2024 Cambridge IAS-Level Mathematics (9709) Practice Paper with Answers

Equating the line and the curve: \( 2x - k = x^2 - 2kx + 3 \). Rearranging into a standard quadratic equation in \( x \): \( x^2 - 2(k+1)x + (k+3) = 0 \). Since the line and the curve intersect at two distinct points, the discriminant must be strictly positive: \( \Delta = [-2(k+1)]^2 - 4(1)(k+3) > 0 \). Expanding the terms gives: \( 4(k^2 + 2k + 1) - 4k - 12 > 0 \) which simplifies to \( 4k^2 + 8k + 4 - 4k - 12 > 0 \), and thus \( 4k^2 + 4k - 8 > 0 \). Dividing the inequality by 4 gives: \( k^2 + k - 2 > 0 \). Factorising the quadratic inequality: \( (k+2)(k-1) > 0 \). Therefore, the set of values of \( k \) is \( k < -2 \) or \( k > 1 \).

M1: Equating line and curve and rearranging into standard quadratic form. A1: Correct quadratic in \( x \): \( x^2 - 2(k+1)x + (k+3) = 0 \) or equivalent. M1: Applying discriminant condition \( b^2 - 4ac > 0 \). A1: Correct simplified quadratic inequality \( k^2 + k - 2 > 0 \). M1: Finding the critical values of \( k \). A2: Correct final range: \( k < -2 \) or \( k > 1 \) (1 mark for each inequality).

(a) Completing the square for \( f(x) \): \( f(x) = 2(x^2 - 4x) + 5 = 2[(x-2)^2 - 4] + 5 = 2(x-2)^2 - 3 \). The vertex of this parabola is at \( x = 2 \). For the function to have an inverse, it must be a one-to-one function. Given that the domain is \( x \ge k \), the smallest value of \( k \) to ensure the function is strictly increasing is \( k = 2 \). (b) To find the inverse function, let \( y = 2(x-2)^2 - 3 \). Rearranging for \( x \) gives: \( y + 3 = 2(x-2)^2 \) which leads to \( (x-2)^2 = \frac{y+3}{2} \). Taking the square root: \( x - 2 = \pm\sqrt{\frac{y+3}{2}} \). Since the domain of \( f \) is \( x \ge 2 \), we choose the positive root: \( x = 2 + \sqrt{\frac{y+3}{2}} \). Thus, \( f^{-1}(x) = 2 + \sqrt{\frac{x+3}{2}} \). The domain of \( f^{-1} \) is the range of \( f \). Since \( x \ge 2 \), the minimum value of \( f(x) \) is \( -3 \), so the range of \( f \) is \( f(x) \ge -3 \). Therefore, the domain of \( f^{-1} \) is \( x \ge -3 \).

(a) M1: For completing the square to identify the coordinate of the vertex. A1: Correctly stating \( k = 2 \). (b) M1: Setting \( y = 2(x-2)^2 - 3 \) and making \( (x-2)^2 \) the subject. M1: Taking the square root and choosing the positive root based on the domain. A1: Correct inverse function \( f^{-1}(x) = 2 + \sqrt{\frac{x+3}{2}} \) in terms of \( x \). M1: Attempting to find the range of \( f \). A1: Correctly stating the domain of \( f^{-1} \) as \( x \ge -3 \).

(a) The midpoint \( M \) of \( AB \) is \( \left(\frac{2+8}{2}, \frac{5-3}{2}\right) = (5, 1) \). The gradient of \( AB \) is \( m_{AB} = \frac{-3 - 5}{8 - 2} = -\frac{8}{6} = -\frac{4}{3} \). The gradient of the perpendicular bisector is the negative reciprocal: \( m_{\perp} = -\frac{1}{m_{AB}} = \frac{3}{4} \). The equation of the perpendicular bisector is \( y - 1 = \frac{3}{4}(x - 5) \), which simplifies to \( 4y - 4 = 3x - 15 \) or \( 3x - 4y - 11 = 0 \). (b) Solve simultaneously with \( y = x - 3 \): Substitute \( y = x - 3 \) into the perpendicular bisector equation: \( 3x - 4(x-3) - 11 = 0 \implies -x + 12 - 11 = 0 \implies x = 1 \). Substitute \( x = 1 \) back: \( y = 1 - 3 = -2 \). Thus, the coordinates of \( C \) are \( (1, -2) \). The distance \( AC \) between \( A(2, 5) \) and \( C(1, -2) \) is: \( AC = \sqrt{(2-1)^2 + (5 - (-2))^2} = \sqrt{1 + 49} = \sqrt{50} = 5\sqrt{2} \).

(a) M1: Finding the midpoint of \( AB \). M1: Finding the gradient of \( AB \) and applying the perpendicular condition. A1: Stating the correct perpendicular bisector equation in any equivalent form. (b) M1: Solving simultaneously to find \( C \). A1: Correct coordinates \( C(1, -2) \). M1: Applying the distance formula to find \( AC \). A1: Stating the correct exact distance \( 5\sqrt{2} \) (or \( \sqrt{50} \) or decimal \( 7.07 \)).

(a) In the right-angled triangle \( OAD \): \( OD = 8 \cos(\frac{\pi}{3}) = 8 \times 0.5 = 4\text{ cm} \), and \( AD = 8 \sin(\frac{\pi}{3}) = 8 \times \frac{\sqrt{3}}{2} = 4\sqrt{3}\text{ cm} \). Since \( OB = 8\text{ cm} \), the length of \( BD \) is \( BD = OB - OD = 8 - 4 = 4\text{ cm} \). The arc length \( AB \) is \( s = r\theta = 8 \times \frac{\pi}{3} = \frac{8\pi}{3}\text{ cm} \). The perimeter of the shaded region is \( \text{Arc } AB + BD + AD = \frac{8\pi}{3} + 4 + 4\sqrt{3}\text{ cm} \). (b) The area of sector \( OAB \) is \( A_{\text{sector}} = \frac{1}{2} r^2 \theta = \frac{1}{2} \times 8^2 \times \frac{\pi}{3} = \frac{32\pi}{3}\text{ cm}^2 \). The area of right-angled triangle \( OAD \) is \( A_{\text{triangle}} = \frac{1}{2} \times OD \times AD = \frac{1}{2} \times 4 \times 4\sqrt{3} = 8\sqrt{3}\text{ cm}^2 \). The area of the shaded region is \( A_{\text{sector}} - A_{\text{triangle}} = \frac{32\pi}{3} - 8\sqrt{3}\text{ cm}^2 \).

(a) M1: Calculating \( OD \) and \( AD \) using trigonometry. M1: Calculating the length of \( BD \) and the arc length \( AB \). A1: Stating the correct perimeter: \( \frac{8\pi}{3} + 4 + 4\sqrt{3} \). (b) M1: Calculating the area of the sector \( OAB \). M1: Calculating the area of the triangle \( OAD \). M1: Subtracting the area of the triangle from the area of the sector. A1: Stating the correct exact area: \( \frac{32\pi}{3} - 8\sqrt{3} \).

(a) Using the identity \( \tan \theta = \frac{\sin \theta}{\cos \theta} \), the equation becomes \( 3 \sin \theta \left(\frac{\sin \theta}{\cos \theta}\right) = 8 \implies 3 \sin^2 \theta = 8 \cos \theta \). Using the identity \( \sin^2 \theta = 1 - \cos^2 \theta \), we obtain \( 3(1 - \cos^2 \theta) = 8 \cos \theta \implies 3 - 3 \cos^2 \theta = 8 \cos \theta \). Rearranging gives \( 3 \cos^2 \theta + 8 \cos \theta - 3 = 0 \). (b) Factorising the quadratic in \( \cos \theta \): \( (3 \cos \theta - 1)(\cos \theta + 3) = 0 \). This gives \( \cos \theta = \frac{1}{3} \) or \( \cos \theta = -3 \). Since \( \cos \theta = -3 \) has no solutions, we only solve \( \cos \theta = \frac{1}{3} \). This yields the primary solution \( \theta = \cos^{-1}(\frac{1}{3}) \approx 70.5^\circ \). The secondary solution in the range is \( 360^\circ - 70.5^\circ = 289.5^\circ \).

(a) M1: Substituting \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). M1: Substituting \( \sin^2 \theta = 1 - \cos^2 \theta \). A1: Reaching the given equation with all steps shown. (b) M1: Attempting to factorise or solve the quadratic. A1: Identifying \( \cos \theta = \frac{1}{3} \) and rejecting \( \cos \theta = -3 \). A1: Obtaining \( 70.5^\circ \) (or equivalent). A1: Obtaining \( 289.5^\circ \) (or equivalent). Deduct 1 mark for extra answers in range.

(a) The terms of the AP are \( T_1 = a \), \( T_3 = a + 2d \), and \( T_7 = a + 6d \). The terms of the GP are \( G_1 = a \), \( G_2 = ar \), and \( G_3 = ar^2 \). Since they are equal: \( a + 2d = ar \implies 2d = a(r-1) \) (Equation 1), and \( a + 6d = ar^2 \implies 6d = a(r^2-1) \) (Equation 2). Dividing Equation 2 by Equation 1 gives: \( \frac{6d}{2d} = \frac{a(r^2-1)}{a(r-1)} \implies 3 = r + 1 \) (since \( r \ne 1 \)). Thus, \( r = 2 \). (b) The sum of the first 5 terms of the GP is \( S_5 = \frac{a(r^5 - 1)}{r-1} \). Since \( r = 2 \), \( S_5 = \frac{a(2^5 - 1)}{2-1} = 31a \). Given \( S_5 = 93 \), \( 31a = 93 \implies a = 3 \). Substitute \( a = 3 \) and \( r = 2 \) into Equation 1 to find \( d \): \( 2d = 3(2-1) = 3 \implies d = 1.5 \). The sum of the first 10 terms of the AP is \( S_{10} = \frac{10}{2}[2a + 9d] = 5[2(3) + 9(1.5)] = 5[6 + 13.5] = 5 \times 19.5 = 97.5 \).

(a) M1: Writing down expressions for terms and setting up equations. M1: Eliminating \( d \) to form an equation in \( r \). A1: Correctly showing \( r = 2 \). (b) M1: Using the GP sum formula for 5 terms and equating to 93 to find \( a \). A1: Correctly finding \( a = 3 \). M1: Substituting to find \( d = 1.5 \) and applying the AP sum formula. A1: Correctly finding \( S_{10} = 97.5 \).

(a) Rewrite the curve as \( y = 16x^{-1} + x^2 \). Differentiating: \( \frac{dy}{dx} = -16x^{-2} + 2x = -\frac{16}{x^2} + 2x \). For a stationary point, \( \frac{dy}{dx} = 0 \implies -\frac{16}{x^2} + 2x = 0 \implies 2x^3 = 16 \implies x^3 = 8 \implies x = 2 \). The y-coordinate is \( y = \frac{16}{2} + 2^2 = 12 \). So the stationary point is \( (2, 12) \). To find its nature, compute the second derivative: \( \frac{d^2y}{dx^2} = 32x^{-3} + 2 = \frac{32}{x^3} + 2 \). At \( x = 2 \), \( \frac{d^2y}{dx^2} = \frac{32}{8} + 2 = 6 > 0 \), so it is a minimum point. (b) Given \( \frac{dx}{dt} = 0.5 \). At \( x = 4 \), \( \frac{dy}{dx} = -\frac{16}{4^2} + 2(4) = -1 + 8 = 7 \). Using the chain rule, the rate of change of the y-coordinate is \( \frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt} = 7 \times 0.5 = 3.5 \) units per second.

(a) M1: Correct differentiation to find \( \frac{dy}{dx} \). M1: Setting \( \frac{dy}{dx} = 0 \) and solving for \( x \). A1: Stating coordinates \( (2, 12) \). A1: Correct second derivative and deduction of minimum nature. (b) M1: Finding the value of \( \frac{dy}{dx} \) at \( x = 4 \). M1: Applying the chain rule formula. A1: Stating the correct rate of change \( 3.5 \).

(a) Writing the curve equation as \( y = 4(2x+1)^{-3/4} \) and differentiating: \( \frac{dy}{dx} = 4 \cdot \left(-\frac{3}{4}\right)(2x+1)^{-7/4} \cdot 2 = -6(2x+1)^{-7/4} \). At \( x = 0 \), \( \frac{dy}{dx} = -6(1)^{-7/4} = -6 \). (b) The volume of revolution is \( V = \pi \int_{0}^{4} y^2 dx \). Here, \( y^2 = 16(2x+1)^{-3/2} \). Integrating this expression: \( \int 16(2x+1)^{-3/2} dx = 16 \left[ \frac{(2x+1)^{-1/2}}{(-1/2) \cdot 2} \right] = -16(2x+1)^{-1/2} = -\frac{16}{\sqrt{2x+1}} \). Substituting the limits: \( V = \pi \left[ -\frac{16}{\sqrt{2x+1}} \right]_{0}^{4} = \pi \left[ -\frac{16}{\sqrt{9}} - \left(-\frac{16}{\sqrt{1}}\right) \right] = \pi \left[ -\frac{16}{3} + 16 \right] = \frac{32\pi}{3} \).

(a) M1: Differentiating with chain rule. A1: Stating the correct derivative \( \frac{dy}{dx} = -6(2x+1)^{-7/4} \). A1: Showing gradient is \( -6 \) at \( x = 0 \). (b) M1: Finding the correct squared expression \( y^2 = 16(2x+1)^{-3/2} \). M1: Integrating to get \( k(2x+1)^{-1/2} \). A1: Correct integration \( -16(2x+1)^{-1/2} \). A1: Correct final volume of \( \frac{32\pi}{3} \) (or equivalent exact form, or decimal \( 33.5 \)).

(a) To find the intersection, equate the curve and the line:
\(kx^2 - 10x + k = 2x - 9\)

Rearranging into standard quadratic form:
\(kx^2 - 12x + (k + 9) = 0\)

For the line and the curve to not intersect, the discriminant of this quadratic equation must be strictly less than 0:
\(b^2 - 4ac < 0\)
\((-12)^2 - 4(k)(k + 9) < 0\)
\(144 - 4k^2 - 36k < 0\)

Divide by 4:
\(36 - k^2 - 9k < 0\)

Rearranging gives:
\(k^2 + 9k - 36 > 0\)

Factorizing the quadratic expression:
\((k + 12)(k - 3) > 0\)

Therefore, the set of values of \(k\) is:
\(k < -12\) or \(k > 3\)

(b) Substituting \(k = -9\) into the rearranged intersection equation:
\(-9x^2 - 12x + (-9 + 9) = 0\)
\(-9x^2 - 12x = 0\)
\(-3x(3x + 4) = 0\)

So, \(x = 0\) or \(x = -\frac{4}{3}\).

Substituting these \(x\)-values back into the linear equation \(y = 2x - 9\):
For \(x = 0\):
\(y = 2(0) - 9 = -9\)

For \(x = -\frac{4}{3}\):
\(y = 2\left(-\frac{4}{3}\right) - 9 = -\frac{8}{3} - \frac{27}{3} = -\frac{35}{3}\)

Thus, the coordinates of the points of intersection are \((0, -9)\) and \((-\frac{4}{3}, -\frac{35}{3})\).

**Part (a):**
* **M1:** Equating the line and curve equations and rearranging into a standard three-term quadratic form \(Ax^2 + Bx + C = 0\).
* **M1:** Attempting to find the discriminant \(b^2 - 4ac\) and setting it \(< 0\).
* **A1:** Obtaining the correct simplified quadratic inequality \(k^2 + 9k - 36 > 0\) (or equivalent).
* **A1:** Correctly solving the inequality to find \(k < -12\) or \(k > 3\).

**Part (b):**
* **M1:** Substituting \(k = -9\) into the quadratic equation and attempting to solve for \(x\).
* **A1:** Finding both correct \(x\)-values: \(x = 0\) and \(x = -\frac{4}{3}\).
* **A1:** Correctly finding both \(y\)-values to give the coordinates \((0, -9)\) and \((-\frac{4}{3}, -\frac{35}{3})\).

(a) First, find the derivative of \(y = 6x^{1/2} - x\):
\(\frac{dy}{dx} = 6 \left(\frac{1}{2}x^{-1/2}\right) - 1 = \frac{3}{\sqrt{x}} - 1\)

At the point \(P(4, 8)\), the \(x\)-coordinate is 4. Substitute \(x = 4\) into the derivative to find the gradient of the tangent, \(m\):
\(m = \frac{3}{\sqrt{4}} - 1 = \frac{3}{2} - 1 = \frac{1}{2}\)

Using the equation of a straight line \(y - y_1 = m(x - x_1)\) with the point \((4, 8)\):
\(y - 8 = \frac{1}{2}(x - 4)\)
\(y - 8 = \frac{1}{2}x - 2\)
\(y = \frac{1}{2}x + 6\)

(b) The region is bounded by the tangent \(y = \frac{1}{2}x + 6\), the curve \(y = 6\sqrt{x} - x\), and the \(y\)-axis (\(x = 0\)). The tangent and the curve meet at \(P\) where \(x = 4\).

Since the tangent lies above the curve on the interval \([0, 4]\), the area of the region is:
\(\text{Area} = \int_{0}^{4} \left( \left(\frac{1}{2}x + 6\right) - \left(6\sqrt{x} - x\right) \right) dx\)
\(\text{Area} = \int_{0}^{4} \left( \frac{3}{2}x + 6 - 6x^{1/2} \right) dx\)

Now, integrate each term with respect to \(x\):
\(\int \left( \frac{3}{2}x + 6 - 6x^{1/2} \right) dx = \left[ \frac{3}{4}x^2 + 6x - 4x^{3/2} \right]_{0}^{4}\)

Evaluate at the upper limit \(x = 4\):
\(\frac{3}{4}(4)^2 + 6(4) - 4(4)^{3/2} = \frac{3}{4}(16) + 24 - 4(8) = 12 + 24 - 32 = 4\)

Evaluate at the lower limit \(x = 0\):
\(0\)

Subtracting the lower limit from the upper limit:
\(\text{Area} = 4 - 0 = 4\)

**Part (a):**
* **M1:** Attempting to differentiate the equation of the curve to find \(\frac{dy}{dx}\).
* **M1:** Substituting \(x = 4\) into their derivative to find the gradient of the tangent.
* **A1:** Correctly finding the equation of the tangent as \(y = \frac{1}{2}x + 6\) (or equivalent standard form).

**Part (b):**
* **M1:** Setting up the correct integral representing the area, \(\int_{0}^{4} (y_{\text{tangent}} - y_{\text{curve}}) dx\).
* **M1:** Integrating the terms correctly, obtaining \(\frac{3}{4}x^2 + 6x - 4x^{3/2}\) (accept unsimplified forms).
* **M1:** Applying the limits 0 and 4 to their integrated expression.
* **A1:** Correctly evaluating the definite integral to get the exact area of \(4\).

(a) Start with the left-hand side (LHS) of the identity:
\(\text{LHS} = \frac{\sin \theta}{1 - \cos \theta} - \frac{\sin \theta}{1 + \cos \theta}\)

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

Expand the numerator and the denominator:
\(\text{Numerator} = \sin \theta + \sin \theta \cos \theta - \sin \theta + \sin \theta \cos \theta = 2 \sin \theta \cos \theta\)
\(\text{Denominator} = 1 - \cos^2 \theta = \sin^2 \theta\) (using the identity \(\sin^2 \theta + \cos^2 \theta = 1\))

Substitute these back:
\(\text{LHS} = \frac{2 \sin \theta \cos \theta}{\sin^2 \theta}\)

Cancel one \(\sin \theta\) from the numerator and denominator:
\(\text{LHS} = \frac{2 \cos \theta}{\sin \theta}\)

Since \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), we have \(\frac{\cos \theta}{\sin \theta} = \frac{1}{\tan \theta}\):
\(\text{LHS} = \frac{2}{\tan \theta} = \text{RHS}\) [Proved]

(b) Using the identity proven in part (a) with \(\theta = 2x\), the given equation:
\(\frac{\sin 2x}{1 - \cos 2x} - \frac{\sin 2x}{1 + \cos 2x} = 3\tan 2x\)

becomes:
\(\frac{2}{\tan 2x} = 3\tan 2x\)

Multiply both sides by \(\tan 2x\) (assuming \(\tan 2x \neq 0\)):
\(2 = 3\tan^2 2x\)
\(\tan^2 2x = \frac{2}{3}\)

Taking the square root:
\(\tan 2x = \pm \sqrt{\frac{2}{3}}\)

Let \(\theta = 2x\). Given \(0^\circ < x < 180^\circ\), the range for \(\theta\) is \(0^\circ < \theta < 360^\circ\).
Find the basic angle \(\alpha\):
\(\alpha = \tan^{-1}\left(\sqrt{\frac{2}{3}}\right) \approx 39.23^\circ\)

For \(\tan 2x = \sqrt{\frac{2}{3}}\)
\(2x = 39.23^\circ \implies x = 19.6^\circ\) (to 1 d.p.)
\(2x = 180^\circ + 39.23^\circ = 219.23^\circ \implies x = 109.6^\circ\) (to 1 d.p.)

For \(\tan 2x = -\sqrt{\frac{2}{3}}\)
\(2x = 180^\circ - 39.23^\circ = 140.77^\circ \implies x = 70.4^\circ\) (to 1 d.p.)
\(2x = 360^\circ - 39.23^\circ = 320.77^\circ \implies x = 160.4^\circ\) (to 1 d.p.)

So the solutions are \(x = 19.6^\circ, 70.4^\circ, 109.6^\circ, 160.4^\circ\).

**Part (a):**
* **M1:** Attempting to combine the two fractions on the LHS over a common denominator \((1 - \cos \theta)(1 + \cos \theta)\).
* **M1:** Using the identity \(1 - \cos^2 \theta = \sin^2 \theta\) in the denominator.
* **A1:** Simplifying the numerator and completing the proof to reach \(\frac{2}{\tan \theta}\) with clear intermediate steps shown.

**Part (b):**
* **M1:** Using the identity from part (a) to obtain the equation \(\frac{2}{\tan 2x} = 3\tan 2x\).
* **M1:** Solving to find \(\tan 2x = \pm \sqrt{\frac{2}{3}}\).
* **A1:** Finding at least two correct values of \(x\) to 1 d.p.
* **A1:** Finding all four correct values of \(x\) (\(19.6^\circ, 70.4^\circ, 109.6^\circ, 160.4^\circ\)) to 1 d.p. and no extra values in the range.

Using the factor theorem, \(p(2) = 0\), which gives \(16 + 4a + 2b - 6 = 0\), simplifying to \(2a + b = -5\). Using the remainder theorem, \(p(-0.5) = -15\), which gives \(2(-1/8) + a(1/4) - b/2 - 6 = -15\), which simplifies to \(a - 2b = -35\). Solving these simultaneous equations gives \(a = -9\) and \(b = 13\). Substituting these values, \(p(x) = 2x^3 - 9x^2 + 13x - 6\). Since \((x - 2)\) is a factor, division yields \(p(x) = (x - 2)(2x^2 - 5x + 3)\). Factorising the quadratic term gives the complete factorisation \(p(x) = (x - 2)(x - 1)(2x - 3)\).

M1: Attempt to use the factor theorem with \(p(2) = 0\). A1: Obtain correct equation \(2a + b = -5\). M1: Attempt to use the remainder theorem with \(p(-0.5) = -15\). A1: Obtain correct equation \(a - 2b = -35\). M1: Solve simultaneous equations to find \(a\) and \(b\). A1: Obtain \(a = -9\) and \(b = 13\). B1: Factorise completely to obtain \((x-2)(x-1)(2x-3)\).

Taking exponentials of both sides gives \(2e^{2x} + 5 = e^{x + \ln 7} = e^x \cdot e^{\ln 7} = 7e^x\). Let \(u = e^x\). The equation becomes \(2u^2 - 7u + 5 = 0\). Factorising this quadratic equation gives \((2u - 5)(u - 1) = 0\), which yields solutions \(u = 2.5\) or \(u = 1\). Since \(u = e^x\), we have \(e^x = 2.5\) or \(e^x = 1\). Solving for \(x\) gives \(x = \ln(2.5)\) and \(x = 0\).

M1: Apply exponential to remove logarithms correctly. A1: Obtain \(2e^{2x} + 5 = 7e^x\). M1: Substitute \(u = e^x\) to obtain a quadratic equation. A1: Solve the quadratic to get \(u = 2.5\) and \(u = 1\). M1: Set up equations \(e^x = 2.5\) and \(e^x = 1\). A1: Obtain \(x = 0\). A1: Obtain \(x = \ln(2.5)\) (or equivalent exact expression).

(i) LHS is \(\frac{1}{\sin 2\theta} + \frac{\cos 2\theta}{\sin 2\theta} = \frac{1 + \cos 2\theta}{\sin 2\theta}\). Using double angle identities, \(1 + \cos 2\theta = 2\cos^2\theta\) and \(\sin 2\theta = 2\sin\theta\cos\theta\). Substituting these gives \(\frac{2\cos^2\theta}{2\sin\theta\cos\theta} = \frac{\cos\theta}{\sin\theta} = \cot\theta\). (ii) The equation is equivalent to \(\cot\theta = 3\tan\theta\). This gives \(\frac{1}{\tan\theta} = 3\tan\theta\), which leads to \(\tan^2\theta = \frac{1}{3}\). Taking the square root gives \(\tan\theta = \pm \frac{1}{\sqrt{3}}\). For \(0^\circ < \theta < 180^\circ\), \(\theta = 30^\circ\) or \(\theta = 150^\circ\).

(i) M1: Rewrite LHS in terms of sine and cosine. M1: Apply double-angle formulae to numerator and denominator. A1: Simplify the fraction to \(\cos\theta / \sin\theta\). A1: Complete proof to show \(\cot\theta\). (ii) M1: Substitute the identity to obtain \(\cot\theta = 3\tan\theta\) and rewrite as \(\tan^2\theta = k\). A1: Obtain \(\theta = 30^\circ\). A1: Obtain \(\theta = 150^\circ\) and no other values in the range.

(i) Using the quotient rule: \(\frac{\text{d}y}{\text{d}x} = \frac{3e^{3x}(2x + 1) - 2e^{3x}}{(2x + 1)^2} = \frac{e^{3x}(6x + 3 - 2)}{(2x + 1)^2} = \frac{e^{3x}(6x + 1)}{(2x + 1)^2}\). (ii) For stationary points, set \(\frac{\text{d}y}{\text{d}x} = 0\). Since \(e^{3x} \neq 0\), we have \(6x + 1 = 0\), giving \(x = -1/6\). The corresponding \(y\)-value is \(\frac{e^{-1/2}}{2(-1/6)+1} = \frac{e^{-1/2}}{2/3} = \frac{3}{2}e^{-1/2}\). To determine the nature, consider the sign of \(\frac{\text{d}y}{\text{d}x}\) near \(x = -1/6\). Since \(e^{3x} > 0\) and \((2x+1)^2 > 0\), the sign depends on \(6x+1\). For \(x < -1/6\), \(6x+1 < 0\) and for \(x > -1/6\), \(6x+1 > 0\). Thus the gradient changes from negative to positive, which shows that \((-1/6, \frac{3}{2}e^{-0.5})\) is a local minimum.

(i) M1: Apply quotient rule or product rule. A1: Obtain correct numerator \(3e^{3x}(2x+1) - 2e^{3x}\). A1: Obtain correct simplified derivative. (ii) M1: Set derivative to zero and solve for \(x\). A1: Obtain \(x = -1/6\). A1: Obtain exact \(y = 1.5e^{-0.5}\). M1: Determine the nature of the stationary point using first derivative sign test or second derivative. A1: Conclude it is a local minimum with correct justification.

(i) The integral is \(\int \frac{1}{2x + 3} \text{d}x = \frac{1}{2} \ln|2x + 3| + c\). (ii) Using the double angle formula, \(\sin^2(\frac{1}{2}x) = \frac{1}{2}(1 - \cos x)\). Substituting this into the integral gives \(\int_{0}^{\pi} \frac{1}{2}(1 - \cos x) \text{d}x = \left[ \frac{1}{2}x - \frac{1}{2}\sin x \right]_{0}^{\pi} = (\frac{\pi}{2} - \frac{1}{2}\sin\pi) - (0 - \frac{1}{2}\sin 0) = \frac{\pi}{2} - 0 - 0 = \frac{\pi}{2}\).

(i) M1: Recognize logarithmic form. A1: Obtain \(\frac{1}{2} \ln|2x + 3| + c\) (must include constant of integration). (ii) M1: Use a double angle identity to rewrite the integrand. A1: Obtain \(\frac{1}{2}(1 - \cos x)\). M1: Integrate to get \(\frac{1}{2}x - \frac{1}{2}\sin x\). M1: Substitute limits \(\pi\) and 0 correctly. A1: Obtain the correct exact value \(\frac{\pi}{2}\) with no errors shown.

(i) Let \(f(x) = x^3 - 5x + 1\). Then \(f(0) = 1 > 0\) and \(f(1) = 1 - 5 + 1 = -3 < 0\). Since \(f(x)\) is continuous and there is a change of sign between 0 and 1, there is a root \(\alpha\) in this interval. (ii) Using \(x_{n+1} = \frac{x_n^3 + 1}{5}\) with \(x_1 = 0.2\): \(x_2 = \frac{0.2^3 + 1}{5} = 0.20160\); \(x_3 = \frac{0.20160^3 + 1}{5} = 0.20164\); \(x_4 = \frac{0.20164^3 + 1}{5} = 0.20164\). Thus, \(\alpha \approx 0.202\). (iii) The boundaries for 0.2016 to 4 d.p. are 0.20155 and 0.20165. \(f(0.20155) = (0.20155)^3 - 5(0.20155) + 1 \approx 0.00043 > 0\). \(f(0.20165) = (0.20165)^3 - 5(0.20165) + 1 \approx -0.000055 < 0\). Since there is a sign change in the interval \([0.20155, 0.20165]\), \(\alpha = 0.2016\) to 4 decimal places.

(i) M1: Calculate values of \(f(0)\) and \(f(1)\). A1: Observe sign change and conclude. (ii) M1: Perform iterations using the correct formula. A1: Obtain correct values of \(x_2\), \(x_3\), \(x_4\) to 5 d.p. A1: State final value as 0.202. (iii) M1: Choose correct boundaries 0.20155 and 0.20165 and evaluate \(f(x)\). A1: Show sign change and make the final conclusion.

(i) Squaring both sides of the inequality gives \((3x - 2)^2 < (x + 4)^2\), which simplifies to \(9x^2 - 12x + 4 < x^2 + 8x + 16\). Rearranging gives \(8x^2 - 20x - 12 < 0\). Dividing by 4 gives \(2x^2 - 5x - 3 < 0\). Factorising yields \((2x + 1)(x - 3) < 0\). The solution is \(-0.5 < x < 3\). (ii) Let \(x = 3^y\). Then \(3^{y+1} = 3 \cdot 3^y = 3x\). Thus, the inequality becomes \(|3x - 2| < |x + 4|\), so the solution from part (i) applies: \(-0.5 < 3^y < 3\). Since \(3^y > 0\) for all real \(y\), the inequality \(3^y > -0.5\) is always satisfied. This leaves \(3^y < 3\), which gives \(y < 1\).

(i) M1: Square both sides or consider two cases to find critical values. A1: Obtain critical values \(x = -0.5\) and \(x = 3\). M1: Solve inequality for inside range. A1: Obtain correct interval \(-0.5 < x < 3\). (ii) M1: Substitute \(x = 3^y\) and recognize relationship. M1: Deduce \(3^y < 3\) (discarding lower limit because \(3^y > 0\)). A1: Obtain final solution \(y < 1\).