2023 Cambridge IGCSE Mathematics (0580) Practice Paper with Answers

\(a)\) Find the derivative of \(y\): \(\frac{dy}{dx} = 6x^2 - 6x - 12\). For stationary points, set \(\frac{dy}{dx} = 0\): \(6(x^2 - x - 2) = 0 \Rightarrow 6(x-2)(x+1) = 0\). Thus, \(x = 2\) or \(x = -1\). When \(x = -1\): \(y = 2(-1)^3 - 3(-1)^2 - 12(-1) + 8 = 15\). So one stationary point is \((-1, 15)\). When \(x = 2\): \(y = 2(2)^3 - 3(2)^2 - 12(2) + 8 = -12\). So the other stationary point is \((2, -12)\). \(b)\) The equation has exactly three distinct real solutions where the horizontal line \(y = k\) intersects the curve three times, which occurs when \(k\) is strictly between the \(y\)-coordinates of the local minimum and local maximum: \(-12 < k < 15\). \(c)\) When \(x = 0\), \(y = 8\). The gradient of the tangent at \(x = 0\) is: \(m = \frac{dy}{dx}\Big|_{x=0} = -12\). The equation of the tangent is: \(y - 8 = -12(x - 0) \Rightarrow 12x + y = 8\). \(d)\) We set the gradient equal to \(-12\): \(6x^2 - 6x - 12 = -12 \Rightarrow 6x^2 - 6x = 0 \Rightarrow 6x(x - 1) = 0\). This gives \(x = 0\) or \(x = 1\). Since \(x = 0\) is the point in part (c), the other \(x\)-coordinate is \(x = 1\).

\(a)\) M1 for \(\frac{dy}{dx} = 6x^2 - 6x - 12\) (at least two terms correct). M1 for setting derivative to 0 and solving to get \(x = 2\) and \(x = -1\). A1 for \((-1, 15)\). A1 for \((2, -12)\). \(b)\) M2 for identifying the boundaries are the \(y\)-coordinates of the stationary points, i.e., \(-12\) and \(15\) (M1 for one boundary correct or writing inequality with incorrect boundaries). A1 for \(-12 < k < 15\). \(c)\) M1 for substituting \(x = 0\) into the original equation to find \(y = 8\). M1 for substituting \(x = 0\) into the derivative to find \(m = -12\). A1 for full correct completion to show \(12x + y = 8\). \(d)\) M1 for setting derivative equal to \(-12\). A1 for \(x = 1\).

\(a)\) Time taken = \(\frac{40}{x}\) hours. \(b)\) The speed for the return journey is \(x - 3\) km/h. Time taken = \(\frac{40}{x-3}\) hours. \(c)\) Convert 40 minutes to hours: \(\frac{40}{60} = \frac{2}{3}\) hours. The return journey is longer, so: \(\frac{40}{x - 3} - \frac{40}{x} = \frac{2}{3}\). Multiply the entire equation by \(3x(x-3)\) to clear denominators: \(120x - 120(x-3) = 2x(x-3) \Rightarrow 120x - 120x + 360 = 2x^2 - 6x \Rightarrow 360 = 2x^2 - 6x\). Divide both sides by 2: \(180 = x^2 - 3x \Rightarrow x^2 - 3x - 180 = 0\). \(d)\) We factorise \(x^2 - 3x - 180 = 0\) as \((x - 15)(x + 12) = 0\). Thus, \(x = 15\) or \(x = -12\). \(e)\) Since speed must be positive, \(x = 15\). The return speed is \(15 - 3 = 12\) km/h. The time taken is \(\frac{40}{12} = 3\frac{1}{3}\) hours, which is 3 hours 20 minutes.

\(a)\) B1 for \(\frac{40}{x}\). \(b)\) B1 for \(\frac{40}{x-3}\). \(c)\) B1 for converting 40 minutes to \(\frac{2}{3}\) hours. M1 for setting up \(\frac{40}{x-3} - \frac{40}{x} = \frac{2}{3}\). M1 for multiplying by \(3x(x-3)\) to clear fractions. M1 for expanding brackets correctly: \(120x - 120x + 360 = 2x^2 - 6x\). A1 for fully correct simplification leading to \(x^2 - 3x - 180 = 0\). \(d)\) M1 for \((x + a)(x + b) = 0\) where \(ab = -180\) or \(a+b = -3\). A1 for correct factors \((x - 15)(x + 12) = 0\). A1 for \(x = 15\) and \(x = -12\). \(e)\) M1 for choosing \(x = 15\) and calculating return time using \(\frac{40}{12}\). A1 for 3 hours 20 minutes.

\(a)(i)\) Setting \(x = 0\) gives \(y = -5\), so the intersection is \((0, -5)\). \(a)(ii)\) Setting \(y = 0\) gives \(x^2 - 4x - 5 = 0 \Rightarrow (x - 5)(x + 1) = 0\), so \(x = 5\) or \(x = -1\). The coordinates are \((5, 0)\) and \((-1, 0)\). \(b)\) Completing the square: \(x^2 - 4x - 5 = (x - 2)^2 - 4 - 5 = (x - 2)^2 - 9\). This gives \(a = 2\) and \(b = 9\). The minimum point of the curve is \((2, -9)\). \(c)\) Equating the curve and line equations: \(x^2 - 4x - 5 = 2x + 2 \Rightarrow x^2 - 6x - 7 = 0\). Factorising: \((x - 7)(x + 1) = 0\), so \(x = 7\) or \(x = -1\). When \(x = -1\), \(y = 2(-1) + 2 = 0\). When \(x = 7\), \(y = 2(7) + 2 = 16\). The coordinates are \((-1, 0)\) and \((7, 16)\). \(d)\) The inequality is satisfied where the curve is below the line, which occurs between their intersection points, so \(-1 < x < 7\).

\(a)(i)\) B1 for \((0, -5)\). \(a)(ii)\) M1 for factorising or solving the quadratic. A1 for \((5, 0)\) and \((-1, 0)\). \(b)\) M1 for \((x-2)^2 \pm c\). A1 for \((x - 2)^2 - 9\). A1 for minimum point \((2, -9)\). \(c)\) M1 for equating curve and line: \(x^2 - 4x - 5 = 2x + 2\). M1 for setting to quadratic form: \(x^2 - 6x - 7 = 0\). A1 for finding \(x = 7\) and \(x = -1\). A1 for both coordinates \((-1, 0)\) and \((7, 16)\). \(d)\) M1 for stating interval bounds based on their \(x\) values in (c). A1 for \(-1 < x < 7\).

(a) Substitute the values of \(x\) into the equation \(y = \frac{8}{x^2} + 2x - 5\):
- For \(x = -2\): \(p = \frac{8}{(-2)^2} + 2(-2) - 5 = 2 - 4 - 5 = -7\).
- For \(x = 1\): \(q = \frac{8}{1^2} + 2(1) - 5 = 8 + 2 - 5 = 5\).
- For \(x = 4\): \(r = \frac{8}{4^2} + 2(4) - 5 = 0.5 + 8 - 5 = 3.5\).

(b) (i) Start with:
\(\frac{8}{x^2} + 2x - 5 = -x + 3\)
Rearrange terms by adding \(x\) and subtracting \(3\) on both sides:
\(\frac{8}{x^2} + 3x - 8 = 0\)
Multiply the entire equation by \(x^2\) (since \(x \neq 0\)):
\(8 + 3x^3 - 8x^2 = 0\)
Rearrange in descending powers of \(x\):
\(3x^3 - 8x^2 + 8 = 0\)

(b) (ii) The intersections of the curve and the line \(y = -x + 3\) are given by the roots of \(3x^3 - 8x^2 + 8 = 0\).
Since \(x = 2\) is one root, we can divide \(3x^3 - 8x^2 + 8\) by \((x - 2)\) to get \(3x^2 - 2x - 4 = 0\).
Solving \(3x^2 - 2x - 4 = 0\) using the quadratic formula:
\(x = \frac{2 \pm \sqrt{(-2)^2 - 4(3)(-4)}}{2(3)} = \frac{2 \pm \sqrt{52}}{6}\)
This gives \(x \approx 1.54\) and \(x \approx -0.87\).

(c) (i) Expressing the curve as \(y = 8x^{-2} + 2x - 5\) and differentiating with respect to \(x\):
\(\frac{dy}{dx} = 8(-2)x^{-3} + 2 = -16x^{-3} + 2 = -\frac{16}{x^3} + 2\)

(c) (ii) Substitute \(x = 2\) into \(\frac{dy}{dx}\):
\(\text{Gradient} = -\frac{16}{2^3} + 2 = -\frac{16}{8} + 2 = -2 + 2 = 0\)

(a)
B1 for p = -7
B1 for q = 5
B1 for r = 3.5

(b)(i)
M1 for collecting terms to get \(\frac{8}{x^2} + 3x - 8 = 0\) or equivalent
M1 for multiplying by \(x^2\) to eliminate the fraction
A1 for obtaining the final cubic equation \(3x^3 - 8x^2 + 8 = 0\) clearly without errors

(b)(ii)
B1 for \(x = 1.54\) (accept range \(1.5\) to \(1.6\))
B1 for \(x = -0.87\) (accept range \(-0.95\) to \(-0.80\))

(c)(i)
M2 for differentiating at least one term correctly: \(8x^{-2} \rightarrow -16x^{-3}\) or \(2x \rightarrow 2\)
A1 for the complete correct derivative: \(-\frac{16}{x^3} + 2\) or \(-16x^{-3} + 2\)

(c)(ii)
B1 for gradient = 0 (allow follow through from their derivative)

(a) (i) Time taken from A to B is \(\frac{\text{Distance}}{\text{Speed}} = \frac{1200}{x}\) hours.
(a) (ii) Return speed is \(x - 80\) km/h. Time taken is \(\frac{1200}{x - 80}\) hours.

(b) Since the total time is 5.5 hours:
\(\frac{1200}{x} + \frac{1200}{x-80} = 5.5\)
Write 5.5 as \(\frac{11}{2}\):
\(\frac{1200}{x} + \frac{1200}{x-80} = \frac{11}{2}\)
Multiply both sides by \(2x(x - 80)\) to clear denominators:
\(2400(x - 80) + 2400x = 11x(x - 80)\)
\(2400x - 192000 + 2400x = 11x^2 - 880x\)
\(4800x - 192000 = 11x^2 - 880x\)
Rearrange terms to form a quadratic equation:
\(11x^2 - 5680x + 192000 = 0\)

(c) Solve using the quadratic formula with \(a = 11\), \(b = -5680\), \(c = 192000\):
\(x = \frac{-(-5680) \pm \sqrt{(-5680)^2 - 4(11)(192000)}}{2(11)}
= \frac{5680 \pm \sqrt{32262400 - 8448000}}{22}
= \frac{5680 \pm \sqrt{23814400}}{22}\)
Since \(\sqrt{23814400} = 4880\):
\(x = \frac{5680 \pm 4880}{22}\)
This yields:
\(x_1 = \frac{10560}{22} = 480\)
\(x_2 = \frac{800}{22} \approx 36.36\)

(d) The average speed \(x\) must be greater than 80 km/h so that the return speed \(x - 80\) is positive. If \(x = 36.36\), the speed on the return journey would be \(36.36 - 80 = -43.64\) km/h, which is physically impossible. Therefore, the average speed from A to B is \(480\) km/h.

(a)(i)
B1 for \(\frac{1200}{x}\)

(a)(ii)
B1 for \(\frac{1200}{x-80}\)

(b)
M1 for writing \(\frac{1200}{x} + \frac{1200}{x-80} = 5.5\) or equivalent
M1 for multiplying by \(x(x-80)\) to clear fractions
M1 for expanding terms to get \(1200(x-80) + 1200x = 5.5x(x-80)\) or equivalent
A1 for reaching \(11x^2 - 5680x + 192000 = 0\) with no algebraic errors

(c)
M1 for correct substitution into quadratic formula: \(\frac{-(-5680) \pm \sqrt{(-5680)^2 - 4(11)(192000)}}{2(11)}\) (allow 1 sign error)
A1 for evaluating the discriminant to \(23814400\) or evaluating its square root to \(4880\)
B1 for \(x = 480\)
B1 for \(x = 36.36\) (accept \(36.4\))

(d)
B1 for explaining that speed must be greater than 80 (or return speed \(x-80\) must be positive, so \(x = 36.36\) is rejected)
B1 for stating the average speed is \(480\) km/h

(a) (i) \(f(5) = \frac{3(5) + 2}{5 - 4} = \frac{17}{1} = 17\).
(a) (ii) First find \(f(6) = \frac{3(6) + 2}{6 - 4} = \frac{20}{2} = 10\).
Then find \(g(10) = 10^2 - 5 = 100 - 5 = 95\).

(b) Set \(y = \frac{3x + 2}{x - 4}\). Swap variables or solve for \(x\):
\(y(x - 4) = 3x + 2\)
\(xy - 4y = 3x + 2\)
\(xy - 3x = 4y + 2\)
Factorise \(x\):
\(x(y - 3) = 4y + 2\)
\(x = \frac{4y + 2}{y - 3}\)
So, \(f^{-1}(x) = \frac{4x + 2}{x - 3}\).

(c) Set \(g(x) = 31\):
\(x^2 - 5 = 31\)
\(x^2 = 36\)
\(x = \pm 6\).

(d) Set \(f(x) = x\):
\(\frac{3x + 2}{x - 4} = x\)
\(3x + 2 = x(x - 4)\)
\(3x + 2 = x^2 - 4x\)
Rearrange into standard quadratic form:
\(x^2 - 7x - 2 = 0\)
Apply the quadratic formula with \(a = 1\), \(b = -7\), \(c = -2\):
\(x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(1)(-2)}}{2(1)} = \frac{7 \pm \sqrt{49 + 8}}{2} = \frac{7 \pm \sqrt{57}}{2}\)
Since \(\sqrt{57} \approx 7.54983\):
\(x_1 = \frac{7 + 7.54983}{2} \approx 7.27\)
\(x_2 = \frac{7 - 7.54983}{2} \approx -0.27\)

(a)(i)
B1 for 17

(a)(ii)
M1 for finding \(f(6) = 10\)
A1 for 95

(b)
M1 for multiplying by \((x - 4)\) to get \(y(x - 4) = 3x + 2\) (or vice-versa with \(x\) and \(y\))
M1 for isolating terms containing \(x\) on one side and factorising: \(x(y - 3) = 4y + 2\)
A1 for \(f^{-1}(x) = \frac{4x + 2}{x - 3}\) (or equivalent)

(c)
M1 for \(x^2 = 36\)
A1 for \(x = 6\) and \(x = -6\) (both solutions required)

(d)
M1 for setting up the equation \(3x + 2 = x(x - 4)\)
M1 for arranging into the quadratic form \(x^2 - 7x - 2 = 0\)
A1 for \(x = \frac{7 \pm \sqrt{57}}{2}\) or substitution into formula
A1 for both \(7.27\) and \(-0.27\) (correct to 2 decimal places)

(a) To find \( a \), substitute \( x = 0.5 \) into the equation:
\( a = (0.5)^2 - \frac{4}{0.5} = 0.25 - 8 = -7.75 \).
To find \( b \), substitute \( x = 1.5 \) into the equation:
\( b = (1.5)^2 - \frac{4}{1.5} = 2.25 - 2.6667 = -0.4167 \approx -0.42 \).

(b) Set the curve equation equal to the line equation:
\( x^2 - \frac{4}{x} = 3x - 4 \)
Multiply the entire equation by \( x \) (where \( x \neq 0 \)) to clear the fraction:
\( x^3 - 4 = 3x^2 - 4x \)
Rearrange all terms to one side:
\( x^3 - 3x^2 + 4x - 4 = 0 \)
Comparing this to \( x^3 + px^2 + qx + r = 0 \), we get:
\( p = -3 \), \( q = 4 \), \( r = -4 \).

(c) Substitute \( x = 2 \) into the equation:
\( 2^3 - 3(2)^2 + 4(2) - 4 = 8 - 12 + 8 - 4 = 0 \).
Since this equals 0, \( x = 2 \) is a solution.
To find other solutions, factorise out \( (x - 2) \) from the cubic expression:
\( x^3 - 3x^2 + 4x - 4 = (x - 2)(x^2 - x + 2) = 0 \).
For the remaining quadratic part, \( x^2 - x + 2 = 0 \), calculate the discriminant:
\( \Delta = (-1)^2 - 4(1)(2) = 1 - 8 = -7 \).
Since the discriminant is negative, there are no other real roots. Therefore, \( x = 2 \) is the only real solution.

M1: for substituting \( x = 0.5 \) to find \( a \) or substituting \( x = 1.5 \) to find \( b \)
A1: for \( a = -7.75 \) and \( b = -0.42 \)
M1: for equating the curve and the line \( x^2 - \frac{4}{x} = 3x - 4 \)
M1: for multiplying by \( x \) to obtain a cubic equation
A1: for correctly identifying \( p = -3 \), \( q = 4 \), \( r = -4 \)
B1: for verifying \( x = 2 \) by substitution (showing \( 8 - 12 + 8 - 4 = 0 \))
M1: for factorising the cubic expression to obtain the quadratic factor \( x^2 - x + 2 \)
M1: for calculating the discriminant \( \Delta = -7 \)
A1: for a clear conclusion that \( \Delta < 0 \) implies no other real solutions

(a) Factorise the numerator and denominator:
Numerator: \( 2x^2 - 8 = 2(x^2 - 4) = 2(x - 2)(x + 2) \)
Denominator: \( x^2 + 5x + 6 = (x + 2)(x + 3) \)
Simplify by cancelling the common factor of \( (x + 2) \):
\( \frac{2(x - 2)(x + 2)}{(x + 2)(x + 3)} = \frac{2(x - 2)}{x + 3} = \frac{2x - 4}{x + 3} \).

(b) Multiply both sides by \( (2 - t) \):
\( w(2 - t) = 3t + 5 \)
Expand the left-hand side:
\( 2w - wt = 3t + 5 \)
Collect terms with \( t \) on one side and other terms on the opposite side:
\( 2w - 5 = wt + 3t \)
Factorise out \( t \):
\( 2w - 5 = t(w + 3) \)
Divide by \( (w + 3) \):
\( t = \frac{2w - 5}{w + 3} \).

(c) Find a common denominator, which is \( (x - 1)(x + 2) \):
\( \frac{3(x + 2) - 2(x - 1)}{(x - 1)(x + 2)} \)
Expand the numerator:
\( 3x + 6 - 2x + 2 = x + 8 \)
Combine to write the final simplified fraction:
\( \frac{x + 8}{(x - 1)(x + 2)} \) or \( \frac{x + 8}{x^2 + x - 2} \).

M1: for factorising numerator as \( 2(x-2)(x+2) \)
M1: for factorising denominator as \( (x+2)(x+3) \)
A1: for final answer \( \frac{2(x - 2)}{x + 3} \) (or equivalent)
M1: for multiplying both sides by \( (2 - t) \) to get \( w(2 - t) = 3t + 5 \)
M1: for expanding to get \( 2w - wt = 3t + 5 \)
M1: for factorising \( t \) to get \( t(w + 3) = 2w - 5 \)
A1: for final answer \( t = \frac{2w - 5}{w + 3} \)
M1: for expressing fractions with a common denominator \( (x - 1)(x + 2) \)
M1: for expanding numerator \( 3(x + 2) - 2(x - 1) \) correctly with signs
A1: for final answer \( \frac{x + 8}{(x - 1)(x + 2)} \) (or equivalent)

(a) Time taken for the first part of the journey is \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{12}{x} \) hours.

(b) Time taken for the second part of the journey is \( \text{Time} = \frac{12}{x - 1} \) hours.

(c) The sum of the times is equal to the total time of 7 hours:
\( \frac{12}{x} + \frac{12}{x - 1} = 7 \)
Multiply through by the common denominator \( x(x - 1) \):
\( 12(x - 1) + 12x = 7x(x - 1) \)
Expand both sides:
\( 12x - 12 + 12x = 7x^2 - 7x \)
\( 24x - 12 = 7x^2 - 7x \)
Rearrange all terms to one side:
\( 7x^2 - 7x - 24x + 12 = 0 \)
\( 7x^2 - 31x + 12 = 0 \) (as required).

(d) Solve by factorising the quadratic equation \( 7x^2 - 31x + 12 = 0 \):
Find two numbers that multiply to \( 7 \times 12 = 84 \) and add up to \( -31 \). These numbers are \( -28 \) and \( -3 \).
\( 7x^2 - 28x - 3x + 12 = 0 \)
\( 7x(x - 4) - 3(x - 4) = 0 \)
\( (7x - 3)(x - 4) = 0 \)
So \( x = \frac{3}{7} \) or \( x = 4 \).

(e) Since the speed on the second part is \( (x - 1) \) km/h, \( x \) must be greater than 1 so that the speed remains positive.
Therefore, \( x = 4 \).
The time taken for the first 12 km is:
\( \frac{12}{4} = 3 \) hours.
In minutes, this is \( 3 \times 60 = 180 \) minutes.

B1: for \( \frac{12}{x} \)
B1: for \( \frac{12}{x - 1} \)
M1: for setting up the equation \( \frac{12}{x} + \frac{12}{x - 1} = 7 \)
M1: for multiplying by the common denominator \( x(x - 1) \) to get \( 12(x - 1) + 12x = 7x(x - 1) \)
A1: for fully showing the algebraic steps to arrive at \( 7x^2 - 31x + 12 = 0 \)
M1: for factorising into \( (7x - 3)(x - 4) = 0 \) (or equivalent use of the quadratic formula)
A1: for finding the roots \( x = 4 \) and \( x = \frac{3}{7} \)
B1: for selecting \( x = 4 \) with a valid reason (e.g., speed \( x - 1 \) must be positive)
M1: for calculating the time taken as \( \frac{12}{4} = 3 \) hours
A1: for converting to 180 minutes

(a) Substitute \(x = -3\), \(1\), and \(3\) into the equation \(y = \frac{x^3}{4} - 3x + 1\): \\
- For \(x = -3\): \(p = \frac{(-3)^3}{4} - 3(-3) + 1 = -6.75 + 9 + 1 = 3.25\) \\
- For \(x = 1\): \(q = \frac{1^3}{4} - 3(1) + 1 = 0.25 - 3 + 1 = -1.75\) \\
- For \(x = 3\): \(r = \frac{3^3}{4} - 3(3) + 1 = 6.75 - 9 + 1 = -1.25\) \\
\\
(b) The turning points of the curve occur where the derivative is zero: \(\frac{dy}{dx} = \frac{3x^2}{4} - 3 = 0 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2\). \\
- At \(x = -2\), the local maximum is \(y = \frac{(-2)^3}{4} - 3(-2) + 1 = 5\). \\
- At \(x = 2\), the local minimum is \(y = \frac{2^3}{4} - 3(2) + 1 = -3\). \\
To have three distinct real solutions, the horizontal line \(y = k\) must intersect the curve at three distinct points, which lies strictly between the local minimum and local maximum. Thus, \(-3 < k < 5\). \\
\\
(c) (i) We want to solve \(\frac{x^3}{4} - 4x - 1 = 0\). Rearrange to isolate the curve's expression on one side: \(\frac{x^3}{4} - 3x + 1 - x - 2 = 0 \Rightarrow \frac{x^3}{4} - 3x + 1 = x + 2\). Thus, the equation of the line is \(y = x + 2\). \\
(ii) The other two solutions are the remaining \(x\)-coordinates of the intersection of \(y = \frac{x^3}{4} - 3x + 1\) and \(y = x + 2\). Estimating these roots gives \(x \approx -3.9\) and \(x \approx 4.1\). \\
\\
(d) Differentiating \(y = \frac{1}{4}x^3 - 3x + 1\) with respect to \(x\) gives: \(\frac{dy}{dx} = \frac{3}{4}x^2 - 3\). Substitute \(x = 3\): \(\text{Gradient} = \frac{3}{4}(3)^2 - 3 = 6.75 - 3 = 3.75\).

(a) B3: 1 mark for each correct value: \(p = 3.25\), \(q = -1.75\), \(r = -1.25\) (or equivalent fractions). \\
(b) M1: for finding or identifying the y-coordinates of the turning points (e.g. 5 or -3 seen). A1: for either boundary correct, i.e. \(k > -3\) or \(k < 5\). A1: for the complete correct inequality range \(-3 < k < 5\). \\
(c) (i) M1: for equating \(\frac{x^3}{4} - 3x + 1 = mx + c\) to \(\frac{x^3}{4} - 4x - 1 = 0\). A1: for \(y = x + 2\). \\
(ii) B1: for \(x = -3.9\) (accept in the range \([-3.9, -3.8]\)). B1: for \(x = 4.1\) (accept in the range \([4.1, 4.2]\)). \\
(d) M1: for derivative of the form \(ax^2 - b\) where \(a \ne 0\) or \(b \ne 0\) (e.g. \(\frac{3}{4}x^2 - 3\)). A1: for \(3.75\) (or equivalent fraction).

(a) (i) Time taken by Printer A = \(\frac{900}{x}\) hours. \\
(ii) Time taken by Printer B = \(\frac{900}{x + 15}\) hours. \\
\\
(b) Since Printer A takes 3 hours longer: \\
\(\frac{900}{x} - \frac{900}{x + 15} = 3\) \\
Multiply by \(x(x + 15)\): \\
\(900(x + 15) - 900x = 3x(x + 15)\) \\
\(900x + 13500 - 900x = 3x^2 + 45x\) \\
\(13500 = 3x^2 + 45x\) \\
Divide by 3: \\
\(4500 = x^2 + 15x \Rightarrow x^2 + 15x - 4500 = 0\). \\
\\
(c) Factorising the quadratic equation: \\
\((x + 75)(x - 60) = 0\) \\
Giving solutions \(x = 60\) or \(x = -75\). \\
\\
(d) Since rate must be positive, \(x = 60\) grams per hour. \\
Printer B's rate = \(60 + 15 = 75\) grams per hour. \\
For a mass of 950 grams, time taken = \(\frac{950}{75} = \frac{38}{3} = 12\frac{2}{3}\) hours. \\
\(\frac{2}{3}\text{ hours} = \frac{2}{3} \times 60 = 40\) minutes. \\
So, the time taken is 12 hours 40 minutes.

(a) (i) B1: for \(\frac{900}{x}\) \\
(ii) B1: for \(\frac{900}{x + 15}\) \\
\\
(b) M1: for setting up the initial equation: \(\frac{900}{x} - \frac{900}{x + 15} = 3\). M1: for multiplying by \(x(x + 15)\) to clear denominators: \(900(x + 15) - 900x = 3x(x + 15)\). A1: for correct expansion: \(900x + 13500 - 900x = 3x^2 + 45x\). A1: for correct division by 3 and rearrangement to the target equation. \\
\\
(c) M1: for attempt to factorise (e.g. \((x+a)(x-b)\) where \(ab=4500\)) or correct use of quadratic formula. A1: for \((x + 75)(x - 60) = 0\). A1: for \(x = 60\) and \(x = -75\). \\
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(d) M1: for using the positive rate of Printer B: \(60 + 15 = 75\). M1: for calculating \(\frac{950}{75}\) (or \(12.67\) hours). A1: for 12 hours 40 minutes.