - x²)/(1 + x²)
- -2/(1 + x²)
- The demand and cost functions of Cyntax Nig. Ltd are p = 22500 - 3q², C = 5000 + 14400q. The marginal revenue function is
- 22500 - 3q²
- -6q
- 22500² - 3q³
- 22500 - 9q²
- The quantity for maximum revenue is
- 49
- 50
- 51
- 52
- The price for maximum revenue is
- 13000
- 14500
- 15000
- 13500
- The profit function is
- 5000 - 8100q + 3q³
- 8100q - 3q³ + 5000
- -8100q - 5000 + 3q³
- 8100q - 5000 - 3q³
- The marginal profit function is
- 8100 - 9q²
- -8100 + 9q²
- 8100 + 9q²
- -8100 - 9q²
- The maximum profit is
- 175000
- 150070
- 157000
- 751000
- The price for maximum profit is
- 19800
- 8100
- 5000
- 22500
- The average cost function is
- 5000q + 14400q²
- 14400q
- 5000q⁻¹ + 14400q²
- 5000q + 7200q²
- The marginal cost function is
- 5000
- 14400
- q² + 1
- 5000q - q²
- At what quantity is the marginal cost a minimum?
- 0
- Any quantity
- 50
- 28
- Find the equation of the tangent at the point (2,2) on the curve y² = 2x.
- x - 2y + 2 = 0
- y - 2x - 2 = 0
- x - 2y - 2 = 0
- y + 2x - 2 = 0
- The equation of the normal at the point (3, 4/3) on xy = 4 is
- 12x - 27y - 65 = 0
- 27x + 12y + 65 = 0
- 12x + 27y + 65 = 0
- 27x - 12y - 65 = 0
- The minimum value of y = x³ - 3x² - 9x + 27 is
- 54
- 0
- 27
- -27
- The maximum value of y = x³ - 3x² - 9x + 27 is
- 36
- 34
- 32
- 30
- The integral ∫ (5x + 3)^0.5 dx evaluates to
- (5x + 3)^1.5/1.5
- (3/2)(5x + 3)^(-0.5)
- (3/10)(5x + 3)^0.5
- (2/15)(5x + 3)^1.5
- ∫₂^{2√3} dx/(x² + 4) =
- π/4
- 1/2 arctan(1/2 x)
- π/3 - π/4
- π/4 - π/3
- The best substitution to evaluate ∫ x√(a² - x²) dx is
- x = a cos θ
- x = a sin θ
- u = a² - x²
- x = a tan θ
- Evaluate ∫₀² √(x/(4 - x)) dx.
- π + 2
- 2 - π
- π - 2
- 2
- Integrate cos² ax with respect to x.
- cot ax + c
- tan² ax + c
- (1/a)cot ax + c
- (-1/a)cot ax + c
- Evaluate ∫ ln x dx.
- Not Possible
- ln x + c
- ln x - x + c
- x ln x - x + c
- The expression for ∫ x e^{3x} dx is
- (1/3)x e^{3x} - (1/9)e^{3x} + c
- (1/3)e^{3x} - (1/9)x e^{3x} + c
- (1/3)x e^{3x} + (1/9)e^{3x} + c
- (-1/3)e^{3x} + (1/9)x e^{3x} + c
- A rule f which assigns to each element x of a set X a unique element y of a set Y is called
- Equation
- Differential
- Function
- Integration
- If y(x) = u(x)v(x)w(x), then dy/dx is
- uvw + uvw + uvw
- does not exist
- uvw + uvw + uvw
- uvw + uvw
- Given that the acceleration of an electron is y = ∫ e^{cos x} ln x arcsin x dx, then the velocity of this electron is
- e^{sin x} ln x arccos x
- ln x sin x
- e^{cos x} e^x
- e^{cos x} ln x arcsin x
- Evaluate ∫_a^a e^{sin x} ln x arccos x dx.
- 0
- 1
- 2
- 2a
- Evaluate ∫ dx/(x ln x).
- ln x + c
- ln(x ln x) + c
- ln(ln x) + c
- x ln x + c
- Integrate x with respect to x.
- -x + c
- x²/2 + c
- x + c
- 1 + c
- Using the substitution u = t - 1/t, the integral ∫ (t² + 1)/(t⁴ + 1) dt reduces to
- ∫ du/(u² + 1)
- ∫ du/(u² + 2)
- ∫ du/(u² + 3)
- ∫ du/(u² + 4)
- Given that y = ln t² and x = 1/t, find dy/dx.
- −2/t³
- −2t
- 1/t³
- 2t²
- A curve has equation y = x³ + 3x² − 9x + p, where p is a constant. The value of p for which the curve has a stationary point on the x-axis is:
- 4
- 5
- 3
- 2
- Find the second derivative of y = f(θ) = sinθ/(1 + cosθ).
- (1 + cosθ)²/sinθ
- sinθ/(1 + sinθ)²
- sinθ/(1 + cosθ)²
- sinθ/(1 + sinθ)
- For the function f(x) = 4x² − 1 if x < 1, and f(x) = 3x + 2 if x ≥ 1, the limit of f(x) as x → 1⁺ is:
- 3
- does not exist
- 5
- 1
- ∫ (5x² + 6)/(x² + 1) dx =
- cos⁻¹(x) + C
- sin⁻¹(x) + C
- tan⁻¹(x) + C
- 5x + tan⁻¹(x) + C
- A point is moving along the curve y = 4√x + 2/√x in such a way that the x-coordinate is increasing at a constant rate of 0.12 units per second. Find the rate of change of the y-coordinate when x = 4.
- 0.12
- 0.015
- 0.75
- 0.105
- Find dy/dx when xy + sin y = 1.
- −y/(x + sin y)
- y/(x + sin y)
- −y/(x + cos y)
- y/(x + cos y)
- Find the equation of the tangent to the curve x² + y² = 5 at (−1, 2).
- x − 2y + 5 = 0
- 2x − y + 4 = 0
- x − 2y + 4 = 0
- 2x − y + 5 = 0
- ∫ (sin(2x) + cos(x/3)) dx =
- −(1/2)cos(2x) + 3 sin(x/3) + C
- cos(x/3) + sin(2x) + C
- cos(x) + sin(x) + C
- cos(2x) + sin(x/3) + C
- ∫ x²/(x − 1) dx =
- ln(x − 1) + C
- x + ln(x − 1)/(x² + 1) + C
- x²/2 + x + ln(x − 1) + C
- x²/2 + x + C
- ∫ (2 − x)/(1 − x) dx =
- ln(1 + x) + C
- ln(x² − 1) + x + C
- ln(1 − x) + C
- x − ln(1 − x) + C
- What is the domain of the function (x + 3)/(x² − 3x)?
- x ∈ (−∞, 3) ∪ (3, 0) ∪ (0, ∞)
- x ∈ (−∞, −3) ∪ (0, 3) ∪ (3, ∞)
- x ∈ (−∞, ∞)
- x ∈ (−∞, 0) ∪ (0, 3) ∪ (3, ∞)
- If f(x) = 1 + x² for x < 2 and f(x) = x³ for x ≥ 2, then the limit of f(x) as x → 2⁻ is:
- 5
- 7
- 8
- 6
- Evaluate the limit: lim (x → 0) [(x + 1)² − 1]/x
- 0
- 1
- does not exist
- 2
- Find the point of discontinuity of p(x) = 3x/(√x − 4).
- 2
- −2
- 16
- 8
- The equation of a curve is y = x³ + px², where p is a positive constant. The nature of the stationary point at the origin is?
- Increasing point
- Point of inflexion
- Minimum point
- Maximum point
- Find the gradient of the tangent to the curve v = sinθ/(cosθ + sinθ) at the point where θ = π/3.
- (2 + √3)/2
- 2/(1 + √3)
- 2 + √3
- 2/(2 + √3)
- The function f is such that f(x) = (3x + 2)³ − 5 for x ≥ 0. Which of the following is true about f?
- f is a decreasing function
- f is an increasing function
- f has a stationary point at (0, 0)
- f has a maximum point at (0, 0)
- ∫₀¹ (2x + 8x²)/(1 + 4x) dx =
- 4
- 3
- 2
- 1
- The region enclosed by part of the curve y = 4x², the lines y = 1, y = 3 and the y-axis is rotated about the y-axis through a complete revolution. Find the volume of the solid of revolution generated.
- π/4
- π/2
- π
- 2π
- ∫ (2 − x)/(1 − x) dx =
- ln(1 − x) + C
- ln(1 + x) + C
- ln(x² − 1) + x + C
- x − ln(1 − x) + C
- Given that y = ln t² and x = 1/t, find dy/dx.
- −2/t³
- 2t²
- −2t
- 1/t³
- Find the second derivative of y = f(θ) = sinθ/(1 + cosθ).
- (1 + cosθ)²/sinθ
- sinθ/(1 + sinθ)
- sinθ/(1 + sinθ)²
- sinθ/(1 + cosθ)²
- The region enclosed by part of the curve y = 4x², the lines y = 1, y = 3 and the y-axis is rotated about the y-axis through a complete revolution. Find the volume of the solid of revolution generated.
- π/2
- π/4
- π
- 2π
- The function f is such that f(x) = (3x + 2)³ − 5 for x ≥ 0. Which of the following is true about f?
- f is a decreasing function
- f has a maximum point at (0, 0)
- f has a stationary point at (0, 0)
- f is an increasing function
- Find the equation of the tangent to the curve x² + y² = 5 at (−1, 2).
- x − 2y + 4 = 0
- 2x − y + 4 = 0
- x − 2y + 5 = 0
- 2x − y + 5 = 0
- Find the point of discontinuity of p(x) = 3x/(√x − 4).
- 8
- −2
- 2
- 16
- A curve has equation y = x³ + 3x² − 9x + p, where p is a constant. The value of p for which the curve has a stationary point on the x-axis is:
- 2
- 3
- 4
- 5
- Find the gradient of the tangent to the curve v = sinθ/(cosθ + sinθ) at the point where θ = π/3.
- (2 + √3)/2
- 2 + √3
- 2/(2 + √3)
- 2/(1 + √3)
- If f(x) = 1 + x² for x < 2 and f(x) = x³ for x ≥ 2, then the limit of f(x) as x → 2⁻ is:
- 5
- 7
- 8
- 6
- ∫ x²/(x − 1) dx =
- ln(x − 1) + C
- x²/2 + x + C
- x²/2 + x + ln(x − 1) + C
- x + ln(x − 1)/(x² + 1) + C
- ∫₀¹ (2x + 8x²)/(1 + 4x) dx =
- 1
- 3
- 2
- 4
- What is the domain of the function (x + 3)/(x² − 3x)?
- x ∈ (−∞, ∞)
- x ∈ (−∞, 0) ∪ (0, 3) ∪ (3, ∞)
- x ∈ (∞, 3) ∪ (3, 0) ∪ (0, ∞)
- x ∈ (−∞, −3) ∪ (0, 3) ∪ (3, ∞)
- The equation of a curve is y = x³ + px², where p is a positive constant. The nature of the stationary point at the origin is?
- Point of inflexion
- Minimum point
- Increasing point
- Maximum point
- A point is moving along the curve y = 4√x + 2/√x in such a way that the x-coordinate is increasing at a constant rate of 0.12 units per second. Find the rate of change of the y-coordinate when x = 4.
- 0.12
- 0.015
- 0.105
- 0.75
- ∫ (5x² + 6)/(x² + 1) dx =
- cos⁻¹(x) + C
- 5x + tan⁻¹(x) + C
- tan⁻¹(x) + C
- sin⁻¹(x) + C
- Find dy/dx when xy + sin y = 1.
- y/(x + sin y)
- −y/(x + cos y)
- −y/(x + sin y)
- y/(x + cos y)
- Evaluate the limit: lim (x → 0) [(x + 1)² − 1]/x
- 2
- 0
- 1
- does not exist
- For the function f(x) = 4x² − 1 if x < 1, and f(x) = 3x + 2 if x ≥ 1, the limit of f(x) as x → 1⁺ is:
- 3
- 5
- does not exist
- 1
- ∫ (sin(2x) + cos(x/3)) dx =
- cos(x) + sin(x) + C
- cos(2x) + sin(x/3) + C
- cos(x/3) + sin(2x) + C
- −(1/2)cos(2x) + 3 sin(x/3) + C
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