2023 Cambridge IGCSE Mathematics - Additional (0606) Practice Paper with Answers

When \(x = 0\), \(y = 0 \cdot e^0 = 0\). The point of contact is \((0, 0)\). To find the gradient of the tangent, we differentiate \(y\) using the product rule: \(\frac{dy}{dx} = 1 \cdot e^{-2x} + x \cdot (-2 e^{-2x}) = e^{-2x}(1 - 2x)\). At \(x = 0\), the gradient of the tangent is \(\frac{dy}{dx} = e^{0}(1 - 0) = 1\). Since the normal is perpendicular to the tangent, the gradient of the normal is \(-\frac{1}{1} = -1\). The equation of the normal is \(y - 0 = -1(x - 0)\), which simplifies to \(y = -x\).

M1: Apply the product rule correctly to differentiate \(y = x e^{-2x}\). A1: Obtain correct derivative \(\frac{dy}{dx} = e^{-2x}(1 - 2x)\). M1: Substitute \(x = 0\) to find the gradient of the tangent, then find the negative reciprocal for the normal gradient. A1.5: State the correct final equation of the normal as \(y = -x\) or \(x + y = 0\).

Using the identity \
\sin^2 x = 1 - \cos^2 x\, we substitute into the equation to obtain: \(4(1 - \cos^2 x) - 5\cos x - 5 = 0\). This simplifies to \(4 - 4\cos^2 x - 5\cos x - 5 = 0 \implies 4\cos^2 x + 5\cos x + 1 = 0\). Factorising the quadratic equation gives \((4\cos x + 1)(\cos x + 1) = 0\). This gives two possible cases: \(\cos x = -0.25\) or \(\cos x = -1\). For \(\cos x = -1\), the solution within the range is \(x = 180^\circ\). For \(\cos x = -0.25\), the reference angle is \(\cos^{-1}(0.25) \approx 75.52^\circ\). Since cosine is negative in the second and third quadrants, \(x = 180^\circ - 75.52^\circ = 104.5^\circ\) and \(x = 180^\circ + 75.52^\circ = 255.5^\circ\). Thus, the solutions are \(x = 104.5^\circ, 180^\circ, 255.5^\circ\).

M1: Substitute \(\sin^2 x = 1 - \cos^2 x\) to write the equation in terms of \(\cos x\). A1: Form the correct quadratic equation \(4\cos^2 x + 5\cos x + 1 = 0\). M1: Factorise or solve the quadratic to find \(\cos x = -0.25\) and \(\cos x = -1\). A1.5: State all three correct angles: \(104.5^\circ\), \(180^\circ\), and \(255.5^\circ\) (rounded to 1 decimal place; deduct 0.5 marks for extra out-of-range values or incorrect rounding).

The formula for the \(n\)-th term of a geometric progression is \(u_n = a r^{n-1}\). We are given \(u_3 = a r^2 = 18\) and \(u_6 = a r^5 = 2.25\). Dividing the sixth term equation by the third term equation gives: \(\frac{a r^5}{a r^2} = \frac{2.25}{18} \implies r^3 = 0.125\). Taking the cube root, we get \(r = 0.5\). Substituting \(r = 0.5\) back into the third term equation gives: \(a (0.5)^2 = 18 \implies 0.25a = 18 \implies a = 72\). Using the sum to infinity formula \(S_\infty = \frac{a}{1 - r}\), we get \(S_\infty = \frac{72}{1 - 0.5} = \frac{72}{0.5} = 144\).

M1: Formulate simultaneous equations for the third and sixth terms and attempt to solve for \(r\) by division. A1: Determine correct values \(r = 0.5\) and \(a = 72\). M1: Substitute their \(a\) and \(r\) values into the sum to infinity formula \(S_\infty = \frac{a}{1-r}\). A1.5: Obtain the correct sum to infinity of \(144\).

Using the product law of logarithms, the equation can be written as \(\log_3((x + 4)(x - 2)) = 3\). Converting this logarithmic equation into its equivalent exponential form gives: \((x + 4)(x - 2) = 3^3 \implies x^2 + 2x - 8 = 27\). Rearranging this quadratic equation yields: \(x^2 + 2x - 35 = 0\). Factorising the quadratic equation gives \((x + 7)(x - 5) = 0\), which leads to \(x = -7\) or \(x = 5\). We must check the validity of these solutions in the original logarithmic terms: for \(x = -7\), the term \(\log_3(x-2)\) becomes \(\log_3(-9)\), which is undefined. Therefore, \(x = -7\) is rejected, and the only valid solution is \(x = 5\).

M1: Apply the addition law of logarithms to combine terms: \(\log_3((x+4)(x-2))\). M1: Rewrite the equation in exponential form: \((x+4)(x-2) = 3^3\). A1: Form and solve the quadratic equation \(x^2 + 2x - 35 = 0\) to obtain \(x = 5\) and \(x = -7\). A1.5: Reject the invalid solution \(x = -7\) and state the final correct answer \(x = 5\).

The perimeter of a sector is given by \(P = 2r + r\theta = 30\), which can be rewritten to express \(r\theta\) as: \(r\theta = 30 - 2r\). The area of a sector is given by \(A = \frac{1}{2}r^2\theta = 50\). This can be rewritten as \(r(r\theta) = 100\). Substituting the expression for \(r\theta\) into this equation yields: \(r(30 - 2r) = 100 \implies 30r - 2r^2 = 100\). Rearranging into standard quadratic form gives: \(2r^2 - 30r + 100 = 0\). Dividing the entire equation by 2, we get \(r^2 - 15r + 50 = 0\). Factorising this quadratic equation yields \((r - 5)(r - 10) = 0\). Therefore, the possible values of \(r\) are \(r = 5\) or \(r = 10\).

M1: State correct equations for both perimeter \(2r + r\theta = 30\) and area \(\frac{1}{2}r^2\theta = 50\). M1: Set up a substitution to eliminate \(\theta\) and obtain an equation in \(r\) only. A1: Correctly simplify to the quadratic equation \(r^2 - 15r + 50 = 0\) (or equivalent). A1.5: Solve to find both correct values: \(r = 5\) and \(r = 10\).

Since the team must contain at least 3 engineers, we must consider three mutually exclusive cases: Case 1: Exactly 3 engineers and 2 technicians. The number of ways is \(\binom{6}{3} \times \binom{4}{2} = 20 \times 6 = 120\). Case 2: Exactly 4 engineers and 1 technician. The number of ways is \(\binom{6}{4} \times \binom{4}{1} = 15 \times 4 = 60\). Case 3: Exactly 5 engineers and 0 technicians. The number of ways is \(\binom{6}{5} \times \binom{4}{0} = 6 \times 1 = 6\). To find the total number of ways, we add the results of the three cases together: \(120 + 60 + 6 = 186\).

M1: Identify the three valid cases and calculate the combinations for at least one case correctly. A1: Calculate the combination counts for at least two cases correctly. M1: Add the number of ways from all three correct mutually exclusive cases. A1.5: State the correct final answer of \(186\).

(i) To find where the curve meets the \( x \)-axis, set \( y = 0 \): \( e^{2x} - 4e^x + 3 = 0 \). Let \( u = e^x \), then \( u^2 - 4u + 3 = 0 \). Factoring gives \( (u - 1)(u - 3) = 0 \), so \( e^x = 1 \) or \( e^x = 3 \). This yields \( x = 0 \) or \( x = \ln 3 \). Thus, the coordinates are \( (0, 0) \) and \( (\ln 3, 0) \). (ii) The area \( A \) is given by \( -\int_{0}^{\ln 3} (e^{2x} - 4e^x + 3) \, dx \) because the curve lies below the \( x \)-axis in this interval. Integrating: \( \int (e^{2x} - 4e^x + 3) \, dx = \left[ \frac{1}{2}e^{2x} - 4e^x + 3x \right] \). Evaluating this expression from \( 0 \) to \( \ln 3 \): at \( x = \ln 3 \), we have \( \frac{1}{2}(9) - 4(3) + 3\ln 3 = 4.5 - 12 + 3\ln 3 = -7.5 + 3\ln 3 \). At \( x = 0 \), we have \( \frac{1}{2}(1) - 4(1) + 0 = -3.5 \). The difference is \( (-7.5 + 3\ln 3) - (-3.5) = 3\ln 3 - 4 \). Since this difference is negative, the area is the absolute value: \( 4 - 3\ln 3 \).

(i) M1: Set \( y = 0 \) and form a quadratic in \( e^x \). A1: Correctly solve for \( e^x = 1 \) and \( e^x = 3 \). A1: Write coordinates \( (0, 0) \) and \( (\ln 3, 0) \). (ii) M1: Attempt integration of the curve equation. A1: Correctly integrate to get \( \frac{1}{2}e^{2x} - 4e^x + 3x \). M1: Substitute limits \( 0 \) and \( \ln 3 \) into their integrated expression. A1: Obtain \( 3\ln 3 - 4 \) or equivalent. A1: Provide the final positive area value \( 4 - 3\ln 3 \).

Divide both sides of the equation by \( \cos(2\theta - \frac{\pi}{4}) \) to get: \( 3 \tan(2\theta - \frac{\pi}{4}) = 2 \implies \tan(2\theta - \frac{\pi}{4}) = \frac{2}{3} \). Let \( x = 2\theta - \frac{\pi}{4} \). Since \( 0 \le \theta \le \pi \), we have \( -\frac{\pi}{4} \le x \le \frac{7\pi}{4} \). The basic angle is \( \tan^{-1}(\frac{2}{3}) \approx 0.5880 \) radians. In the interval, the solutions for \( x \) are: \( x = 0.5880 \) (Quadrant 1) and \( x = \pi + 0.5880 \approx 3.7296 \) (Quadrant 3). Now solve for \( \theta \): 1) \( 2\theta - \frac{\pi}{4} = 0.5880 \implies 2\theta = 0.5880 + 0.7854 = 1.3734 \implies \theta \approx 0.687 \) radians. 2) \( 2\theta - \frac{\pi}{4} = 3.7296 \implies 2\theta = 3.7296 + 0.7854 = 4.5150 \implies \theta \approx 2.26 \) radians.

M1: Divide to obtain \( \tan(2\theta - \frac{\pi}{4}) = \frac{2}{3} \). M1: Find basic angle in radians \( \approx 0.588 \). M1: Identify the correct quadrants and intervals. A1: Find one correct value of \( 2\theta - \frac{\pi}{4} \). A1: Find the second correct value of \( 2\theta - \frac{\pi}{4} \). A1: Calculate \( \theta = 0.687 \). A1: Calculate \( \theta = 2.26 \).

(i) The 2nd, 5th, and 14th terms of the AP are \( T_2 = a + d \), \( T_5 = a + 4d \), and \( T_{14} = a + 13d \). Since these form a GP, \( (a+4d)^2 = (a+d)(a+13d) \). Expanding both sides: \( a^2 + 8ad + 16d^2 = a^2 + 14ad + 13d^2 \). Simplifying gives: \( 3d^2 - 6ad = 0 \implies 3d(d - 2a) = 0 \). Since \( d \neq 0 \), we must have \( d = 2a \). (ii) The 10th term of the AP is \( T_{10} = a + 9d = 57 \). Substituting \( d = 2a \): \( a + 18a = 57 \implies 19a = 57 \implies a = 3 \). (iii) The first term of the GP is \( A = T_2 = a + d = 3 + 6 = 9 \). The common ratio is \( r = 3 \). The sum of the first 15 terms is \( S_{15} = \frac{A(r^{15}-1)}{r-1} = \frac{9(3^{15}-1)}{3-1} = 4.5(3^{15}-1) \).

(i) M1: Express the three terms of the AP. M1: Set up the geometric progression equation \( (a+4d)^2 = (a+d)(a+13d) \). A1: Expand and simplify to obtain \( 3d^2 - 6ad = 0 \). A1: Conclude \( d = 2a \) since \( d \neq 0 \). (ii) M1: Use formula for the 10th term of AP. A1: Correctly solve for \( a = 3 \). (iii) M1: Find first term \( A = 9 \) and common ratio \( r = 3 \) of GP. M1: Apply GP sum formula for \( n = 15 \). A1: Provide final exact form \( 4.5(3^{15}-1) \) or equivalent.

(i) \( \overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (11\mathbf{i} - 2\mathbf{j}) - (3\mathbf{i} + 4\mathbf{j}) = 8\mathbf{i} - 6\mathbf{j} \). The magnitude is \( |\overrightarrow{AB}| = \sqrt{8^2 + (-6)^2} = 10 \). The unit vector is \( \frac{1}{10}(8\mathbf{i} - 6\mathbf{j}) = 0.8\mathbf{i} - 0.6\mathbf{j} \). (ii) \( \overrightarrow{AC} = \frac{3}{4}(8\mathbf{i} - 6\mathbf{j}) = 6\mathbf{i} - 4.5\mathbf{j} \). The position vector of \( C \) is \( \overrightarrow{OC} = \overrightarrow{OA} + \overrightarrow{AC} = (3\mathbf{i} + 4\mathbf{j}) + (6\mathbf{i} - 4.5\mathbf{j}) = 9\mathbf{i} - 0.5\mathbf{j} \). (iii) Let \( \overrightarrow{OD} = p\mathbf{i} + q\mathbf{j} \). Since \( \overrightarrow{OD} \) is perpendicular to \( \overrightarrow{AB} \), their dot product is zero: \( 8p - 6q = 0 \implies p = \frac{3}{4}q \). The magnitude of \( \overrightarrow{OD} \) is 10, so \( p^2 + q^2 = 100 \). Substituting \( p \): \( (\frac{3}{4}q)^2 + q^2 = 100 \implies \frac{25}{16}q^2 = 100 \implies q^2 = 64 \implies q = \pm 8 \). If \( q = 8 \), \( p = 6 \); if \( q = -8 \), \( p = -6 \). Thus, the position vectors of \( D \) are \( 6\mathbf{i} + 8\mathbf{j} \) and \( -6\mathbf{i} - 8\mathbf{j} \).

(i) M1: Calculate \( \overrightarrow{AB} = 8\mathbf{i} - 6\mathbf{j} \). M1: Divide by magnitude to get unit vector. A1: Correct unit vector \( 0.8\mathbf{i} - 0.6\mathbf{j} \). (ii) M1: Compute \( \overrightarrow{AC} \). A1: Correct position vector \( 9\mathbf{i} - 0.5\mathbf{j} \). (iii) M1: Use perpendicularity condition \( 8p - 6q = 0 \). M1: Use magnitude condition \( p^2 + q^2 = 100 \). A1: Solve to find \( q = \pm 8 \) and \( p = \pm 6 \). A1: Correct two vectors.

From the first equation, \( \log_3(2x + y) = 2 \implies 2x + y = 3^2 = 9 \implies y = 9 - 2x \). From the second equation, \( \log_3 x + \log_3(y - 2) = 1 \implies \log_3(x(y - 2)) = 1 \implies x(y - 2) = 3 \). Substitute \( y = 9 - 2x \) into this equation: \( x((9 - 2x) - 2) = 3 \implies x(7 - 2x) = 3 \implies 2x^2 - 7x + 3 = 0 \). Factoring the quadratic: \( (2x - 1)(x - 3) = 0 \). This gives two potential solutions for \( x \): \( x = 3 \) or \( x = 0.5 \). Find the corresponding \( y \) values: If \( x = 3 \), then \( y = 9 - 2(3) = 3 \). If \( x = 0.5 \), then \( y = 9 - 2(0.5) = 8 \). Check validity of solutions for the logarithmic expressions: For \( (3, 3) \): \( 2x+y = 9 > 0 \), \( x = 3 > 0 \), \( y-2 = 1 > 0 \). (Valid) For \( (0.5, 8) \): \( 2x+y = 9 > 0 \), \( x = 0.5 > 0 \), \( y-2 = 6 > 0 \). (Valid) Thus, both solutions are valid.

M1: Eliminate logarithm from the first equation to get \( 2x + y = 9 \). M1: Use laws of logarithms on the second equation to get \( x(y - 2) = 3 \). M1: Substitute \( y = 9 - 2x \) to form a quadratic equation. A1: Correct quadratic equation \( 2x^2 - 7x + 3 = 0 \). A1: Solve quadratic to get \( x = 3 \) and \( x = 0.5 \). A1: Find corresponding \( y \) values \( y = 3 \) and \( y = 8 \). B1: Verify that both pairs satisfy the domain of logarithmic functions.

(i) The perimeter of a sector is given by \( P = 2r + r\theta \). We are given \( 2r + r\theta = 24 \). Thus, \( r\theta = 24 - 2r \implies \theta = \frac{24}{r} - 2 \). (ii) The area of a sector is given by \( A = \frac{1}{2}r^2\theta \). Substituting \( \theta = \frac{24-2r}{r} \) into this formula: \( A = \frac{1}{2}r^2 \left(\frac{24-2r}{r}\right) = \frac{1}{2}r(24-2r) = 12r - r^2 \). (iii) Set \( A = 32 \): \( 12r - r^2 = 32 \implies r^2 - 12r + 32 = 0 \). Factoring the quadratic: \( (r-4)(r-8) = 0 \). This gives two possible values for \( r \): \( r = 4 \) or \( r = 8 \). Using \( \theta = \frac{24}{r} - 2 \): If \( r = 4 \), then \( \theta = \frac{24}{4} - 2 = 4 \) radians. If \( r = 8 \), then \( \theta = \frac{24}{8} - 2 = 1 \) radian.

(i) M1: Set up the perimeter equation \( 2r + r\theta = 24 \). A1: Correctly express \( \theta = \frac{24}{r} - 2 \). (ii) M1: Use the area of a sector formula \( A = \frac{1}{2}r^2\theta \). A1: Complete verification to show \( A = 12r - r^2 \). (iii) M1: Equate area to 32 to get a quadratic equation in \( r \). A1: Correctly solve for \( r = 4 \) and \( r = 8 \). A1: Calculate \( \theta = 4 \) corresponding to \( r = 4 \). A1: Calculate \( \theta = 1 \) corresponding to \( r = 8 \).

From \(\log_3(x-2y) = 2\), we have \(x-2y = 3^2 = 9\), which gives \(x = 2y + 9\). From \(\log_3 x + \log_3 y = \log_3 35\), using the laws of logarithms we get \(\log_3(xy) = \log_3 35\), which simplifies to \(xy = 35\). Substituting \(x = 2y + 9\) into \(xy = 35\) gives \((2y+9)y = 35\), which expands and rearranges to the quadratic equation \(2y^2 + 9y - 35 = 0\). Factoring this quadratic gives \((2y-5)(y+7) = 0\), which yields \(y = 2.5\) or \(y = -7\). Since the argument of a logarithm must be positive, \(y\) must be greater than zero, so we reject \(y = -7\) and keep \(y = 2.5\). Substituting \(y = 2.5\) back into \(x = 2y + 9\) gives \(x = 14\). Thus, the solution is \(x = 14\) and \(y = 2.5\).

M1 for converting logarithmic equation to index form to get \(x-2y=9\).
M1 for applying the product law of logarithms to get \(xy=35\).
M1 for substituting one equation into the other to obtain a quadratic equation in one variable.
A1 for obtaining the correct quadratic equation \(2y^2+9y-35=0\) or equivalent.
M1 for solving the quadratic equation to find two roots and rejecting the negative root.
A1.25 for both correct final values \(x = 14\) and \(y = 2.5\).

To find the equation of the curve, we integrate the gradient function: \(y = \int (10(2x-3)^4 - 160) dx\). Integrating term by term gives: \(y = \frac{10(2x-3)^5}{5 \times 2} - 160x + C\), which simplifies to \(y = (2x-3)^5 - 160x + C\). We are given that the curve passes through the point \((1, 5)\). Substituting \(x = 1\) and \(y = 5\) into the equation: \(5 = (2(1)-3)^5 - 160(1) + C \implies 5 = (-1)^5 - 160 + C \implies 5 = -1 - 160 + C \implies C = 166\). Therefore, the equation of the curve is \(y = (2x-3)^5 - 160x + 166\).

M1 for attempting to integrate \(10(2x-3)^4\).
A1 for correct integration of this term to get \((2x-3)^5\) (or equivalent).
A1 for correct integration of \(-160\) to get \(-160x\).
M1 for substituting \(x=1\) and \(y=5\) into their integrated expression containing a constant of integration \(C\).
A1 for finding \(C = 166\).
A1.25 for the correct final equation \(y = (2x-3)^5 - 160x + 166\).

Using the trigonometric identity \(\sec^2 \theta = 1 + \tan^2 \theta\), we substitute this into the equation to get: \(3(1 + \tan^2 \theta) - 5 \tan \theta - 5 = 0 \implies 3\tan^2 \theta - 5\tan \theta - 2 = 0\). Let \(u = \tan \theta\), then we have the quadratic equation \(3u^2 - 5u - 2 = 0\). Factoring the quadratic gives \((3u+1)(u-2) = 0\), which yields \(\tan \theta = 2\) or \(\tan \theta = -\frac{1}{3}\). For \(\tan \theta = 2\), the basic angle is \(\tan^{-1}(2) \approx 63.43^\circ\). Since \(0^\circ \le \theta \le 180^\circ\) and \(\tan \theta\) is positive, \(\theta = 63.4^\circ\). For \(\tan \theta = -\frac{1}{3}\), the basic angle is \(\tan^{-1}(\frac{1}{3}) \approx 18.43^\circ\). Since \(\tan \theta\) is negative, \(\theta = 180^\circ - 18.43^\circ = 161.6^\circ\). Therefore, the solutions in the given range are \(\theta = 63.4^\circ\) and \(\theta = 161.6^\circ\).

M1 for using the identity \(\sec^2 \theta = 1 + \tan^2 \theta\).
A1 for forming the correct quadratic equation \(3\tan^2 \theta - 5\tan \theta - 2 = 0\).
M1 for solving the quadratic equation to get \(\tan \theta = 2\) or \(\tan \theta = -1/3\).
A1 for finding \(\theta = 63.4^\circ\).
M1 for finding the second angle in the correct quadrant.
A1.25 for both correct angles \(\theta = 63.4^\circ\) and \(\theta = 161.6^\circ\) rounded to 1 decimal place.

Using the binomial theorem, the expansion of \((1 + ax)^n\) is: \((1 + ax)^n = 1 + n(ax) + \frac{n(n-1)}{2}(ax)^2 + \dots = 1 + anx + \frac{n(n-1)a^2}{2}x^2 + \dots\). Comparing the coefficients of \(x\) and \(x^2\) with the given expansion: 1) \(an = -30\), 2) \(\frac{n(n-1)a^2}{2} = 405\). From the first equation, we can express \(a\) in terms of \(n\): \(a = -\frac{30}{n}\). Substituting this into the second equation: \(\frac{n(n-1)}{2} \left(-\frac{30}{n}\right)^2 = 405 \implies \frac{n(n-1)}{2} \cdot \frac{900}{n^2} = 405 \implies \frac{450(n-1)}{n} = 405\). Multiplying both sides by \(n\) gives: \(450n - 450 = 405n \implies 45n = 450 \implies n = 10\). Substituting \(n = 10\) back into \(a = -\frac{30}{n}\) gives \(a = -\frac{30}{10} = -3\). Thus, \(a = -3\) and \(n = 10\).

M1 for writing down the terms of the expansion in terms of \(a\) and \(n\).
A1 for setting up the two simultaneous equations: \(an = -30\) and \(\frac{n(n-1)a^2}{2} = 405\).
M1 for attempting to eliminate one variable (e.g., substituting \(a = -30/n\)).
A1 for obtaining a linear or quadratic equation in \(n\) (e.g., \(450(n-1) = 405n\)).
A1 for finding \(n = 10\).
A1.25 for finding \(a = -3\).

(a) Using the quotient rule with \(u = \mathrm{e}^{2x}\) and \(v = 2x - 1\):
\(\frac{\mathrm{d}u}{\mathrm{d}x} = 2\mathrm{e}^{2x}\)
\(\frac{\mathrm{d}v}{\mathrm{d}x} = 2\)

\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{v \frac{\mathrm{d}u}{\mathrm{d}x} - u \frac{\mathrm{d}v}{\mathrm{d}x}}{v^2} = \frac{(2x-1)(2\mathrm{e}^{2x}) - (\mathrm{e}^{2x})(2)}{(2x-1)^2}\)

\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2\mathrm{e}^{2x}(2x-1-1)}{(2x-1)^2} = \frac{4(x-1)\mathrm{e}^{2x}}{(2x-1)^2}\).

(b) At the stationary point, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\(\frac{4(x-1)\mathrm{e}^{2x}}{(2x-1)^2} = 0 \implies 4(x-1)\mathrm{e}^{2x} = 0\).

Since \(\mathrm{e}^{2x} \ne 0\), we have \(x - 1 = 0 \implies x = 1\).

Substituting \(x = 1\) into the original equation:
\(y = \frac{\mathrm{e}^{2(1)}}{2(1) - 1} = \mathrm{e}^2\).

So the exact coordinates are \((1, \mathrm{e}^2)\).

(c) To find the nature, we look at the sign of \(\frac{\mathrm{d}y}{\mathrm{d}x}\) on either side of \(x = 1\):

For \(x = 0.9\): \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{4(0.9-1)\mathrm{e}^{1.8}}{(2(0.9)-1)^2} = \frac{-0.4\mathrm{e}^{1.8}}{0.64} < 0\).

For \(x = 1.1\): \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{4(1.1-1)\mathrm{e}^{2.2}}{(2(1.1)-1)^2} = \frac{0.4\mathrm{e}^{2.2}}{1.44} > 0\).

Since the gradient changes from negative to positive as \(x\) increases through 1, the stationary point is a minimum.

(Alternatively, using the second derivative:
\(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{(2x-1)^2[4\mathrm{e}^{2x} + 8(x-1)\mathrm{e}^{2x}] - 4(x-1)\mathrm{e}^{2x}[4(2x-1)]}{(2x-1)^4}\).
At \(x = 1\):
\(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{(1)^2[4\mathrm{e}^2 + 0] - 0}{1^4} = 4\mathrm{e}^2 > 0\).
Since \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} > 0\), it is a minimum point.)

Part (a)
M1: For applying the quotient rule correctly.
A1: For correct differentiation of both parts and simplification to the required form.

Part (b)
M1: For setting dy/dx = 0 and solving for x.
A1: For x = 1 and y = e^2.

Part (c)
M1: For a valid method to determine the nature of the stationary point (either sign change of gradient or second derivative).
A1: For showing the gradient is negative before and positive after (or second derivative > 0) and concluding it is a minimum.

(a) Using the double-angle identities \(\cos 2\theta = 1 - 2\sin^2\theta\) and \(\sin 2\theta = 2\sin\theta\cos\theta\):

LHS \(= \frac{1 - (1 - 2\sin^2\theta)}{2\sin\theta\cos\theta} = \frac{2\sin^2\theta}{2\sin\theta\cos\theta} = \frac{\sin\theta}{\cos\theta} = \tan\theta\). (proven)

(b) Using the identity from part (a) with \(\theta = 2x\):

\(\frac{1 - \cos 4x}{\sin 4x} = \tan 2x\).

The equation becomes:

\(\tan 2x = \sqrt{3}\).

For \(0^\circ \le x \le 180^\circ\), we have \(0^\circ \le 2x \le 360^\circ\).

The basic angle for \(\tan 2x = \sqrt{3}\) is \(60^\circ\).

Since tangent is positive in the 1st and 3rd quadrants:

\(2x = 60^\circ\) or \(2x = 180^\circ + 60^\circ = 240^\circ\).

Solving for \(x\):

\(x = 30^\circ\) or \(x = 120^\circ\).

Part (a)
M1: For using cos(2θ) = 1 - 2sin^2(θ) (or equivalent) or sin(2θ) = 2sin(θ)cos(θ).
A1: For fully correct algebraic simplification to tan(θ).

Part (b)
M1: For recognizing the relationship with part (a) to write tan(2x) = √3.
M1: For finding at least one correct value for 2x or finding the basic angle 60°.
A1: For x = 30°.
A1: For x = 120° and no other values in the range.

(a) In terms of \(a\) and \(d\):
1st term: \(T_1 = a\)
4th term: \(T_4 = a+3d\)
5th term: \(T_5 = a+4d\)

Since these terms are in geometric progression, we have:
\(\frac{a+3d}{a} = \frac{a+4d}{a+3d}\)
\((a+3d)^2 = a(a+4d)\)
\(a^2 + 6ad + 9d^2 = a^2 + 4ad\)
\(2ad + 9d^2 = 0 \implies d(2a + 9d) = 0\).

Since \(d \ne 0\), we have \(2a + 9d = 0 \implies a = -4.5d\) (shown).

Using the 2nd term of the arithmetic progression:
\(T_2 = a + d = 7\).

Substitute \(a = -4.5d\):
\(-4.5d + d = 7 \implies -3.5d = 7 \implies d = -2\).

Then \(a = -4.5(-2) = 9\).

(b) The sum of the first 25 terms of the AP:
\(S_{n} = \frac{n}{2}[2a + (n-1)d]\)
\(S_{25} = \frac{25}{2}[2(9) + 24(-2)] = 12.5[18 - 48] = 12.5[-30] = -375\).

(c) The terms of the GP are:
First term, \(A = T_1 = a = 9\)
Second term, \(T_4 = a + 3d = 9 + 3(-2) = 3\)

So the common ratio \(r = \frac{3}{9} = \frac{1}{3}\).

The sum to infinity is:
\(S_{\infty} = \frac{A}{1-r} = \frac{9}{1 - 1/3} = \frac{9}{2/3} = 13.5\).

Part (a)
M1: For setting up the ratio of GP terms: (a+3d)/a = (a+4d)/(a+3d).
A1: For deriving a = -4.5d.
M1: For using a+d = 7 with a = -4.5d to find d.
A1: For d = -2 and a = 9.

Part (b)
M1: For applying the correct AP sum formula.
A1: For -375.

Part (c)
M1: For identifying the first term of the GP (A = 9) and finding the common ratio (r = 1/3).
A1: For applying the sum to infinity formula to get 13.5 (or 27/2).

(a) Total number of people is \(6 + 7 = 13\).

The number of ways to choose 5 people from 13 is:
\(\binom{13}{5} = \frac{13 \times 12 \times 11 \times 10 \times 9}{5 \times 4 \times 3 \times 2 \times 1} = 1287\).

(b) For at least 3 women:
Case 1: 3 women and 2 men
Number of ways = \(\binom{7}{3} \times \binom{6}{2} = 35 \times 15 = 525\).

Case 2: 4 women and 1 man
Number of ways = \(\binom{7}{4} \times \binom{6}{1} = 35 \times 6 = 210\).

Case 3: 5 women and 0 men
Number of ways = \(\binom{7}{5} \times \binom{6}{0} = 21 \times 1 = 21\).

Total number of ways = \(525 + 210 + 21 = 756\).

(c) Let the particular man be \(M\) and the particular woman be \(W\).

The number of committees where both \(M\) and \(W\) are present:
Since both \(M\) and \(W\) are selected, we must choose 3 more people from the remaining \(13 - 2 = 11\) people.

Number of ways = \(\binom{11}{3} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\).

The number of ways they do not serve together is:
Total committees - Committees with both \(M\) and \(W\)
\(= 1287 - 165 = 1122\).

(Alternatively, we can sum the cases where neither is chosen, only \(M\) is chosen, or only \(W\) is chosen:
\(\binom{11}{5} + \binom{11}{4} + \binom{11}{4} = 462 + 330 + 330 = 1122\).)

Part (a)
M1: For 13C5.
A1: For 1287.

Part (b)
M1: For considering three cases (3W 2M, 4W 1M, 5W) and finding the combinations for at least one case.
A1: For all three correct individual products (525, 210, 21).
A1: For 756.

Part (c)
M1: For calculating the number of ways containing both restricted people: 11C3 = 165.
M1: For subtracting this from their total in part (a).
A1: For 1122.

(a) To find the intersection points, set the equations equal to each other:
\(\sqrt{3x + 4} = \frac{1}{2}x + 2\)

Square both sides:
\(3x + 4 = \left(\frac{1}{2}x + 2\right)^2\)
\(3x + 4 = \frac{1}{4}x^2 + 2x + 4\)

Multiply by 4:
\(12x + 16 = x^2 + 8x + 16\)
\(x^2 - 4x = 0\)
\(x(x - 4) = 0\)

So, \(x = 0\) or \(x = 4\).

When \(x = 0\): \(y = \frac{1}{2}(0) + 2 = 2\). Point is \((0, 2)\).
When \(x = 4\): \(y = \frac{1}{2}(4) + 2 = 4\). Point is \((4, 4)\).

The intersection points are \((0, 2)\) and \((4, 4)\).

(b) For \(0 \le x \le 4\), the curve lies above the line. The area \(A\) is given by:
\(A = \int_{0}^{4} \left( \sqrt{3x + 4} - \left(\frac{1}{2}x + 2\right) \right) \mathrm{d}x\)

Let's integrate each part:
\(\int \sqrt{3x + 4} \mathrm{d}x = \int (3x + 4)^{1/2} \mathrm{d}x = \frac{(3x+4)^{3/2}}{\frac{3}{2} \times 3} = \frac{2}{9}(3x+4)^{3/2}\)
\(\int \left(\frac{1}{2}x + 2\right) \mathrm{d}x = \frac{1}{4}x^2 + 2x\)

Evaluate from 0 to 4:
\(\left[ \frac{2}{9}(3x+4)^{3/2} - \frac{1}{4}x^2 - 2x \right]_0^4\)

At \(x = 4\):
\(\frac{2}{9}(3(4)+4)^{3/2} - \frac{1}{4}(4)^2 - 2(4) = \frac{2}{9}(16)^{3/2} - 4 - 8 = \frac{2}{9}(64) - 12 = \frac{128}{9} - \frac{108}{9} = \frac{20}{9}\)

At \(x = 0\):
\(\frac{2}{9}(4)^{3/2} - 0 - 0 = \frac{2}{9}(8) = \frac{16}{9}\)

Area = \(\frac{20}{9} - \frac{16}{9} = \frac{4}{9}\).

Part (a)
M1: For equating the curve and the line and squaring both sides.
M1: For forming and solving the quadratic equation x^2 - 4x = 0.
A1: For both points (0, 2) and (4, 4).

Part (b)
M1: For integrating (3x+4)^{1/2} to obtain k(3x+4)^{3/2}.
A1: For the correct integrated term 2/9(3x+4)^{3/2}.
M1: For finding the area under the line or integrating the line term: 1/4x^2 + 2x.
M1: For substituting the limits 4 and 0 into their integrated expression.
A1: For the correct final exact area of 4/9 (or 0.444).

(a) Using the laws of logarithms:
\(\log_2(x+3) + \log_2(x-3) = \log_2((x+3)(x-3))\)

So, \(\log_2(x^2 - 9) = 4\)

Convert to exponential form:
\(x^2 - 9 = 2^4\)
\(x^2 - 9 = 16\)
\(x^2 = 25 \implies x = 5\) or \(x = -5\).

We must check the validity of these solutions:
For \(x = 5\), both \(x+3 = 8 > 0\) and \(x-3 = 2 > 0\), which are valid.
For \(x = -5\), \(x+3 = -2 < 0\), so \(\log_2(x+3)\) is undefined.

Thus, the only solution is \(x = 5\).

(b) For the first equation, express both sides with base 3:
\(3^x \cdot (3^2)^y = 3^5\)
\(3^{x+2y} = 3^5 \implies x + 2y = 5\). (Equation 1)

For the second equation, convert to exponential form:
\(\log_2(x - y) = 1 \implies x - y = 2^1 \implies x - y = 2\). (Equation 2)

Subtract Equation 2 from Equation 1:
\((x + 2y) - (x - y) = 5 - 2\)
\(3y = 3 \implies y = 1\).

Substitute \(y = 1\) into Equation 2:
\(x - 1 = 2 \implies x = 3\).

So the solution is \(x = 3, y = 1\).

Part (a)
M1: For applying the product rule of logarithms: log_2((x+3)(x-3)).
M1: For removing logarithms to get x^2 - 9 = 2^4.
A1: For x = 5 and explicitly rejecting x = -5.

Part (b)
M1: For converting the first equation to the linear form x + 2y = 5.
M1: For converting the second equation to the linear form x - y = 2.
M1: For solving the system of simultaneous linear equations.
A1: For x = 3 and y = 1.