ESAT Mathematics 2: Differentiation and integration, set 4

• Answer all questions. No calculator.
• Each question has exactly one correct answer.
• This is a stretch set: several questions combine two specification points, and the quickest route is rarely the one that grinds through every term.

1. 11 mark

   Let f(x) = sqrt(x) (x - 6) for x > 0. Find f'(9).

   1. A 5/2
   2. B 1/2
   3. C 7/2
   4. D -3/2
2. 21 mark

   The curve y = x^3 - 3x^2 - 9x + 5 has tangents at two points that are parallel to the line y = 15x + 2. Find the sum of the x-coordinates of these two points.

   1. A 2
   2. B 6
   3. C -2
   4. D 1
3. 31 mark

   For x > 0, let f(x) = x^2 + 54/x. Find the set of values of x for which f is a decreasing function.

   1. A x > 3
   2. B 0 < x < 3 sqrt(3)
   3. C 0 < x < 9
   4. D 0 < x < 3
4. 41 mark

   The curve f(x) = 2x^3 - 3x^2 - 12x + 5 has two stationary points. Find the value of f(x) at its local maximum.

   1. A -15
   2. B 12
   3. C 18
   4. D 1
5. 51 mark

   Evaluate the definite integral of (x^3 - 8)/(x - 2) dx from x = 3 to x = 4.

   1. A 55/3
   2. B 70/3
   3. C 143/4
   4. D 28/3
6. 61 mark

   The curve y = x^3 - x^2 - 2x crosses the x-axis at x = -1, x = 0 and x = 2. Find the area enclosed between the curve and the x-axis for -1 <= x <= 2.

   1. A 37/12
   2. B 9/4
   3. C 5/12
   4. D 8/3
7. 71 mark

   Given that the definite integral of 2x dx from x = 2 to x = k equals 32, and that k > 2, find k.

   1. A 4 sqrt(2)
   2. B 2 sqrt(5)
   3. C 6
   4. D 2 sqrt(7)
8. 81 mark

   Given that the definite integral of f(x) dx from x = 1 to x = 3 equals 6, and the definite integral of f(x) dx from x = 3 to x = 8 equals 11, find the definite integral of f(x) dx from x = 8 to x = 1.

   1. A 17
   2. B -17
   3. C 5
   4. D -11
9. 91 mark

   A concave-up function f(x) (f''(x) > 0 throughout) takes the values f(0) = 3, f(2) = 7, f(4) = 15, f(6) = 27. Use the trapezium rule with all three strips to estimate the definite integral of f(x) dx from x = 0 to x = 6, and state whether this estimate is an overestimate or an underestimate of the true value.

   1. A 74, an overestimate
   2. B 148, an overestimate
   3. C 74, an underestimate
   4. D 52, an overestimate
10. 101 mark

    A curve y = f(x) satisfies dy/dx = 6x^2 + kx - 5, where k is a constant. The gradient of the curve at the point where x = 2 is 21. Given also that the curve passes through the point (2, 10), find f(x).

    1. A f(x) = 2x^3 + (1/2)x^2 - 5x + 10
    2. B f(x) = 6x^3 + (1/2)x^2 - 5x - 30
    3. C f(x) = 2x^3 + (7/2)x^2 - 5x - 10
    4. D f(x) = 2x^3 + (1/2)x^2 - 5x + 2
11. 111 mark

    Find the x-intercept of the normal to the curve y = x^3 - 3x + 4 at the point where x = 2.

    1. A 4/3
    2. B -52
    3. C 56
    4. D 8/3
12. 121 mark

    The number of bacteria in a culture t hours after the start of an experiment is modelled by N(t) = -2t^3 + 24t^2 + 100, for 0 <= t <= 12. Find the greatest value of the rate of change of N, dN/dt, over this interval.

    1. A 356
    2. B 96
    3. C 288
    4. D 0
13. 131 mark

    Evaluate the definite integral of (x^5 + 2x^4 - 3x^2)/x^2 dx from x = 1 to x = 2.

    1. A 65/12
    2. B 17/12
    3. C -65/12
    4. D 59/4
14. 141 mark

    The curve f(x) = x^3 - 6x^2 + kx - 3 has a stationary point at x = 5. Find the x-coordinate of the curve's other stationary point, and determine whether it is a maximum or a minimum.

    1. A x = -3, a local maximum
    2. B x = 1, a local maximum
    3. C x = -1, a local minimum
    4. D x = -1, a local maximum
15. 151 mark

    The function g(x) satisfies g(-x) = g(x) for every real x (g is even). Given that the definite integral of g(x) dx from 0 to 4 equals 9, find the definite integral of g(x) dx from -4 to 4.

    1. A 36
    2. B 18
    3. C 0
    4. D 9

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: C

   1. Expand f(x) = sqrt(x)(x-6) as a sum of powers of x before differentiating: f(x) = x^(1/2)(x-6) = x^(3/2) - 6x^(1/2).
   2. Differentiate term by term: d/dx(x^(3/2)) = (3/2)x^(1/2), and d/dx(-6x^(1/2)) = -6 x (1/2) x^(-1/2) = -3x^(-1/2).
   3. So f'(x) = (3/2)x^(1/2) - 3x^(-1/2).
   4. At x = 9: x^(1/2) = 3 and x^(-1/2) = 1/3, so f'(9) = (3/2)(3) - 3(1/3) = 4.5 - 1 = 3.5.
   5. So the answer is C, 7/2.
   • Why not A: Differentiates -6x^(1/2) but forgets to also halve the coefficient the way the exponent rule requires, so it stays -6x^(-1/2) instead of becoming -3x^(-1/2), giving f'(x) = (3/2)x^(1/2) - 6x^(-1/2) and f'(9) = 4.5 - 2 = 5/2.
   • Why not B: Reaches for the product rule on sqrt(x)(x-6) without expanding first, and keeps only one of the two required terms: (x-6) times the derivative of sqrt(x), giving (x-6)/(2 sqrt(x)), which is 3/6 = 1/2 at x = 9.
   • Why not D: Expands sqrt(x)(x-6) incorrectly as x^(3/2) - 6x, forgetting that the second term should also carry a square root (sqrt(x) times -6 is -6x^(1/2), not -6x), and differentiates that wrong expression to get (3/2)x^(1/2) - 6, which is 4.5 - 6 = -3/2 at x = 9.

2. Question 2Answer: A

   1. Differentiate: dy/dx = 3x^2 - 6x - 9.
   2. A tangent parallel to y = 15x + 2 has gradient 15, so set 3x^2 - 6x - 9 = 15, giving 3x^2 - 6x - 24 = 0.
   3. Rather than factorising, use the sum-of-roots shortcut for ax^2 + bx + c = 0: the two x-coordinates sum to -b/a.
   4. Here a = 3 and b = -6, so the sum is -(-6)/3 = 6/3 = 2.
   5. (Check by factorising: x^2 - 2x - 8 = (x-4)(x+2) = 0, so x = 4 or x = -2, and indeed 4 + (-2) = 2.)
   6. So the answer is A, 2.
   • Why not B: Differentiates x^3 as x^2, forgetting to bring the power 3 down as a multiplying factor, so the resulting quadratic is x^2 - 6x - 24 = 0 instead of 3x^2 - 6x - 24 = 0, changing the sum-of-roots formula -b/a from 6/3 = 2 to 6/1 = 6.
   • Why not C: Recalls the sum-of-roots formula for ax^2 + bx + c = 0 as b/a rather than -b/a, dropping the negative sign, and computes -6/3 = -2 instead of -(-6)/3 = 2.
   • Why not D: Differentiates -3x^2 but forgets to multiply by its exponent 2, treating d/dx(-3x^2) as -3x instead of -6x, which changes the quadratic to 3x^2 - 3x - 24 = 0 and gives a sum of roots of 3/3 = 1 instead of 6/3 = 2.

3. Question 3Answer: D

   1. Differentiate: f'(x) = 2x - 54x^(-2) = 2x - 54/x^2.
   2. f is decreasing where f'(x) < 0: 2x < 54/x^2.
   3. Since x > 0, x^2 is positive, so multiplying both sides by x^2 keeps the inequality direction: 2x^3 < 54, so x^3 < 27.
   4. This gives x < 3 (the real cube root of 27 is 3), and combined with the domain x > 0, the function is decreasing for 0 < x < 3.
   5. So the answer is D.
   • Why not A: Correctly reaches x^3 < 27, but then wrongly reverses the inequality when cubing back (thinking that clearing a squared term flips the direction the way a negative multiplier would), concluding x > 3 instead of x < 3.
   • Why not B: Differentiates 54/x with the wrong power, treating d/dx(54x^(-1)) as -54x^(-1) instead of -54x^(-2) (forgetting to reduce the power by one), which gives 2x < 54/x rather than 2x < 54/x^2, and so x^2 < 27, x < sqrt(27) = 3 sqrt(3).
   • Why not C: Correctly reaches x^3 < 27, but then miscalculates the cube root, treating it as 27 divided by 3 (giving 9) rather than the number whose cube is 27 (which is 3).

4. Question 4Answer: B

   1. Differentiate: f'(x) = 6x^2 - 6x - 12 = 6(x^2 - x - 2) = 6(x-2)(x+1).
   2. Stationary points at x = 2 and x = -1.
   3. Differentiate again: f''(x) = 12x - 6. At x = 2, f''(2) = 18 > 0, a minimum. At x = -1, f''(-1) = -18 < 0, a maximum.
   4. So the local maximum is at x = -1. Evaluate: f(-1) = 2(-1)^3 - 3(-1)^2 - 12(-1) + 5 = -2 - 3 + 12 + 5 = 12.
   5. So the answer is B, 12.
   • Why not A: Correctly finds both stationary points, x = 2 and x = -1, and correctly evaluates f at both, but misapplies the second derivative test, treating the point where f''(x) > 0 as the maximum rather than the minimum, and so reports f(2) = -15 instead of f(-1) = 12.
   • Why not C: Correctly identifies x = -1 as the local maximum, but makes a squaring error evaluating (-1)^2, treating it as -1 instead of 1, which flips the sign of the -3x^2 term and gives 18 instead of 12.
   • Why not D: Mis-factorises f'(x) = 6x^2 - 6x - 12 as 6(x-1)(x+2) instead of 6(x-2)(x+1), giving wrong stationary points x = 1 and x = -2, and then assumes without checking that the more negative x-value is the maximum, evaluating f(-2) = 1.

5. Question 5Answer: B

   1. The numerator factorises using the difference of cubes: x^3 - 8 = x^3 - 2^3 = (x - 2)(x^2 + 2x + 4).
   2. For x != 2 (which holds throughout 3 <= x <= 4), this cancels with the denominator, so the integrand simplifies to x^2 + 2x + 4 before integrating.
   3. Integrate term by term: the antiderivative is F(x) = x^3/3 + x^2 + 4x.
   4. Evaluate at x = 4: F(4) = 64/3 + 16 + 16 = 64/3 + 32 = 160/3.
   5. Evaluate at x = 3: F(3) = 9 + 9 + 12 = 30 = 90/3.
   6. The definite integral is F(4) - F(3) = 160/3 - 90/3 = 70/3.
   7. So the answer is B.
   • Why not A: Correctly simplifies the integrand to x^2 + 2x + 4, but treats the linear term 2x as already integrated and carries it forward unchanged instead of applying the power rule to raise it to x^2, so uses the antiderivative x^3/3 + 2x + 4x instead of x^3/3 + x^2 + 4x.
   • Why not C: Attempts to integrate the original fraction without dividing out (x - 2) first, treating the numerator x^3 - 8 as though it could be integrated on its own while ignoring the denominator entirely, and evaluates the definite integral of x^3 - 8 dx from x = 3 to x = 4 instead.
   • Why not D: Divides x^3 - 8 by (x - 2) using the wrong sign in the difference-of-cubes identity, writing a^3 - b^3 = (a - b)(a^2 - ab + b^2) instead of (a - b)(a^2 + ab + b^2), and so integrates x^2 - 2x + 4 instead of x^2 + 2x + 4.

6. Question 6Answer: A

   1. The curve is positive on (-1,0) and negative on (0,2) (check x = -0.5: (-0.5)^3 - (-0.5)^2 - 2(-0.5) = -0.125 - 0.25 + 1 = 0.625 > 0; check x = 1: 1 - 1 - 2 = -2 < 0), so the definite integral over the whole interval would let the two parts cancel. The area must be found by splitting at the interior root x = 0.
   2. Antiderivative: F(x) = x^4/4 - x^3/3 - x^2. F(-1) = 1/4 - (-1/3) - 1 = 1/4 + 1/3 - 1 = -5/12. F(0) = 0. F(2) = 4 - 8/3 - 4 = -8/3.
   3. Area on (-1,0): F(0) - F(-1) = 0 - (-5/12) = 5/12 (positive, matching the curve being above the axis there).
   4. Area on (0,2): the integral is F(2) - F(0) = -8/3, and since area must be positive, this contributes |-8/3| = 8/3.
   5. Total enclosed area = 5/12 + 8/3 = 5/12 + 32/12 = 37/12, so the answer is A.
   • Why not B: Computes the definite integral of (x^3 - x^2 - 2x) directly from -1 to 2 without splitting at the interior root x = 0, so the positive region on (-1,0) and the negative contribution on (0,2) partly cancel: F(2) - F(-1) = -8/3 - (-5/12) = -9/4, and taking the size of this single combined value as the area gives 9/4 instead of the true total.
   • Why not C: Correctly finds the area of the region between x = -1 and x = 0 as 5/12, but forgets that the interval also includes the region between x = 0 and x = 2, reporting only the first piece as the whole area.
   • Why not D: Correctly finds the area of the region between x = 0 and x = 2 as 8/3, but forgets that the interval also includes the region between x = -1 and x = 0, reporting only the second piece as the whole area.

7. Question 7Answer: C

   1. The antiderivative of 2x is F(x) = x^2.
   2. By the Fundamental Theorem of Calculus, the definite integral from 2 to k is F(k) - F(2) = k^2 - 4.
   3. Set this equal to 32: k^2 - 4 = 32, so k^2 = 36.
   4. Since k > 2, take the positive root: k = 6 (rejecting k = -6).
   5. So the answer is C.
   • Why not A: Treats the definite integral as simply F(k) rather than F(k) - F(2), forgetting to subtract the antiderivative's value at the lower limit, and solves k^2 = 32 to get k = 4 sqrt(2) instead of k^2 - 4 = 32.
   • Why not B: Integrates 2x by raising the power to x^2 but then forgets to divide by the new power, so uses the antiderivative 2x^2 instead of x^2, and solves 2k^2 - 8 = 32 to get k^2 = 20 and k = 2 sqrt(5).
   • Why not D: Adds the value of the antiderivative at the lower limit instead of subtracting it, computing k^2 + 4 = 32 instead of k^2 - 4 = 32, giving k = 2 sqrt(7).

8. Question 8Answer: B

   1. The intervals [1,3] and [3,8] are contiguous, so the integral from 1 to 8 combines them by addition: integral from 1 to 8 = 6 + 11 = 17.
   2. Reversing the limits of a definite integral negates its value: the integral from 8 to 1 is the negative of the integral from 1 to 8.
   3. So the integral from 8 to 1 = -17.
   4. So the answer is B.
   • Why not A: Correctly combines the two contiguous ranges, 6 + 11 = 17, but forgets that going from x = 8 to x = 1 is the reverse direction of x = 1 to x = 8, and so omits the sign change that reversing the limits requires.
   • Why not C: Reverses the direction of only one of the two pieces, negating the [1,3] integral to -6 while leaving the [3,8] integral unreversed at 11, and adds these to get 5 instead of reversing the whole combined [1,8] integral.
   • Why not D: Uses only the [3,8] integral, correctly reversed to -11, but forgets to also include the [1,3] piece in the combined range from 8 to 1.

9. Question 9Answer: A

   1. The strip width is h = 2. The trapezium rule for 3 strips is: integral is approximately h/2 x [y0 + y3 + 2(y1 + y2)].
   2. Here y0 = 3, y1 = 7, y2 = 15, y3 = 27, so the estimate is 1 x [3 + 27 + 2(7+15)] = 1 x [30 + 44] = 74.
   3. Since f''(x) > 0 (concave up) throughout, each chord joining two of the given points lies above the curve, so each trapezium's area is greater than the true area of the strip it covers.
   4. So the trapezium rule overestimates the true integral here, and the answer is A: 74, an overestimate.
   • Why not B: Uses the correct concavity reasoning, but forgets to multiply by the leading factor h/2 in the trapezium rule, using h = 2 on its own instead, which doubles the correct estimate to 148.
   • Why not C: Correctly computes the trapezium estimate as 74, but reverses the concavity rule: for a concave-up curve the chord joining two points lies above the curve, so the trapezium rule overestimates the true area, not underestimates it.
   • Why not D: Uses the correct concavity reasoning, but omits the factor of 2 on the interior ordinates in the trapezium formula, computing h/2 x (y0+y1+y2+y3) = 1 x (3+7+15+27) = 52 instead of doubling the interior terms.

10. Question 10Answer: D

    1. Use the gradient condition to find k: at x = 2, dy/dx = 6(2)^2 + k(2) - 5 = 24 + 2k - 5 = 19 + 2k. Set this equal to 21: 19 + 2k = 21, so k = 1.
    2. So dy/dx = 6x^2 + x - 5. Integrate to find f(x): f(x) = 2x^3 + x^2/2 - 5x + C.
    3. Use the point (2, 10) to find C: f(2) = 2(8) + 4/2 - 5(2) + C = 16 + 2 - 10 + C = 8 + C.
    4. Set this equal to 10: 8 + C = 10, so C = 2.
    5. So f(x) = 2x^3 + (1/2)x^2 - 5x + 2, and the answer is D.
    • Why not A: Correctly finds k = 1 and correctly integrates to f(x) = 2x^3 + (1/2)x^2 - 5x + C, but then skips solving for C from the given point, setting C equal to the point's y-coordinate directly (C = 10) instead of substituting to get 8 + C = 10 and C = 2.
    • Why not B: Correctly finds k = 1, but integrates 6x^2 by raising the power to x^3 without dividing by the new power 3, using 6x^3 instead of 2x^3 as the antiderivative of that term, and finds a different constant of integration to match.
    • Why not C: Miscalculates 6(2)^2 as 6(2) = 12 instead of 6(4) = 24 when solving for k (forgetting to square the 2 before multiplying by 6), obtaining k = 7 instead of k = 1, and carries this wrong value through the rest of the solution.

11. Question 11Answer: C

    1. At x = 2: y = (2)^3 - 3(2) + 4 = 8 - 6 + 4 = 6, so the point is (2, 6).
    2. Differentiate: dy/dx = 3x^2 - 3. At x = 2, the tangent gradient is 3(4) - 3 = 9.
    3. The normal is perpendicular to the tangent, so its gradient is the negative reciprocal of 9, which is -1/9.
    4. The normal line is y - 6 = (-1/9)(x - 2). Find the x-intercept by setting y = 0: -6 = (-1/9)(x-2), so multiplying both sides by -9 gives 54 = x - 2, so x = 56.
    5. So the answer is C.
    • Why not A: Uses the tangent gradient (9) directly instead of taking the negative reciprocal for the normal, writing the tangent line y - 6 = 9(x-2) and finding its x-intercept, 4/3, instead of the normal's.
    • Why not B: Takes the reciprocal of the tangent gradient but forgets to negate it, using 1/9 instead of -1/9 as the normal's gradient, giving the line y - 6 = (1/9)(x-2) and an x-intercept of -52.
    • Why not D: Takes the negative of the tangent gradient but forgets to also take the reciprocal, using -9 instead of -1/9 as the normal's gradient, giving the line y - 6 = -9(x-2) and an x-intercept of 8/3.

12. Question 12Answer: B

    1. Differentiate once to get the rate of change: dN/dt = -6t^2 + 48t.
    2. This is itself a downward-opening quadratic in t (the coefficient of t^2 is negative), so it is greatest at its own stationary point: differentiate again and set to zero: d(dN/dt)/dt = -12t + 48 = 0, giving t = 4.
    3. Since the coefficient of t^2 in dN/dt is negative, t = 4 gives a maximum of dN/dt, not a minimum.
    4. Evaluate the rate at t = 4: dN/dt = -6(4)^2 + 48(4) = -96 + 192 = 96.
    5. So the answer is B, 96.
    • Why not A: Correctly finds that dN/dt is greatest at t = 4, but then evaluates N(4), the population itself, rather than dN/dt at t = 4, the rate being asked for, giving 356 instead of 96.
    • Why not C: Differentiates N(t) incorrectly, treating d/dt(-2t^3) as -2t^2 rather than -6t^2 (forgetting to bring down the exponent 3 as a multiplying factor), which shifts the maximising time to t = 12 and the resulting value to 288.
    • Why not D: Confuses maximising the rate of change with finding when the population itself is stationary, solving dN/dt = 0 instead of d(dN/dt)/dt = 0, and finds t = 8, where the rate has in fact fallen back to zero rather than reached its peak.

13. Question 13Answer: A

    1. Simplify each term of the numerator by dividing by x^2 before integrating: x^5/x^2 = x^3, 2x^4/x^2 = 2x^2, -3x^2/x^2 = -3. So the integrand is x^3 + 2x^2 - 3.
    2. Integrate: F(x) = x^4/4 + (2/3)x^3 - 3x.
    3. Evaluate at x = 2: F(2) = 16/4 + (2/3)(8) - 6 = 4 + 16/3 - 6 = -2 + 16/3 = 10/3.
    4. Evaluate at x = 1: F(1) = 1/4 + 2/3 - 3 = 3/12 + 8/12 - 36/12 = -25/12.
    5. The definite integral is F(2) - F(1) = 10/3 - (-25/12) = 40/12 + 25/12 = 65/12.
    6. So the answer is A.
    • Why not B: Divides the first two terms of the numerator by x^2 correctly, giving x^3 and 2x^2, but fails to divide the last term at all, leaving it as -3x^2 instead of -3, and integrates the wrong simplified expression x^3 - x^2.
    • Why not C: Correctly simplifies and integrates, but subtracts the limits in the wrong order, computing F(1) - F(2) instead of F(2) - F(1), giving the negative of the correct answer.
    • Why not D: Simplifies the integrand correctly to x^3 + 2x^2 - 3, but integrates 2x^2 by raising the power to x^3 without dividing by the new power 3, keeping the antiderivative as 2x^3 instead of (2/3)x^3.

14. Question 14Answer: D

    1. Differentiate: f'(x) = 3x^2 - 12x + k. Since x = 5 is a stationary point, f'(5) = 0: 3(25) - 12(5) + k = 75 - 60 + k = 15 + k = 0, so k = -15.
    2. So f'(x) = 3x^2 - 12x - 15 = 3(x^2 - 4x - 5) = 3(x-5)(x+1), giving stationary points at x = 5 and x = -1.
    3. Differentiate again: f''(x) = 6x - 12. At x = -1, f''(-1) = -6 - 12 = -18 < 0, so this is a local maximum.
    4. So the answer is D: x = -1, a local maximum.
    • Why not A: Differentiates -6x^2 incorrectly as -6x, forgetting to double the coefficient, so uses f'(x) = 3x^2 - 6x + k instead of 3x^2 - 12x + k when solving for k, obtaining k = -45 instead of k = -15 and a different (and wrong) second stationary point, x = -3.
    • Why not B: Correctly finds k = -15 and correctly factorises f'(x) = 3(x-5)(x+1), but misreads the root of the factor (x+1) as x = 1 instead of x = -1.
    • Why not C: Correctly finds the other stationary point at x = -1, but misapplies the second derivative test there, treating f''(-1) < 0 as indicating a minimum rather than a maximum.

15. Question 15Answer: B

    1. Since g is even, its graph is symmetric about the y-axis, so the area under it on [-4,0] equals the area on [0,4]: the definite integral of g(x) dx from -4 to 0 also equals 9.
    2. Combining these two contiguous, equal-sized ranges: the integral from -4 to 4 = integral from -4 to 0 + integral from 0 to 4 = 9 + 9 = 18.
    3. So the answer is B.
    • Why not A: Correctly applies the even-function symmetry to double the given value, but then also scales by the ratio of the interval lengths (8/4 = 2) as if that were a separate adjustment still needed, compounding the two effects into 9 x 2 x 2 = 36 instead of just 9 x 2 = 18.
    • Why not C: Treats g as an odd function instead of even, assuming the integral over the negative half cancels the integral over the positive half, giving a total of 0 regardless of the given value.
    • Why not D: Forgets that the negative side of the interval contributes its own equal share, and reports the one-sided value 9 as if it already were the total over the whole symmetric interval.