2025 Cambridge IAS-Level Mathematics (9709) Practice Paper with Answers

To find the points of intersection, we equate the equations of the line and the curve: \(x^2 + kx - 1 = 2kx - 5\). Rearranging this gives the quadratic equation: \(x^2 - kx + 4 = 0\). For the line and curve not to intersect, this quadratic equation must have no real roots. Therefore, the discriminant must be less than zero: \(b^2 - 4ac < 0\). Substituting the coefficients: \((-k)^2 - 4(1)(4) < 0\), which simplifies to \(k^2 - 16 < 0\). Solving this inequality gives \(-4 < k < 4\).

M1: Equates the line and curve equations and rearranges into a 3-term quadratic in \(x\).
A1: Correct quadratic equation \(x^2 - kx + 4 = 0\).
M1: Uses the condition \(b^2 - 4ac < 0\).
A1: Correct final interval \(-4 < k < 4\).

Let the first term be \(a\) and the common ratio be \(r\). The sum of the first two terms is given by \(a + ar = 15\), which factors to \(a(1+r) = 15\). The sum to infinity is given by \(\frac{a}{1-r} = 27\), which gives \(a = 27(1-r)\). Substituting this expression for \(a\) into the first equation: \(27(1-r)(1+r) = 15\). This simplifies to \(27(1-r^2) = 15\), so \(1-r^2 = \frac{15}{27} = \frac{5}{9}\). This gives \(r^2 = \frac{4}{9}\). Since all terms are positive, both \(a > 0\) and \(r > 0\), so \(r = \frac{2}{3}\). Substituting \(r = \frac{2}{3}\) back into \(a = 27(1-r)\) gives \(a = 27\left(1 - \frac{2}{3}\right) = 9\).

M1: Sets up two equations using the sum of the first two terms and the sum to infinity.
M1: Eliminates one variable (either \(a\) or \(r\)) to form an equation in one variable.
A1: Solves to find the correct value of \(r = \frac{2}{3}\) (rejecting the negative root).
A1: Obtains \(a = 9\).

Using the identity \(\cos^2 \theta = 1 - \sin^2 \theta\), we substitute this into the equation: \(3(1 - \sin^2 \theta) + 8 \sin \theta - 7 = 0\). Expanding and simplifying: \(3 - 3 \sin^2 \theta + 8 \sin \theta - 7 = 0 \implies -3 \sin^2 \theta + 8 \sin \theta - 4 = 0\). Multiplying by \(-1\) gives: \(3 \sin^2 \theta - 8 \sin \theta + 4 = 0\). Factoring this quadratic in \(\sin \theta\) yields: \((3 \sin \theta - 2)(\sin \theta - 2) = 0\). This gives \(\sin \theta = \frac{2}{3}\) or \(\sin \theta = 2\) (which has no solution). For \(\sin \theta = \frac{2}{3}\), the principal value is \(\theta = \sin^{-1}\left(\frac{2}{3}\right) \approx 41.8^\circ\). The second solution in the range is \(180^\circ - 41.8^\circ = 138.2^\circ\).

M1: Uses the identity \(\cos^2 \theta = 1 - \sin^2 \theta\) to form a quadratic in \(\sin \theta\).
A1: Correct quadratic equation \(3 \sin^2 \theta - 8 \sin \theta + 4 = 0\).
M1: Solves the quadratic to find \(\sin \theta = \frac{2}{3}\) and finds the principal angle.
A1: Both angles \(41.8^\circ\) and \(138.2^\circ\) correct (rounded to 1 decimal place), and no extra solutions in range.

The perimeter of a sector is given by \(P = 2r + r\theta\), so \(2r + r\theta = 30\), which gives \(\theta = \frac{30 - 2r}{r}\). The area of a sector is given by \(A = \frac{1}{2}r^2\theta\), so \(\frac{1}{2}r^2\theta = 50\). Substituting the expression for \(\theta\) into the area equation: \(\frac{1}{2}r^2\left(\frac{30 - 2r}{r}\right) = 50\). This simplifies to \(\frac{1}{2}r(30 - 2r) = 50 \implies 15r - r^2 = 50\). Rearranging into a standard quadratic form: \(r^2 - 15r + 50 = 0\). Factoring the quadratic: \((r - 5)(r - 10) = 0\). This gives \(r = 5\) or \(r = 10\).

M1: Writes down two correct equations for the perimeter and area in terms of \(r\) and \(\theta\).
M1: Eliminates \(\theta\) to form a quadratic equation in \(r\).
A1: Correct quadratic equation \(r^2 - 15r + 50 = 0\) (or equivalent).
A1: Both values \(r = 5\) and \(r = 10\) correctly identified.

First, find the \(y\)-coordinate at \(x = 1\): \(y = \frac{8}{\sqrt{3(1) + 1}} = \frac{8}{\sqrt{4}} = 4\). To find the gradient of the tangent, we find the derivative of \(y = 8(3x + 1)^{-\frac{1}{2}}\). Using the chain rule: \(\frac{dy}{dx} = 8 \left(-\frac{1}{2}\right)(3x+1)^{-\frac{3}{2}} \times 3 = -12(3x+1)^{-\frac{3}{2}}\). Evaluating the derivative at \(x = 1\): \(\frac{dy}{dx} = -12(4)^{-\frac{3}{2}} = -12 \times \frac{1}{8} = -\frac{3}{2}\). The equation of the tangent is \(y - 4 = -\frac{3}{2}(x - 1)\). Multiplying by 2: \(2y - 8 = -3x + 3\). Rearranging gives \(3x + 2y - 11 = 0\).

M1: Finds the correct \(y\)-coordinate of \(4\) and differentiates \(y\) using chain rule.
A1: Correct derivative expression \(-12(3x+1)^{-\frac{3}{2}}\).
M1: Substitutes \(x=1\) to find the gradient and uses it to set up the equation of the line.
A1: Correct equation in the specified integer form: \(3x + 2y - 11 = 0\) (or any integer multiple).

(a) Completing the square: \( f(x) = 2(x^2 - 6x) + 13 = 2[(x-3)^2 - 9] + 13 = 2(x-3)^2 - 18 + 13 = 2(x-3)^2 - 5 \). So \( a = 2 \), \( b = 3 \), and \( c = -5 \). (b) For the function to have an inverse, it must be one-to-one. The line of symmetry of the quadratic is at \( x = 3 \). Since the domain is \( x \le k \), the largest value of \( k \) is 3. (c) Let \( y = 2(x-3)^2 - 5 \). Since \( x \le 3 \), we have \( x-3 \le 0 \). Rearranging for \( x \): \( y + 5 = 2(x-3)^2 \implies \frac{y+5}{2} = (x-3)^2 \). Taking the negative square root because \( x \le 3 \): \( x - 3 = -\sqrt{\frac{y+5}{2}} \implies x = 3 - \sqrt{\frac{y+5}{2}} \). Replacing variables gives \( f^{-1}(x) = 3 - \sqrt{\frac{x+5}{2}} \). The domain of \( f^{-1} \) is the range of \( f \). For \( x \le 3 \), the minimum value of \( f(x) \) is \( -5 \), so the range is \( f(x) \ge -5 \). Hence, the domain of \( f^{-1}(x) \) is \( x \ge -5 \).

(a) M1 for attempting to complete the square to get \( a(x-b)^2 + c \). A1 for \( 2(x-3)^2 \). A1 for \( -5 \). (b) B1 for \( k = 3 \). (c) M1 for attempting to make \( x \) the subject. M1 for choosing the negative square root due to the domain. A1 for \( f^{-1}(x) = 3 - \sqrt{\frac{x+5}{2}} \). B1.6 for stating the domain as \( x \ge -5 \).

(a) The terms of the AP are: \( A_1 = a \), \( A_3 = a + 2d \), \( A_9 = a + 8d \). The terms of the GP are: \( g_1 = a \), \( g_2 = ar \), \( g_3 = ar^2 \). Since they are equal: \( ar = a + 2d \implies 2d = a(r - 1) \implies d = \frac{a(r-1)}{2} \) (Equation 1). \( ar^2 = a + 8d \implies 8d = a(r^2 - 1) \) (Equation 2). Substitute \( d \) from Equation 1 into Equation 2: \( 8 \left( \frac{a(r-1)}{2} \right) = a(r^2 - 1) \implies 4a(r-1) = a(r-1)(r+1) \). Since \( d \ne 0 \), we must have \( a \ne 0 \) and \( r \ne 1 \). Dividing both sides by \( a(r-1) \) yields: \( 4 = r + 1 \implies r = 3 \). (b) The sum of the first 3 terms of the GP is \( a + ar + ar^2 = 26 \). Substitute \( r = 3 \): \( a + 3a + 9a = 26 \implies 13a = 26 \implies a = 2 \). Substitute \( a = 2 \) and \( r = 3 \) into Equation 1: \( d = \frac{2(3-1)}{2} = 2 \). (c) The sum of the first 20 terms of the AP is: \( S_{20} = \frac{20}{2} [2a + 19d] = 10 [2(2) + 19(2)] = 10 [4 + 38] = 10 \times 42 = 420 \).

(a) B1 for expressing AP terms correctly. M1 for expressing \( d \) in terms of \( a \) and \( r \). M1 for substituting and setting up the equation for \( r \). A1 for showing \( r = 3 \) convincingly. (b) M1 for using the GP sum formula. A1 for obtaining both \( a = 2 \) and \( d = 2 \). (c) M1 for using the AP sum formula with \( n = 20 \). A1.6 for 420.

(a) Combine the fractions on the LHS: \( \frac{\sin^2\theta + (1+\cos\theta)^2}{\sin\theta(1+\cos\theta)} = \frac{\sin^2\theta + 1 + 2\cos\theta + \cos^2\theta}{\sin\theta(1+\cos\theta)} \). Since \( \sin^2\theta + \cos^2\theta = 1 \), we have: \( \frac{1 + 1 + 2\cos\theta}{\sin\theta(1+\cos\theta)} = \frac{2 + 2\cos\theta}{\sin\theta(1+\cos\theta)} = \frac{2(1+\cos\theta)}{\sin\theta(1+\cos\theta)} = \frac{2}{\sin\theta} \). (b) Substituting the identity into the equation: \( \frac{2}{\sin\theta} = 3\tan\theta \implies \frac{2}{\sin\theta} = 3\frac{\sin\theta}{\cos\theta} \). Multiply both sides by \( \sin\theta\cos\theta \): \( 2\cos\theta = 3\sin^2\theta \). Substitute \( \sin^2\theta = 1 - \cos^2\theta \): \( 2\cos\theta = 3(1 - \cos^2\theta) \implies 3\cos^2\theta + 2\cos\theta - 3 = 0 \). Let \( x = \cos\theta \), then solve \( 3x^2 + 2x - 3 = 0 \): \( x = \frac{-2 \pm \sqrt{4 - 4(3)(-3)}}{6} = \frac{-2 \pm \sqrt{40}}{6} = \frac{-1 \pm \sqrt{10}}{3} \). So, \( \cos\theta \approx 0.7208 \) or \( \cos\theta \approx -1.387 \) (no solution as \( \cos\theta \ge -1 \)). Solving \( \cos\theta = 0.7208 \): \( \theta = \cos^{-1}(0.7208) \approx 43.9^\circ \). The other solution in the given interval is \( 360^\circ - 43.9^\circ = 316.1^\circ \).

(a) M1 for finding a common denominator and expressing the numerator correctly. M1 for applying \( \sin^2\theta + \cos^2\theta = 1 \). A1 for correct simplification to the RHS. (b) M1 for replacing LHS with \( \frac{2}{\sin\theta} \). M1 for substituting \( \tan\theta = \frac{\sin\theta}{\cos\theta} \) and obtaining a quadratic equation in \( \cos\theta \). A1 for \( 3\cos^2\theta + 2\cos\theta - 3 = 0 \). M1 for solving the quadratic correctly. A1 for \( 43.9^\circ \). A0.6 for \( 316.1^\circ \) (allow answers rounding to 1 d.p.).

(a)(i) When \( k = 1 \), the curve is \( y = x^2 - 3x + 2 \). Equating to the line \( y = x - 1 \): \( x^2 - 3x + 2 = x - 1 \implies x^2 - 4x + 3 = 0 \implies (x-1)(x-3) = 0 \). This gives \( x = 1 \) or \( x = 3 \). The points of intersection are \( (1, 0) \) and \( (3, 2) \). (a)(ii) The curve lies strictly above the line when \( x^2 - 3x + 2 > x - 1 \implies x^2 - 4x + 3 > 0 \). The critical values are 1 and 3, so the solution set is \( x < 1 \) or \( x > 3 \). (b) Equating the general curve and the line: \( kx^2 - 3x + k + 1 = x - 1 \implies kx^2 - 4x + (k+2) = 0 \). For no intersection, the discriminant must be strictly negative: \( b^2 - 4ac < 0 \implies (-4)^2 - 4(k)(k+2) < 0 \implies 16 - 4k^2 - 8k < 0 \). Divide by \(-4\) (and reverse the inequality): \( k^2 + 2k - 4 > 0 \). The roots of \( k^2 + 2k - 4 = 0 \) are \( k = \frac{-2 \pm \sqrt{4 - 4(1)(-4)}}{2} = -1 \pm \sqrt{5} \). Thus, \( k < -1 - \sqrt{5} \) or \( k > -1 + \sqrt{5} \).

(a)(i) M1 for equating the line and curve with \( k = 1 \). A1 for \( x = 1, 3 \). A1 for coordinates \( (1, 0) \) and \( (3, 2) \). (a)(ii) M1 for establishing the quadratic inequality. A1 for correct ranges \( x < 1 \) or \( x > 3 \). (b) M1 for setting up the quadratic in terms of \( k \). M1 for using \( b^2 - 4ac < 0 \). A1 for obtaining \( k^2 + 2k - 4 > 0 \). A0.6 for the final range \( k < -1 - \sqrt{5} \) or \( k > -1 + \sqrt{5} \).

(a) Write \( y = x^2 - 2x + 8x^{-1} \). Differentiating once: \( \frac{dy}{dx} = 2x - 2 - 8x^{-2} = 2x - 2 - \frac{8}{x^2} \). Differentiating a second time: \( \frac{d^2y}{dx^2} = 2 + 16x^{-3} = 2 + \frac{16}{x^3} \). (b) Set \( \frac{dy}{dx} = 0 \): \( 2x - 2 - \frac{8}{x^2} = 0 \implies 2x^3 - 2x^2 - 8 = 0 \implies x^3 - x^2 - 4 = 0 \). Testing \( x = 2 \) gives \( 2^3 - 2^2 - 4 = 0 \), so \( x = 2 \) is a root. Factoring out \( (x-2) \) yields \( (x-2)(x^2 + x + 2) = 0 \). Since \( x^2 + x + 2 = 0 \) has no real roots (discriminant is \( -7 < 0 \)), \( x = 2 \) is the only real solution. When \( x = 2 \), \( y = 2^2 - 2(2) + \frac{8}{2} = 4 - 4 + 4 = 4 \). Thus, the coordinates are \( (2, 4) \). (c) Substitute \( x = 2 \) into the second derivative: \( \frac{d^2y}{dx^2} = 2 + \frac{16}{2^3} = 2 + 2 = 4 \). Since \( \frac{d^2y}{dx^2} = 4 > 0 \), the stationary point is a minimum.

(a) M1 for differentiating at least one term correctly. A1 for correct \( \frac{dy}{dx} \). A1 for correct \( \frac{d^2y}{dx^2} \). (b) M1 for setting \( \frac{dy}{dx} = 0 \). M1 for solving the cubic equation to find \( x = 2 \). A1 for \( x = 2 \). A0.6 for finding \( y = 4 \). (c) M1 for evaluating the second derivative at their stationary value. A1 for concluding it is a Minimum based on a positive value.

(a) Integrate \( \frac{dy}{dx} \) to find \( y \): \( y = \int 6(3x+1)^{-1/2} \, dx = 6 \cdot \frac{(3x+1)^{1/2}}{\frac{1}{2} \cdot 3} + C = 4(3x+1)^{1/2} + C \). Substituting the point \( (1, 8) \): \( 8 = 4\sqrt{3(1) + 1} + C \implies 8 = 8 + C \implies C = 0 \). Thus, the equation of the curve is \( y = 4\sqrt{3x+1} \). (b) The area under the curve between \( x = 1 \) and \( x = 5 \) is: \( A = \int_{1}^{5} 4(3x+1)^{1/2} \, dx = \left[ 4 \cdot \frac{(3x+1)^{3/2}}{\frac{3}{2} \cdot 3} \right]_{1}^{5} = \left[ \frac{8}{9}(3x+1)^{3/2} \right]_{1}^{5} \). Evaluating this at the limits: \( A = \frac{8}{9}(3(5)+1)^{3/2} - \frac{8}{9}(3(1)+1)^{3/2} = \frac{8}{9}(16)^{3/2} - \frac{8}{9}(4)^{3/2} = \frac{8}{9}(64) - \frac{8}{9}(8) = \frac{512}{9} - \frac{64}{9} = \frac{448}{9} \).

(a) M1 for attempting to integrate to obtain \( k(3x+1)^{1/2} \). A1 for obtaining \( y = 4(3x+1)^{1/2} + C \). M1 for substituting the point \( (1, 8) \) to find \( C \). A1 for the correct equation. (b) M1 for attempting to integrate \( 4(3x+1)^{1/2} \) to obtain \( m(3x+1)^{3/2} \). A1 for correct integral expression \( \frac{8}{9}(3x+1)^{3/2} \). M1 for substituting both limits 1 and 5. A1.6 for obtaining \( \frac{448}{9} \) (or equivalent exact fraction/decimal rounding to 49.8).

Using algebraic long division: First, divide the leading term \(2x^3\) by \(x^2\) to get the first term of the quotient, \(2x\). Multiply the divisor \(x^2 - x - 2\) by \(2x\) to obtain \(2x^3 - 2x^2 - 4x\). Subtracting this from the original polynomial \(2x^3 - 3x^2 - 8x + 15\) gives \(-x^2 - 4x + 15\). Next, divide \(-x^2\) by \(x^2\) to get the next term of the quotient, \(-1\). Multiply the divisor by \(-1\) to obtain \(-x^2 + x + 2\). Subtracting this from \(-x^2 - 4x + 15\) gives the remainder \(-5x + 13\). Therefore, the quotient is \(2x - 1\) and the remainder is \(-5x + 13\).

M1: For attempting algebraic long division or equating coefficients up to finding a linear quotient. A1: For obtaining the correct quotient \(2x - 1\). A1: For obtaining the correct remainder \(-5x + 13\).

Using the laws of logarithms, we can write the right-hand side of the equation as: \(2\ln y - \ln 3 = \ln(y^2) - \ln 3 = \ln\left(\frac{y^2}{3}\right)\). Equating this to the left-hand side, we have: \(\ln(2y - 3) = \ln\left(\frac{y^2}{3}\right)\). Since the logarithmic function is one-to-one, we can remove the logarithms: \(2y - 3 = \frac{y^2}{3}\). Multiplying both sides by 3 gives: \(6y - 9 = y^2\), which rearranges to the quadratic equation: \(y^2 - 6y + 9 = 0\). Factoring the quadratic, we get: \((y - 3)^2 = 0\), which yields \(y = 3\). Checking the validity of the solution, we require the arguments of the logarithms in the original equation to be positive: \(2y - 3 = 3 > 0\) and \(y = 3 > 0\). Both conditions are satisfied, so the only valid solution is \(y = 3\).

M1: For applying logarithm laws to combine the right-hand side into a single logarithm. M1: For removing logarithms correctly to form a quadratic equation. A1: For solving the quadratic to get \(y = 3\) and verifying its validity.

We use the trigonometric identity \(\sec^2 \theta = 1 + \tan^2 \theta\) to rewrite the equation as: \(1 + \tan^2 \theta + 2\tan \theta = 4\). Rearranging terms into a quadratic in \(\tan \theta\), we get: \(\tan^2 \theta + 2\tan \theta - 3 = 0\). Factoring this quadratic expression gives: \((\tan \theta + 3)(\tan \theta - 1) = 0\), which yields \(\tan \theta = 1\) or \(\tan \theta = -3\). For the interval \(0^\circ \le \theta \le 180^\circ\): 1) If \(\tan \theta = 1\), the solution is \(\theta = 45^\circ\). 2) If \(\tan \theta = -3\), the solution is \(\theta = 180^\circ - \arctan(3) \approx 108.4^\circ\). Therefore, the solutions in the given interval are \(\theta = 45^\circ\) and \(\theta = 108.4^\circ\).

M1: For using \(\sec^2 \theta = 1 + \tan^2 \theta\) to form a quadratic in \(\tan \theta\). M1: For solving the quadratic to find \(\tan \theta = 1\) and \(\tan \theta = -3\). A1: For finding \(\theta = 45^\circ\). A1: For finding \(\theta = 108.4^\circ\) (accept 108.4 or 108.43; round to 1 decimal place as per standard instruction).

(i) We start from the left-hand side of the identity:
\(\cos 3x + \cos x = \cos(2x + x) + \cos(2x - x)\)

Using the compound angle formulas:
\(\cos(2x + x) = \cos 2x \cos x - \sin 2x \sin x\)
\(\cos(2x - x) = \cos 2x \cos x + \sin 2x \sin x\)

Adding these two expressions gives:
\(\cos 3x + \cos x = 2\cos 2x \cos x\) (as required).

(ii) Using the identity from part (i), we can rewrite the equation:
\(\cos 3x + \cos x = 2\cos 2x \implies 2\cos 2x \cos x = 2\cos 2x\)

Rearranging this:
\(2\cos 2x \cos x - 2\cos 2x = 0\)

Factoring out \(2\cos 2x\):
\(2\cos 2x(\cos x - 1) = 0\) (as required).

(iii) To solve \(2\cos 2x(\cos x - 1) = 0\) in the interval \(0 \le x \le \pi\):

Either \(\cos 2x = 0\):
For \(0 \le x \le \pi\), we have \(0 \le 2x \le 2\pi\).
\(2x = \frac{\pi}{2}, \frac{3\pi}{2} \implies x = \frac{\pi}{4}, \frac{3\pi}{4}\).

Or \(\cos x - 1 = 0 \implies \cos x = 1\):
For \(0 \le x \le \pi\), this gives:
\(x = 0\).

Thus, the solutions are \(x = 0, \frac{\pi}{4}, \frac{3\pi}{4}\).

(i)
M1: For attempting to use compound angle formulas on \(\cos(2x+x)\) and \(\cos(2x-x)\).
A1: For correct expansion of both terms.
A1: For completing the proof with no errors.

(ii)
M1: For substituting the identity from (i) into the equation.
A1: For rearranging and factorising to obtain the given form.

(iii)
M1: For setting \(\cos 2x = 0\) and solving for \(x\) in the given range.
A1: For obtaining \(x = \frac{\pi}{4}\) and \(x = \frac{3\pi}{4}\).
M1: For setting \(\cos x = 1\) and solving to get \(x = 0\).
A1: For obtaining all three correct solutions and no extras in the range.

(i) Differentiating \(x\) and \(y\) with respect to \(t\):
\(\frac{\text{d}x}{\text{d}t} = 4e^{2t} - 1\)
\(\frac{\text{d}y}{\text{d}t} = 3e^{3t} + 2\)

Using the chain rule:
\(\frac{\text{d}y}{\text{d}x} = \frac{\text{d}y/\text{d}t}{\text{d}x/\text{d}t} = \frac{3e^{3t} + 2}{4e^{2t} - 1}\).

(ii) When \(t = 0\):
\(x = 2e^0 - 0 = 2\)
\(y = e^0 + 2(0) = 1\)

Gradient of the tangent, \(m = \frac{\text{d}y}{\text{d}x} = \frac{3(1) + 2}{4(1) - 1} = \frac{5}{3}\).

Using the equation of a straight line:
\(y - 1 = \frac{5}{3}(x - 2)\)
\(3(y - 1) = 5(x - 2) \implies 3y - 3 = 5x - 10 \implies 5x - 3y - 7 = 0\).

(iii) The gradient of the curve is zero when \(\frac{\text{d}y}{\text{d}t} = 0\).
Here, \(\frac{\text{d}y}{\text{d}t} = 3e^{3t} + 2\).
Since \(e^{3t} > 0\) for all real values of \(t\), we have \(3e^{3t} + 2 > 2\).
Therefore, \(\frac{\text{d}y}{\text{d}t}\) can never be equal to zero, which means the gradient of the curve is never zero.

(i)
M1: For attempting to differentiate \(x\) and \(y\) with respect to \(t\).
A1: For correct derivatives \(\frac{\text{d}x}{\text{d}t} = 4e^{2t} - 1\) and \(\frac{\text{d}y}{\text{d}t} = 3e^{3t} + 2\).
A1: For correct expression of \(\frac{\text{d}y}{\text{d}x}\).

(ii)
M1: For finding the coordinates \((2, 1)\) at \(t = 0\).
M1: For finding the numerical value of the gradient \(m = \frac{5}{3}\) at \(t = 0\).
M1: For attempting to find the equation of the line using their point and gradient.
A1.7: For correct final equation \(5x - 3y - 7 = 0\) (or any integer multiple thereof).

(iii)
M1: For setting their numerator \(\frac{\text{d}y}{\text{d}t} = 0\) or stating that gradient is zero when \(\frac{\text{d}y}{\text{d}t} = 0\).
A1: For a complete explanation showing that \(3e^{3t} + 2 > 0\) for all real \(t\).

(i) We know the double-angle formula for cosine is:
\(\cos 2\theta = 1 - 2\sin^2 \theta\)

Letting \(\theta = 3x\):
\(\cos 6x = 1 - 2\sin^2 3x\)

Rearranging to make \(\sin^2 3x\) the subject:
\(2\sin^2 3x = 1 - \cos 6x \implies \sin^2 3x = \frac{1}{2}(1 - \cos 6x)\) (as required).

(ii) Substituting the identity into the integrand:
\(4\sin^2 3x + e^{4x} = 4 \cdot \frac{1}{2}(1 - \cos 6x) + e^{4x} = 2(1 - \cos 6x) + e^{4x} = 2 - 2\cos 6x + e^{4x}\).

Now we perform the integration:
\(\int_{0}^{\pi/12} (2 - 2\cos 6x + e^{4x}) \text{d}x = \left[ 2x - \frac{2}{6}\sin 6x + \frac{1}{4}e^{4x} \right]_{0}^{\pi/12}\)
\(= \left[ 2x - \frac{1}{3}\sin 6x + \frac{1}{4}e^{4x} \right]_{0}^{\pi/12}\)

Evaluating at the upper limit \(x = \frac{\pi}{12}\):
\(2\left(\frac{\pi}{12}\right) - \frac{1}{3}\sin\left(6 \cdot \frac{\pi}{12}\right) + \frac{1}{4}e^{4(\pi/12)}\)
\(= \frac{\pi}{6} - \frac{1}{3}\sin\left(\frac{\pi}{2}\right) + \frac{1}{4}e^{\pi/3}\)
\(= \frac{\pi}{6} - \frac{1}{3}(1) + \frac{1}{4}e^{\pi/3} = \frac{\pi}{6} - \frac{1}{3} + \frac{1}{4}e^{\pi/3}\).

Evaluating at the lower limit \(x = 0\):
\(2(0) - \frac{1}{3}\sin(0) + \frac{1}{4}e^{0} = 0 - 0 + \frac{1}{4} = \frac{1}{4}\).

Subtracting the lower limit from the upper limit:
\(\left(\frac{\pi}{6} - \frac{1}{3} + \frac{1}{4}e^{\pi/3}\right) - \frac{1}{4} = \frac{\pi}{6} - \frac{7}{12} + \frac{1}{4}e^{\pi/3}\).

(i)
M1: For attempting to use the double-angle identity \(\cos 2\theta = 1 - 2\sin^2 \theta\) with \(\theta = 3x\).
A1: For correctly completing the proof.

(ii)
M1: For substituting the identity from (i) into the integral to express the integrand as \(2 - 2\cos 6x + e^{4x}\).
M1: For integrating the trigonometric part to obtain \(2x - k\sin 6x\) (where \(k \ne 0\)).
A1: For correct term \(-\frac{1}{3}\sin 6x\).
M1: For integrating \(e^{4x}\) to obtain \(\frac{1}{4}e^{4x}\).
M1: For substituting the limits correctly (must show working for both limits, especially at \(x=0\)).
A1: For correct value at the upper limit: \(\frac{\pi}{6} - \frac{1}{3} + \frac{1}{4}e^{\pi/3}\).
A1.7: For the correct final exact answer: \(\frac{\pi}{6} - \frac{7}{12} + \frac{1}{4}e^{\pi/3}\).

(i) Let \(f(x) = x^3 - 5x - 3\).
We evaluate the function at the boundary values:
\(f(2) = (2)^3 - 5(2) - 3 = 8 - 10 - 3 = -5\)
\(f(3) = (3)^3 - 5(3) - 3 = 27 - 15 - 3 = 9\)

Since there is a change of sign between \(f(2) < 0\) and \(f(3) > 0\), and the function is continuous, the equation \(f(x) = 0\) must have at least one root \(\alpha\) in the interval \(2 < \alpha < 3\).

(ii) Starting with the equation:
\(x^3 - 5x - 3 = 0\)
\(x^3 = 5x + 3\)

Since \(x \ne 0\) (as \(2 < \alpha < 3\)), we can divide both sides by \(x\):
\(x^2 = 5 + \frac{3}{x}\)

Taking the positive square root (since \(\alpha > 0\)):
\(x = \sqrt{5 + \frac{3}{x}}\) (as required).

(iii) Applying the iterative formula \(x_{n+1} = \sqrt{5 + \frac{3}{x_n}}\) with \(x_1 = 2.5\):
\(x_2 = \sqrt{5 + \frac{3}{2.5}} = \sqrt{6.2} \approx 2.48998\)
\(x_3 = \sqrt{5 + \frac{3}{2.48998}} \approx 2.49095\)
\(x_4 = \sqrt{5 + \frac{3}{2.49095}} \approx 2.49086\)
\(x_5 = \sqrt{5 + \frac{3}{2.49086}} \approx 2.49086\)

Since \(x_4\) and \(x_5\) round to the same value to 3 decimal places, we can conclude that:
\(\alpha \approx 2.491\) (to 3 decimal places).

(i)
M1: For attempting to evaluate the function \(f(x) = x^3 - 5x - 3\) at both \(x = 2\) and \(x = 3\).
A1: For getting \(f(2) = -5\) and \(f(3) = 9\) and giving a conclusion based on the sign change.

(ii)
M1: For adding \(5x + 3\) to both sides and dividing by \(x\).
A1: For taking the square root to obtain the given form.

(iii)
M1: For substituting \(x_1 = 2.5\) into the iterative formula to find \(x_2\).
A1: For correct value of \(x_2 = 2.48998\).
A1: For correct values of \(x_3 = 2.49095\) and \(x_4 = 2.49086\).
M1: For continuing the iteration to check convergence.
A1.7: For final answer of \(2.491\) correct to 3 decimal places.