2023 Cambridge IAS-Level Mathematics (9709) 模擬試題連答案詳解

The general term in the expansion is given by \(\binom{9}{r} x^{9-r} \left(\frac{a}{x^2}\right)^r = \binom{9}{r} a^r x^{9-3r}\). For the term to be independent of \(x\), we require the exponent of \(x\) to be zero: \(9 - 3r = 0 \Rightarrow r = 3\). The term independent of \(x\) is therefore \(\binom{9}{3} a^3 = 84a^3\). We are given that this term is \(145152\), so \(84a^3 = 145152 \Rightarrow a^3 = 1728 \Rightarrow a = 12\).

M1: Identify the term where the power of \(x\) is 0 (i.e., \(r=3\)).
A1: Correctly calculate the coefficient \(\binom{9}{3} = 84\).
M1: Set up the equation \(84a^3 = 145152\).
A1.5: Solve for \(a\) to obtain \(a = 12\).

To find the points of intersection, we equate the equations: \(x^2 - 4x + 12 = 2x + k \Rightarrow x^2 - 6x + (12 - k) = 0\). For the line and the curve not to intersect, the discriminant of this quadratic equation must be negative: \(\Delta < 0 \Rightarrow (-6)^2 - 4(1)(12-k) < 0 \Rightarrow 36 - 48 + 4k < 0 \Rightarrow -12 + 4k < 0 \Rightarrow 4k < 12 \Rightarrow k < 3\).

M1: Equate the line and the curve to form a single quadratic equation.
A1: Correctly simplify to \(x^2 - 6x + (12-k) = 0\).
M1: State and apply the condition \(\Delta < 0\) for no intersection.
A1.5: Solve the inequality to obtain \(k < 3\).

First, express \(\text{f}(x)\) in completed square form: \(\text{f}(x) = 2(x^2 - 6x) + 13 = 2[(x-3)^2 - 9] + 13 = 2(x-3)^2 - 5\). Now, let \(y = 2(x-3)^2 - 5\). Rearranging to make \(x\) the subject: \(y + 5 = 2(x-3)^2 \Rightarrow \frac{y+5}{2} = (x-3)^2\). Since the domain of \(\text{f}\) is \(x \le 3\), we must choose the negative square root: \(x - 3 = -\sqrt{\frac{y+5}{2}} \Rightarrow x = 3 - \sqrt{\frac{y+5}{2}}\). Thus, \(\text{f}^{-1}(x) = 3 - \sqrt{\frac{x+5}{2}}\). The domain of \(\text{f}^{-1}(x)\) is the range of \(\text{f}(x)\). Since \(x \le 3\), the minimum value of \(\text{f}(x)\) occurs at \(x = 3\) where \(\text{f}(3) = -5\). Therefore, the range of \(\text{f}\) is \(\text{f}(x) \ge -5\), which means the domain of \(\text{f}^{-1}\) is \(x \ge -5\).

M1: Attempt to complete the square for \(2x^2 - 12x + 13\).
A1: Obtain the correct completed square form \(2(x-3)^2 - 5\).
M1: Attempt to make \(x\) the subject of the equation.
A1: Correctly identify the negative square root to get \(\text{f}^{-1}(x) = 3 - \sqrt{\frac{x+5}{2}}\).
B0.5: State the correct domain \(x \ge -5\).

The perimeter of the sector is given by \(P = 2r + r\theta = 32 \Rightarrow \theta = \frac{32-2r}{r}\). The area of the sector is given by \(A = \frac{1}{2}r^2\theta = 60\). Substituting the expression for \(\theta\) into the area equation: \(\frac{1}{2}r^2\left(\frac{32-2r}{r}\right) = 60 \Rightarrow \frac{1}{2}r(32-2r) = 60 \Rightarrow 16r - r^2 = 60 \Rightarrow r^2 - 16r + 60 = 0\). Factoring the quadratic: \((r-6)(r-10) = 0\), which gives \(r = 6\) or \(r = 10\). Both values of \(r\) yield valid positive angles \(\theta\).

M1: Use the perimeter formula to express \(\theta\) in terms of \(r\).
M1: Substitute into the area formula to obtain an equation in terms of \(r\).
A1: Form a correct quadratic equation, e.g., \(r^2 - 16r + 60 = 0\).
M1: Solve the quadratic equation.
A0.5: State the correct values \(r = 6\) and \(r = 10\).

To find the intersection points, substitute \(y = 2x - 1\) into the circle equation: \(x^2 + (2x-1)^2 - 8x + 6(2x-1) - 11 = 0 \Rightarrow x^2 + (4x^2 - 4x + 1) - 8x + 12x - 6 - 11 = 0 \Rightarrow 5x^2 - 16 = 0\). The x-coordinates of the points of intersection are the roots of this equation: \(x^2 = \frac{16}{5} \Rightarrow x_1 = -\frac{4}{\sqrt{5}}\) and \(x_2 = \frac{4}{\sqrt{5}}\). The x-coordinate of the midpoint \(M\) of \(AB\) is \(x_M = \frac{x_1 + x_2}{2} = 0\). Substituting \(x_M = 0\) into the line equation \(y = 2x - 1\) gives \(y_M = 2(0) - 1 = -1\). Thus, the coordinates of the midpoint are \((0, -1)\).

M1: Substitute the equation of the line into the equation of the circle.
A1: Correctly simplify to a quadratic equation, e.g., \(5x^2 - 16 = 0\).
M1: Find the x-coordinate of the midpoint (using the sum of roots or by calculating individual roots).
A1: Substitute the midpoint's x-coordinate into the line equation to find the y-coordinate.
A0.5: State the final coordinates \((0, -1)\).

The volume of revolution about the x-axis is given by \(V = \pi \int_{a}^{b} y^2 \, \text{d}x\). Here, \(y^2 = (\sqrt{3x + 1})^2 = 3x + 1\). Thus, the volume is \(V = \pi \int_{1}^{5} (3x + 1) \, \text{d}x = \pi \left[ \frac{3}{2}x^2 + x \right]_{1}^{5}\). Evaluating this at the limits: At \(x = 5\), \(\frac{3}{2}(25) + 5 = \frac{75}{2} + 5 = 42.5\). At \(x = 1\), \(\frac{3}{2}(1) + 1 = 2.5\). Therefore, the volume is \(V = \pi (42.5 - 2.5) = 40\pi\).

M1: Apply the volume of revolution formula with correct limits.
A0.5: Identify \(y^2 = 3x+1\).
M1: Integrate \(3x+1\) to obtain \(\frac{3}{2}x^2 + x\).
A1: Apply limits correctly.
A1: Obtain the exact value \(40\pi\).

(i) The tenth, sixth, and third terms of the AP are given by:
\(u_{10} = a + 9d\)
\(u_6 = a + 5d\)
\(u_3 = a + 2d\)

These terms are the first three terms of a GP, so:
\(g_1 = a + 9d\)
\(g_2 = a + 5d\)
\(g_3 = a + 2d\)

Since they form a GP, their common ratio \(r\) is constant, giving:
\(\frac{g_2}{g_1} = \frac{g_3}{g_2} \implies \frac{a+5d}{a+9d} = \frac{a+2d}{a+5d}\)

Multiplying both sides by the denominators:
\((a+5d)^2 = (a+9d)(a+2d)\)
\(a^2 + 10ad + 25d^2 = a^2 + 11ad + 18d^2\)

Rearranging the terms yields:
\(7d^2 = ad\)

Since \(d \neq 0\), we can divide both sides by \(d\) to obtain:
\(a = 7d\).

Now, substitute \(a = 7d\) into the expressions for \(g_1\) and \(g_2\) to find the common ratio \(r\):
\(g_1 = 7d + 9d = 16d\)
\(g_2 = 7d + 5d = 12d\)
\(r = \frac{g_2}{g_1} = \frac{12d}{16d} = \frac{3}{4}\).

(ii)(a) The sum of the first \(n\) terms of an AP is given by:
\(S_n = \frac{n}{2} [2a + (n-1)d]\)

For \(n = 15\) and \(S_{15} = 240\):
\(\frac{15}{2} [2a + 14d] = 240\)
\(15(a + 7d) = 240\)
\(a + 7d = 16\)

Substituting \(a = 7d\) into this equation:
\(7d + 7d = 16 \implies 14d = 16 \implies d = \frac{8}{7}\).

Then, \(a = 7d = 7\left(\frac{8}{7}\right) = 8\).

(ii)(b) The first term of the GP is:
\(g_1 = 16d = 16\left(\frac{8}{7}\right) = \frac{128}{7}\).

Using the sum to infinity formula with \(r = \frac{3}{4}\):
\(S_{\infty} = \frac{g_1}{1 - r} = \frac{\frac{128}{7}}{1 - \frac{3}{4}} = \frac{\frac{128}{7}}{\frac{1}{4}} = \frac{512}{7}\).

(i)
M1: For writing down the expressions for \(u_3\), \(u_6\), and \(u_{10}\) in terms of \(a\) and \(d\).
M1: For setting up the GP ratio equation and expanding it to obtain a quadratic relation.
A1: For showing \(a = 7d\) with clear, correct working.
A1: For finding \(r = \frac{3}{4}\) (or 0.75).

(ii)(a)
M1: For using the sum formula of an AP with \(n=15\) and setting it equal to 240.
M1: For substituting \(a = 7d\) into their equation and solving for \(d\) or \(a\).
A1: For both correct values: \(a = 8\) and \(d = \frac{8}{7}\).

(ii)(b)
M1: For finding the first term of the GP, \(g_1 = 16d = \frac{128}{7}\), using their value of \(d\).
M1: For using the sum to infinity formula \(S_{\infty} = \frac{g_1}{1-r}\) with their \(g_1\) and \(r = \frac{3}{4}\).
A1: For the correct exact value \​\(\frac{512}{7}\) (or \(73\frac{1}{7}\)).

(i) We complete the square for \(\mathrm{f}(x) = 2x^2 - 12x + 13\):
\(\mathrm{f}(x) = 2(x^2 - 6x) + 13\)
\(\mathrm{f}(x) = 2\left[(x-3)^2 - 3^2\right] + 13\)
\(\mathrm{f}(x) = 2(x-3)^2 - 18 + 13\)
\(\mathrm{f}(x) = 2(x-3)^2 - 5\).
Thus, \(a = 2\), \(b = 3\), and \(c = -5\).

(ii) The vertex of this quadratic function is located at \(x = 3\). For \(\mathrm{f}\) to have an inverse, it must be a one-to-one function. Since the domain is restricted to \(x \le k\), the largest value of \(k\) that keeps the function one-to-one (occupying only the left half of the parabola) is \(k = 3\).

(iii) Let \(y = \mathrm{f}(x)\):
\(y = 2(x-3)^2 - 5\)
\(y + 5 = 2(x-3)^2\)
\(\frac{y+5}{2} = (x-3)^2\)

Since the domain of \(\mathrm{f}\) is \(x \le 3\), we take the negative square root:
\(x - 3 = -\sqrt{\frac{y+5}{2}}\)
\(x = 3 - \sqrt{\frac{y+5}{2}}\).

Swapping variables gives:
\(\mathrm{f}^{-1}(x) = 3 - \sqrt{\frac{x+5}{2}}\).

The domain of \(\mathrm{f}^{-1}\) is the range of \(\mathrm{f}\). Since \(x \le 3\), the minimum value of \(\mathrm{f}(x)\) is \(-5\), so the range of \(\mathrm{f}\) is \(\mathrm{f}(x) \ge -5\).
Therefore, the domain of \(\mathrm{f}^{-1}\) is \(x \ge -5\).

(iv) To solve \(\mathrm{f}^{-1}(x) = 1\):
By the definition of an inverse function, \(\mathrm{f}^{-1}(x) = 1 \iff x = \mathrm{f}(1)\).
Since \(1 \le 3\), we can evaluate:
\(\mathrm{f}(1) = 2(1)^2 - 12(1) + 13 = 2 - 12 + 13 = 3\).

Alternatively, using the expression for the inverse:
\(3 - \sqrt{\frac{x+5}{2}} = 1\)
\(\sqrt{\frac{x+5}{2}} = 2\)
\(\frac{x+5}{2} = 4 \implies x + 5 = 8 \implies x = 3\).

(i)
M1: For factoring out 2 from the quadratic and linear terms: \(2(x^2 - 6x) + \dots\)
M1: For completing the square inside the bracket: \(2(x-3)^2 + \dots\)
A1: For the correct expression: \(2(x-3)^2 - 5\).

(ii)
B1: For \(k = 3\) (accept \(k \le 3\)).

(iii)
M1: For attempting to make \(x\) the subject of \(y = 2(x-3)^2 - 5\).
M1: For correctly identifying the need for the negative square root due to \(x \le 3\).
A1: For \(\mathrm{f}^{-1}(x) = 3 - \sqrt{\frac{x+5}{2}}\).
B1: For stating the domain as \(x \ge -5\).

(iv)
M1: For either setting their \(\mathrm{f}^{-1}(x) = 1\) and attempting to solve, or stating/using \(x = \mathrm{f}(1)\).
A1: For \(x = 3\).

(i) Substitute \(y = 5 - x\) into the equation of the circle:
\(x^2 + (5-x)^2 - 10x - 8(5-x) + 31 = 0\)
\(x^2 + (25 - 10x + x^2) - 10x - 40 + 8x + 31 = 0\)
\(2x^2 - 12x + 16 = 0\)

Divide the equation by 2:
\(x^2 - 6x + 8 = 0\)
\((x-2)(x-4) = 0\)
This yields \(x = 2\) or \(x = 4\).

For \(x = 2\):
\(y = 5 - 2 = 3 \implies A(2, 3)\).
For \(x = 4\):
\(y = 5 - 4 = 1 \implies B(4, 1)\).

Thus, the coordinates of \(A\) and \(B\) are \(A(2, 3)\) and \(B(4, 1)\) (in any order).

(ii) First, find the midpoint \(M\) of the chord \(AB\):
\(M = \left(\frac{2+4}{2}, \frac{3+1}{2}\right) = (3, 2)\).

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

The gradient of the perpendicular bisector, \(m_{\perp}\), is:
\(m_{\perp} = -\frac{1}{m_{AB}} = 1\).

Using the point-gradient form with \(M(3, 2)\) and gradient \(1\):
\(y - 2 = 1(x - 3) \implies y = x - 1\).

(iii) To find the centre, complete the square on the circle equation \(x^2 + y^2 - 10x - 8y + 31 = 0\):
\((x-5)^2 - 25 + (y-4)^2 - 16 + 31 = 0\)
\((x-5)^2 + (y-4)^2 = 10\).

So the centre of the circle is \(C(5, 4)\).

To verify that the perpendicular bisector \(y = x - 1\) passes through the centre \(C(5, 4)\), substitute \(x = 5\) and \(y = 4\):
\(4 = 5 - 1\), which is true.
Hence, the perpendicular bisector passes through the centre of the circle.

(i)
M1: For substituting \(y = 5 - x\) into the circle equation.
M1: For simplifying to a 3-term quadratic in \(x\).
A1: For correct x-values: \(x = 2\) and \(x = 4\) (or y-values: \(y = 3\) and \(y = 1\)).
A1: For both points \(A(2, 3)\) and \(B(4, 1)\) correctly identified.

(ii)
M1: For finding the midpoint of \(AB\).
M1: For finding the gradient of \(AB\) and taking its negative reciprocal.
A1: For the correct equation of the perpendicular bisector: \(y = x - 1\) (or equivalent).

(iii)
M1: For completing the square on the circle equation to find the coordinates of the centre.
A1: For identifying the centre as \((5, 4)\).
A1: For substituting \((5, 4)\) into their perpendicular bisector equation and concluding that it lies on the line.

(i) To find the equation of the curve, integrate the derivative:
\(y = \int \left(12(2x-1)^{-2} - 3\right) \mathrm{d}x\)
\(y = \frac{12(2x-1)^{-1}}{-1 \times 2} - 3x + c\)
\(y = -\frac{6}{2x-1} - 3x + c\).

Substitute the coordinates of the point \((2, 5)\) into the integrated equation:
\(5 = -\frac{6}{2(2)-1} - 3(2) + c\)
\(5 = -2 - 6 + c \implies c = 13\).

So, the equation of the curve is:
\(y = -\frac{6}{2x-1} - 3x + 13\).

(ii) Stationary points occur when \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\):
\(\frac{12}{(2x-1)^2} - 3 = 0\)
\(\frac{12}{(2x-1)^2} = 3\)
\((2x-1)^2 = 4\)
\(2x - 1 = \pm 2\).

If \(2x - 1 = 2 \implies 2x = 3 \implies x = 1.5\).
If \(2x - 1 = -2 \implies 2x = -1 \implies x = -0.5\).

Now, calculate the corresponding \(y\)-coordinates:
For \(x = 1.5\):
\(y = -\frac{6}{2(1.5)-1} - 3(1.5) + 13 = -3 - 4.5 + 13 = 5.5\).
So one stationary point is \((1.5, 5.5)\).

For \(x = -0.5\):
\(y = -\frac{6}{2(-0.5)-1} - 3(-0.5) + 13 = 3 + 1.5 + 13 = 17.5\).
So the second stationary point is \((-0.5, 17.5)\).

(iii) Differentiate \(\frac{\mathrm{d}y}{\mathrm{d}x} = 12(2x-1)^{-2} - 3\):
\(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 12 \times (-2)(2x-1)^{-3} \times 2 = -\frac{48}{(2x-1)^3}\).

Test the stationary points:
For \(x = 1.5\):
\(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -\frac{48}{(2(1.5)-1)^3} = -\frac{48}{8} = -6\).
Since \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} < 0\), the point \((1.5, 5.5)\) is a **maximum** point.

For \(x = -0.5\):
\(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -\frac{48}{(2(-0.5)-1)^3} = -\frac{48}{-8} = 6\).
Since \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} > 0\), the point \((-0.5, 17.5)\) is a **minimum** point.

(i)
M1: For integrating \(12(2x-1)^{-2} - 3\), with at least one term correct.
A1: For the correct integration: \(-\frac{6}{2x-1} - 3x\) (must include dividing by 2).
M1: For substituting \((2, 5)\) to find the constant \(c\).
A1: For the correct equation: \(y = -\frac{6}{2x-1} - 3x + 13\) (or equivalent).

(ii)
M1: For setting \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) and solving for \(x\).
A1: For both \(x = 1.5\) and \(x = -0.5\).
A1: For both points \((1.5, 5.5)\) and \((-0.5, 17.5)\) correct.

(iii)
M1: For differentiating \(\frac{\mathrm{d}y}{\mathrm{d}x}\) to find \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\).
A1: For \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -\frac{48}{(2x-1)^3}\).
A1: For evaluating at both \(x\) values and correctly concluding \((1.5, 5.5)\) is a maximum and \((-0.5, 17.5)\) is a minimum.

(i) Express the LHS with a common denominator:
\(\text{LHS} = \frac{\sin \theta}{1 - \cos \theta} + \frac{1 - \cos \theta}{\sin \theta}\)
\(\text{LHS} = \frac{\sin^2 \theta + (1 - \cos \theta)^2}{\sin \theta (1 - \cos \theta)}\)

Expand the numerator:
\(\sin^2 \theta + (1 - 2\cos \theta + \cos^2 \theta) = \sin^2 \theta + \cos^2 \theta + 1 - 2\cos \theta\)

Using the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\):
\(\text{Numerator} = 1 + 1 - 2\cos \theta = 2 - 2\cos \theta = 2(1 - \cos \theta)\).

Substitute the simplified numerator back:
\(\text{LHS} = \frac{2(1 - \cos \theta)}{\sin \theta (1 - \cos \theta)}\)

Cancelling the common factor \((1 - \cos \theta)\) (where \(\cos \theta \neq 1\)) yields:
\(\text{LHS} = \frac{2}{\sin \theta}\). (proven)

(ii) Using the identity proven in part (i), we can rewrite the equation as:
\(\frac{2}{\sin \theta} = 3\tan \theta\)
\(\frac{2}{\sin \theta} = \frac{3\sin \theta}{\cos \theta}\)

Multiplying both sides by \(\sin \theta \cos \theta\) (noting \(\sin \theta \neq 0\)):
\(2\cos \theta = 3\sin^2 \theta\)

Substitute \(\sin^2 \theta = 1 - \cos^2 \theta\):
\(2\cos \theta = 3(1 - \cos^2 \theta)\)
\(3\cos^2 \theta + 2\cos \theta - 3 = 0\).

Let \(u = \cos \theta\):
\(3u^2 + 2u - 3 = 0\).

Using the quadratic formula:
\(u = \frac{-2 \pm \sqrt{2^2 - 4(3)(-3)}}{6} = \frac{-2 \pm \sqrt{40}}{6} = \frac{-1 \pm \sqrt{10}}{3}\).

This gives two values:
1. \(\cos \theta = \frac{-1 + \sqrt{10}}{3} \approx 0.7208\).
2. \(\cos \theta = \frac{-1 - \sqrt{10}}{3} \approx -1.387\).

Since \(-1.387 < -1\), the second case has no real solutions.
For \(\cos \theta = 0.7208\):
The principal angle is:
\(\theta = \cos^{-1}(0.7208) \approx 43.9^{\circ}\).

Since cosine is positive in the first and fourth quadrants, the solutions in the range \(0^{\circ} \le \theta \le 360^{\circ}\) are:
\(\theta_1 = 43.9^{\circ}\)
\(\theta_2 = 360^{\circ} - 43.9^{\circ} = 316.1^{\circ}\).

(i)
M1: For putting the two fractions over a common denominator.
M1: For expanding \((1-\cos\theta)^2\) correctly.
M1: For using \(\sin^2\theta + \cos^2\theta = 1\) to simplify the numerator.
A1: For obtaining \(\frac{2}{\sin\theta}\) with clear and complete working.

(ii)
M1: For rewriting the equation as \(\frac{2}{\sin\theta} = 3\tan\theta\).
M1: For writing \(\tan\theta = \frac{\sin\theta}{\cos\theta}\) and obtaining \(2\cos\theta = 3\sin^2\theta\).
M1: For using \(\sin^2\theta = 1 - \cos^2\theta\) to form a quadratic in \(\cos\theta\).
A1: For the correct quadratic equation: \(3\cos^2\theta + 2\cos\theta - 3 = 0\).
M1: For solving the quadratic equation to get \(\cos\theta \approx 0.7208\) and discarding the other root.
A1: For both correct solutions \(\theta = 43.9^{\circ}\) and \(\theta = 316.1^{\circ}\) (and no extra values in the range).

To solve the equation:

\[\ln(5x - 2) - 2\ln(x) = \ln(2)\]

First, we use the power law of logarithms to rewrite the second term:

\[2\ln(x) = \ln(x^2)\]

Substituting this back into the equation gives:

\[\ln(5x - 2) - \ln(x^2) = \ln(2)\]

Next, apply the subtraction law of logarithms:

\[\ln\left(\frac{5x - 2}{x^2}\right) = \ln(2)\]

Taking the exponential of both sides, we eliminate the logarithms:

\[\frac{5x - 2}{x^2} = 2\]

Multiply both sides by \(x^2\):

\[5x - 2 = 2x^2\]

Rearrange this into a quadratic equation:

\[2x^2 - 5x + 2 = 0\]

Factorize the quadratic expression:

\[(2x - 1)(x - 2) = 0\]

This yields two possible solutions:

\[x = \frac{1}{2} \quad \text{or} \quad x = 2\]

We must check if both solutions are valid for the original logarithmic equation.
- For \(x = 0.5\), the argument \(5x - 2 = 0.5 > 0\) and \(x = 0.5 > 0\), so it is a valid solution.
- For \(x = 2\), the argument \(5x - 2 = 8 > 0\) and \(x = 2 > 0\), so it is also a valid solution.

Thus, the solutions are \(x = 2\) and \(x = 0.5\).

M1: Apply power law of logarithms to write \(2\ln(x)\) as \(\ln(x^2)\).
M1: Apply subtraction law to combine logarithms into a single logarithmic term.
A1: Obtain the correct quadratic equation \(2x^2 - 5x + 2 = 0\) (or equivalent).
M1: Solve the quadratic equation to find two values of \(x\).
A1: State both correct solutions \(x = 2\) and \(x = 0.5\) and show awareness of validity (both arguments must be positive).

Given the equation:

\[4\sin(2\theta) = 3\sec\theta\]

We can use the double-angle identity for sine, \(\sin(2\theta) = 2\sin\theta\cos\theta\), and the definition of secant, \(\sec\theta = \frac{1}{\cos\theta}\):

\[4(2\sin\theta\cos\theta) = \frac{3}{\cos\theta}\]

\[8\sin\theta\cos\theta = \frac{3}{\cos\theta}\]

Since \(\sec\theta\) is defined, \(\cos\theta \neq 0\). We can multiply both sides by \(\cos\theta\):

\[8\sin\theta\cos^2\theta = 3\]

Now use the identity \(\cos^2\theta = 1 - \sin^2\theta\) to write the equation in terms of \(\sin\theta\):

\[8\sin\theta(1 - \sin^2\theta) = 3\]

\[8\sin\theta - 8\sin^3\theta = 3\]

Rearrange to form a cubic equation:

\[8\sin^3\theta - 8\sin\theta + 3 = 0\]

Let \(u = \sin\theta\). The equation becomes:

\[8\u^3 - 8u + 3 = 0\]

By trying simple rational values, we find that \(u = \frac{1}{2}\) is a root because:

\[8\left(\frac{1}{2}\right)^3 - 8\left(\frac{1}{2}\right) + 3 = 1 - 4 + 3 = 0\]

Thus, \((2u - 1)\) is a factor of the polynomial.
Dividing the cubic by \((2u - 1)\) gives:

\[(2u - 1)(4u^2 + 2u - 3) = 0\]

This yields two cases:

**Case 1:** \(2u - 1 = 0 \Rightarrow u = \frac{1}{2}\)

\[\sin\theta = \frac{1}{2}\]

For \(0^\circ < \theta < 360^\circ\), the solutions are:

\[\theta = 30.0^\circ, 150.0^\circ\]

**Case 2:** \(4u^2 + 2u - 3 = 0\)

Using the quadratic formula:

\[u = \frac{-2 \pm \sqrt{2^2 - 4(4)(-3)}}{2(4)} = \frac{-2 \pm \sqrt{52}}{8} = \frac{-1 \pm \sqrt{13}}{4}\]

Calculating the values:

\[u_1 = \frac{-1 + \sqrt{13}}{4} \approx 0.65139\]

\[u_2 = \frac{-1 - \sqrt{13}}{4} \approx -1.15139\]

Since \(\sin\theta\) must lie in the range \([-1, 1]\), the root \(u_2 \approx -1.15139\) is rejected.

For \(\sin\theta \approx 0.65139\):
- The basic angle is \(\sin^{-1}(0.65139) \approx 40.64^\circ\).
- Since \(\sin\theta > 0\), the solutions in the interval \(0^\circ < \theta < 360^\circ\) are:

\[\theta = 40.6^\circ, 180^\circ - 40.6^\circ = 139.4^\circ\]

Combining all valid solutions, we obtain:

\[\theta = 30.0^\circ, 40.6^\circ, 139.4^\circ, 150.0^\circ\]

M1: Use the identities \(\sin(2\theta) = 2\sin\theta\cos\theta\) and \(\sec\theta = \frac{1}{\cos\theta}\) to rewrite the equation in terms of sines and cosines.
M1: Use the identity \(\cos^2\theta = 1 - \sin^2\theta\) to form a cubic equation in \(\sin\theta\).
A1: Obtain the correct cubic equation \(8\sin^3\theta - 8\sin\theta + 3 = 0\) (or equivalent).
M1: Solve the cubic equation to find \(\sin\theta = 0.5\) and \(\sin\theta = \frac{-1+\sqrt{13}}{4}\) (or 0.651), clearly rejecting the invalid root.
A1: Find all four correct angles: \(\theta = 30.0^\circ, 40.6^\circ, 139.4^\circ, 150.0^\circ\) (accept 30° and 150°).

(i) Let \(f(x) = x^3 - 3x^2 + 5\).

Evaluate \(f(x)\) at the endpoints of the interval \([-1.5, -1.1]\):

\[f(-1.5) = (-1.5)^3 - 3(-1.5)^2 + 5 = -3.375 - 6.75 + 5 = -5.125\]

\[f(-1.1) = (-1.1)^3 - 3(-1.1)^2 + 5 = -1.331 - 3.63 + 5 = 0.039\]

Since there is a change of sign between \(f(-1.5) < 0\) and \(f(-1.1) > 0\), and the function \(f(x)\) is continuous on this interval, there must be a root \(\alpha\) lying between \(-1.5\) and \(-1.1\).

(ii) Using the iterative formula \(x_{n+1} = -\sqrt{\frac{5}{3-x_n}}\) with \(x_1 = -1.2\):

\[x_2 = -\sqrt{\frac{5}{3 - (-1.2)}} = -\sqrt{\frac{5}{4.2}} \approx -1.0911\]

\[x_3 = -\sqrt{\frac{5}{3 - (-1.0911)}} = -\sqrt{\frac{5}{4.0911}} \approx -1.1055\]

\[x_4 = -\sqrt{\frac{5}{3 - (-1.1055)}} = -\sqrt{\frac{5}{4.1055}} \approx -1.1036\]

\[x_5 = -\sqrt{\frac{5}{3 - (-1.1036)}} = -\sqrt{\frac{5}{4.1036}} \approx -1.1038\]

\[x_6 = -\sqrt{\frac{5}{3 - (-1.1038)}} = -\sqrt{\frac{5}{4.1038}} \approx -1.1038\]

Since successive iterations round to \(-1.10\) to 2 decimal places, we conclude that \(\alpha \approx -1.10\).

M1: Evaluate \(f(-1.5)\) and \(f(-1.1)\) with at least one correct calculation and state the conclusion based on sign change.
A1: Obtain correct values \(f(-1.5) = -5.13\) (or \(-5.125\)) and \(f(-1.1) = 0.039\) (or \(0.04\)) with a fully correct explanation.
M1: Apply the iterative formula at least twice, showing correct substitution of previous values.
A1: Obtain the correct sequence of values \(x_2 \approx -1.0911\), \(x_3 \approx -1.1055\), \(x_4 \approx -1.1036\), \(x_5 \approx -1.1038\) (accept minor rounding differences in the 4th decimal place).
A1: State the final root \(\alpha = -1.10\) correct to 2 decimal places, following sufficient convergence.

**(i)** To find the intersection points with the \(x\)-axis, we set \( y = 0 \):
\( 3^{2x+1} - 10(3^x) + 3 = 0 \)

Since \( 3^{2x+1} = 3 \cdot (3^x)^2 \), we can substitute \( u = 3^x \):
\( 3u^2 - 10u + 3 = 0 \)

Factorising the quadratic equation:
\( (3u - 1)(u - 3) = 0 \)

This gives \( u = \frac{1}{3} \) or \( u = 3 \).

Substituting back \( u = 3^x \):
- For \( 3^x = \frac{1}{3} \), we have \( x = -1 \).
- For \( 3^x = 3 \), we have \( x = 1 \).

So, the \(x\)-coordinates of the intersection points are \( x = -1 \) and \( x = 1 \).

**(ii)** To find the gradient, we differentiate \( y = 3^{2x+1} - 10(3^x) + 3 \) with respect to \( x \):
Using the rule \( \frac{d}{dx}(a^{f(x)}) = a^{f(x)} \ln(a) f'(x) \):
\( \frac{dy}{dx} = 2 \cdot 3^{2x+1} \ln 3 - 10
\cdot 3^x \ln 3 \)

We can factorise this as:
\( \frac{dy}{dx} = (2 \cdot 3^{2x+1} - 10 \cdot 3^x) \ln 3 \)

Now, substitute the \(x\)-coordinates found in part (i):
- At \( x = -1 \):
\( \frac{dy}{dx} = (2 \cdot 3^{-1} - 10 \cdot 3^{-1}) \ln 3 = \left(\frac{2}{3} - \frac{10}{3}\right) \ln 3 = -\frac{8}{3} \ln 3 \)
- At \( x = 1 \):
\( \frac{dy}{dx} = (2 \cdot 3^3 - 10 \cdot 3^1) \ln 3 = (54 - 30) \ln 3 = 24
\ln 3 \)

**(i)**
M1: For substituting \( u = 3^x \) to obtain a quadratic equation in \( u \).
A1: For the correct quadratic equation \( 3u^2 - 10u + 3 = 0 \).
M1: For solving the quadratic equation to find two positive values for \( u \).
A1: For finding both correct \(x\)-coordinates: \( x = -1 \) and \( x = 1 \).

**(ii)**
M1: For attempting to differentiate an exponential term of the form \( a^{kx} \) to obtain \( k a^{kx} \ln a \).
A1: For the correct derivative \( \frac{dy}{dx} = 2 \cdot 3^{2x+1} \ln 3 - 10 \cdot 3^x \ln 3 \).
M1: For substituting either of their \(x\)-values from part (i) into their derivative.
A1: For the correct exact gradient at \( x = -1 \), which is \( -\frac{8}{3} \ln 3 \) (or equivalent).
A0.75: For the correct exact gradient at \( x = 1 \), which is \( 24 \ln 3 \).

**(i)** Using the trigonometric identity \( \sec^2 \theta = 1 + \tan^2 \theta \), we substitute this into the equation:
\( 3(1 + \tan^2 \theta) + 2 \tan \theta = 8 \)

Expand the brackets:
\( 3 + 3 \tan^2 \theta + 2 \tan \theta = 8 \)

Subtract 8 from both sides:
\( 3 \tan^2 \theta + 2 \tan \theta - 5 = 0 \) (as required).

**(ii)** Let \( \theta = 2x + 0.1 \). The equation becomes \( 3 \tan^2 \theta + 2 \tan \theta - 5 = 0 \).

Factorising the quadratic:
\( (3 \tan \theta + 5)(\tan \theta - 1) = 0 \)

This gives two cases:
- **Case 1**: \( \tan \theta = 1 \)
- **Case 2**: \( \tan \theta = -\frac{5}{3} \)

Given \( 0 \le x \le \pi \), we have \( 0 \le 2x \le 2\pi \), so \( 0.1 \le 2x + 0.1 \le 2\pi + 0.1 \), which is approximately \( 0.1 \le \theta \le 6.383 \).

- For **Case 1**: \( \tan \theta = 1 \)
The basic angle is \( \tan^{-1}(1) = \frac{\pi}{4} \approx 0.7854 \).
Since \( \tan \theta > 0 \), the solutions in the interval are:
\( \theta = \frac{\pi}{4} \approx 0.7854 \)
\( \theta = \pi + \frac{\pi}{4} = \frac{5\pi}{4} \approx 3.9270 \)

Now solve for \( x \):
\( 2x + 0.1 = 0.7854 \implies 2x = 0.6854 \implies x \approx 0.343 \)
\( 2x + 0.1 = 3.9270 \implies 2x = 3.8270 \implies x \approx 1.91 \)

- For **Case 2**: \( \tan \theta = -\frac{5}{3} \)
The basic angle is \( \tan^{-1}\left(\frac{5}{3}\right) \approx 1.0304 \).
Since \( \tan \theta < 0 \), the solutions in the interval are:
\( \theta = \pi - 1.0304 \approx 2.1112 \)
\( \theta = 2\pi - 1.0304 \approx 5.2528 \)

Now solve for \( x \):
\( 2x + 0.1 = 2.1112 \implies 2x = 2.0112 \implies x \approx 1.01 \)
\( 2x + 0.1 = 5.2528 \implies 2x = 5.1528 \implies x \approx 2.58 \)

**(i)**
M1: For using the identity \( \sec^2 \theta = 1 + \tan^2 \theta \).
A1: For correct substitution to obtain \( 3(1 + \tan^2 \theta) + 2 \tan \theta = 8 \).
A1: For obtaining the given equation \( 3 \tan^2 \theta + 2 \tan \theta - 5 = 0 \) with no errors shown.

**(ii)**
M1: For factorising or solving the quadratic to find two values for \( \tan(2x + 0.1) \).
A1: For obtaining \( \tan(2x+0.1) = 1 \) and \( \tan(2x+0.1) = -\frac{5}{3} \).
M1: For finding at least two values of \( 2x+0.1 \) using inverse tangent (in radians).
A1: For finding any two of the four correct values of \( x \) (0.343, 1.01, 1.91, 2.58).
A1: For finding the remaining two correct values of \( x \).
A0.75: For writing all four final answers correct to 3 significant figures and no extra values in the range.

**(i)** Let \( f(x) = e^{-x} + 2x^2 - 3 \).
Evaluate the function at \( x = 1.1 \) and \( x = 1.2 \):
\( f(1.1) = e^{-1.1} + 2(1.1)^2 - 3 = 0.33287 + 2.42 - 3 = -0.24713 < 0 \)
\( f(1.2) = e^{-1.2} + 2(1.2)^2 - 3 = 0.30119 + 2.88 - 3 = 0.18119 > 0 \)

Since \( f(x) \) is continuous and there is a change of sign between \( x = 1.1 \) and \( x = 1.2 \), the root \( \alpha \) must lie in this interval.

**(ii)** Let the initial value be \( x_1 = 1.15 \).
Using the formula \( x_{n+1} = \sqrt{\frac{3 - e^{-x_n}}{2}} \):
\( x_2 = \sqrt{\frac{3 - e^{-1.15}}{2}} \approx 1.15831 \)
\( x_3 = \sqrt{\frac{3 - e^{-1.15831}}{2}} \approx 1.15934 \)
\( x_4 = \sqrt{\frac{3 - e^{-1.15934}}{2}} \approx 1.15947 \)
\( x_5 = \sqrt{\frac{3 - e^{-1.15947}}{2}} \approx 1.15949 \)
\( x_6 = \sqrt{\frac{3 - e^{-1.15949}}{2}} \approx 1.15949 \)

Since successive iterations converge to \( 1.15949 \), we have \( \alpha = 1.159 \) correct to 3 decimal places.

**(iii)** To rearrange the equation:
\( e^{-x} + 2x^2 - 3 = 0 \)
\( e^{-x} = 3 - 2x^2 \)

Taking the reciprocal of both sides:
\( e^x = \frac{1}{3 - 2x^2} \)

Taking the natural logarithm of both sides:
\( x = \ln \left( \frac{1}{3 - 2x^2} \right) \) (as required).

If we use the initial value \( x_0 = 1.5 \):
The expression inside the logarithm is \( \frac{1}{3 - 2x^2} \).
Evaluating \( 3 - 2x^2 \) at \( x = 1.5 \):
\( 3 - 2(1.5)^2 = 3 - 2(2.25) = 3 - 4.5 = -1.5 \)

Since this value is negative, the argument of the logarithm is negative (specifically, \( -\frac{2}{3} \)). Because the natural logarithm of a negative number is undefined in the set of real numbers, this iterative formula cannot be used with \( x_0 = 1.5 \).

**(i)**
M1: For attempting to evaluate \( f(x) = e^{-x} + 2x^2 - 3 \) at \( 1.1 \) and \( 1.2 \).
A1: For correct values showing a sign change (e.g. \( -0.247 \) and \( 0.181 \)) and a concluding statement.

**(ii)**
M1: For attempting to use the iterative formula with an appropriate starting value in the interval \( (1.1, 1.2) \).
A1: For obtaining at least two correct intermediate iterations to 5 decimal places.
A1: For showing that iterations converge to \( 1.15949 \) (or showing \( x_5 \approx 1.15949 \) and \( x_6 \approx 1.15949 \)).
A0.75: For concluding \( \alpha = 1.159 \) to 3 decimal places.

**(iii)**
M1: For attempting to isolate \( e^{-x} \) and taking natural logarithms of both sides.
A1: For showing all steps clearly to arrive at \( x = \ln \left( \frac{1}{3 - 2x^2}
\right) \).
A1: For substituting \( x = 1.5 \) to find that \( 3 - 2x^2 = -1.5 < 0 \) and stating that the logarithm of a negative number is undefined.

**(i)** We differentiate the equation \( y^3 - 3xy + x^3 = 3 \) implicitly with respect to \( x \):
Using the chain rule on \( y^3 \) gives \( 3y^2 \frac{dy}{dx} \).
Using the product rule on \( -3xy \) gives \( -3y - 3x \frac{dy}{dx} \).
Differentiating \( x^3 \) gives \( 3x^2 \).
Differentiating the constant 3 gives 0.

This gives:
\( 3y^2 \frac{dy}{dx} - 3y - 3x \frac{dy}{dx} + 3x^2 = 0 \)

Divide the entire equation by 3:
\( y^2 \frac{dy}{dx} - y - x \frac{dy}{dx} + x^2 = 0 \)

Group the \( \frac{dy}{dx} \) terms:
\( (y^2 - x) \frac{dy}{dx} = y - x^2 \)

Divide by \( y^2 - x \):
\( \frac{dy}{dx} = \frac{y - x^2}{y^2 - x} \).

**(ii)** For the tangent to be parallel to the \(x\)-axis, the gradient \( \frac{dy}{dx} \) must be 0:
\( \frac{y - x^2}{y^2 - x} = 0 \implies y - x^2 = 0 \implies y = x^2 \) (provided the denominator \( y^2 - x \neq 0 \)).

Substitute \( y = x^2 \) back into the original curve equation:
\( (x^2)^3 - 3x(x^2) + x^3 = 3 \)
\( x^6 - 3x^3 + x^3 = 3 \)
\( x^6 - 2x^3 - 3 = 0 \)

Let \( u = x^3 \). This is a quadratic equation in \( u \):
\( u^2 - 2u - 3 = 0 \)

Factorising gives:
\( (u - 3)(u + 1) = 0 \)

So \( u = 3 \) or \( u = -1 \).

- If \( u = -1 \):
\( x^3 = -1 \implies x = -1 \)
Since \( y = x^2 \), we have \( y = (-1)^2 = 1 \).
So, one point is \( (-1, 1) \).
Check denominator: \( y^2 - x = 1^2 - (-1) = 2 \neq 0 \). This point is valid.

- If \( u = 3 \):
\( x^3 = 3 \implies x = 3^{1/3} \)
Since \( y = x^2 \), we have \( y = (3^{1/3})^2 = 3^{2/3} \).
So, another point is \( (3^{1/3}, 3^{2/3}) \).
Check denominator: \( y^2 - x = 3^{4/3} - 3^{1/3} = 3^{1/3}(3 - 1) = 2 \cdot 3^{1/3} \neq 0 \). This point is valid.

**(i)**
M1: For attempting to differentiate implicit terms (introducing \( \frac{dy}{dx} \) correctly for at least one term).
A1: For correct differentiation of the product term \( -3xy \).
A1: For the fully correct differentiated equation \( 3y^2 \frac{dy}{dx} - 3y - 3x \frac{dy}{dx} + 3x^2 = 0 \) (or equivalent).
A1: For isolating \( \frac{dy}{dx} \) to get \( \frac{dy}{dx} = \frac{y - x^2}{y^2 - x} \) (or equivalent).

**(ii)**
M1: For setting the numerator of \( \frac{dy}{dx} \) to 0 to get \( y = x^2 \).
M1: For substituting \( y = x^2 \) into the original equation to obtain a quadratic equation in \( x^3 \) (e.g. \( x^6 - 2x^3 - 3 = 0 \)).
M1: For solving the quadratic in \( x^3 \) to find \( x^3 = 3 \) and \( x^3 = -1 \).
A1: For finding the exact coordinates of the first point: \( (-1, 1) \).
A0.75: For finding the exact coordinates of the second point: \( (3^{1/3}, 3^{2/3}) \) or equivalent.