TMUA Paper 1: Differentiation, set 2

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

1. 11 mark

   Given that y = 5x^3 - 2x^(-2) + 7 for x != 0, find dy/dx.

   1. A 15x^2 - 4x^(-3)
   2. B 15x^2 + 4x^(-3)
   3. C 15x^2 + 4x^(-2)
   4. D 15x^2 + 4x^(-3) + 7
2. 21 mark

   A shop's daily profit, P pounds, when it sells x items is modeled by P = 60x - (1/2)x^2 - 300 for 0 <= x <= 100. Find the rate of change of P with respect to x when x = 40.

   1. A 1300
   2. B 40
   3. C 100
   4. D 20
3. 31 mark

   Given that f'(x) = 6x^2 - 4x + 1, find f''(2).

   1. A 20
   2. B 17
   3. C 8
   4. D 24
4. 41 mark

   Given that y = (2x - 1)(3x + 4)/x for x != 0, find dy/dx in simplified form.

   1. A 6 - 4/x^2
   2. B 12x + 4/x^2
   3. C 6 + 4/x^2
   4. D 6
5. 51 mark

   Find the values of x at which the tangent to the curve y = x^3 - 12x + 5 is parallel to the line y = 15x + 2.

   1. A x = 1 or x = -1
   2. B x = 3 sqrt(3) or x = -3 sqrt(3)
   3. C x = 3 only
   4. D x = 3 or x = -3
6. 61 mark

   Find the equation of the normal to the curve y = x^2 - 4x + 7 at the point where x = 1.

   1. A y = (1/2)x + 7/2
   2. B y = -2x + 6
   3. C y = -(1/2)x + 9/2
   4. D y = (1/2)x - 7/2
7. 71 mark

   For which set of values of x is the function f(x) = 2x^3 - 3x^2 - 12x + 4 strictly increasing?

   1. A -1 < x < 2
   2. B x < -1 or x > 2
   3. C x < -2 or x > 1
   4. D x = -1 or x = 2 only
8. 81 mark

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

   1. A (1/2)x^(-1/2)
   2. B (3/2)x^(3/2) - (3/2)x^(-1/2)
   3. C (3/2)x^(1/2) - (3/2)x^(-1/2)
   4. D (3/2)x^(1/2) - 3x^(-1/2)
9. 91 mark

   The curve y = x^3 - 6x^2 + 9x + 1 has two stationary points, at x = a and x = b. Find the value of a + b.

   1. A 4
   2. B -4
   3. C 3
   4. D 6
10. 101 mark

    A stone is thrown vertically upwards, and its height, h metres, after t seconds is given by h = 36t - 6t^2 for 0 <= t <= 6. At what time t is the stone's velocity zero?

    1. A t = 6
    2. B t = -3
    3. C t = 0
    4. D t = 3
11. 111 mark

    Find the x-coordinate of the point on the curve y = x^2 - 5x + 3 at which the tangent is parallel to the x-axis.

    1. A x = 5
    2. B x = -5/2
    3. C x = 5/2
    4. D x = -13/4
12. 121 mark

    For which values of the constant k is the function f(x) = x^3 + kx + 1 strictly increasing for all real values of x?

    1. A k >= 0
    2. B k > 0
    3. C k < 0
    4. D k > 3

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: B

   1. Differentiate term by term using the power rule for a negative exponent.
   2. d/dx(5x^3) = 15x^2.
   3. d/dx(-2x^(-2)) = -2 x (-2) x^(-3) = 4x^(-3), since multiplying the two negative numbers -2 and -2 gives a positive coefficient.
   4. d/dx(7) = 0, since the derivative of a constant is always zero.
   5. So dy/dx = 15x^2 + 4x^(-3), which is option B.
   • Why not A: Makes a sign error multiplying the coefficient -2 by the exponent -2, treating the two negatives as giving -4 instead of +4, so writes -4x^(-3) instead of +4x^(-3).
   • Why not C: Multiplies the coefficient -2 by the exponent -2 correctly to get 4, but forgets to reduce the power by one, leaving the term as 4x^(-2) instead of 4x^(-3).
   • Why not D: Carries the constant term +7 through unchanged into the derivative, forgetting that the derivative of a constant is zero.

2. Question 2Answer: D

   1. The rate of change of P with respect to x is dP/dx, found by differentiating term by term.
   2. d/dx(60x) = 60, d/dx(-(1/2)x^2) = -x, d/dx(-300) = 0, so dP/dx = 60 - x.
   3. Substitute x = 40: dP/dx = 60 - 40 = 20.
   4. So the rate of change at x = 40 is 20 pounds per item, option D.
   • Why not A: Evaluates P at x = 40 directly (P(40) = 2400 - 800 - 300 = 1300) instead of differentiating first, confusing the total profit with its rate of change.
   • Why not B: Differentiates -(1/2)x^2 as -(1/2)x, forgetting to multiply the coefficient by the original power 2, so uses dP/dx = 60 - (1/2)x and gets 60 - 20 = 40 at x = 40.
   • Why not C: Makes a sign error differentiating -(1/2)x^2, writing dP/dx = 60 + x instead of 60 - x, and gets 60 + 40 = 100 at x = 40.

3. Question 3Answer: A

   1. f'(x) is already given, so f''(x) is found by differentiating f'(x) once more with respect to x.
   2. Differentiate term by term: d/dx(6x^2) = 12x, d/dx(-4x) = -4, d/dx(1) = 0.
   3. So f''(x) = 12x - 4.
   4. Substitute x = 2: f''(2) = 12(2) - 4 = 24 - 4 = 20, which is option A.
   • Why not B: Evaluates the given f'(x) at x = 2 (f'(2) = 24 - 8 + 1 = 17) instead of differentiating again, confusing the first derivative with the second.
   • Why not C: Differentiates 6x^2 incorrectly as 6x, forgetting to multiply the coefficient by the power 2, giving f''(x) = 6x - 4 and f''(2) = 12 - 4 = 8.
   • Why not D: Drops the derivative of the linear term -4x, treating it as though it were already a constant contributing 0, giving f''(x) = 12x and f''(2) = 24.

4. Question 4Answer: C

   1. Expand the product first, since the expression needs simplifying before differentiating: (2x - 1)(3x + 4) = 6x^2 + 8x - 3x - 4 = 6x^2 + 5x - 4.
   2. Divide every term of the numerator by x: y = 6x + 5 - 4/x, valid since x != 0.
   3. Differentiate term by term: d/dx(6x) = 6, d/dx(5) = 0, d/dx(-4x^(-1)) = 4x^(-2) = 4/x^2.
   4. Combine the terms: dy/dx = 6 + 4/x^2, which is option C.
   • Why not A: Makes a sign error differentiating the -4/x term, treating d/dx(-4x^(-1)) as -4x^(-2) instead of +4x^(-2), missing that multiplying two negative numbers gives a positive coefficient.
   • Why not B: Forgets to divide the x^2 term of the numerator by x, working instead from y = 6x^2 + 5 - 4/x, which gives d/dx(6x^2) = 12x rather than the correctly divided 6x term differentiating to 6.
   • Why not D: Divides the -4 constant term of the numerator by x correctly when simplifying, but then treats the resulting -4/x as though it had already vanished to a plain constant, so drops it entirely before differentiating and gets only dy/dx = 6.

5. Question 5Answer: D

   1. Two lines (or a tangent and a line) are parallel exactly when they have the same gradient, and the line y = 15x + 2 has gradient 15.
   2. Differentiate the curve: dy/dx = 3x^2 - 12.
   3. Set the gradient equal to 15: 3x^2 - 12 = 15, so 3x^2 = 27, so x^2 = 9.
   4. Taking the square root of both sides gives two solutions: x = 3 or x = -3, which is option D.
   • Why not A: Makes a sign error differentiating -12x as +12x, using dy/dx = 3x^2 + 12 and solving 3x^2 + 12 = 15 to get x^2 = 1.
   • Why not B: Sets up 3x^2 - 12 = 15 correctly but forgets to divide both sides by the coefficient 3, taking the square root of 27 directly instead of first isolating x^2 = 9.
   • Why not C: Correctly reaches x^2 = 9 but forgets that a squared variable equal to a positive number has two roots, reporting only the positive one.

6. Question 6Answer: A

   1. Find the point on the curve at x = 1: y = 1^2 - 4(1) + 7 = 1 - 4 + 7 = 4, so the point is (1, 4).
   2. Differentiate to find the tangent's gradient: dy/dx = 2x - 4. At x = 1: dy/dx = 2 - 4 = -2.
   3. The normal is perpendicular to the tangent, so its gradient is the negative reciprocal of -2, which is 1/2.
   4. Use y - y1 = m(x - x1) with (x1, y1) = (1, 4) and m = 1/2: y - 4 = (1/2)(x - 1), so y = (1/2)x - 1/2 + 4 = (1/2)x + 7/2, which is option A.
   • Why not B: Uses the tangent's own gradient (-2) to write the line through the point, instead of taking the normal's gradient, which is the negative reciprocal of the tangent's.
   • Why not C: Takes the reciprocal of the tangent gradient -2 to get -1/2, but forgets to also flip its sign, so uses gradient -1/2 instead of the correct +1/2 for the normal.
   • Why not D: Uses the correct normal gradient of 1/2, but evaluates the point on the curve incorrectly as (1, -3) by omitting the +7 constant term, instead of the true point (1, 4).

7. Question 7Answer: B

   1. Differentiate: f'(x) = 6x^2 - 6x - 12 = 6(x^2 - x - 2) = 6(x - 2)(x + 1).
   2. The function is strictly increasing where f'(x) > 0. Since the coefficient of x^2 in f'(x) is positive, the parabola f'(x) is positive outside its roots and negative between them.
   3. The roots of f'(x) are x = 2 and x = -1, so f'(x) > 0 when x < -1 or x > 2, which is option B.
   • Why not A: Solves the inequality with the sign reversed, giving the interval where f is strictly decreasing rather than where it is strictly increasing.
   • Why not C: Factorises x^2 - x - 2 = 0 as (x + 2)(x - 1) instead of the correct (x - 2)(x + 1), swapping which root is positive and which is negative, and gets critical points -2 and 1 instead of 2 and -1.
   • Why not D: Gives the critical points where f'(x) = 0, rather than the interval of x over which f'(x) > 0.

8. Question 8Answer: C

   1. Expand the product first, since it needs simplifying before differentiating: sqrt(x)(x - 3) = x^(1/2) x x - 3x^(1/2) = x^(3/2) - 3x^(1/2).
   2. Differentiate term by term using the power rule: d/dx(x^(3/2)) = (3/2)x^(1/2).
   3. d/dx(-3x^(1/2)) = -3 x (1/2) x^(-1/2) = -(3/2)x^(-1/2).
   4. Combine the terms: dy/dx = (3/2)x^(1/2) - (3/2)x^(-1/2), which is option C.
   • Why not A: Differentiates each factor separately and multiplies the results together (d/dx(sqrt(x)) x d/dx(x - 3) = (1/2)x^(-1/2) x 1), instead of expanding the product into a sum before differentiating.
   • Why not B: Multiplies the first term's coefficient by its exponent correctly, but forgets to reduce the power by one, leaving x^(3/2) unchanged instead of reducing it to x^(1/2).
   • Why not D: Reduces the power on the second term correctly to x^(-1/2), but forgets to multiply its coefficient -3 by the original exponent 1/2, leaving the coefficient as -3 instead of -3/2.

9. Question 9Answer: A

   1. Stationary points occur where dy/dx = 0. Differentiate: dy/dx = 3x^2 - 12x + 9.
   2. Set this equal to zero and divide by 3: x^2 - 4x + 3 = 0, which factorises as (x - 1)(x - 3) = 0.
   3. So the stationary points are at x = 1 and x = 3, meaning a = 1 and b = 3 (in some order).
   4. So a + b = 1 + 3 = 4, which is option A.
   • Why not B: Makes a sign error differentiating -6x^2, using dy/dx = 3x^2 + 12x + 9 instead of 3x^2 - 12x + 9, and solves x^2 + 4x + 3 = 0 to get roots -1 and -3.
   • Why not C: Finds the product of the two x-coordinates (1 x 3 = 3) rather than their sum.
   • Why not D: Adds the y-coordinates of the two stationary points (y(1) = 5 and y(3) = 1, giving 6) instead of adding their x-coordinates.

10. Question 10Answer: D

    1. Velocity is the rate of change of height with time, that is, dh/dt.
    2. Differentiate: dh/dt = 36 - 12t.
    3. Set the velocity to zero: 36 - 12t = 0, so 12t = 36, so t = 3.
    4. So the stone's velocity is zero (at its greatest height) at t = 3 seconds, which is option D.
    • Why not A: Solves h = 0 instead of dh/dt = 0, finding the time the stone lands back on the ground rather than the time its velocity is zero.
    • Why not B: Makes a sign error differentiating -6t^2, using dh/dt = 36 + 12t instead of 36 - 12t, and solves 36 + 12t = 0.
    • Why not C: Assumes the stone's velocity is zero at the instant it is thrown, misreading the initial condition instead of solving dh/dt = 0.

11. Question 11Answer: C

    1. A tangent parallel to the x-axis has gradient zero, so this is asking for the point where dy/dx = 0.
    2. Differentiate: dy/dx = 2x - 5.
    3. Set this equal to zero: 2x - 5 = 0, so 2x = 5, so x = 5/2.
    4. So the tangent is parallel to the x-axis at x = 5/2, which is option C.
    • Why not A: Correctly rearranges to 2x = 5, but then forgets to divide both sides by the coefficient 2, giving x = 5 instead of x = 5/2.
    • Why not B: Makes a sign error rearranging 2x - 5 = 0, writing 2x = -5 instead of 2x = 5.
    • Why not D: Correctly locates the turning point at x = 5/2, but then reports its y-coordinate (y = 25/4 - 25/2 + 3 = -13/4) instead of the x-coordinate the question asks for.

12. Question 12Answer: A

    1. For f(x) to be strictly increasing for all real x, its derivative must satisfy f'(x) >= 0 everywhere, with f'(x) = 0 only at isolated points rather than on a whole interval (otherwise f would be constant there).
    2. Differentiate: f'(x) = 3x^2 + k.
    3. Since 3x^2 >= 0 for every real x, with equality only at the single point x = 0, the smallest possible value of f'(x) is k, attained only at x = 0.
    4. If k >= 0, then f'(x) >= 0 everywhere and is zero only at the single point x = 0 (when k = 0), which is the isolated zero gradient a simple cubic like y = x^3 has at its point of inflexion, so f is still strictly increasing overall. If k < 0, f'(x) is negative on a whole interval around x = 0, so f decreases there and cannot be strictly increasing for all real x.
    5. So the condition is k >= 0, which is option A.
    • Why not B: Requires f'(x) > 0 strictly at every x, wrongly excluding the boundary case k = 0. When k = 0, f(x) = x^3 + 1 has f'(x) = 3x^2, which is zero only at the single point x = 0; a derivative that vanishes at just one isolated point does not stop the function being strictly increasing overall, the same qualitative fact that makes y = x^3 an increasing curve through its point of inflexion at the origin.
    • Why not C: Reverses the inequality, wrongly reasoning that a negative k is needed to keep 3x^2 + k from growing too large, when in fact a negative k makes f'(x) negative for x near 0, so f decreases on that interval instead of increasing everywhere.
    • Why not D: Mistakes the coefficient 3 in f'(x) = 3x^2 + k for a threshold that k itself must exceed, adding an unnecessary and incorrect restriction.