2024 Cambridge IGCSE Mathematics - Additional (0606) Practice Paper with Answers

1. Differentiate the equation with respect to \(x\) using the chain rule: \(\frac{dy}{dx} = 4(3x - 1)^3 \cdot 3 - 8 = 12(3x - 1)^3 - 8\). 2. Set the gradient of the tangent to 4: \(12(3x - 1)^3 - 8 = 4\), which simplifies to \(12(3x - 1)^3 = 12\), so \((3x - 1)^3 = 1\). 3. Taking the cube root gives \(3x - 1 = 1\), which solves to \(x = \frac{2}{3}\). 4. Substitute \(x = \frac{2}{3}\) back into the curve equation to find the \(y\)-coordinate: \(y = (3(\frac{2}{3}) - 1)^4 - 8(\frac{2}{3}) = (2 - 1)^4 - \frac{16}{3} = 1 - \frac{16}{3} = -\frac{13}{3}\). Therefore, the coordinates of the point are \((\frac{2}{3}, -\frac{13}{3})\).

M1: Differentiates to find \(\frac{dy}{dx} = k(3x-1)^3 - 8\) where \(k\) is a constant. M1: Sets their derivative equal to 4 and solves for \(x\). A1: Obtains the correct coordinates \((\frac{2}{3}, -\frac{13}{3})\) or equivalent exact fractions.

Let the first term be \(a\) and the common ratio be \(r\) where \(r > 0\). The third term is \(ar^2 = 16\), so \(a = \frac{16}{r^2}\). The sum of the first three terms is \(a + ar + ar^2 = 28\). Substituting \(a\) into this equation gives \(\frac{16}{r^2} + \frac{16}{r} + 16 = 28\). Dividing by 4 and multiplying by \(r^2\) yields the quadratic equation \(4 + 4r + 4r^2 = 7r^2\), which simplifies to \(3r^2 - 4r - 4 = 0\). Factorising this gives \((3r + 2)(r - 2) = 0\). Since \(r\) must be positive, we find \(r = 2\). Substituting this back gives \(a = \frac{16}{2^2} = 4\).

M1: Sets up the equations \(a(1+r+r^2) = 28\) and \(ar^2 = 16\). M1: Eliminates \(a\) to obtain a quadratic in \(r\) and solves for positive \(r\) to find \(r = 2\). A1: Obtains the correct first term \(a = 4\).

1. Substitute \(\cos^2 \theta = 1 - \sin^2 \theta\) into the equation: \(3(1 - \sin^2 \theta) + 5 \sin \theta - 1 = 0\), which simplifies to \(3\sin^2 \theta - 5 \sin \theta - 2 = 0\). 2. Factorising the quadratic gives \((3\sin \theta + 1)(\sin \theta - 2) = 0\). This yields \(\sin \theta = -\frac{1}{3}\) or \(\sin \theta = 2\) (which has no solution). 3. For \(\sin \theta = -\frac{1}{3}\), the basic angle in the range is \(\alpha = \sin^{-1}(\frac{1}{3}) \approx 19.47^\circ\). Since \(\sin \theta\) is negative, the solutions lie in the third and fourth quadrants: \(\theta = 180^\circ + 19.47^\circ = 199.5^\circ\) and \(\theta = 360^\circ - 19.47^\circ = 340.5^\circ\) (both correct to 1 decimal place).

M1: Uses \(\cos^2 \theta = 1 - \sin^2 \theta\) to form a three-term quadratic in \(\sin \theta\) and attempts to solve. M1: Identifies \(\sin \theta = -\frac{1}{3}\). A1: Obtains both correct angles \(199.5^\circ\) and \(340.5^\circ\) (rounded to 1 d.p. as per standard angle rules, accept 199 or 341 if 3 s.f. is used).

The committee of 4 students must contain at least 2 girls. Since there are only 3 girls in total, the committee can contain either 2 girls or 3 girls. Case 1: 2 girls and 2 boys. Number of selections = \(\binom{3}{2} \times \binom{5}{2} = 3 \times 10 = 30\). Case 2: 3 girls and 1 boy. Number of selections = \(\binom{3}{3} \times \binom{5}{1} = 1 \times 5 = 5\). Total number of ways = \(30 + 5 = 35\).

M1: Identifies and considers the two valid cases (2 girls and 2 boys) AND (3 girls and 1 boy). M1: Calculates the combinations for at least one case correctly (e.g., \(\binom{3}{2} \times \binom{5}{2} = 30\)). A1: Obtains the correct total of 35.

1. Combine the logarithms using the product rule: \(\log_4((x + 3)(x - 3)) = 2\), which simplifies to \(\log_4(x^2 - 9) = 2\). 2. Rewrite in exponential form: \(x^2 - 9 = 4^2\), so \(x^2 - 9 = 16\). 3. Solve for \(x\): \(x^2 = 25\), which gives \(x = 5\) or \(x = -5\). Since the argument of the logarithm must be positive, we must have \(x > 3\), which means we reject the negative solution \(x = -5\). Thus, the only solution is \(x = 5\).

M1: Applies logarithm laws to obtain \(\log_4(x^2 - 9) = 2\) or equivalent. M1: Converts to exponential form correctly to get \(x^2 - 9 = 16\) and solves for \(x^2\). A1: Obtains the final answer \(x = 5\) only (explicitly rejecting \(x = -5\)).

(a) To find the points of intersection, we equate the equations: \(3x^2 - x^3 = 2x\). Rearranging gives \(x^3 - 3x^2 + 2x = 0\). Factorising out \(x\): \(x(x^2 - 3x + 2) = 0\), which further factorises to \(x(x-1)(x-2) = 0\). Thus, the \(x\)-coordinates of the intersection points are \(x = 0\), \(x = 1\), and \(x = 2\). (b) The enclosed area is split into two regions: Region 1 from \(x = 0\) to \(x = 1\) and Region 2 from \(x = 1\) to \(x = 2\). For Region 1, the line is above the curve: \(\text{Area}_1 = \int_0^1 (2x - (3x^2 - x^3)) \, dx = \int_0^1 (x^3 - 3x^2 + 2x) \, dx = \left[ \frac{x^4}{4} - x^3 + x^2 \right]_0^1 = \left(\frac{1}{4} - 1 + 1\right) - 0 = \frac{1}{4}\). For Region 2, the curve is above the line: \(\text{Area}_2 = \int_1^2 ((3x^2 - x^3) - 2x) \, dx = \int_1^2 (-x^3 + 3x^2 - 2x) \, dx = \left[ -\frac{x^4}{4} + x^3 - x^2 \right]_1^2 = \left(-\frac{16}{4} + 8 - 4\right) - \left(-\frac{1}{4} + 1 - 1\right) = 0 - \left(-\frac{1}{4}\right) = \frac{1}{4}\). The total area is \(\text{Area}_1 + \text{Area}_2 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}\) or \(0.5\).

(a) M1: Equating the two equations to form a cubic equation. M1: Attempting to factorise the cubic equation. A1: Identifying \(x=0\) and \(x=1\). A1: Identifying \(x=2\). (b) M1: Recognising the need to split the area into two intervals: 0 to 1 and 1 to 2. M1: Attempting to integrate \(\pm(x^3 - 3x^2 + 2x)\). A1: Obtaining correct integrated term: \(\frac{x^4}{4} - x^3 + x^2\) (or signs reversed). M1: Substituting limits 0 and 1 into their integrated expression. M1: Substituting limits 1 and 2 into their integrated expression. A1: Finding both individual areas as \(0.25\) (or equivalent). A1: Obtaining the correct total area of \(0.5\).

(a) Let \(y = (x-2)(2x+3)^{1/2}\). Using the product rule: \(\frac{dy}{dx} = (1)\sqrt{2x+3} + (x-2) \cdot \frac{1}{2}(2x+3)^{-1/2} \cdot 2 = \sqrt{2x+3} + \frac{x-2}{\sqrt{2x+3}}\). Combining these terms over a common denominator: \(\frac{dy}{dx} = \frac{(2x+3) + (x-2)}{\sqrt{2x+3}} = \frac{3x+1}{\sqrt{2x+3}}\). Thus, \(a = 3\) and \(b = 1\). (b) For the stationary point, set \(\frac{dy}{dx} = 0\), which gives \(3x + 1 = 0\), so \(x = -\frac{1}{3}\). Substituting \(x = -\frac{1}{3}\) back into the equation of the curve: \(y = \left(-\frac{1}{3} - 2\right)\sqrt{2\left(-\frac{1}{3}\right)+3} = -\frac{7}{3}\sqrt{\frac{7}{3}} = -\frac{7\sqrt{21}}{9} \approx -3.56\). To find the nature, test the sign of \(\frac{dy}{dx}\) on either side of \(x = -\frac{1}{3}\): for \(x < -\frac{1}{3}\) (e.g., \(x = -1\)), \(\frac{dy}{dx} = \frac{-2}{1} < 0\); for \(x > -\frac{1}{3}\) (e.g., \(x = 0\)), \(\frac{dy}{dx} = \frac{1}{\sqrt{3}} > 0\). Since the gradient changes from negative to positive, the stationary point is a local minimum.

(a) M1: For attempting to use the product rule. A1: For correctly differentiating \(\sqrt{2x+3}\) to get \(\frac{1}{\sqrt{2x+3}}\). A1: For obtaining \(\sqrt{2x+3} + \frac{x-2}{\sqrt{2x+3}}\). M1: For attempting to combine over a common denominator. A1: For obtaining \(\frac{3x+1}{\sqrt{2x+3}}\). (b) M1: Setting their numerator to 0. A1: Correctly finding \(x = -\frac{1}{3}\). M1: Substituting their \(x\)-value into the original equation. A1: Correctly finding the exact \(y\)-coordinate \(-\frac{7\sqrt{21}}{9}\) or decimal equivalent \(-3.56\) (to 3 s.f.). M1: A valid method to determine the nature (e.g., first derivative test or sign table of gradient). A1: Concluding it is a local minimum with correct justification.

(a) We are given: \(T_1 = a = G_1 = a\), \(T_3 = a + 2d = G_2 = ar\), and \(T_{11} = a + 10d = G_3 = ar^2\). From these equations, we can express the common differences: \(a(r-1) = 2d\) (Equation 1) and \(a(r^2-1) = 10d\) (Equation 2). Dividing Equation 2 by Equation 1 (assuming \(d \neq 0\)): \(\frac{a(r^2-1)}{a(r-1)} = \frac{10d}{2d} \implies r+1 = 5 \implies r = 4\). Substitute \(r = 4\) back into Equation 1: \(a(4-1) = 2d \implies 3a = 2d \implies d = 1.5a\). Using the sum formula for the arithmetic progression: \(S_{11} = \frac{11}{2}(2a + 10d) = 187 \implies 11(a+5d) = 187 \implies a+5d=17\). Substituting \(d = 1.5a\): \(a + 5(1.5a) = 17 \implies 8.5a = 17 \implies a = 2\). Using \(a = 2\), we find \(d = 1.5(2) = 3\). (b) The geometric progression has first term \(a = 2\) and common ratio \(r = 4\). The sum of the first 8 terms is given by: \(S_8 = \frac{a(r^8 - 1)}{r - 1} = \frac{2(4^8 - 1)}{4 - 1} = \frac{2(65536 - 1)}{3} = \frac{131070}{3} = 43690\).

(a) M1: Writing expressions for \(T_3\) and \(T_{11}\) in terms of \(a\) and \(d\). M1: Equating these to the geometric progression terms, i.e., \(a + 2d = ar\) and \(a + 10d = ar^2\). M1: Eliminating \(a\) and \(d\) to find \(r\). A1: Correctly obtaining \(r = 4\). M1: Using the sum formula for AP: \(\frac{11}{2}(2a + 10d) = 187\). A1: Showing clearly that \(d = 3\). A1: Finding \(a = 2\). (b) M1: Identifying the geometric progression parameters \(a=2\) and \(r=4\). M1: Applying the geometric sum formula \(S_n = \frac{a(r^n - 1)}{r - 1}\) for \(n=8\). A1: Correctly evaluating \(4^8 = 65536\). A1: Obtaining \(43690\).

(a) Starting with the Left Hand Side (LHS): \(\frac{\sin \theta}{1 + \cos \theta} + \frac{1 + \cos \theta}{\sin \theta} = \frac{\sin^2 \theta + (1 + \cos \theta)^2}{\sin \theta (1 + \cos \theta)} = \frac{\sin^2 \theta + 1 + 2\cos \theta + \cos^2 \theta}{\sin \theta (1 + \cos \theta)}\). Since \(\sin^2 \theta + \cos^2 \theta = 1\), the numerator becomes \(1 + 1 + 2\cos \theta = 2 + 2\cos \theta = 2(1 + \cos \theta)\). Thus, the expression simplfies to \(\frac{2(1 + \cos \theta)}{\sin \theta (1 + \cos \theta)} = \frac{2}{\sin \theta}\) (LHS = RHS). (b) Using the identity, we can rewrite the equation as: \(\frac{2}{\sin 2x} = 5\sin 2x\), which gives \(2 = 5\sin^2 2x \implies \sin^2 2x = 0.4 \implies \sin 2x = \pm \sqrt{0.4}\). For \(0^\circ \le x \le 180^\circ\), we have \(0^\circ \le 2x \le 360^\circ\). Case 1: \(\sin 2x = \sqrt{0.4} \approx 0.63246 \implies 2x = 39.23^\circ\) or \(140.77^\circ \implies x = 19.6^\circ\) or \(70.4^\circ\). Case 2: \(\sin 2x = -\sqrt{0.4} \approx -0.63246 \implies 2x = 219.23^\circ\) or \(320.77^\circ \implies x = 109.6^\circ\) or \(160.4^\circ\). Thus, the solutions in the range are \(x = 19.6^\circ, 70.4^\circ, 109.6^\circ, 160.4^\circ\).

(a) M1: Combining the fractions over a common denominator. A1: Expanding the numerator correctly. M1: Using the identity \(\sin^2\theta + \cos^2\theta = 1\). A1: Factorising and cancelling to obtain \(\frac{2}{\sin\theta}\). (b) M1: Using the identity to rewrite the equation as \(\frac{2}{\sin 2x} = 5\sin 2x\). A1: Rearranging to \(\sin^2 2x = 0.4\). M1: Taking both positive and negative roots. M1: Finding the principal angle (approx \(39.2^\circ\)). A1: Obtaining \(x = 19.6^\circ\) and \(70.4^\circ\). A1: Obtaining \(x = 109.6^\circ\) and \(160.4^\circ\). A1: Correct answers with no extra solutions in range.

(a)(i) Total members is \(7 + 5 = 12\). The number of ways to select 5 members with no restrictions is: \(\binom{12}{5} = \frac{12!}{5!7!} = 792\). (a)(ii) For the committee to have more men than women, the possibilities are: 3 men and 2 women: \(\binom{7}{3} \times \binom{5}{2} = 35 \times 10 = 350\) ways. 4 men and 1 woman: \(\binom{7}{4} \times \binom{5}{1} = 35 \times 5 = 175\) ways. 5 men and 0 women: \(\binom{7}{5} \times \binom{5}{0} = 21 \times 1 = 21\) ways. Total number of ways = \(350 + 175 + 21 = 546\). (b) The total number of ways 5 members can sit in a row is \(5! = 120\). Let the 2 tallest members be treated as a single block. The number of ways to arrange this block and the remaining 3 members is \(4! = 24\). Within the block, the 2 tallest members can be arranged in \(2! = 2\) ways. So, the number of arrangements where they sit next to each other is \(24 \times 2 = 48\). The number of arrangements where they do not sit next to each other is \(120 - 48 = 72\).

(a)(i) M1: For showing \(\binom{12}{5}\). A1: \(792\). (a)(ii) M1: Identifying the three cases (3M 2W, 4M 1W, 5M 0W). M1: Calculating products correctly. A1: Finding individual case values (350, 175, 21). A1: \(546\). (b) M1: Finding total unrestricted arrangements (\(5! = 120\)). M1: Treating the 2 tallest members as a single block. A1: Correctly finding arrangements with them together (\(4! \times 2! = 48\)). M1: Subtracting their 'together' arrangements from the total. A1: \(72\).

(a) The area of the shaded region is found by subtracting the area of triangle \(OAP\) from the area of sector \(OAB\). The area of sector \(OAB\) is \(\frac{1}{2} r^2 \theta = \frac{1}{2} (12^2) \theta = 72\theta\). In the right-angled triangle \(OAP\), \(AP\) is perpendicular to \(OB\), so \(OP = 12\cos\theta\) and \(AP = 12\sin\theta\). The area of triangle \(OAP\) is \(\frac{1}{2} \times OP \times AP = \frac{1}{2} (12\cos\theta)(12\sin\theta) = 72\sin\theta\cos\theta\). Using the identity \(\sin 2\theta = 2\sin\theta\cos\theta\), we have Area of triangle \(OAP = 36\sin 2\theta\). Thus, the area of the shaded region is \(72\theta - 36\sin 2\theta\). (b)(i) When \(\theta = \frac{\pi}{6}\): Arc length \(AB = r\theta = 12 \left(\frac{\pi}{6}\right) = 2\pi\). Length \(AP = 12\sin\left(\frac{\pi}{6}\right) = 6\). Length \(OP = 12\cos\left(\frac{\pi}{6}\right) = 6\sqrt{3}\), so \(PB = OB - OP = 12 - 6\sqrt{3}\). Total perimeter is \(AB + AP + PB = 2\pi + 6 + 12 - 6\sqrt{3} = 2\pi + 18 - 6\sqrt{3}\). (b)(ii) Substituting \(\theta = \frac{\pi}{6}\) into the area formula: \(\text{Area} = 72\left(\frac{\pi}{6}\right) - 36\sin\left(2 \cdot \frac{\pi}{6}\right) = 12\pi - 36\sin\left(\frac{\pi}{3}\right) = 12\pi - 36\left(\frac{\sqrt{3}}{2}\right) = 12\pi - 18\sqrt{3}\).

(a) M1: For sector area \(72\theta\). M1: Expressing \(OP\) and \(AP\) in terms of \(\theta\). A1: For triangle area \(72\sin\theta\cos\theta\). A1: Correctly applying double-angle formula to obtain \(36\sin 2\theta\) and finishing the proof. (b)(i) M1: Finding arc length \(AB = 2\pi\). M1: Finding length \(AP = 6\). M1: Finding length \(PB = 12 - 6\sqrt{3}\). A1: Summing and getting exact perimeter \(2\pi + 18 - 6\sqrt{3}\). (b)(ii) M1: Substituting \(\theta = \frac{\pi}{6}\) into their formula from part (a). A1: Using the exact value \(\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}\). A1: Correctly obtaining \(12\pi - 18\sqrt{3}\).