2023 Cambridge IAS-Level Mathematics (9709) Practice Paper with Answers

Equating the equations of the line and the curve:
\( mx - 5 = x^2 - 3x - 1 \)

Rearranging into standard quadratic form:
\( x^2 - (3 + m)x + 4 = 0 \)

Since the line is a tangent to the curve, the discriminant of this quadratic equation must be zero:
\( b^2 - 4ac = 0 \)
\( (-(3 + m))^2 - 4(1)(4) = 0 \)
\( (3 + m)^2 - 16 = 0 \)

Solving for \( m \):
\( 3 + m = \pm 4 \)

This gives:
\( 3 + m = 4 \implies m = 1 \)
\( 3 + m = -4 \implies m = -7 \)

M1: Equating the line and the curve, rearranging into quadratic form, and setting the discriminant \( b^2 - 4ac = 0 \) (or equivalent method).
A1: For obtaining both correct values: \( m = 1 \) and \( m = -7 \).

For the function \( \mathrm{f} \) to have an inverse, it must be a one-to-one function.

The graph of the quadratic function \( y = 2x^2 - 12x + 13 \) is a parabola that opens upwards. The turning point (vertex) represents the boundary where the function changes from decreasing to increasing.

We can find the \( x \)-coordinate of the vertex by completing the square:
\( \mathrm{f}(x) = 2(x^2 - 6x) + 13 \)
\( \mathrm{f}(x) = 2[(x - 3)^2 - 9] + 13 \)
\( \mathrm{f}(x) = 2(x - 3)^2 - 5 \)

Alternatively, we can find the vertex by differentiating and setting the derivative to zero:
\( \mathrm{f}'(x) = 4x - 12 = 0 \implies x = 3 \)

In either case, the vertex is at \( x = 3 \). For the domain \( x \ge k \), the function is one-to-one if and only if \( k \ge 3 \).

Thus, the least value of the constant \( k \) is \( 3 \).

M1: Attempting to find the \( x \)-coordinate of the vertex of the quadratic (e.g., by completing the square, using the formula \( x = -\frac{b}{2a} \), or differentiating and setting to 0).
A1: Correctly stating the least value \( k = 3 \).

To find the points of intersection, we set the equations equal to each other: \(2kx - 3 = x^2 + (k-2)x + 1\). Rearranging this into a standard quadratic equation gives: \(x^2 + (k-2-2k)x + 4 = 0\), which simplifies to \(x^2 - (k+2)x + 4 = 0\). For the line and curve not to intersect, this quadratic equation must have no real roots. Therefore, the discriminant must be less than zero: \(b^2 - 4ac < 0\). Here, \(a = 1\), \(b = -(k+2)\), and \(c = 4\). So, \((-(k+2))^2 - 4(1)(4) < 0\), which simplifies to \((k+2)^2 - 16 < 0\). Solving this inequality: \((k+2)^2 < 16 \implies -4 < k+2 < 4 \implies -6 < k < 2\).

M1: Set the equations equal to each other and attempt to form a quadratic equation in \(x\).
A1: Obtain the correct quadratic equation \(x^2 - (k+2)x + 4 = 0\) (or equivalent).
M1: Use the discriminant condition \(b^2 - 4ac < 0\) with their coefficients.
M1: Solve the quadratic inequality for \(k\).
A1: Obtain the correct range \(-6 < k < 2\).

First, we express \(\mathrm{f}(x)\) in completed square form: \(\mathrm{f}(x) = 2(x^2 - 6x) + 11 = 2((x-3)^2 - 9) + 11 = 2(x-3)^2 - 7\). The curve is a parabola with a minimum point at \((3, -7)\). For the function to be one-to-one, we must restrict the domain to only one side of the line of symmetry \(x = 3\). Since the domain is defined as \(x \le k\), we must have \(k \le 3\). Thus, the largest value of \(k\) is \(3\). For this value of \(k\), the domain is \(x \le 3\). To find the inverse function, we set \(y = 2(x-3)^2 - 7\) and rearrange to make \(x\) the subject: \(y+7 = 2(x-3)^2 \implies (x-3)^2 = \frac{y+7}{2}\). Taking the square root, and noting that since \(x \le 3\) we must choose the negative square root: \(x-3 = -\sqrt{\frac{y+7}{2}} \implies x = 3 - \sqrt{\frac{y+7}{2}}\). Therefore, the inverse function is \(\mathrm{f}^{-1}(x) = 3 - \sqrt{\frac{x+7}{2}}\).

M1: Attempt to complete the square on \(2x^2 - 12x + 11\).
A1: State \(k = 3\) (or identify the line of symmetry \(x = 3\)).
M1: Set \(y = 2(x-3)^2 - 7\) and attempt to make \(x\) (or \(y\)) the subject.
A1: Identify the correct sign for the square root (negative because of the domain \(x \le 3\)).
A1: Obtain correct inverse function \(\mathrm{f}^{-1}(x) = 3 - \sqrt{\frac{x+7}{2}}\) (must be in terms of \(x\)).

First, find the midpoint of \(AB\): \(M = \left(\frac{2+8}{2}, \frac{5+(-3)}{2}\right) = (5, 1)\). Next, find the gradient of \(AB\): \(m_{AB} = \frac{-3-5}{8-2} = \frac{-8}{6} = -\frac{4}{3}\). The gradient of the perpendicular bisector is the negative reciprocal: \(m = -\frac{1}{m_{AB}} = \frac{3}{4}\). Now, find the equation of the perpendicular bisector using the midpoint \((5, 1)\) and gradient \(\frac{3}{4}\): \(y - 1 = \frac{3}{4}(x - 5) \implies 4y - 4 = 3x - 15 \implies 3x - 4y - 11 = 0\). The perpendicular bisector meets the \(y\)-axis at point \(C\), where \(x = 0\). Substituting \(x = 0\) into the equation gives: \(3(0) - 4y - 11 = 0 \implies -4y = 11 \implies y = -\frac{11}{4} = -2.75\). Thus, the coordinates of \(C\) are \((0, -2.75)\).

B1: Find the correct midpoint \((5, 1)\) of \(AB\).
M1: Find the gradient of \(AB\) and state the perpendicular gradient as \(\frac{3}{4}\).
M1: Form the equation of the perpendicular bisector passing through their midpoint with their perpendicular gradient.
M1: Substitute \(x = 0\) to find the \(y\)-intercept.
A1: Obtain the correct coordinates \((0, -2.75)\) (or equivalent fractional form \((0, -\frac{11}{4})\)).

The area of a sector is given by \(A = \frac{1}{2}r^2\theta\). Since the area is \(18\), we have: \(\frac{1}{2}r^2\theta = 18 \implies r^2\theta = 36 \implies \theta = \frac{36}{r^2}\). The perimeter of a sector is given by \(P = 2r + r\theta\). Since the perimeter is \(18\), we have: \(2r + r\theta = 18\). Substituting \(\theta = \frac{36}{r^2}\) into the perimeter equation: \(2r + r\left(\frac{36}{r^2}\right) = 18 \implies 2r + \frac{36}{r} = 18\). Multiplying the entire equation by \(r\) gives: \(2r^2 + 36 = 18r \implies 2r^2 - 18r + 36 = 0\). Dividing by 2: \(r^2 - 9r + 18 = 0\). Factoring the quadratic: \((r-3)(r-6) = 0\). Thus, the two possible values of \(r\) are \(3\) and \(6\).

B1: Write down the correct area equation \(\frac{1}{2}r^2\theta = 18\).
B1: Write down the correct perimeter equation \(2r + r\theta = 18\).
M1: Eliminate \(\theta\) to obtain an equation in terms of \(r\) only.
M1: Form and solve a three-term quadratic equation in \(r\).
A1: Obtain the correct values \(r = 3\) and \(r = 6\).

The formula for the \(n\)-th term of an arithmetic progression is \(u_n = a + (n-1)d\). Therefore, the 3rd term is: \(a + 2d = 18\) (Equation 1). The formula for the sum of the first \(n\) terms is \(S_n = \frac{n}{2}(2a + (n-1)d)\). Therefore, the sum of the first 8 terms is: \(S_8 = \frac{8}{2}(2a + 7d) = 4(2a + 7d) = 180 \implies 2a + 7d = 45\) (Equation 2). We can solve Equation 1 and Equation 2 simultaneously. From Equation 1, \(2a + 4d = 36\). Subtracting this from Equation 2: \((2a + 7d) - (2a + 4d) = 45 - 36 \implies 3d = 9 \implies d = 3\). Substituting \(d = 3\) back into Equation 1: \(a + 2(3) = 18 \implies a = 12\). Now, find the 15th term: \(u_{15} = a + 14d = 12 + 14(3) = 12 + 42 = 54\).

B1: Write down the correct equation for the 3rd term: \(a + 2d = 18\).
M1: Use the sum formula to write an equation for the sum of the first 8 terms and simplify to obtain \(2a + 7d = 45\) (or equivalent).
M1: Solve the simultaneous equations to find \(a\) and \(d\).
A1: Obtain \(a = 12\) and \(d = 3\).
A1: Obtain the correct 15th term, which is \(54\).

Use the identity \(\sin^2 \theta = 1 - \cos^2 \theta\) to write the equation in terms of \(\cos \theta\): \(3(1 - \cos^2 \theta) - 5 \cos \theta - 1 = 0 \implies 3 - 3 \cos^2 \theta - 5 \cos \theta - 1 = 0 \implies 3 \cos^2 \theta + 5 \cos \theta - 2 = 0\). Let \(y = \cos \theta\). The equation becomes \(3y^2 + 5y - 2 = 0\), which factorises to \((3y - 1)(y + 2) = 0\). This gives \(y = \frac{1}{3}\) or \(y = -2\). Since \(-1 \le \cos \theta \le 1\), \(\cos \theta = -2\) has no solutions. For \(\cos \theta = \frac{1}{3}\): The basic angle is \(\theta = \cos^{-1}\left(\frac{1}{3}\right) \approx 70.53^\circ\). In the interval \(0^\circ \le \theta \le 360^\circ\), \(\cos \theta\) is positive in the 1st and 4th quadrants: \(\theta = 70.5^\circ\) (to 1 d.p.) and \(\theta = 360^\circ - 70.53^\circ = 289.5^\circ\) (to 1 d.p.).

M1: Use the identity \(\sin^2 \theta = 1 - \cos^2 \theta\) to form a quadratic equation in \(\cos \theta\).
A1: Obtain correct quadratic equation \(3 \cos^2 \theta + 5 \cos \theta - 2 = 0\).
M1: Factorise/solve the quadratic equation to get \(\cos \theta = \frac{1}{3}\) (and identify that \(\cos \theta = -2\) yields no solutions).
A1: Obtain first angle \(70.5^\circ\) (accept \(70.5\) or \(70.6\) due to rounding).
A1: Obtain second angle \(289.5^\circ\) (accept \(289.5\) or \(289.4\)). Deduct 1 mark overall if extra solutions in the range are given.

Rewrite the equation of the curve as \(y = 8x^{-1} + 2x\). First, find the first derivative: \(\frac{\mathrm{d}y}{\mathrm{d}x} = -8x^{-2} + 2 = -\frac{8}{x^2} + 2\). At stationary points, \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0 \implies -\frac{8}{x^2} + 2 = 0 \implies \frac{8}{x^2} = 2 \implies x^2 = 4 \implies x = \pm 2\). When \(x = 2\), \(y = \frac{8}{2} + 2(2) = 4 + 4 = 8\). The coordinate is \((2, 8)\). When \(x = -2\), \(y = \frac{8}{-2} + 2(-2) = -4 - 4 = -8\). The coordinate is \((-2, -8)\). Now find the second derivative to determine their nature: \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{\mathrm{d}}{\mathrm{d}x}(-8x^{-2} + 2) = 16x^{-3} = \frac{16}{x^3}\). For \(x = 2\): \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{16}{8} = 2 > 0\), which means \((2, 8)\) is a local minimum point. For \(x = -2\): \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{16}{-8} = -2 < 0\), which means \((-2, -8)\) is a local maximum point.

M1: Differentiate the curve equation to find \(\frac{\mathrm{d}y}{\mathrm{d}x}\).
A1: Set the derivative equal to 0 and find \(x = \pm 2\).
M1: Find the corresponding \(y\)-coordinates \(8\) and \(-8\).
M1: Find the second derivative \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\) and substitute \(x\) values to determine the nature of the stationary points.
A1: Correctly identify \((2, 8)\) as a minimum point and \((-2, -8)\) as a maximum point.

To find the equation of the curve, we integrate the derivative: \(\mathrm{f}(x) = \int 6(3x+1)^{-1/2} \, \mathrm{d}x\). Using the reverse chain rule for integration: \(\mathrm{f}(x) = 6 \cdot \frac{(3x+1)^{1/2}}{(1/2) \cdot 3} + C\), where the division by 3 comes from the derivative of the inside function \(3x+1\). Simplifying the expression: \(\mathrm{f}(x) = 6 \cdot \frac{2}{3}(3x+1)^{1/2} + C = 4(3x+1)^{1/2} + C = 4\sqrt{3x+1} + C\). We are given that the curve passes through the point \((1, 8)\), so we substitute \(x = 1\) and \(y = 8\) into our equation: \(8 = 4\sqrt{3(1)+1} + C \implies 8 = 4\sqrt{4} + C \implies 8 = 4(2) + C \implies 8 = 8 + C \implies C = 0\). Therefore, the equation of the curve is \(y = 4\sqrt{3x+1}\).

M1: Write the integral as \(\int 6(3x+1)^{-1/2} \, \mathrm{d}x\).
A1: Perform the integration to obtain \(4(3x+1)^{1/2}\) (award M1 for \(k(3x+1)^{1/2}\) and A1 for correct coefficient \(4\)).
M1: Include the constant of integration \(C\) and substitute the coordinates \((1, 8)\).
A1: Solve for \(C\) to find \(C = 0\).
A1: State the final equation of the curve \(y = 4\sqrt{3x+1}\) (or \(\mathrm{f}(x) = 4\sqrt{3x+1}\)).

(a) To find the points of intersection, we equate the equations of the curve and the line:
\( (2x - 3)^2 = 2x + 3 \)

Expand the left-hand side:
\( 4x^2 - 12x + 9 = 2x + 3 \)

Rearrange into standard quadratic form:
\( 4x^2 - 14x + 6 = 0 \)

Divide by 2:
\( 2x^2 - 7x + 3 = 0 \)

Factorise the quadratic:
\( (2x - 1)(x - 3) = 0 \)

This gives the x-coordinates of the intersection points:
\( x = \frac{1}{2} \) and \( x = 3 \)

Substitute these x-coordinates back into the line equation to find the corresponding y-coordinates:
For \( x = \frac{1}{2} \):
\( y = 2\left(\frac{1}{2}\right) + 3 = 4 \)

For \( x = 3 \):
\( y = 2(3) + 3 = 9 \)

So, the coordinates of \( A \) and \( B \) are \( (\frac{1}{2}, 4) \) and \( (3, 9) \).

(b) The area \( A \) of the enclosed region is given by the integral of the upper function (the line) minus the lower function (the curve) between their points of intersection:
\( A = \int_{1/2}^{3} \left[ (2x + 3) - (2x - 3)^2 \right] dx \)

Simplify the integrand:
\( A = \int_{1/2}^{3} \left[ (2x + 3) - (4x^2 - 12x + 9) \right] dx \)
\( A = \int_{1/2}^{3} \left[ -4x^2 + 14x - 6 \right] dx \)

Integrate each term:
\( \left[ -\frac{4}{3}x^3 + 7x^2 - 6x \right]_{1/2}^{3} \)

Evaluate the integral at the upper limit \( x = 3 \):
\( -\frac{4}{3}(3)^3 + 7(3)^2 - 6(3) = -36 + 63 - 18 = 9 \)

Evaluate the integral at the lower limit \( x = \frac{1}{2} \):
\( -\frac{4}{3}\left(\frac{1}{2}\right)^3 + 7\left(\frac{1}{2}\right)^2 - 6\left(\frac{1}{2}\right) = -\frac{1}{6} + \frac{7}{4} - 3 = \frac{-2 + 21 - 36}{12} = -\frac{17}{12} \)

Subtract the lower limit value from the upper limit value:
\( A = 9 - \left(-\frac{17}{12}\right) = 9 + \frac{17}{12} = \frac{125}{12} \)

(a)
* **M1**: Equating the line and the curve and attempting to solve the resulting quadratic equation.
* **A1**: Finding correct x-coordinates \( x = \frac{1}{2} \) and \( x = 3 \).
* **A1**: Finding correct y-coordinates, giving points \( (\frac{1}{2}, 4) \) and \( (3, 9) \).

(b)
* **M1**: Formulating the integral of the difference of the functions, \( \int (2x + 3 - (2x - 3)^2) dx \) or equivalent, with their limits from part (a).
* **A1**: Obtaining the correct integrated expression \( -\frac{4}{3}x^3 + 7x^2 - 6x \) (ignore signs for this mark if limits are swapped).
* **M1**: Substituting their limits \( 3 \) and \( \frac{1}{2} \) correctly into their integrated expression.
* **A1**: Obtaining the correct final area of \( \frac{125}{12} \) (or \( 10.4 \) or \( 10\frac{5}{12} \)).

Using the laws of logarithms, \(\ln(2x - 3) + \ln(x + 1) = \ln((2x - 3)(x + 1))\) and \(2\ln x = \ln(x^2)\). Equating the arguments of the logarithms gives \((2x - 3)(x + 1) = x^2\), which simplifies to \(2x^2 - x - 3 = x^2\) and hence \(x^2 - x - 3 = 0\). Using the quadratic formula, we find the roots \(x = \frac{1 \pm \sqrt{13}}{2}\). Since the term \(\ln(2x - 3)\) is only defined when \(2x - 3 > 0\) (or \(x > 1.5\)), the negative root \(x = \frac{1 - \sqrt{13}}{2} \approx -1.30\) must be rejected. Therefore, the only valid solution is \(x = \frac{1 + \sqrt{13}}{2}\), which to 3 significant figures is \(x = 2.30\).

M1: Apply addition law of logarithms to combine left-hand side into a single log term. M1: Apply power law of logarithms to the right-hand side. A1: Obtain the correct quadratic equation \(x^2 - x - 3 = 0\). M1: Solve the quadratic equation to find two algebraic roots. B1: State the domain restriction \(x > 1.5\) (or equivalent) and explicitly reject the negative root. A1: Obtain the final answer \(x = 2.30\).

Using the compound angle formulae, we expand both sides: \(\cos\theta \cos\frac{\pi}{6} - \sin\theta \sin\frac{\pi}{6} = 3\left(\sin\theta \cos\frac{\pi}{6} - \cos\theta \sin\frac{\pi}{6}\right)\). Substituting the exact values \(\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}\) and \(\sin\frac{\pi}{6} = \frac{1}{2}\), we get \(\frac{\sqrt{3}}{2}\cos\theta - \frac{1}{2}\sin\theta = 3\left(\frac{\sqrt{3}}{2}\sin\theta - \frac{1}{2}\cos\theta\right)\). Multiplying by 2 to clear denominators yields \(\sqrt{3}\cos\theta - \sin\theta = 3\sqrt{3}\sin\theta - 3\cos\theta\). Rearranging terms gives \((\sqrt{3} + 3)\cos\theta = (3\sqrt{3} + 1)\sin\theta\). Dividing both sides by \(\cos\theta\) and \(3\sqrt{3} + 1\), we obtain \(\tan\theta = \frac{\sqrt{3} + 3}{3\sqrt{3} + 1}\). Evaluating this fraction gives \(\tan\theta \approx 0.76370\). Finding the inverse tangent in the range \(0 \le \theta \le \pi\) gives \(\theta \approx 0.652\) radians.

M1: Expand both sides using the correct compound angle formulae. M1: Substitute the correct exact values for sine and cosine of \(\pi/6\). M1: Rearrange the equation to group \(\sin\theta\) and \(\cos\theta\) terms. M1: Convert the equation to the form \(\tan\theta = k\). A1: Obtain \(\tan\theta = \frac{\sqrt{3} + 3}{3\sqrt{3} + 1}\) or decimal equivalent. A1: Obtain final answer \(\theta = 0.652\) (allow 0.651 to 0.653).

For part (i), using the double angle identity \(\sin 2\theta = 2\sin\theta\cos\theta\), we square both sides to get \(\sin^2 2\theta = 4\sin^2\theta\cos^2\theta\). Thus, \(8\sin^2\theta\cos^2\theta = 2(4\sin^2\theta\cos^2\theta) = 2\sin^2 2\theta\). Using the identity \(\cos 4\theta = 1 - 2\sin^2 2\theta\), we can rewrite this as \(1 - \cos 4\theta\). Hence, \(8\sin^2\theta\cos^2\theta \equiv 1 - \cos 4\theta\). For part (ii), the integral is \(\int_{\pi/12}^{\pi/4} 16\sin^2 x \cos^2 x \, dx = \int_{\pi/12}^{\pi/4} 2(1 - \cos 4x) \, dx = \int_{\pi/12}^{\pi/4} (2 - 2\cos 4x) \, dx\). Integrating terms gives \(\left[ 2x - \frac{1}{2}\sin 4x \right]_{\pi/12}^{\pi/4}\). Evaluating at the upper limit \(x = \pi/4\) gives \(2(\pi/4) - \frac{1}{2}\sin(\pi) = \frac{\pi}{2}\). Evaluating at the lower limit \(x = \pi/12\) gives \(2(\pi/12) - \frac{1}{2}\sin(\pi/3) = \frac{\pi}{6} - \frac{\sqrt{3}}{4}\). Subtracting the lower limit value from the upper limit value gives \(\frac{\pi}{2} - \left(\frac{\pi}{6} - \frac{\sqrt{3}}{4}\right) = \frac{\pi}{3} + \frac{\sqrt{3}}{4}\).

M1: Rewrite the LHS in terms of \(\sin 2\theta\). A1: Correctly use double angle cosine formula to complete the proof of the identity. M1: Express the integrand using the identity as \(2(1 - \cos 4x)\). A1: Correctly integrate to obtain \(2x - \frac{1}{2}\sin 4x\). M1: Substitute the limits \(\pi/4\) and \(\pi/12\) into their integrated expression. A1: Obtain the exact value \(\frac{\pi}{3} + \frac{\sqrt{3}}{4}\) in simplest form.

We differentiate using the quotient rule: \(\frac{dy}{dx} = \frac{3e^{3x}(2x + 1) - 2e^{3x}}{(2x + 1)^2} = \frac{e^{3x}(6x + 3 - 2)}{(2x + 1)^2} = \frac{e^{3x}(6x + 1)}{(2x + 1)^2}\). Setting \(\frac{dy}{dx} = 0\) to find the stationary point, we have \(e^{3x}(6x + 1) = 0\). Since \(e^{3x}\) is never zero, we have \(6x + 1 = 0 \implies x = -\frac{1}{6}\). Substituting \(x = -\frac{1}{6}\) into the equation of the curve gives \(y = \frac{e^{-1/2}}{2(-1/6) + 1} = \frac{e^{-1/2}}{2/3} = \frac{3}{2\sqrt{e}}\). To find its nature, we examine the sign of \(\frac{dy}{dx}\) near \(x = -1/6\). For \(-0.5 < x < -1/6\), \(6x + 1 < 0\), so \(\frac{dy}{dx} < 0\). For \(x > -1/6\), \(6x + 1 > 0\), so \(\frac{dy}{dx} > 0\). Because the gradient changes from negative to positive, the stationary point is a local minimum.

M1: Apply quotient rule or product rule correctly. A1: Obtain correct simplified derivative \(\frac{dy}{dx} = \frac{e^{3x}(6x + 1)}{(2x + 1)^2}\). M1: Set derivative to 0 and solve for \(x\). A1: Find the correct x-coordinate \(x = -\frac{1}{6}\). A1: Find the correct y-coordinate \(y = \frac{3}{2\sqrt{e}}\) (or equivalent exact form). B1: Correctly justify that the stationary point is a local minimum.

(i) To find the stationary point, we differentiate \(y\) with respect to \(x\):
\(\frac{dy}{dx} = -8e^{-2x} + 2\).
Setting \(\frac{dy}{dx} = 0\):
\(-8e^{-2x} + 2 = 0 \implies e^{-2x} = \frac{1}{4}\).
Taking the natural logarithm of both sides:
\(-2x = \ln\left(\frac{1}{4}\right) = -\ln(4) = -2\ln(2) \implies x = \ln 2\).

Substitute \(x = \ln 2\) back into the equation of the curve to find the \(y\)-coordinate:
\(y = 4e^{-2\ln 2} + 2\ln 2 - 5 = 4\left(\frac{1}{4}\right) + 2\ln 2 - 5 = 1 + 2\ln 2 - 5 = 2\ln 2 - 4\).
Thus, the stationary point is \((\ln 2, 2\ln 2 - 4)\).

To find the nature of the stationary point, we find the second derivative:
\(\frac{d^2y}{dx^2} = 16e^{-2x}\).
Since \(16e^{-2x} > 0\) for all real values of \(x\), the second derivative at \(x = \ln 2\) is positive (specifically, \(16 e^{-2\ln 2} = 4 > 0\)). Therefore, the stationary point is a minimum.

(ii) Let \(f(x) = 4e^{-2x} + 2x - 5\).
Evaluate \(f(x)\) at the boundaries:
\(f(2) = 4e^{-4} + 2(2) - 5 = 4e^{-4} - 1 \approx 4(0.018316) - 1 = -0.9267\) (which is negative).
\(f(2.5) = 4e^{-5} + 2(2.5) - 5 = 4e^{-5} \approx 4(0.006738) = 0.0270\) (which is positive).
Since there is a change of sign between \(x = 2\) and \(x = 2.5\) and the function is continuous, there must be at least one root \(\alpha\) in the interval \(2 < \alpha < 2.5\).

(iii) Set \(4e^{-2x} + 2x - 5 = 0\):
\(2x = 5 - 4e^{-2x}\)
Divide both sides by 2:
\(x = 2.5 - 2e^{-2x}\).

(iv) Using the iterative formula \(x_{n+1} = 2.5 - 2e^{-2x_n}\) with \(x_1 = 2.4\):
\(x_2 = 2.5 - 2e^{-2(2.4)} = 2.5 - 2e^{-4.8} \approx 2.48354\)
\(x_3 = 2.5 - 2e^{-2(2.48354)} = 2.5 - 2e^{-4.96708} \approx 2.48607\)
\(x_4 = 2.5 - 2e^{-2(2.48607)} = 2.5 - 2e^{-4.97214} \approx 2.48615\)
\(x_5 = 2.5 - 2e^{-2(2.48615)} = 2.5 - 2e^{-4.97230} \approx 2.48615\)

Thus, the value of \(\alpha\) correct to 3 decimal places is 2.486.

(i)
M1: Attempt to differentiate \(y\) to find \(\frac{dy}{dx}\) containing an exponential term.
A1: Obtain correct derivative \(-8e^{-2x} + 2\) and set to 0 to find \(x = \ln 2\) (or equivalent exact form).
A1: Obtain exact \(y\)-coordinate \(2\ln 2 - 4\) (or \(\ln 4 - 4\)).
B1: Correctly show that the stationary point is a minimum using \(\frac{d^2y}{dx^2}\) or sign table.

(ii)
M1: Evaluate \(f(2)\) and \(f(2.5)\) with at least one correct numerical value (to 2 s.f. or more).
A1: State correct values with correct signs, and make a conclusion mentioning sign change.

(iii)
B1: Complete algebraic steps showing the rearrangement clearly.

(iv)
M1: Attempt to calculate iterations using the given formula starting with \(x_1 = 2.4\).
A1: Obtain \(x_2 = 2.48354\) (or better).
A1: Obtain correct values for remaining iterations to 5 d.p., establishing convergence.
A1: Conclude \(\alpha = 2.486\) to 3 d.p.

(i) Recall the double-angle identity for cosine:
\(\cos(2x) = 2\cos^2(x) - 1\).
Rearranging this gives:
\(\cos^2(x) = \frac{1}{2}\cos(2x) + \frac{1}{2}\).
Substituting this back into the expression for \(y\):
\(y = \sin(2x) + \cos^2(x) = \sin(2x) + \frac{1}{2}\cos(2x) + \frac{1}{2}\).

(ii) The exact area of region \(R\) is given by the definite integral:
\(\text{Area} = \int_{0}^{\frac{\pi}{4}} \left( \sin(2x) + \frac{1}{2}\cos(2x) + \frac{1}{2} \right) dx\).
Integrating term by term:
\(\int \sin(2x) dx = -\frac{1}{2}\cos(2x)\)
\(\int \frac{1}{2}\cos(2x) dx = \frac{1}{4}\sin(2x)\)
\(\int \frac{1}{2} dx = \frac{1}{2}x\).
So,
\(\text{Area} = \left[ -\frac{1}{2}\cos(2x) + \frac{1}{4}\sin(2x) + \frac{1}{2}x \right]_{0}^{\frac{\pi}{4}}\).

Substitute the upper limit \(x = \frac{\pi}{4}\):
\(-\frac{1}{2}\cos\left(\frac{\pi}{2}\right) + \frac{1}{4}\sin\left(\frac{\pi}{2}\right) + \frac{1}{2}\left(\frac{\pi}{4}\right) = 0 + \frac{1}{4}(1) + \frac{\pi}{8} = \frac{1}{4} + \frac{\pi}{8}\).

Substitute the lower limit \(x = 0\):
\(-\frac{1}{2}\cos(0) + \frac{1}{4}\sin(0) + 0 = -\frac{1}{2}(1) + 0 = -\frac{1}{2}\).

Subtract the lower limit value from the upper limit value:
\(\text{Area} = \left( \frac{1}{4} + \frac{\pi}{8} \right) - \left( -\frac{1}{2} \right) = \frac{3}{4} + \frac{\pi}{8}\).

(iii) The width of each interval is \(h = \frac{\frac{\pi}{4} - 0}{2} = \frac{\pi}{8}\).
The boundaries are \(x_0 = 0\), \(x_1 = \frac{\pi}{8}\), and \(x_2 = \frac{\pi}{4}\).
Evaluate the original function \(y = \sin(2x) + \cos^2(x)\) at these points:
\(y_0 = \sin(0) + \cos^2(0) = 1\).
\(y_1 = \sin\left(\frac{\pi}{4}\right) + \cos^2\left(\frac{\pi}{8}\right) = \frac{\sqrt{2}}{2} + \frac{2 + \sqrt{2}}{4} = \frac{1}{2} + \frac{3\sqrt{2}}{4} \approx 1.56066\).
\(y_2 = \sin\left(\frac{\pi}{2}\right) + \cos^2\left(\frac{\pi}{4}\right) = 1 + \frac{1}{2} = 1.5\).

Using the trapezium rule:
\(\text{Area} \approx \frac{1}{2}h [y_0 + y_2 + 2y_1] = \frac{\pi}{16} [1 + 1.5 + 2(1.56066)] = \frac{\pi}{16} [2.5 + 3.12132] = \frac{\pi}{16} \times 5.62132 \approx 1.1037\).
To 3 significant figures, the approximate area is 1.10.

(iv) The exact area is \(\frac{3}{4} + \frac{\pi}{8} \approx 1.14270\).
Using the unrounded approximate value of 1.10372:
\(\text{Percentage Error} = \frac{|1.14270 - 1.10372|}{1.14270} \times 100\% = \frac{0.03898}{1.14270} \times 100\% \approx 3.41\%\).

If using the rounded approximate value of 1.10:
\(\text{Percentage Error} = \frac{|1.14270 - 1.10|}{1.14270} \times 100\% \approx 3.74\%\).

(i)
M1: State or use the identity \(\cos^2(x) = \frac{1}{2}\cos(2x) + \frac{1}{2}\).
A1: Combine terms correctly to show the given identity.

(ii)
M1: Attempt integration of \(\sin(2x)\) and \(\cos(2x)\) terms, resulting in forms \(a \cos(2x)\) and \(b \sin(2x)\).
A1: Obtain the correct integrated expression \(-\frac{1}{2}\cos(2x) + \frac{1}{4}\sin(2x) + \frac{1}{2}x\).
M1: Substitute the limits \(0\) and \(\frac{\pi}{4}\) correctly into their integrated expression.
A1: Obtain the correct exact value \(\frac{3}{4} + \frac{\pi}{8}\) (or equivalent single fraction).

(iii)
M1: Use the trapezium rule structure with correct \(h = \frac{\pi}{8}\) (or equivalent).
A1: Calculate correct ordinate values \(y_0 = 1\), \(y_1 \approx 1.56\) (or exact), \(y_2 = 1.5\).
A1: Obtain the approximate area of 1.10 (or 1.104).

(iv)
M1: Attempt to find the percentage error using their values.
A1: Obtain correct percentage error of 3.41% (from unrounded values) or 3.74% (from rounded value 1.10).