2024 Cambridge IAS-Level Mathematics (9709) 模擬試題連答案詳解

Let the first term of the arithmetic progression be \(a\) and the common difference be \(d\).

The 1st, 3rd, and 11th terms of the arithmetic progression are given by:
\(u_1 = a\)
\(u_3 = a + 2d\)
\(u_{11} = a + 10d\)

Since these three terms form a geometric progression, the ratio between consecutive terms is constant:
\(\frac{a + 2d}{a} = \frac{a + 10d}{a + 2d}\)

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

Subtracting \(a^2\) and \(4ad\) from both sides:
\(4d^2 = 6ad\)
\(4d^2 - 6ad = 0\)
\(2d(2d - 3a) = 0\)

Since the arithmetic progression has a non-zero common difference, \(d \neq 0\). Therefore:
\(2d = 3a \implies d = 1.5a\)

We can find the common ratio \(r\) of the geometric progression by using the first two terms:
\(r = \frac{a + 2d}{a}\)

Substituting \(d = 1.5a\):
\(r = \frac{a + 2(1.5a)}{a} = \frac{a + 3a}{a} = \frac{4a}{a} = 4\)

(Note: \(a\) cannot be \(0\), otherwise \(d\) would also be \(0\)).

M1: Set up the correct terms of the arithmetic progression in terms of \(a\) and \(d\).
M1: Formulate the geometric progression equation: \((a + 2d)^2 = a(a + 10d)\).
A1: Correctly solve the equation to find \(2d = 3a\) or \(d = 1.5a\) (rejecting \(d=0\)).
A1: Correctly deduce that the common ratio \(r = 4\).

First, find the midpoint \(M\) of the line segment \(AB\):
\(M = \left(\frac{1+5}{2}, \frac{3+11}{2}\right) = (3, 7)\)

Next, find the gradient of the line segment \(AB\):
\(m_{AB} = \frac{11 - 3}{5 - 1} = \frac{8}{4} = 2\)

The gradient of the perpendicular bisector \(L\) is the negative reciprocal of \(m_{AB}\):
\(m_L = -\frac{1}{2}\)

Now, find the equation of the line \(L\) passing through \(M(3, 7)\) with gradient \(-\frac{1}{2}\):
\(y - 7 = -\frac{1}{2}(x - 3)\)
\(2(y - 7) = -(x - 3)\)
\(2y - 14 = -x + 3\)
\(2y = -x + 17\)

We need to find the intersection of this line with the line \(y = x + 1\). Substitute \(y = x + 1\) into the equation of \(L\):
\(2(x + 1) = -x + 17\)
\(2x + 2 = -x + 17\)
\(3x = 15\)
\(x = 5\)

Substitute \(x = 5\) back into \(y = x + 1\) to find \(y\):
\(y = 5 + 1 = 6\)

Thus, the coordinates of the point of intersection are \((5, 6)\).

M1: Find the midpoint \((3, 7)\) and the gradient of \(AB\) as \(2\).
M1: Determine the equation of the perpendicular bisector \(L\) (using gradient \(-\frac{1}{2}\) and the midpoint).
A1: Obtain the correct equation of \(L\), e.g., \(2y = -x + 17\) or \(y = -0.5x + 8.5\).
A1: Solve simultaneously with \(y = x + 1\) to obtain the correct coordinates \((5, 6)\).

First, find the \(x\)-coordinates of the points of intersection by setting the equations equal to each other:
\(x^2 - 4x + 6 = x + 2\)
\(x^2 - 5x + 4 = 0\)

Factorising the quadratic equation:
\((x - 1)(x - 4) = 0\)

So the points of intersection occur at \(x = 1\) and \(x = 4\).

The area \(A\) of the enclosed region is given by the integral of the upper function (the line) minus the lower function (the curve) from \(x = 1\) to \(x = 4\):
\(A = \int_{1}^{4} ((x + 2) - (x^2 - 4x + 6)) \, dx\)
\(A = \int_{1}^{4} (-x^2 + 5x - 4) \, dx\)

Now, integrate each term with respect to \(x\):
\(A = \left[ -\frac{x^3}{3} + \frac{5x^2}{2} - 4x \right]_{1}^{4}\)

Substitute the upper limit \(x = 4\):
\(\left( -\frac{4^3}{3} + \frac{5(4^2)}{2} - 4(4) \right) = -\frac{64}{3} + 40 - 16 = -\frac{64}{3} + 24 = \frac{8}{3}\)

Substitute the lower limit \(x = 1\):
\(\left( -\frac{1^3}{3} + \frac{5(1^2)}{2} - 4(1) \right) = -\frac{1}{3} + \frac{5}{2} - 4 = -\frac{11}{6}\)

Subtract the lower limit value from the upper limit value:
\(A = \frac{8}{3} - \left( -\frac{11}{6} \right) = \frac{16}{6} + \frac{11}{6} = \frac{27}{6} = 4.5\)

Thus, the area of the enclosed region is \(4.5\).

M1: Set up the equation \(x^2 - 4x + 6 = x + 2\) and solve to find the limits \(x = 1\) and \(x = 4\).
M1: Integrate the difference of the functions, showing at least two terms integrated correctly.
A1: Obtain the correct integrated expression: \(-\frac{x^3}{3} + \frac{5x^2}{2} - 4x\) (or equivalent).
A1: Apply limits correctly and evaluate the exact area to get \(4.5\) (or \(\frac{9}{2}\)).

(a) The center \(M\) of the circle is the midpoint of \(AB\):
\(M = \left(\frac{1+7}{2}, \frac{3+11}{2}\right) = (4, 7)\).
The radius \(r\) is the distance from \(M\) to \(A\):
\(r^2 = (4-1)^2 + (7-3)^2 = 3^2 + 4^2 = 25\).
Thus, the equation of the circle is:
\((x-4)^2 + (y-7)^2 = 25\).

(b) Substituting \(P(7, 3)\) into the circle equation:
\((7-4)^2 + (3-7)^2 = 3^2 + (-4)^2 = 9 + 16 = 25\).
This is equal to \(25\), so \(P\) lies on the circle.
The gradient of the radius \(MP\) is:
\(m_{MP} = \frac{3-7}{7-4} = -\frac{4}{3}\).
The tangent is perpendicular to the radius, so its gradient is:
\(m = \frac{3}{4}\).
The equation of the tangent at \(P\) is:
\(y - 3 = \frac{3}{4}(x - 7)\)
\(4y - 12 = 3x - 21\)
\(3x - 4y = 9\).

(c) To find \(Q\) on the \(x\)-axis, set \(y = 0\):
\(3x = 9 \implies x = 3\), so \(Q\) has coordinates \((3, 0)\).
To find \(R\) on the \(y\)-axis, set \(x = 0\):
\(-4y = 9 \implies y = -2.25\), so \(R\) has coordinates \((0, -2.25)\).
The area of the triangle \(OQR\) is:
\(\text{Area} = \frac{1}{2} \times OQ \times OR = \frac{1}{2} \times 3 \times 2.25 = 3.375\).

(a)
M1: For attempting to find the midpoint of \(AB\) and calculating the radius or diameter.
A1: Correct center at \((4, 7)\) and radius squared as \(25\).
A1: Correct equation of the circle \((x-4)^2 + (y-7)^2 = 25\) (or equivalent).

(b)
B1: For verifying that \(P(7, 3)\) satisfies the equation of the circle.
M1: For finding the gradient of the radius \(MP\) and utilizing \(m_1 m_2 = -1\) to find the gradient of the tangent.
A1: Correct equation of the tangent in the required form: \(3x - 4y = 9\).

(c)
M1: For finding the coordinates of the intercepts \(Q\) and \(R\) by setting \(y = 0\) and \(x = 0\).
A1: Correct values \(x = 3\) and \(y = -2.25\).
A1: Correct area of the triangle: \(3.375\) (or \(\frac{27}{8}\)).

(a) Complete the square:
\(2x^2 - 12x + 11 = 2(x^2 - 6x) + 11 = 2[(x-3)^2 - 9] + 11 = 2(x-3)^2 - 18 + 11 = 2(x-3)^2 - 7\).

(b) The quadratic curve has its vertex (minimum) at \(x = 3\). For a function to have an inverse, it must be one-to-one. Since the domain is defined as \(x \le a\), the largest possible value of \(a\) is \(3\).

(c) Set \(y = 2(x-3)^2 - 7\):
\(y + 7 = 2(x-3)^2\)
\(\frac{y+7}{2} = (x-3)^2\)
Since \(x \le 3\), we take the negative square root:
\(x - 3 = -\sqrt{\frac{y+7}{2}}\)
\(x = 3 - \sqrt{\frac{y+7}{2}}\)
Thus, \(f^{-1}(x) = 3 - \sqrt{\frac{x+7}{2}}\).

(d) We want to solve \(fg(x) = 11\):
\(2(g(x)-3)^2 - 7 = 11\)
\(2(g(x)-3)^2 = 18\)
\((g(x)-3)^2 = 9\)
\(g(x) - 3 = \pm 3\)
So, \(g(x) = 6\) or \(g(x) = 0\).
But since the domain of \(f\) is \(x \le 3\), we must have \(g(x) \le 3\). Hence, we reject \(g(x) = 6\), leaving \(g(x) = 0\).
Since \(g(x) = x - 2\), this gives:
\(x - 2 = 0 \implies x = 2\).
This value of \(x\) lies within the domain of \(g\) (since \(2 \le 5\)), so the only solution is \(x = 2\).

(a)
M1: For attempting to complete the square, identifying \(p = 2\) and \(q = 3\).
A1: Correctly obtaining \(2(x-3)^2 - 7\).

(b)
B1: State \(a = 3\).

(c)
M1: Correctly rearrange their equation to make \((x-3)^2\) the subject.
M1: For choosing the negative square root due to the domain \(x \le 3\).
A1: Correct inverse function: \(f^{-1}(x) = 3 - \sqrt{\frac{x+7}{2}}\).

(d)
M1: Set up the equation \(2(g(x)-3)^2 - 7 = 11\) (or equivalent using the expanded form).
A1: For obtaining \(g(x) = 0\) and rejecting \(g(x) = 6\).
A1: Correct solution \(x = 2\).

(a) From the given information:
2nd term of GP: \(ar\)
3rd term of AP: \(a + 2d\)
So, \(ar = a + 2d \implies 2d = a(r - 1) \implies d = \frac{a(r-1)}{2}\).

3rd term of GP: \(ar^2\)
11th term of AP: \(a + 10d\)
So, \(ar^2 = a + 10d \implies 10d = a(r^2 - 1) = a(r-1)(r+1)\).

Since \(d \ne 0\) and \(r > 1\), we can divide the second equation by the first:
\(\frac{10d}{2d} = \frac{a(r-1)(r+1)}{a(r-1)}\)
\(5 = r + 1 \implies r = 4\).

(b) Given \(r = 4\), we substitute this back to find \(d\):
\(d = \frac{a(4-1)}{2} = 1.5a\).
The sum of the first 4 terms of the AP is:
\(S_4 = \frac{4}{2}(2a + 3d) = 2(2a + 3(1.5a)) = 2(2a + 4.5a) = 13a\).
We are given \(S_4 = 78\):
\(13a = 78 \implies a = 6\).

(c) Using \(a = 6\) and \(r = 4\), the sum of the first 6 terms of the GP is:
\(S_6 = \frac{a(r^6 - 1)}{r - 1} = \frac{6(4^6 - 1)}{4 - 1} = \frac{6(4096 - 1)}{3} = 2(4095) = 8190\).

(a)
M1: For writing down the equations for both given relationships: \(ar = a + 2d\) and \(ar^2 = a + 10d\).
M1: For expressing \(d\) in terms of \(a\) and \(r\) from both equations (or substituting one into another).
M1: For setting up a division or factorization of the equations to eliminate \(a\) and \(d\).
A1: For obtaining \(r = 4\) clearly with no errors.

(b)
M1: For writing the AP sum formula and substituting \(d = 1.5a\) (or equivalent).
A1: For finding \(a = 6\).

(c)
M1: For utilizing the GP sum formula \(S_n = \frac{a(r^n - 1)}{r-1}\) with \(n = 6\).
A1: For substituting \(a = 6\) and \(r = 4\) correctly.
A1: Correct answer: \(8190\).

(a) First, we find the length of the components of the perimeter of the shaded region, which are the arc \(AB\), the line segment \(AC\), and the line segment \(BC\).
- Arc length \(AB = r\theta = 8 \times \frac{1}{3}\pi = \frac{8\pi}{3}\text{ cm}\).
- In the right-angled triangle \(OAC\):
\(AC = 8 \sin\left(\frac{1}{3}\pi\right) = 8 \times \frac{\sqrt{3}}{2} = 4\sqrt{3}\text{ cm}\).
\(OC = 8 \cos\left(\frac{1}{3}\pi\right) = 8 \times \frac{1}{2} = 4\text{ cm}\).
- The length of \(BC\) is:
\(BC = OB - OC = 8 - 4 = 4\text{ cm}\).
- Therefore, the perimeter of the shaded region is:
\(P = \text{Arc } AB + AC + BC = \frac{8\pi}{3} + 4\sqrt{3} + 4\text{ cm}\).

(b) The area of the shaded region is found by subtracting the area of the triangle \(OAC\) from the area of the sector \(OAB\).
- Area of sector \(OAB = \frac{1}{2} r^2 \theta = \frac{1}{2} \times 8^2 \times \frac{1}{3}\pi = \frac{32\pi}{3}\text{ cm}^2\).
- Area of right-angled triangle \(OAC = \frac{1}{2} \times OC \times AC = \frac{1}{2} \times 4 \times 4\sqrt{3} = 8\sqrt{3}\text{ cm}^2\).
- Area of the shaded region is:
\(\text{Area} = \frac{32\pi}{3} - 8\sqrt{3}\text{ cm}^2\).

(a)
M1: For calculating the arc length \(AB\) using \(r\theta\).
B1: For finding the length of \(AC = 4\sqrt{3}\) using exact trigonometric values.
M1: For finding \(OC\) and calculating \(BC = 8 - OC = 4\).
A1: Correct perimeter: \(4 + 4\sqrt{3} + \frac{8\pi}{3}\).

(b)
M1: For calculating the area of the sector \(OAB\).
A1: Correct sector area: \(\frac{32\pi}{3}\).
M1: For calculating the area of the triangle \(OAC\).
A1: Correct triangle area: \(8\sqrt{3}\).
A1: Correct shaded area: \(\frac{32\pi}{3} - 8\sqrt{3}\).

(a) Using the identity \(\tan \theta = \frac{\sin \theta}{\cos \theta}\):
\(3 \sin \theta \left(\frac{\sin \theta}{\cos \theta}\right) = 5 - \cos \theta\)
\(\frac{3 \sin^2 \theta}{\cos \theta} = 5 - \cos \theta\)
\(3 \sin^2 \theta = 5 \cos \theta - \cos^2 \theta\)

Now, substitute \(\sin^2 \theta = 1 - \cos^2 \theta\):
\(3(1 - \cos^2 \theta) = 5 \cos \theta - \cos^2 \theta\)
\(3 - 3 \cos^2 \theta = 5 \cos \theta - \cos^2 \theta\)
\(0 = 2 \cos^2 \theta + 5 \cos \theta - 3\).
Thus, \(2 \cos^2 \theta + 5 \cos \theta - 3 = 0\).

(b) Let \(\theta = 2x\). Since \(0^\circ \le x \le 180^\circ\), we have \(0^\circ \le 2x \le 360^\circ\).
Solving the quadratic equation from (a):
\(2 \cos^2 \theta + 5 \cos \theta - 3 = 0\)
\((2 \cos \theta - 1)(\cos \theta + 3) = 0\)

So, \(\cos \theta = \frac{1}{2}\) or \(\cos \theta = -3\).
Since \(\cos \theta = -3\) has no solutions, we only solve \(\cos \theta = \frac{1}{2}\).
For \(0^\circ \le \theta \le 360^\circ\):
\(\theta = 60^\circ\) or \(\theta = 300^\circ\).

Since \(\theta = 2x\):
\(2x = 60^\circ \implies x = 30^\circ\)
\(2x = 300^\circ \implies x = 150^\circ\).
Both solutions are within the interval \(0^\circ \le x \le 180^\circ\).

(a)
M1: For substituting \(\tan \theta = \frac{\sin \theta}{\cos \theta}\).
M1: For using \(\sin^2 \theta = 1 - \cos^2 \theta\) to eliminate the sine term.
A1: Correctly completing the algebraic steps to show the given quadratic equation.

(b)
M1: For factorizing the quadratic equation to get \(\cos 2x = \frac{1}{2}\) (and indicating no solution for \(\cos 2x = -3\)).
A1: Correctly stating \(\cos 2x = \frac{1}{2}\).
M1: For identifying the two angles for \(2x\) in the range \(0^\circ \le 2x \le 360^\circ\) (e.g. \(60^\circ\) and \(300^\circ\)).
A1: Correct answer \(x = 30^\circ\).
A1: Correct answer \(x = 150^\circ\).
B1: Rejection of any out-of-range solutions, or no extra incorrect answers within the range.

(a) We rewrite the equation as \(y = 12x^{-1} + x^2 - x\).
Differentiating with respect to \(x\):
\(\frac{dy}{dx} = -12x^{-2} + 2x - 1 = -\frac{12}{x^2} + 2x - 1\).

Differentiating again:
\(\frac{d^2y}{dx^2} = 24x^{-3} + 2 = \frac{24}{x^3} + 2\).

(b) At a stationary point, \(\frac{dy}{dx} = 0\):
\(-\frac{12}{x^2} + 2x - 1 = 0\)
Multiply by \(x^2\) (since \(x > 0\)):
\(-12 + 2x^3 - x^2 = 0 \implies 2x^3 - x^2 - 12 = 0\).
Let \(P(x) = 2x^3 - x^2 - 12\). We can test \(x = 2\):
\(2(2)^3 - (2)^2 - 12 = 16 - 4 - 12 = 0\).
So \(x = 2\) is a root. Factorizing out \((x-2)\):
\(2x^3 - x^2 - 12 = (x - 2)(2x^2 + 3x + 6) = 0\).
The quadratic factor is \(2x^2 + 3x + 6\). Its discriminant is:
\(\Delta = 3^2 - 4(2)(6) = 9 - 48 = -39 < 0\).
Since the discriminant is negative, \(2x^2 + 3x + 6 = 0\) has no real roots. Thus, \(x = 2\) is the only real root, and hence there is only one stationary point.
Substituting \(x = 2\) back into the original curve equation:
\(y = \frac{12}{2} + 2^2 - 2 = 6 + 4 - 2 = 8\).
The coordinates of the stationary point are \((2, 8)\).

(c) Evaluate \(\frac{d^2y}{dx^2}\) at \(x = 2\):
\(\frac{d^2y}{dx^2} = \frac{24}{2^3} + 2 = \frac{24}{8} + 2 = 3 + 2 = 5\).
Since \(\frac{d^2y}{dx^2} > 0\), the stationary point \((2, 8)\) is a local minimum.

(a)
M1: For attempting to differentiate \(\frac{12}{x}\) or \(x^2-x\).
A1: Correct first derivative: \(-\frac{12}{x^2} + 2x - 1\).
A1: Correct second derivative: \(\frac{24}{x^3} + 2\).

(b)
M1: Setting their \(\frac{dy}{dx} = 0\) and attempting to solve, resulting in a cubic equation.
A1: Verifying \(x = 2\) is a root and finding the coordinates \((2, 8)\).
M1: Factorizing the cubic to show \((x-2)(2x^2 + 3x + 6) = 0\) and analyzing the discriminant of the quadratic factor.
A1: Correctly showing there are no other real solutions.

(c)
M1: Substituting \(x=2\) into their second derivative expression.
A1: Getting a positive value (e.g. \(5\)) and concluding it is a minimum.

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

Square both sides:
\(9x = (x + 2)^2\)
\(9x = x^2 + 4x + 4\)
\(x^2 - 5x + 4 = 0\)

Factorize:
\((x - 1)(x - 4) = 0\)
So, \(x = 1\) or \(x = 4\).

Substituting back to find \(y\):
When \(x = 1\), \(y = 3\sqrt{1} = 3\).
When \(x = 4\), \(y = 3\sqrt{4} = 6\).
So the intersection points are \((1, 3)\) and \((4, 6)\).

(b) Since \(3\sqrt{x} \ge x + 2\) for \(1 \le x \le 4\), the area of the bounded region is given by the integral:
\(\text{Area} = \int_{1}^{4} \left(3x^{1/2} - (x + 2)\right) dx\)

Integrate each term with respect to \(x\):
\(\int \left(3x^{1/2} - x - 2\right) dx = \left[ 3 \times \frac{2}{3} x^{3/2} - \frac{1}{2} x^2 - 2x \right]_{1}^{4} = \left[ 2x^{3/2} - \frac{1}{2} x^2 - 2x \right]_{1}^{4}\)

Evaluate at the upper limit \(x = 4\):
\(2(4)^{3/2} - \frac{1}{2}(4)^2 - 2(4) = 2(8) - 8 - 8 = 16 - 16 = 0\).

Evaluate at the lower limit \(x = 1\):
\(2(1)^{3/2} - \frac{1}{2}(1)^2 - 2(1) = 2 - 0.5 - 2 = -0.5\).

Subtract the lower limit value from the upper limit value:
\(\text{Area} = 0 - (-0.5) = 0.5\).

(a)
M1: For setting up the equation \(3\sqrt{x} = x + 2\) and attempting to solve (e.g., squaring both sides).
A1: Correct quadratic equation \(x^2 - 5x + 4 = 0\) and roots \(x = 1, 4\).
A1: Correct intersection points \((1, 3)\) and \((4, 6)\).

(b)
M1: For setting up the integral \(\int_{1}^{4} \left(3x^{1/2} - x - 2\right) dx\).
M1: For integrating \(3x^{1/2}\) correctly to \(2x^{3/2}\).
M1: For integrating \(-x-2\) correctly to \(-\frac{1}{2}x^2 - 2x\).
M1: For substituting limits \(4\) and \(1\) into their integrated expression.
A1: Correct calculation of upper limit as \(0\) and lower limit as \(-0.5\).
A1: Correct final area: \(0.5\) (or \(\frac{1}{2}\)).

Let \(u = 5^x\). Since \(5^{2x} = (5^x)^2 = u^2\) and \(5^{x+1} = 5 \times 5^x = 5u\), we can rewrite the equation as:
\(u^2 - 5u + 4 = 0\)

Factorising the quadratic equation:
\((u - 1)(u - 4) = 0\)

So, \(u = 1\) or \(u = 4\).

Case 1: \(5^x = 1\)
\(x = 0\)

Case 2: \(5^x = 4\)
Taking natural logarithms of both sides:
\(\ln(5^x) = \ln 4\)
\(x \ln 5 = \ln 4\)
\(x = \frac{\ln 4}{\ln 5}\)

Therefore, the exact solutions are \(x = 0\) and \(x = \frac{\ln 4}{\ln 5}\).

**M1**: Substitute \(u = 5^x\) to form a quadratic equation \(u^2 - 5u + 4 = 0\) (or equivalent).
**A1**: Obtain the correct roots \(u = 1\) and \(u = 4\).
**A1**: Find the solution \(x = 0\).
**A1**: Find the exact solution \(x = \frac{\ln 4}{\ln 5}\) (or \(\log_5 4\)).

We use the trigonometric identity:
\(\cot^2 \theta = \mathrm{cosec}^2 \theta - 1\)

Substituting this into the equation:
\((\mathrm{cosec}^2 \theta - 1) + 3\ \mathrm{cosec}\ \theta + 3 = 0\)
\(\mathrm{cosec}^2 \theta + 3\ \mathrm{cosec}\ \theta + 2 = 0\)

Letting \(u = \mathrm{cosec}\ \theta\), we have:
\(u^2 + 3u + 2 = 0\)
\((u + 1)(u + 2) = 0\)

So, \(u = -1\) or \(u = -2\), which means:
\(\sin \theta = -1\) or \(\sin \theta = -\frac{1}{2}\)

For the interval \(0^\circ < \theta < 360^\circ\):
- From \(\sin \theta = -1\), we get:
\(\theta = 270^\circ\)
- From \(\sin \theta = -\frac{1}{2}\), the reference angle is \(30^\circ\). Since sine is negative in the third and fourth quadrants:
\(\theta = 180^\circ + 30^\circ = 210^\circ\)
\(\theta = 360^\circ - 30^\circ = 330^\circ\)

Thus, the solutions are \(\theta = 210^\circ, 270^\circ, 330^\circ\).

**M1**: Use the identity \(\cot^2 \theta = \mathrm{cosec}^2 \theta - 1\) to obtain a quadratic equation in terms of \(\mathrm{cosec}\ \theta\) or \(\sin \theta\).
**A1**: Solve the quadratic equation correctly to find \(\sin \theta = -1\) and \(\sin \theta = -\frac{1}{2}\).
**A1**: Find at least one correct value of \(\theta\) (e.g. \(270^\circ\)).
**A1**: Obtain all three correct solutions: \(\theta = 210^\circ, 270^\circ, 330^\circ\) and no other solutions in the range.

Let \( y = 2^x \). Then \( 2^{2x+1} = 2 \cdot (2^x)^2 = 2y^2 \). The given equation can be written as: \( 3(2y^2) - 17y + 10 = 0 \), which simplifies to \( 6y^2 - 17y + 10 = 0 \). Factorising the quadratic equation yields: \( (3y - 2)(2y - 5) = 0 \). Thus, we have the solutions: \( y = \frac{2}{3} \) or \( y = \frac{5}{2} \). Since \( y = 2^x \), we solve for \( x \): 1) \( 2^x = \frac{2}{3} \Rightarrow x = \log_2\left(\frac{2}{3}\right) = \frac{\ln(2/3)}{\ln 2} \). 2) \( 2^x = \frac{5}{2} \Rightarrow x = \log_2\left(\frac{5}{2}\right) = \frac{\ln(5/2)}{\ln 2} \).

M1: Substitute \( y = 2^x \) and express \( 2^{2x+1} \) as \( 2y^2 \).
A1: Obtain the correct quadratic equation \( 6y^2 - 17y + 10 = 0 \).
M1: Attempt to solve the quadratic equation to find two values for \( y \).
A1: Obtain \( y = \frac{2}{3} \) and \( y = \frac{5}{2} \).
M1: Apply logarithmic laws to solve \( 2^x = y \).
A1: Obtain the correct final exact expressions for both values of \( x \).

(i) Let \( 4\sin 2\theta + 3\cos 2\theta = R\sin(2\theta + \alpha) = R\sin 2\theta \cos \alpha + R\cos 2\theta \sin \alpha \). Comparing coefficients gives \( R\cos\alpha = 4 \) and \( R\sin\alpha = 3 \). Thus, \( R = \sqrt{4^2 + 3^2} = 5 \) and \( \tan \alpha = \frac{3}{4} \), which gives \( \alpha \approx 36.87^\circ \). So, \( 4\sin 2\theta + 3\cos 2\theta = 5\sin(2\theta + 36.87^\circ) \).

(ii) The equation is \( 5\sin(2\theta + 36.87^\circ) = 2 \), so \( \sin(2\theta + 36.87^\circ) = 0.4 \). Since \( 0^\circ \le \theta \le 180^\circ \), we have \( 36.87^\circ \le 2\theta + 36.87^\circ \le 396.87^\circ \). The basic angle is \( \sin^{-1}(0.4) \approx 23.58^\circ \). Thus, we solve: 1) \( 2\theta + 36.87^\circ = 180^\circ - 23.58^\circ = 156.42^\circ \Rightarrow 2\theta = 119.55^\circ \Rightarrow \theta \approx 59.8^\circ \). 2) \( 2\theta + 36.87^\circ = 360^\circ + 23.58^\circ = 383.58^\circ \Rightarrow 2\theta = 346.71^\circ \Rightarrow \theta \approx 173.4^\circ \).

Part (i):
B1: State \( R = 5 \).
M1: Use \( \tan\alpha = \frac{3}{4} \) to find \( \alpha \).
A1: Obtain \( \alpha \approx 36.87^\circ \) (or 36.9).

Part (ii):
M1: Set up the equation \( 5\sin(2\theta + 36.87^\circ) = 2 \) and find the basic angle.
A1: Obtain one correct value for the argument \( 2\theta + 36.87^\circ \) (either \( 156.42^\circ \) or \( 383.58^\circ \)).
A1: Obtain \( \theta = 59.8^\circ \).
A1: Obtain \( \theta = 173.4^\circ \).

We rewrite the integrand by performing algebraic division: \( \frac{4x+1}{2x+1} = \frac{2(2x+1) - 1}{2x+1} = 2 - \frac{1}{2x+1} \). Integrating term by term gives: \( \int_{1}^{3} \left( 2 - \frac{1}{2x+1} \right) dx = \left[ 2x - \frac{1}{2}\ln|2x+1| \right]_{1}^{3} \). Substituting the upper and lower limits: At \( x = 3 \): \( 2(3) - \frac{1}{2}\ln(2(3)+1) = 6 - \frac{1}{2}\ln 7 \). At \( x = 1 \): \( 2(1) - \frac{1}{2}\ln(2(1)+1) = 2 - \frac{1}{2}\ln 3 \). Subtracting the values: \( \left( 6 - \frac{1}{2}\ln 7 \right) - \left( 2 - \frac{1}{2}\ln 3 \right) = 4 - \frac{1}{2}(\ln 7 - \ln 3) \). Using the subtraction law for logarithms: \( = 4 - \frac{1}{2}\ln\left(\frac{7}{3}\right) = 4 - \ln\left(\left(\frac{7}{3}\right)^{1/2}\right) = 4 - \ln\left(\sqrt{\frac{7}{3}}\right) \). This is the required form.

M1: Perform division to write the integrand in the form \( A + \frac{B}{2x+1} \).
A1: Obtain \( 2 - \frac{1}{2x+1} \).
M1: Integrate the expression to obtain \( 2x - k\ln|2x+1| \) where \( k \neq 0 \).
A1: Obtain the correct integral \( 2x - \frac{1}{2}\ln|2x+1| \).
M1: Substitute the limits of \( 3 \) and \( 1 \) correctly into their integrated expression.
A1: Apply logarithmic properties to simplify to the final given exact form.

(i) Differentiating both parametric components with respect to \( t \): \( \frac{dx}{dt} = 2t + 2 = 2(t+1) \). Using the chain rule/quotient rule for \( y \): \( \frac{dy}{dt} = \frac{1}{t+1} - \frac{1}{(t+1)^2} = \frac{(t+1) - 1}{(t+1)^2} = \frac{t}{(t+1)^2} \). Applying the parametric chain rule: \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\frac{t}{(t+1)^2}}{2(t+1)} = \frac{t}{2(t+1)^3} \).

(ii) For a stationary point, \( \frac{dy}{dx} = 0 \). Thus, \( \frac{t}{2(t+1)^3} = 0 \Rightarrow t = 0 \). This is valid since \( t > -1 \). Substituting \( t = 0 \) into the parametric equations: \( x = 0^2 + 2(0) = 0 \) and \( y = \ln(1) + \frac{1}{1} = 1 \). Hence, the coordinates of the stationary point are \( (0, 1) \).

(iii) At \( t = 0 \), the curve has the point \( (0, 1) \) and gradient \( \frac{dy}{dx} = 0 \). The equation of the tangent line is \( y - 1 = 0(x - 0) \Rightarrow y = 1 \).

Part (i):
M1: Differentiate \( x \) with respect to \( t \).
M1: Differentiate \( y \) with respect to \( t \).
A1: Obtain correct derivative \( \frac{dy}{dt} = \frac{t}{(t+1)^2} \).
A1: Obtain correct simplified expression for \( \frac{dy}{dx} = \frac{t}{2(t+1)^3} \).

Part (ii):
M1: Set their \( \frac{dy}{dx} = 0 \) and solve for \( t \).
A1: State coordinates \( (0, 1) \).

Part (iii):
M1: Use their coordinates and gradient at \( t=0 \) to find the tangent equation.
A1: State correct tangent equation \( y = 1 \).

(i) Let \( f(x) = x^3 + 2x^2 - 7x - 5 \). Evaluating at the endpoints: \( f(2) = (2)^3 + 2(2)^2 - 7(2) - 5 = 8 + 8 - 14 - 5 = -3 \). \( f(2.5) = (2.5)^3 + 2(2.5)^2 - 7(2.5) - 5 = 15.625 + 12.5 - 17.5 - 5 = 5.625 \). Since there is a change of sign and \( f(x) \) is continuous, there must be a root of \( f(x) = 0 \) in the interval \( 2 < x < 2.5 \).

(ii) We rearrange \( x^3 + 2x^2 - 7x - 5 = 0 \) as follows: \( x^3 + 2x^2 = 7x + 5 \Rightarrow x^2(x + 2) = 7x + 5 \Rightarrow x^2 = \frac{7x+5}{x+2} \). Since we are looking for a root in the interval where \( x > 0 \), we take the positive square root: \( x = \sqrt{\frac{7x+5}{x+2}} \).

(iii) Using \( x_{n+1} = ∑\sqrt{\frac{7x_n+5}{x_n+2}} \) with \( x_1 = 2.2 \):
\( x_2 = \sqrt{\frac{7(2.2)+5}{2.2+2}} = \sqrt{\frac{20.4}{4.2}} \approx 2.2039 \)
\( x_3 = \sqrt{\frac{7(2.2039)+5}{2.2039+2}} \approx 2.2043 \)
\( x_4 = \sqrt{\frac{7(2.2043)+5}{2.2043+2}} \approx 2.2043 \)
Comparing successive iterations, we see the root is \( 2.20 \) to 2 decimal places.

Part (i):
M1: Evaluate \( f(2) \) and \( f(2.5) \).
A1: Obtain the values \( -3 \) and \( 5.625 \) and conclude with a reference to the sign change.

Part (ii):
M1: Factorise \( x^2 \) from \( x^3 + 2x^2 \) or show equivalent division step.
A1: Complete the algebraic rearrangement to show \( x = \sqrt{\frac{7x+5}{x+2}} \).

Part (iii):
M1: Calculate the value of \( x_2 \) using the formula.
A1: Obtain \( x_2 \approx 2.2039 \) and \( x_3 \approx 2.2043 \).
A1: Conclude that the root is \( 2.20 \) correct to 2 decimal places after reaching a stable iteration.