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Admissions tests / TMUA / Paper 1 / Differentiation

Test standard. 12 questions, 12 marks, about 45 minutes.

TMUA Paper 1: Differentiation, set 1

The derivative as a gradient, differentiating powers of x, tangents and normals, stationary points and their nature, and increasing and decreasing functions.

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  • Answer all questions. No calculator.
  • Each question has exactly one correct answer.
  1. 11 mark

    Given that y = x^4 - 3x^2 + 5, what is the value of d2y/dx2 when x = 2?

    1. A 42
    2. B 20
    3. C 48
    4. D 54
  2. 21 mark

    The number of bacteria, N, in a laboratory culture after t hours is modeled by N = 200 + 30t^2 - 2t^3 for 0 <= t <= 10. Find the rate of change of N with respect to t when t = 3.

    1. A 416
    2. B 36
    3. C 162
    4. D 126
  3. 31 mark

    A curve has equation y = f(x). At the point where x = a, it is given that f'(a) = 0 and f''(a) < 0. Which of the following best describes the point (a, f(a))?

    1. A A local minimum of the curve.
    2. B A point where the gradient itself is negative.
    3. C A local maximum of the curve.
    4. D A point where the function is strictly increasing.
  4. 41 mark

    Differentiate y = (x^2 + 3)(x - 2) with respect to x.

    1. A 3x^2 - 4x + 3
    2. B 3x^2 - 4x - 6
    3. C x^2 - 4x + 3
    4. D 3x^2 + 3
  5. 51 mark

    Given that y = (2x^3 - 5x)/x for x != 0, find dy/dx.

    1. A 4x - 5
    2. B 2x
    3. C 4x
    4. D 6x^2 - 5
  6. 61 mark

    Given that y = 3 sqrt(x) - 4/sqrt(x) for x > 0, find dy/dx.

    1. A (3/2)x^(-1/2) - 2x^(-3/2)
    2. B (3/2)x^(1/2) + 2x^(-1/2)
    3. C (3/2)x^(-1/2) + 2x^(-3/2)
    4. D 3x^(-1/2) - 4x^(-3/2)
  7. 71 mark

    Given that y = (x + 1)^2 / x for x > 0, express dy/dx in simplified form.

    1. A 1 + 1/x^2
    2. B 1
    3. C 2x - 1/x^2
    4. D 1 - 1/x^2
  8. 81 mark

    A particle's displacement from a fixed point after t seconds is s(t). Which of the following expressions represents the particle's acceleration at time t?

    1. A ds/dt
    2. B s(t) x t
    3. C d2s/dt2
    4. D 1/(ds/dt)
  9. 91 mark

    Given that f(x) = x^3 + 2x^2 - 5x + 1, find the value of f'(1) + f''(1).

    1. A 12
    2. B 2
    3. C 10
    4. D 9
  10. 101 mark

    Find the equation of the tangent to the curve y = x^3 - 2x^2 + 1 at the point where x = 2.

    1. A y = 4x - 7
    2. B y = -(1/4)x + 3/2
    3. C y = 4x - 8
    4. D y = 8x - 15
  11. 111 mark

    The curve y = x^3 - 3x^2 - 9x + 2 has two stationary points. What is the y-coordinate of the local minimum?

    1. A 7
    2. B 3
    3. C -25
    4. D -27
  12. 121 mark

    For which set of values of x is the function f(x) = x^3 - 12x + 5 strictly decreasing?

    1. A x < -2 or x > 2
    2. B 0 < x < 4
    3. C x = -2 and x = 2 only
    4. D -2 < x < 2

Worked solutions

Every question below carries the reasoning, not just the answer. The official material for this test publishes a correct option letter and nothing else.

  1. Question 1Answer: A

    1. Differentiate y = x^4 - 3x^2 + 5 once: dy/dx = 4x^3 - 6x.
    2. Differentiate again to get the second derivative: d2y/dx2 = 12x^2 - 6.
    3. Substitute x = 2: d2y/dx2 = 12(2^2) - 6 = 12(4) - 6 = 48 - 6 = 42.
    4. So d2y/dx2 at x = 2 is 42, which is option A.
    • Why not B: Evaluates only the first derivative dy/dx = 4x^3 - 6x at x = 2, giving 32 - 12 = 20, confusing the first derivative with the second.
    • Why not C: Drops the -3x^2 term when differentiating, effectively taking the second derivative of x^4 + 5 alone (12x^2 = 48 at x = 2) and losing the middle term's contribution entirely.
    • Why not D: Misreads the sign of the -3x^2 term as +3x^2 throughout, giving d2y/dx2 = 12x^2 + 6, which equals 54 at x = 2 instead of 42.

  2. Question 2Answer: D

    1. dN/dt is the rate of change of the bacteria count with respect to time, found by differentiating N with respect to t.
    2. Differentiate term by term: d/dt(200) = 0, d/dt(30t^2) = 60t, d/dt(-2t^3) = -6t^2, so dN/dt = 60t - 6t^2.
    3. Substitute t = 3: dN/dt = 60(3) - 6(3^2) = 180 - 6(9) = 180 - 54 = 126.
    4. So the rate of change at t = 3 is 126 bacteria per hour, option D.
    • Why not A: Evaluates N at t = 3 directly (N(3) = 200 + 270 - 54 = 416) instead of differentiating first, confusing the population itself with its rate of change.
    • Why not B: Differentiates 30t^2 as 30t, forgetting to multiply the coefficient by the power 2, so uses dN/dt = 30t - 6t^2 and gets 90 - 54 = 36 at t = 3.
    • Why not C: Differentiates -2t^3 as -2t^2, forgetting to multiply the coefficient by the power 3, so uses dN/dt = 60t - 2t^2 and gets 180 - 18 = 162 at t = 3.

  3. Question 3Answer: C

    1. A stationary point occurs where f'(a) = 0, so the tangent to the curve at (a, f(a)) is horizontal.
    2. The second derivative f''(a) measures how the gradient itself is changing: f''(a) < 0 means the gradient is decreasing through zero, from positive just before a to negative just after.
    3. A gradient that changes from positive to negative describes a curve that rises then falls, which is the shape of a local maximum.
    4. So (a, f(a)) is a local maximum, option C.
    • Why not A: Reverses the standard second derivative test, treating f''(a) < 0 as indicating a minimum rather than a maximum.
    • Why not B: Confuses the condition on f''(a), which describes how the gradient is changing, with a condition on f'(a) itself, ignoring that f'(a) = 0 was already given.
    • Why not D: Misreads the stationary condition f'(a) = 0 as though it meant f'(a) > 0, confusing 'stationary' with 'increasing'.

  4. Question 4Answer: A

    1. Expand the product first, since it needs simplifying before differentiating: (x^2 + 3)(x - 2) = x^3 - 2x^2 + 3x - 6.
    2. Differentiate term by term: d/dx(x^3) = 3x^2, d/dx(-2x^2) = -4x, d/dx(3x) = 3, d/dx(-6) = 0.
    3. Combine the terms: dy/dx = 3x^2 - 4x + 3, which is option A.
    • Why not B: Carries the constant term -6 through unchanged, forgetting that the derivative of a constant is zero.
    • Why not C: Forgets to multiply by the exponent when differentiating x^3, applying only the power-reduction step so x^3 becomes x^2 instead of 3x^2.
    • Why not D: Omits the -2x^2 term when differentiating, dropping it during the term-by-term process.

  5. Question 5Answer: C

    1. Simplify before differentiating by dividing each term of the numerator by x: (2x^3 - 5x)/x = 2x^2 - 5, valid since x != 0.
    2. Differentiate the simplified expression term by term: d/dx(2x^2) = 4x, d/dx(-5) = 0.
    3. So dy/dx = 4x, which is option C.
    • Why not A: Divides only the first term of the numerator by x, leaving the second term as -5x instead of -5, so ends up differentiating y = 2x^2 - 5x.
    • Why not B: Differentiates 2x^2 by reducing the power to x^1 but forgetting to multiply by the original exponent 2, giving 2x instead of 4x.
    • Why not D: Differentiates the numerator 2x^3 - 5x directly, ignoring the division by x altogether.

  6. Question 6Answer: C

    1. Rewrite the roots as fractional powers: y = 3x^(1/2) - 4x^(-1/2).
    2. Differentiate term by term using the power rule: d/dx(3x^(1/2)) = 3 x (1/2) x^(-1/2) = (3/2)x^(-1/2).
    3. d/dx(-4x^(-1/2)) = -4 x (-1/2) x^(-3/2) = 2x^(-3/2), since multiplying the two negative numbers gives a positive coefficient.
    4. Combine the terms: dy/dx = (3/2)x^(-1/2) + 2x^(-3/2), which is option C.
    • Why not A: Makes a sign error treating -4 x (-1/2) as -2 instead of +2, mishandling the multiplication of two negative numbers.
    • Why not B: Forgets to reduce the power by 1 during differentiation, multiplying by the original exponent but leaving the power unchanged.
    • Why not D: Reduces each exponent by 1 correctly but forgets to multiply the coefficients 3 and -4 by the original exponent 1/2, leaving them unchanged.

  7. Question 7Answer: D

    1. Expand and simplify before differentiating: (x + 1)^2 / x = (x^2 + 2x + 1)/x = x + 2 + 1/x, valid for x > 0.
    2. Rewrite 1/x as x^(-1) and differentiate term by term: d/dx(x) = 1, d/dx(2) = 0, d/dx(x^(-1)) = -x^(-2).
    3. Combine the terms: dy/dx = 1 - x^(-2) = 1 - 1/x^2, which is option D.
    • Why not A: Differentiates the 1/x term with the wrong sign, treating d/dx(x^(-1)) as +x^(-2) instead of -x^(-2).
    • Why not B: Treats the 1/x term as though it were a constant, since it is not written as an explicit power of x, so differentiates it to 0 along with the +2.
    • Why not C: Divides the x^2 term of the numerator by x incorrectly, leaving it as x^2 instead of reducing it to x, as though dividing by x does not lower the power of that term.

  8. Question 8Answer: C

    1. Velocity is the rate of change of displacement with time, that is, ds/dt.
    2. Acceleration is the rate of change of velocity with time, that is, the derivative of ds/dt with respect to t again.
    3. Differentiating displacement twice with respect to t gives the second derivative d2s/dt2, which is option C.
    • Why not A: This is the velocity, the first derivative of displacement, not the acceleration.
    • Why not B: Multiplies displacement by time, which has no meaning as a rate of change of anything.
    • Why not D: Takes the reciprocal of the velocity, confusing a rate of change with its inverse.

  9. Question 9Answer: A

    1. Differentiate once: f'(x) = 3x^2 + 4x - 5. At x = 1: f'(1) = 3 + 4 - 5 = 2.
    2. Differentiate again: f''(x) = 6x + 4. At x = 1: f''(1) = 6 + 4 = 10.
    3. Add the two values: f'(1) + f''(1) = 2 + 10 = 12, which is option A.
    • Why not B: Computes f'(1) = 2 correctly but forgets to add f''(1), giving only the first derivative's value.
    • Why not C: Computes f''(1) = 10 correctly but forgets to add f'(1), giving only the second derivative's value.
    • Why not D: Finds f'(1) = 2 correctly, but when finding f''(x) differentiates 3x^2 as 3x instead of 6x, forgetting to multiply by the exponent 2, giving f''(1) = 7 and a total of 9 instead of 12.

  10. Question 10Answer: A

    1. Find the point on the curve at x = 2: y = 2^3 - 2(2^2) + 1 = 8 - 8 + 1 = 1, so the point is (2, 1).
    2. Differentiate to find the gradient function: dy/dx = 3x^2 - 4x. At x = 2: dy/dx = 3(4) - 4(2) = 12 - 8 = 4, so the tangent's gradient is 4.
    3. Use y - y1 = m(x - x1) with (x1, y1) = (2, 1) and m = 4: y - 1 = 4(x - 2), so y = 4x - 8 + 1 = 4x - 7, which is option A.
    • Why not B: Uses the normal's gradient, the negative reciprocal of the tangent's gradient, instead of the tangent's own gradient.
    • Why not C: Evaluates y at x = 2 incorrectly, omitting the +1 constant term and using the point (2, 0) instead of the true point (2, 1).
    • Why not D: Miscalculates the gradient by differentiating -2x^2 as -2x instead of -4x, forgetting to multiply the coefficient by the power 2.

  11. Question 11Answer: C

    1. Differentiate: dy/dx = 3x^2 - 6x - 9 = 3(x^2 - 2x - 3) = 3(x - 3)(x + 1), so the stationary points are at x = 3 and x = -1.
    2. Classify using the second derivative: d2y/dx2 = 6x - 6. At x = 3: 18 - 6 = 12 > 0, so x = 3 is a local minimum. At x = -1: -6 - 6 = -12 < 0, so x = -1 is a local maximum.
    3. Evaluate y at the minimum, x = 3: y = 3^3 - 3(3^2) - 9(3) + 2 = 27 - 27 - 27 + 2 = -25, which is option C.
    • Why not A: Reports the y-coordinate of the local maximum instead of the local minimum, confusing which stationary point is which.
    • Why not B: Gives the x-coordinate of the minimum, x = 3, rather than the y-coordinate that the question asks for.
    • Why not D: Correctly identifies x = 3 as the minimum but omits the final constant term (+2) when evaluating y(3), giving 27 - 27 - 27 instead of 27 - 27 - 27 + 2.

  12. Question 12Answer: D

    1. Differentiate: f'(x) = 3x^2 - 12.
    2. The function is strictly decreasing where f'(x) < 0: 3x^2 - 12 < 0, so x^2 < 4.
    3. Since x^2 < 4 means x lies between the negative and positive square roots of 4, this gives -2 < x < 2, which is option D.
    • Why not A: Solves the inequality with the sign reversed, finding where f is increasing rather than where it is decreasing.
    • Why not B: Misdifferentiates the term -12x as though it stayed -12x rather than becoming the constant -12, giving the wrong derivative 3x^2 - 12x and the wrong critical points 0 and 4.
    • Why not C: Gives the critical points where f'(x) = 0 rather than the interval where f'(x) < 0.

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