2025 Cambridge IGCSE Mathematics - Additional (0606) 模擬試題連答案詳解

At \(x = 4\), \(y = \frac{2(4)+1}{4-3} = 9\). The point of contact is \((4, 9)\). Using the quotient rule to differentiate \(y = \frac{2x+1}{x-3}\): \(\frac{dy}{dx} = \frac{2(x-3) - (2x+1)(1)}{(x-3)^2}\) which simplifies to \(\frac{dy}{dx} = \frac{2x - 6 - 2x - 1}{(x-3)^2} = \frac{-7}{(x-3)^2}\). At \(x = 4\), the gradient of the tangent is \(\frac{dy}{dx} = \frac{-7}{(4-3)^2} = -7\). Therefore, the gradient of the normal, \(m\), is \(m = -\frac{1}{-7} = \frac{1}{7}\). The equation of the normal is \(y - 9 = \frac{1}{7}(x - 4)\), which simplifies to \(7y - 63 = x - 4\), and rearranging gives \(x - 7y + 59 = 0\).

M1: Attempts to find the y-coordinate at \(x = 4\). A1: Correctly identifies the point \((4, 9)\). M1: Applies quotient rule to find \(\frac{dy}{dx}\). A1: Obtains \(\frac{dy}{dx} = \frac{-7}{(x-3)^2}\). M1: Finds gradient of normal as the negative reciprocal of the tangent gradient. A1: Correct final equation in the required form \(x - 7y + 59 = 0\).

First, we check if the curve intersects the \(x\)-axis in the interval \([\ln 2, \ln 3]\). Since \(e^{2x} - 4 = 0 \implies e^{2x} = 4 \implies 2x = \ln 4 \implies x = \ln 2\), the region lies entirely on or above the \(x\)-axis for \(x \ge \ln 2\). Area \(A = \int_{\ln 2}^{\ln 3} (e^{2x} - 4) \, dx = \left[ \frac{1}{2}e^{2x} - 4x \right]_{\ln 2}^{\ln 3}\). Evaluating at the upper limit \(x = \ln 3\) gives \(\frac{1}{2}e^{2\ln 3} - 4\ln 3 = \frac{1}{2}(9) - 4\ln 3 = \frac{9}{2} - 4\ln 3\). Evaluating at the lower limit \(x = \ln 2\) gives \(\frac{1}{2}e^{2\ln 2} - 4\ln 2 = \frac{1}{2}(4) - 4\ln 2 = 2 - 4\ln 2\). Subtracting the lower limit evaluation from the upper limit evaluation gives \(A = \left(\frac{9}{2} - 4\ln 3\right) - (2 - 4\ln 2) = \frac{9}{2} - 2 - 4\ln 3 + 4\ln 2 = \frac{5}{2} + 4\ln 2 - 4\ln 3\).

M1: Formulates the definite integral from \(\ln 2\) to \(\ln 3\). A1: Integrates correctly to obtain \(\frac{1}{2}e^{2x} - 4x\). M1: Substitutes the limits \(\ln 3\) and \(\ln 2\) into their integrated expression. A1: Correctly simplifies \(e^{2\ln 3}\) to \(9\) and \(e^{2\ln 2}\) to \(4\). M1: Subtracts lower limit evaluation from upper limit evaluation. A1: Correctly simplifies to the final exact area \(\frac{5}{2} + 4\ln 2 - 4\ln 3\) or equivalent.

We differentiate \(y = x^2 \ln x\) using the product rule: \(\frac{dy}{dx} = 2x \ln x + x^2 \left(\frac{1}{x}\right) = 2x \ln x + x = x(2\ln x + 1)\). Set \(\frac{dy}{dx} = 0\) to find stationary points: \(x(2\ln x + 1) = 0\). Since \(x > 0\), we have \(2\ln x + 1 = 0 \implies \ln x = -\frac{1}{2} \implies x = e^{-1/2}\). Substituting \(x = e^{-1/2}\) back into the original equation gives \(y = \left(e^{-1/2}\right)^2 \ln\left(e^{-1/2}\right) = e^{-1} \left(-\frac{1}{2}\right) = -\frac{1}{2e}\). So the stationary point is \(\left(e^{-1/2}, -\frac{1}{2e}\right)\). To determine the nature, find the second derivative: \(\frac{d^2y}{dx^2} = \frac{d}{dx}(2x\ln x + x) = 2\ln x + 2(1) + 1 = 2\ln x + 3\). Substituting \(x = e^{-1/2}\) gives \(\frac{d^2y}{dx^2} = 2\left(-\frac{1}{2}\right) + 3 = 2\). Since \(\frac{d^2y}{dx^2} > 0\), the stationary point is a local minimum.

M1: Applies product rule to find \(\frac{dy}{dx}\). A1: Correctly finds \(\frac{dy}{dx} = x(2\ln x + 1)\). M1: Sets \(\frac{dy}{dx} = 0\) and solves for \(x\). A1: Obtains \(x = e^{-1/2}\) and \(y = -\frac{1}{2e}\). M1: Finds the second derivative \(\frac{d^2y}{dx^2}\). A1: Evaluates second derivative to confirm it is positive and concludes it is a minimum.

Since the terms form a geometric progression, the ratio between consecutive terms is constant: \(\frac{k-3}{k-1} = \frac{k-4}{k-3}\). Cross-multiplying gives \((k-3)^2 = (k-1)(k-4) \implies k^2 - 6k + 9 = k^2 - 5k + 4\). Solving for \(k\) gives \(-6k + 9 = -5k + 4 \implies -k = -5 \implies k = 5\). The terms of the progression are: First term \(a = 5 - 1 = 4\), Second term \(ar = 5 - 3 = 2\), Third term \(ar^2 = 5 - 4 = 1\). The common ratio is \(r = \frac{2}{4} = \frac{1}{2}\). Since \(|r| < 1\), the sum to infinity exists and is given by \(S_\infty = \frac{a}{1-r} = \frac{4}{1 - 1/2} = 8\).

M1: Sets up the equation using the common ratio property. M1: Expands and solves for \(k\). A1: Correctly finds \(k = 5\). A1: Finds the first term \(a = 4\) and common ratio \(r = 1/2\). M1: Applies the sum to infinity formula. A1: Obtains \(S_\infty = 8\).

First, expand \((1 + 3x)^6\) using the Binomial Theorem up to the term in \(x^2\): \((1 + 3x)^6 = \binom{6}{0}(1)^6 + \binom{6}{1}(1)^5(3x) + \binom{6}{2}(1)^4(3x)^2 + \dots = 1 + 6(3x) + 15(9x^2) + \dots = 1 + 18x + 135x^2 + \dots\). Now substitute this back into the original expression: \((2 - x)(1 + 18x + 135x^2 + \dots)\). To find the coefficient of \(x^2\), collect the terms that result in \(x^2\): \(2 \times (135x^2) + (-x) \times (18x) = 270x^2 - 18x^2 = 252x^2\). Therefore, the coefficient of \(x^2\) is \(252\).

M1: Identifies the terms needed from the binomial expansion of \((1 + 3x)^6\). A1: Correctly expands to obtain \(18x\). A1: Correctly expands to obtain \(135x^2\). M1: Multiplies out the terms of \((2 - x)(1 + 18x + 135x^2)\) that yield an \(x^2\) term. M1: Performs the arithmetic combination \(2(135) - 18\). A1: Correctly calculates the final coefficient as \(252\).

(a) Since \(2x + 3\) is a factor, \(P\left(-\frac{3}{2}\right) = 0 \implies 2\left(-\frac{3}{2}\right)^3 - \left(-\frac{3}{2}\right)^2 + a\left(-\frac{3}{2}\right) + b = 0 \implies -\frac{27}{4} - \frac{9}{4} - \frac{3a}{2} + b = 0 \implies -9 - \frac{3a}{2} + b = 0 \implies -3a + 2b = 18\) [Equation 1]. Since dividing by \(x - 1\) leaves a remainder of \(-20\), \(P(1) = -20 \implies 2(1)^3 - (1)^2 + a(1) + b = -20 \implies 2 - 1 + a + b = -20 \implies a + b = -21\) [Equation 2]. From Equation 2, \(b = -21 - a\). Substitute into Equation 1: \(-3a + 2(-21 - a) = 18 \implies -3a - 42 - 2a = 18 \implies -5a = 60 \implies a = -12\). Using \(b = -21 - a\), we get \(b = -21 - (-12) = -9\). (b) Substitute \(a = -12\) and \(b = -9\) into \(P(x)\) to get \(P(x) = 2x^3 - x^2 - 12x - 9\). Since \(2x + 3\) is a factor, we write \(P(x) = (2x + 3)(x^2 + cx + d)\). By matching coefficients, \(2d = -9 \implies d = -3\), and matching the \(x^2\) term gives \(2c + 3 = -1 \implies 2c = -4 \implies c = -2\). So \(P(x) = (2x + 3)(x^2 - 2x - 3)\). Factorising the quadratic part gives \(x^2 - 2x - 3 = (x - 3)(x + 1)\). Thus, the complete factorisation is \(P(x) = (2x + 3)(x - 3)(x + 1)\).

M1: Applies factor theorem with \(x = -3/2\) to set up an equation in \(a\) and \(b\). M1: Applies remainder theorem with \(x = 1\) to set up a second equation. A1: Correctly solves the simultaneous equations to find \(a = -12\) and \(b = -9\). M1: Attempts algebraic division or coefficient matching to find the quadratic factor. A1: Correctly obtains the quadratic factor \(x^2 - 2x - 3\). A1: Factorises the quadratic factor completely to get \((2x + 3)(x - 3)(x + 1)\).

Using the identity \(\cos^2 \theta = 1 - \sin^2 \theta\), substitute into the equation: \(2(1 - \sin^2 \theta) + 3\sin \theta - 3 = 0 \implies 2 - 2\sin^2 \theta + 3\sin \theta - 3 = 0 \implies 2\sin^2 \theta - 3\sin \theta + 1 = 0\). Let \(u = \sin \theta\). The quadratic is \(2u^2 - 3u + 1 = 0 \implies (2u - 1)(u - 1) = 0\), which gives \(u = \frac{1}{2}\) or \(u = 1\). Case 1: \(\sin \theta = \frac{1}{2}\). For \(0^\circ \le \theta \le 360^\circ\), the solutions are \(\theta = 30^\circ\) and \(\theta = 180^\circ - 30^\circ = 150^\circ\). Case 2: \(\sin \theta = 1\). For \(0^\circ \le \theta \le 360^\circ\), the solution is \(\theta = 90^\circ\). Combining all solutions, we get \(\theta = 30^\circ, 90^\circ, 150^\circ\).

M1: Uses \(\cos^2 \theta = 1 - \sin^2 \theta\) to form a quadratic equation in terms of \(\sin \theta\). A1: Obtains the correct quadratic equation \(2\sin^2 \theta - 3\sin \theta + 1 = 0\). M1: Factorises or solves the quadratic equation to find \(\sin \theta = 1/2\) and \(\sin \theta = 1\). A1: Solves \(\sin \theta = 1\) to get \(\theta = 90^\circ\). M1: Solves \(\sin \theta = 1/2\) to find both angles. A1: Correctly identifies all three solutions: \(30^\circ, 90^\circ, 150^\circ\).

From the first equation: \(\log_2(x - y) = 2 \implies x - y = 2^2 \implies x - y = 4 \implies x = y + 4\) [Equation 1]. From the second equation, applying the product rule of logarithms: \(\log_2(xy) = \log_2 12 \implies xy = 12\) [Equation 2]. Substitute Equation 1 into Equation 2: \((y + 4)y = 12 \implies y^2 + 4y - 12 = 0\). Factorising the quadratic gives \((y + 6)(y - 2) = 0\), which yields \(y = -6\) or \(y = 2\). Since \(\log_2 y\) is only defined for \(y > 0\), we reject \(y = -6\), leaving \(y = 2\). Substitute \(y = 2\) back into Equation 1 to find \(x\): \(x = 2 + 4 = 6\). This gives the valid positive solutions \(x = 6\) and \(y = 2\).

M1: Rewrites the first log equation without logarithms as \(x - y = 4\). M1: Applies product rule of logarithms to the second equation to get \(xy = 12\). M1: Eliminates one variable to form a quadratic equation. A1: Correctly factorises and finds solutions \(y = 2\) and \(y = -6\). M1: Discards the negative solution \(y = -6\) explaining the domain restriction. A1: Correctly finds the final pair \(x = 6\) and \(y = 2\).

First, find the \(y\)-coordinate at \(x = 1\): \(y = (3(1)+1)\ln(1) = 4(0) = 0\). Next, differentiate using the product rule: \(\frac{dy}{dx} = \frac{d}{dx}(3x+1) \cdot \ln(x) + (3x+1) \cdot \frac{d}{dx}(\ln(x)) = 3\ln(x) + \frac{3x+1}{x}\). Substitute \(x = 1\) to find the gradient of the tangent: \(\frac{dy}{dx} = 3\ln(1) + \frac{3(1)+1}{1} = 0 + 4 = 4\). The gradient of the normal is the negative reciprocal of the tangent gradient, which is \(-\frac{1}{4}\). The equation of the normal is given by \(y - 0 = -\frac{1}{4}(x - 1)\). Multiplying by 4 gives \(4y = -(x - 1) \implies 4y = -x + 1 \implies x + 4y - 1 = 0\).

M1: Attempt to differentiate using the product rule. A1: Correct derivative of \(\frac{dy}{dx} = 3\ln(x) + \frac{3x+1}{x}\). B1: Correctly state that when \(x = 1\), \(y = 0\). M1: Correct use of normal gradient relationship, obtaining gradient of \(-\frac{1}{4}\). M1: Attempt to write down equation of line using their point and normal gradient. A1: Correct final equation in the required form: \(x + 4y - 1 = 0\) (or any integer multiple).

The second term of a geometric progression is given by \(ar = 6\). The sum to infinity is given by \(\frac{a}{1-r} = 25\). From the first equation, we express \(a = \frac{6}{r}\). Substituting this into the second equation gives \(\frac{6/r}{1-r} = 25 \implies \frac{6}{r(1-r)} = 25 \implies 25r(1-r) = 6 \implies 25r^2 - 25r + 6 = 0\). This quadratic equation can be factorized as \((5r-2)(5r-3) = 0\), which gives the solutions \(r = \frac{2}{5}\) or \(r = \frac{3}{5}\). When \(r = \frac{2}{5}\), \(a = \frac{6}{2/5} = 15\). When \(r = \frac{3}{5}\), \(a = \frac{6}{3/5} = 10\). Both values of \(r\) are positive and less than 1, so both pairs are valid.

B1: Write down \(ar = 6\). B1: Write down \(\frac{a}{1-r} = 25\). M1: Eliminate one variable to form a quadratic equation in either \(r\) or \(a\). A1: Obtain a correct quadratic equation, e.g., \(25r^2 - 25r + 6 = 0\). M1: Solve their quadratic equation to find two values of \(r\) (or \(a\)). A1: Obtain both correct pairs: \(a = 15, r = \frac{2}{5}\) and \(a = 10, r = \frac{3}{5}\).

Using the identity \(\cos^2(x) = 1 - \sin^2(x)\), the equation becomes \(2(1 - \sin^2(x)) + 3\sin(x) - 3 = 0\). Expanding and simplifying gives \(2 - 2\sin^2(x) + 3\sin(x) - 3 = 0 \implies 2\sin^2(x) - 3\sin(x) + 1 = 0\). Factorizing this quadratic equation yields \((2\sin(x) - 1)(\sin(x) - 1) = 0\), which gives \(\sin(x) = \frac{1}{2}\) or \(\sin(x) = 1\). For \(\sin(x) = \frac{1}{2}\), the solutions in the given range are \(x = 30^\circ\) and \(x = 150^\circ\). For \(\sin(x) = 1\), the solution is \(x = 90^\circ\). Thus, the complete set of solutions is \(x = 30^\circ, 90^\circ, 150^\circ\).

M1: Use the identity \(\cos^2(x) = 1 - \sin^2(x)\) to form an equation in terms of \(\sin(x)\) only. A1: Obtain the correct quadratic equation \(2\sin^2(x) - 3\sin(x) + 1 = 0\). M1: Factorize or solve their quadratic equation to find values for \(\sin(x)\). A1: Obtain both \(\sin(x) = \frac{1}{2}\) and \(\sin(x) = 1\). A1: Find \(x = 30^\circ, 150^\circ\). A1: Find \(x = 90^\circ\) with no other incorrect values in range.

By the factor theorem, since \(x-2\) is a factor, \(p(2) = 0\). This gives \(2(2)^3 + a(2)^2 + b(2) - 6 = 0 \implies 16 + 4a + 2b - 6 = 0 \implies 4a + 2b = -10 \implies 2a + b = -5\) (Equation 1). By the remainder theorem, since the remainder when divided by \(x+1\) is \(-12\), \(p(-1) = -12\). This gives \(2(-1)^3 + a(-1)^2 + b(-1) - 6 = -12 \implies -2 + a - b - 6 = -12 \implies a - b = -4\) (Equation 2). Solving Equations 1 and 2 simultaneously: adding the two equations gives \(3a = -9 \implies a = -3\). Substituting \(a = -3\) into Equation 2 gives \(-3 - b = -4 \implies b = 1\). Therefore, \(a = -3\) and \(b = 1\).

M1: Apply the factor theorem to obtain an equation in \(a\) and \(b\) by setting \(p(2) = 0\). A1: Obtain the simplified equation \(2a + b = -5\) (or equivalent). M1: Apply the remainder theorem to obtain an equation in \(a\) and \(b\) by setting \(p(-1) = -12\). A1: Obtain the simplified equation \(a - b = -4\) (or equivalent). M1: Solve their simultaneous equations to find values for \(a\) and \(b\). A1: Correctly find both \(a = -3\) and \(b = 1\).

(a) By similar triangles, the radius \(r\) of the water surface at depth \(h\) satisfies:
\(\frac{r}{h} = \frac{4}{12} = \frac{1}{3} \implies r = \frac{h}{3}\).
Substituting this into the formula for the volume of a cone:
\(V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{h}{3}\right)^2 h = \frac{\pi}{27} h^3\).

(b) Differentiating \(V\) with respect to \(h\):
\(\frac{dV}{dh} = \frac{\pi}{9} h^2\).
When \(h = 6\):
\(\frac{dV}{dh} = \frac{\pi}{9} (6)^2 = 4\pi\).
Using the chain rule:
\(\frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt}\)
\(3 = 4\pi \times \frac{dh}{dt} \implies \frac{dh}{dt} = \frac{3}{4\pi} \approx 0.239\text{ cm s}^{-1}\).

(c) The surface area is \(A = \pi r^2 = \pi \left(\frac{h}{3}\right)^2 = \frac{\pi}{9} h^2\).
Differentiating with respect to \(h\):
\(\frac{dA}{dh} = \frac{2\pi}{9} h\).
When \(h = 6\):
\(\frac{dA}{dh} = \frac{12\pi}{9} = \frac{4\pi}{3}\).
Using the chain rule:
\(\frac{dA}{dt} = \frac{dA}{dh} \times \frac{dh}{dt} = \frac{4\pi}{3} \times \frac{3}{4\pi} = 1\text{ cm}^2\text{s}^{-1}\).

(a)
M1: Expresses \(r\) in terms of \(h\) using similar triangles.
A1: Obtains \(V = \frac{\pi}{27}h^3\) with clear working.

(b)
M1: Differentiates to find \(\frac{dV}{dh} = \frac{\pi}{9}h^2\).
M1: Applies chain rule \(\frac{dh}{dt} = \frac{dV}{dt} \div \frac{dV}{dh}\) substituting \(h=6\).
A1: Correct exact value \(\frac{3}{4\pi}\) or \(0.239\) (3 s.f.).

(c)
M1: Expresses \(A = \frac{\pi}{9}h^2\) and differentiates to get \(\frac{dA}{dh} = \frac{2\pi}{9}h\).
M1: Applies chain rule \(\frac{dA}{dt} = \frac{dA}{dh} \times \frac{dh}{dt}\) with their values.
A1: Correct answer \(1\).

(a) Equating the expressions for \(y\):
\(8x - x^2 = 2x + 5\)
\(x^2 - 6x + 5 = 0\)
\((x-1)(x-5) = 0\)
So, \(x = 1\) or \(x = 5\).
Substituting back to find the coordinates:
When \(x = 1\), \(y = 2(1) + 5 = 7\).
When \(x = 5\), \(y = 2(5) + 5 = 15\).
Intersection points are \((1, 7)\) and \((5, 15)\).

(b) The area is given by:
\(\int_{1}^{5} (8x - x^2 - (2x + 5)) \, dx = \int_{1}^{5} (6x - x^2 - 5) \, dx\)
Integrating terms:
\(\left[ 3x^2 - \frac{1}{3}x^3 - 5x \right]_{1}^{5}\)
Substituting upper limit \(x = 5\):
\(3(25) - \frac{125}{3} - 25 = 50 - 41.67 = \frac{25}{3}\)
Substituting lower limit \(x = 1\):
\(3(1) - \frac{1}{3} - 5 = -2 - \frac{1}{3} = -\frac{7}{3}\)
Subtracting:
\(\frac{25}{3} - \left(-\frac{7}{3}\right) = \frac{32}{3} = 10\frac{2}{3} \approx 10.7\).

(a)
M1: Equates curve and line to form a quadratic equation.
A1: Solves the quadratic to find \(x = 1\) and \(x = 5\).
A1: Obtains both points of intersection \((1, 7)\) and \((5, 15)\).

(b)
M1: Writes down the correct integral expression with limits.
M1: Integrates at least two terms of their quadratic expression correctly.
A1: Fully correct integrated expression: \(3x^2 - \frac{1}{3}x^3 - 5x\).
M1: Sustitutes limits of 5 and 1 into their integrated expression.
A1: Correct exact answer \(\frac{32}{3}\) or \(10.7\) (3 s.f.).

(a) For both progressions, \(a = 4\).
3rd term of AP: \(a + 2d = 4 + 2d\).
3rd term of GP: \(a r^2 = 4r^2\).
Since they are equal:
\(4 + 2d = 4r^2 \implies 2d = 4r^2 - 4 \implies d = 2r^2 - 2\).

(b) The sum of the first 12 terms of the AP is:
\(S_{12} = \frac{12}{2} [2a + 11d] = 213\)
\(6 [2(4) + 11d] = 213\)
\(8 + 11d = 35.5\)
\(11d = 27.5 \implies d = 2.5\).
Using the relationship from (a):
\(2.5 = 2r^2 - 2\)
\(2r^2 = 4.5\)
\(r^2 = 2.25\)
Since \(r > 0\), \(r = \sqrt{2.25} = 1.5\).

(c) For the GP, \(a = 4\) and \(r = 1.5\).
The sum of the first 5 terms is:
\(S_5 = \frac{a(r^5 - 1)}{r - 1} = \frac{4(1.5^5 - 1)}{1.5 - 1} = \frac{4(7.59375 - 1)}{0.5} = 8(6.59375) = 52.75\).

(a)
M1: Expresses both 3rd terms in terms of \(d\) and \(r\).
A1: Equates and simplifies to obtain \(d = 2r^2 - 2\).

(b)
M1: Uses \(S_{12} = 213\) with \(a = 4\) to set up equation in \(d\).
A1: Solves to find \(d = 2.5\).
M1: Equates \(d\) to \(2r^2 - 2\) and attempts to solve for \(r\).
A1: Obtains \(r = 1.5\) (accepting only positive root).

(c)
M1: Uses geometric series sum formula with \(a = 4\), \(r = 1.5\), and \(n = 5\).
A1: Obtains \(52.75\).

(a) Using the Remainder Theorem:
\(P(1) = 2(1)^3 + p(1)^2 + q(1) + 6 = -6 \implies p + q + 8 = -6 \implies p + q = -14\)
And,
\(P(-3) = 2(-3)^3 + p(-3)^2 + q(-3) + 6 = -42\)
\(-54 + 9p - 3q + 6 = -42 \implies 9p - 3q - 48 = -42 \implies 9p - 3q = 6 \implies 3p - q = 2\).
Solving simultaneously:
Add: \((p+q) + (3p-q) = -14 + 2 \implies 4p = -12 \implies p = -3\).
Then: \(-3 + q = -14 \implies q = -11\).

(b) Using the values \(p = -3\) and \(q = -11\):
\(P(x) = 2x^3 - 3x^2 - 11x + 6\).
By the Factor Theorem, if \(2x-1\) is a factor, then \(P(0.5) = 0\):
\(P(0.5) = 2(0.5)^3 - 3(0.5)^2 - 11(0.5) + 6\)
\(P(0.5) = 0.25 - 0.75 - 5.5 + 6 = 0\).
Since \(P(0.5) = 0\), \(2x-1\) is a factor of \(P(x)\).

(c) Dividing \(2x^3 - 3x^2 - 11x + 6\) by \(2x-1\):
\(2x^3 - 3x^2 - 11x + 6 = (2x - 1)(x^2 - x - 6)\).
Factorising the quadratic:
\(x^2 - x - 6 = (x - 3)(x + 2)\).
So, fully factorised:
\(P(x) = (2x - 1)(x - 3)(x + 2)\).

(a)
M1: Substitutes \(x=1\) and equates to \(-6\).
M1: Substitutes \(x=-3\) and equates to \(-42\).
A1: Correctly finds \(p = -3\).
A1: Correctly finds \(q = -11\).

(b)
M1: Evaluates \(P(0.5)\) using their coefficients.
A1: Obtains 0 and concludes that \(2x-1\) is a factor.

(c)
M1: Performs algebraic division or inspection to obtain quadratic factor \(x^2 - x - 6\).
A1: Correct fully factorised form: \((2x - 1)(x - 3)(x + 2)\).

(a) Using the trigonometric identity \( \sec^2 x = 1 + \tan^2 x \):
\(3(1 + \tan^2 x) + 5\tan x - 7 = 0\)
\(3 + 3\tan^2 x + 5\tan x - 7 = 0\)
\(3\tan^2 x + 5\tan x - 4 = 0\).

(b) Let \(y = \tan x\).
\(3y^2 + 5y - 4 = 0\)
Using the quadratic formula:
\(y = \frac{-5 \pm \sqrt{5^2 - 4(3)(-4)}}{2(3)} = \frac{-5 \pm \sqrt{73}}{6}\)
\(y \approx 0.5907\) or \(y \approx -2.2573\).

Case 1: \(\tan x = 0.5907\)
Basic angle \(\alpha = \tan^{-1}(0.5907) \approx 30.57^{\circ}\).
Since \(\tan x > 0\), \(x\) lies in Quadrants I and III:
\(x = 30.6^{\circ}\)
\(x = 180^{\circ} + 30.57^{\circ} = 210.6^{\circ}\).

Case 2: \(\tan x = -2.2573\)
Basic angle \(\alpha = \tan^{-1}(2.2573) \approx 66.11^{\circ}\).
Since \(\tan x < 0\), \(x\) lies in Quadrants II and IV:
\(x = 180^{\circ} - 66.11^{\circ} = 113.9^{\circ}\)
\(x = 360^{\circ} - 66.11^{\circ} = 293.9^{\circ}\).

(a)
M1: Substitutes \(\sec^2 x = 1 + \tan^2 x\) into equation.
A1: Simplifies to the given quadratic in \(\tan x\).

(b)
M1: Uses quadratic formula (or completing the square) to solve the quadratic equation.
A1: Obtains \(\tan x = 0.5907\) and \(\tan x = -2.2573\) (or exact equivalents).
M1: Finds basic angle for at least one case.
A1: Correctly finds both Quadrant I & III solutions: \(x = 30.6^{\circ}\) and \(210.6^{\circ}\) (accept 30.6 and 211).
A1: Correctly finds Quadrant II solution: \(x = 113.9^{\circ}\) (accept 114).
A1: Correctly finds Quadrant IV solution: \(x = 293.9^{\circ}\) (accept 294).

(a) Since \(x \ge 0\), we have \(3x^2 \ge 0\), and therefore \(g(x) = 3x^2 - 1 \ge -1\).
The range of \(g\) is \(g(x) \ge -1\).

(b) The range of \(g\) is \([-1, \infty)\), which is fully contained within the domain of \(f\), which is \([-4, \infty)\).

(c) \(fg(x) = f(g(x)) = 2\sqrt{(3x^2 - 1) + 4} = 2\sqrt{3x^2 + 3}\).
Domain of \(fg\) is the domain of \(g\), which is \(x \ge 0\).
Range of \(fg\): Since \(x \ge 0\), the minimum value of \(3x^2 + 3\) is 3.
Thus, \(fg(x) \ge 2\sqrt{3}\).

(d) Let \(y = 2\sqrt{x + 4}\) for \(x \ge -4\).
\(\frac{y}{2} = \sqrt{x + 4}\)
\(\frac{y^2}{4} = x + 4 \implies x = \frac{y^2}{4} - 4\).
So, \(f^{-1}(x) = \frac{x^2}{4} - 4\) (or \(\frac{x^2-16}{4}\)).
The domain of \(f^{-1}\) is the range of \(f\).
Since the domain of \(f\) is \(x \ge -4\), its range is \(f(x) \ge 0\).
Therefore, the domain of \(f^{-1}\) is \(x \ge 0\).

(a)
B1: States range \(g(x) \ge -1\).

(b)
B1: Explains that the range of \(g\) is a subset of the domain of \(f\).

(c)
M1: Formulates \(fg(x) = 2\sqrt{3x^2 + 3}\).
A1: Correctly states domain \(x \ge 0\).
A1: Correctly states range \(fg(x) \ge 2\sqrt{3}\) (or \(\ge 3.46\)).

(d)
M1: Rearranges equation to make \(x\) the subject.
A1: Correct expression \(f^{-1}(x) = \frac{x^2}{4} - 4\) (or equivalent).
A1: States correct domain \(x \ge 0\).

(a) From the ratios given:
\(\overrightarrow{OP} = \frac{2}{3}\mathbf{a}\) and \(\overrightarrow{OQ} = \frac{1}{4}\mathbf{b}\).
Therefore:
\(\overrightarrow{AQ} = \overrightarrow{AO} + \overrightarrow{OQ} = -\mathbf{a} + \frac{1}{4}\mathbf{b}\),
\(\overrightarrow{BP} = \overrightarrow{BO} + \overrightarrow{OP} = -\mathbf{b} + \frac{2}{3}\mathbf{a}\).

(b)
(i) \(\overrightarrow{OX} = \overrightarrow{OA} + \overrightarrow{AX} = \mathbf{a} + \lambda \overrightarrow{AQ} = \mathbf{a} + \lambda \left(-\mathbf{a} + \frac{1}{4}\mathbf{b}\right) = (1 - \lambda)\mathbf{a} + \frac{1}{4}\lambda \mathbf{b}\).
(ii) \(\overrightarrow{OX} = \overrightarrow{OB} + \overrightarrow{BX} = \mathbf{b} + \mu \overrightarrow{BP} = \mathbf{b} + \mu \left(-\mathbf{b} + \frac{2}{3}\mathbf{a}\right) = \frac{2}{3}\mu \mathbf{a} + (1 - \mu)\mathbf{b}\).

(c) Since \(\mathbf{a}\) and \(\mathbf{b}\) are non-parallel vectors, we can equate the coefficients:
For \(\mathbf{a}\): \(1 - \lambda = \frac{2}{3}\mu\)
For \(\mathbf{b}\): \(\frac{1}{4}\lambda = 1 - \mu \implies \lambda = 4 - 4\mu\).
Substitute \(\lambda = 4 - 4\mu\) into the first equation:
\(1 - (4 - 4\mu) = \frac{2}{3}\mu\)
\(-3 + 4\mu = \frac{2}{3}\mu\)
Multiply by 3:
\(-9 + 12\mu = 2\mu \implies 10\mu = 9 \implies \mu = 0.9\).
Substitute back to find \(\lambda\):
\(\lambda = 4 - 4(0.9) = 0.4\).

(a)
B1: Correct vector \(\overrightarrow{AQ} = -\mathbf{a} + \frac{1}{4}\mathbf{b}\).
B1: Correct vector \(\overrightarrow{BP} = -\mathbf{b} + \frac{2}{3}\mathbf{a}\).

(b)
(i) M1: Uses vector addition pathway \(\overrightarrow{OA} + \lambda \overrightarrow{AQ}\).
A1: Obtains \((1 - \lambda)\mathbf{a} + \frac{1}{4}\lambda \mathbf{b}\).
(ii) M1: Uses vector addition pathway \(\overrightarrow{OB} + \mu \overrightarrow{BP}\).
A1: Obtains \(\frac{2}{3}\mu \mathbf{a} + (1 - \mu)\mathbf{b}\).

(c)
M1: Equates coefficients of both \(\mathbf{a}\) and \(\mathbf{b}\) to form simultaneous equations.
A1: Correct values \(\lambda = 0.4\) and \(\mu = 0.9\) (or equivalent fractions).

(a) Rearranging and completing the square:
\(x^2 - 8x + y^2 + 4y = 5\)
\((x-4)^2 - 16 + (y+2)^2 - 4 = 5\)
\((x-4)^2 + (y+2)^2 = 25\).
Thus, the center is \((4, -2)\) and the radius is \(\sqrt{25} = 5\).

(b) Substitute \(x = 7\) and \(y = 2\) into the original equation:
\(7^2 + 2^2 - 8(7) + 4(2) - 5 = 49 + 4 - 56 + 8 - 5 = 0\).
Since the LHS equals the RHS, the point \(P(7, 2)\) lies on the circle.

(c) The gradient of the radius from the center \(C(4, -2)\) to \(P(7, 2)\) is:
\(m_{\text{radius}} = \frac{2 - (-2)}{7 - 4} = \frac{4}{3}\).
The tangent is perpendicular to the radius, so its gradient is:
\(m_{\text{tangent}} = -\frac{3}{4}\).
The equation of the tangent is:
\(y - 2 = -\frac{3}{4}(x - 7)\)
\(4y - 8 = -3x + 21\)
\(3x + 4y - 29 = 0\) (or \(y = -0.75x + 7.25\)).

(a)
M1: Completes the square for either \(x\) or \(y\).
A1: Writes correct completed square form: \((x-4)^2 + (y+2)^2 = 25\).
A1: States center \((4, -2)\) and radius \(5\).

(b)
B1: Verifies \(P\) lies on circle by substitution showing both sides equal.

(c)
M1: Finds gradient of radius \(CP\) to be \(\frac{4}{3}\).
M1: Determines perpendicular gradient to be \(-\frac{3}{4}\).
M1: Forms equation of line using their gradient and coordinates of \(P\).
A1: Obtains \(3x + 4y - 29 = 0\) (or equivalent form).

**(a)**
First, find the \(y\)-coordinate at \(x = 4\):
\(y = \frac{16}{\sqrt{2(4) + 1}} = \frac{16}{3}\)

Now, rewrite the equation of the curve to differentiate:
\(y = 16(2x + 1)^{-\frac{1}{2}}\)

Using the chain rule to find \(\frac{dy}{dx}\):
\(\frac{dy}{dx} = 16 \left(-\frac{1}{2}\right) (2x + 1)^{-\frac{3}{2}} \cdot 2 = -16(2x + 1)^{-\frac{3}{2}}\)

At \(x = 4\):
\(\frac{dy}{dx} = -16(2(4) + 1)^{-\frac{3}{2}} = -16(9)^{-\frac{3}{2}} = -16 \cdot \frac{1}{27} = -\frac{16}{27}\)

The gradient of the normal is the negative reciprocal of the tangent gradient:
\(m_{\text{normal}} = -\frac{1}{-\frac{16}{27}} = \frac{27}{16}\)

The equation of the normal at \(\left(4, \frac{16}{3}\right)\) is:
\(y - \frac{16}{3} = \frac{27}{16}(x - 4)\)

Multiplying by 48 to clear fractions:
\(48y - 256 = 81x - 324\)
\(81x - 48y - 68 = 0\)

In gradient-intercept form:
\(y = \frac{27}{16}x - \frac{17}{12}\)

**(b)**
The area of the region is given by the definite integral:
\(\text{Area} = \int_{0}^{4} \frac{16}{\sqrt{2x + 1}} \, dx = \int_{0}^{4} 16(2x + 1)^{-\frac{1}{2}} \, dx\)

Integrate using the reverse chain rule:
\(\int 16(2x + 1)^{-\frac{1}{2}} \, dx = \left[ 16 \cdot \frac{(2x + 1)^{\frac{1}{2}}}{\frac{1}{2} \cdot 2} \right] = \left[ 16\sqrt{2x + 1} \right]\)

Evaluate between the limits \(0\) and \(4\):
\(\text{Area} = \left[ 16\sqrt{2x + 1} \right]_{0}^{4} = 16\sqrt{2(4) + 1} - 16\sqrt{2(0) + 1}\)
\(\text{Area} = 16\sqrt{9} - 16\sqrt{1} = 16(3) - 16(1) = 48 - 16 = 32\)

**(a)**
* **M1**: For an attempt to differentiate, resulting in a form \(k(2x+1)^{-\frac{3}{2}}\).
* **A1**: Correct derivative \(\frac{dy}{dx} = -16(2x+1)^{-\frac{3}{2}}\).
* **M1**: For finding the gradient of the normal from their gradient of the tangent, and using their point \(\left(4, \frac{16}{3}\right)\) to form the equation of a straight line.
* **A1**: Correct equation of the normal, e.g. \(y = \frac{27}{16}x - \frac{17}{12}\) or \(81x - 48y - 68 = 0\) or any equivalent form.

**(b)**
* **M1**: For attempting to integrate, resulting in a form \(c(2x+1)^{\frac{1}{2}}\).
* **A1**: Correct integrated term \(16(2x+1)^{\frac{1}{2}}\) or equivalent.
* **M1**: For substituting the limits \(0\) and \(4\) into their integrated expression and subtracting.
* **A1**: For the correct area of \(32\).

**(a)**
The \(n\)-th term of an arithmetic progression (AP) is given by \(u_n = a + (n-1)d\).

The third, fifth, and eleventh terms of this AP are:
\(u_3 = a + 2d\)
\(u_5 = a + 4d\)
\(u_{11} = a + 10d\)

Since these three terms are the first three terms of a geometric progression (GP), they share a common ratio:
\(\frac{a + 4d}{a + 2d} = \frac{a + 10d}{a + 4d}\)

Cross-multiplying yields:
\((a + 4d)^2 = (a + 2d)(a + 10d)\)

Expanding both sides:
\(a^2 + 8ad + 16d^2 = a^2 + 12ad + 20d^2\)

Subtract \(a^2\) from both sides:
\(8ad + 16d^2 = 12ad + 20d^2\)

Rearranging to group terms:
\(16d^2 - 20d^2 = 12ad - 8ad\)
\(-4d^2 = 4ad\)

Since \(d \neq 0\), we can divide both sides by \(4d\):
\(-d = a \implies a = -d\)

**(b)**
Substitute \(a = -d\) into the first two terms of the GP:
\(\text{First term } (u_3) = a + 2d = -d + 2d = d\)
\(\text{Second term } (u_5) = a + 4d = -d + 4d = 3d\)

The common ratio \(r\) is:
\(r = \frac{\text{Second term}}{\text{First term}} = \frac{3d}{d} = 3\)

**(c)**
The 8th term of the AP is given by:
\(u_8 = a + 7d\)

Substitute \(a = -d\):
\(u_8 = -d + 7d = 6d\)

We are given \(u_8 = 42\):
\(6d = 42 \implies d = 7\)

Thus, the first term of the GP is:
\(G_1 = d = 7\)

The sum of the first 6 terms of this GP (with \(G_1 = 7\) and \(r = 3\)) is:
\(S_6 = \frac{G_1(r^6 - 1)}{r - 1}\)
\(S_6 = \frac{7(3^6 - 1)}{3 - 1} = \frac{7(729 - 1)}{2} = \frac{7 \cdot 728}{2} = 7 \cdot 364 = 2548\)

**(a)**
* **M1**: For expressing the three terms in terms of \(a\) and \(d\), and setting up a correct equation using the GP property, e.g., \((a + 4d)^2 = (a + 2d)(a + 10d)\).
* **A1**: For correct expansion of both sides: \(a^2 + 8ad + 16d^2 = a^2 + 12ad + 20d^2\).
* **A1**: For simplifying to show \(a = -d\) clearly, with no errors seen.

**(b)**
* **M1**: For substituting \(a = -d\) into at least two terms of the GP to express them in terms of \(d\).
* **A1**: For obtaining the correct common ratio \(r = 3\).

**(c)**
* **M1**: For setting \(a + 7d = 42\) and using \(a = -d\) to solve for \(d\).
* **M1**: For substituting their \(d\) (which is the first term of the GP) and \(r = 3\) into the GP sum formula \(S_6 = \frac{a(r^6 - 1)}{r - 1}\).
* **A1**: For the correct final answer of \(2548\).