A Level Pure Mathematics 2 (Edexcel 2023)

✏ Working:

First, find the first derivative \( f'(x) \) using the power rule \( \frac{d}{dx}(ax^n) = anx^{n-1} \):

\[ f'(x) = 3x^2 + 2(2x) – 8 = 3x^2 + 4x – 8 \]

Now, differentiate again to find \( f”(x) \):

\[ f”(x) = 2(3x) + 4 = 6x + 4 \]

✓ (2 marks)

✏ Working:

(i) Solve \( f”(x) = 0 \):

\[ 6x + 4 = 0 \] \[ 6x = -4 \] \[ x = -\frac{4}{6} = -\frac{2}{3} \]

(ii) For \( f(x) \) to be concave, we need \( f”(x) \leq 0 \):

\[ 6x + 4 \leq 0 \] \[ 6x \leq -4 \] \[ x \leq -\frac{2}{3} \]

✓ (2 marks)

✏ Working:

(i) For \( u_2 \), let \( n = 1 \):

\[ u_2 = u_1 + 7\cos\left(\frac{1\pi}{2}\right) – 5(-1)^1 \] \[ u_2 = 35 + 7(0) – 5(-1) \] \[ u_2 = 35 + 0 + 5 = 40 \]

(ii) For \( u_3 \), let \( n = 2 \):

\[ u_3 = u_2 + 7\cos\left(\frac{2\pi}{2}\right) – 5(-1)^2 \] \[ u_3 = 40 + 7\cos(\pi) – 5(1) \] \[ u_3 = 40 + 7(-1) – 5 = 40 – 7 – 5 = 28 \]

For \( u_4 \), let \( n = 3 \):

\[ u_4 = u_3 + 7\cos\left(\frac{3\pi}{2}\right) – 5(-1)^3 \] \[ u_4 = 28 + 7(0) – 5(-1) \] \[ u_4 = 28 + 0 + 5 = 33 \]

✓ (3 marks)

✏ Working:

(i) Since the period is 4, \( u_5 = u_1 = 35 \).

(ii) We need the sum of the first 25 terms. Since the pattern repeats every 4 terms, we check how many full cycles are in 25:

\[ 25 = 6 \times 4 + 1 \]

So we have 6 full cycles plus the first term repeated once.

Sum of one cycle (\( u_1 + u_2 + u_3 + u_4 \)):

\[ S_4 = 35 + 40 + 28 + 33 = 136 \]

Total sum:

\[ \sum_{r=1}^{25} u_r = 6(136) + u_1 \] \[ = 816 + 35 = 851 \]

✓ (3 marks)

✏ Working:

Combine the LHS:

\[ \log_2((x+3)(x+10)) \]

Rewrite the RHS:

\[ 2 + \log_2(x^2) \] \[ \log_2(4) + \log_2(x^2) = \log_2(4x^2) \]

Equate the arguments (remove logs):

\[ (x+3)(x+10) = 4x^2 \]

Expand brackets:

\[ x^2 + 10x + 3x + 30 = 4x^2 \] \[ x^2 + 13x + 30 = 4x^2 \]

Rearrange to form quadratic:

\[ 3x^2 – 13x – 30 = 0 \]

✓ (3 marks)

✏ Working:

(i) Solve \( 3x^2 – 13x – 30 = 0 \):

Factor pairs of \( 3 \times -30 = -90 \) that add to -13 are -18 and +5.

\[ 3x^2 – 18x + 5x – 30 = 0 \] \[ 3x(x – 6) + 5(x – 6) = 0 \] \[ (3x + 5)(x – 6) = 0 \]

Roots: \( x = 6 \) or \( x = -\frac{5}{3} \)

(ii) Check validity:

If \( x = -\frac{5}{3} \), then \( \log_2 x \) is \( \log_2(-\frac{5}{3}) \), which is undefined.

Therefore, \( x = -\frac{5}{3} \) is not a solution.

✓ (2 marks)

✏ Working:

At \( t = 0 \), \( H = 85 \):

\[ 85 = Ae^{-B(0)} + 30 \] \[ 85 = A(1) + 30 \] \[ A = 55 \]

✓ (1 mark)

✏ Working:

Differentiate \( H = 55e^{-Bt} + 30 \) with respect to \( t \):

\[ \frac{dH}{dt} = 55(-B)e^{-Bt} = -55Be^{-Bt} \]

At \( t = 0 \), the rate is cooling at 7.5°C/min, so \( \frac{dH}{dt} = -7.5 \):

\[ -7.5 = -55B e^{0} \] \[ -7.5 = -55B \] \[ B = \frac{7.5}{55} = \frac{3}{22} \approx 0.13636… \]

To 3 decimal places, \( B = 0.136 \).

The complete equation is:

\[ H = 55e^{-0.136t} + 30 \]

✓ (3 marks)

✏ Working:

At \( P(3, -10) \), \( x = 3 \) and \( \frac{dy}{dx} = 0 \).

Substitute these into the gradient equation:

\[ 0 = 2(3)^3 – 9(3)^2 + 5(3) + k \] \[ 0 = 2(27) – 9(9) + 15 + k \] \[ 0 = 54 – 81 + 15 + k \] \[ 0 = -12 + k \] \[ k = 12 \]

✓ (2 marks)

✏ Working:

Integrate \( \frac{dy}{dx} \):

\[ y = \int (2x^3 – 9x^2 + 5x + 12) \, dx \] \[ y = \frac{2x^4}{4} – \frac{9x^3}{3} + \frac{5x^2}{2} + 12x + c \] \[ y = \frac{1}{2}x^4 – 3x^3 + \frac{5}{2}x^2 + 12x + c \]

Use point \( P(3, -10) \) to find \( c \):

\[ -10 = \frac{1}{2}(3)^4 – 3(3)^3 + \frac{5}{2}(3)^2 + 12(3) + c \] \[ -10 = \frac{81}{2} – 81 + \frac{45}{2} + 36 + c \] \[ -10 = 40.5 – 81 + 22.5 + 36 + c \] \[ -10 = 18 + c \] \[ c = -28 \]

The y-intercept occurs when \( x = 0 \):

\[ y = 0 – 0 + 0 + 0 + (-28) \] \[ y = -28 \]

Coordinates: \( (0, -28) \)

✓ (3 marks)

✏ Working:

\[ \vec{AD} = \vec{OD} – \vec{OA} = (22\mathbf{i} + 24\mathbf{j}) – (12\mathbf{i}) = 10\mathbf{i} + 24\mathbf{j} \] \[ \vec{BC} = \vec{OC} – \vec{OB} = (50\mathbf{i} + 136\mathbf{j}) – (16\mathbf{j}) = 50\mathbf{i} + 120\mathbf{j} \]

Comparing the two:

\[ \vec{BC} = 5(10\mathbf{i} + 24\mathbf{j}) = 5 \vec{AD} \]

Since \( \vec{BC} \) is a scalar multiple of \( \vec{AD} \), they are parallel.

✓ (2 marks)

✏ Working:

Calculate lengths:

\[ |\vec{AD}| = \sqrt{10^2 + 24^2} = \sqrt{100+576} = \sqrt{676} = 26 \text{ m} \] \[ |\vec{BC}| = \sqrt{50^2 + 120^2} = 5 \times 26 = 130 \text{ m} \]

Now find \( \vec{AB} \) and \( \vec{CD} \):

\[ \vec{AB} = 16\mathbf{j} – 12\mathbf{i} = -12\mathbf{i} + 16\mathbf{j} \] \[ |\vec{AB}| = \sqrt{(-12)^2 + 16^2} = \sqrt{144+256} = \sqrt{400} = 20 \text{ m} \] \[ \vec{CD} = (22\mathbf{i} + 24\mathbf{j}) – (50\mathbf{i} + 136\mathbf{j}) = -28\mathbf{i} – 112\mathbf{j} \] \[ |\vec{CD}| = \sqrt{(-28)^2 + (-112)^2} = \sqrt{784 + 12544} = \sqrt{13328} \approx 115.447 \text{ m} \]

Total distance for 1 lap = \( 26 + 130 + 20 + 115.447 = 291.447 \) m.

Distance for 2 laps = \( 2 \times 291.447 = 582.89 \) m = \( 0.58289 \) km.

Time = 5 minutes = \( \frac{5}{60} \) hours.

\[ \text{Speed} = \frac{0.58289}{5/60} = 0.58289 \times 12 \approx 6.99 \text{ km/h} \]

✓ (4 marks)

✏ Working:

Differentiate each term w.r.t \( x \):

• \( x^3 \to 3x^2 \)
• \( 2xy \to \) (Product Rule: \( u=2x, v=y \)) \( \to 2y + 2x\frac{dy}{dx} \)
• \( 3y^2 \to 6y\frac{dy}{dx} \)
• \( 47 \to 0 \)

Combine:

\[ 3x^2 + 2y + 2x\frac{dy}{dx} + 6y\frac{dy}{dx} = 0 \]

Group terms with \( \frac{dy}{dx} \):

\[ \frac{dy}{dx}(2x + 6y) = -3x^2 – 2y \] \[ \frac{dy}{dx} = \frac{-3x^2 – 2y}{2x + 6y} \]

✓ (4 marks)

✏ Working:

Substitute \( x = -2, y = 5 \) into \( \frac{dy}{dx} \):

\[ m_T = \frac{-3(-2)^2 – 2(5)}{2(-2) + 6(5)} \] \[ m_T = \frac{-12 – 10}{-4 + 30} = \frac{-22}{26} = \frac{-11}{13} \]

Gradient of normal \( m_N \):

\[ m_N = \frac{-1}{m_T} = \frac{13}{11} \]

Equation of line passing through \( (-2, 5) \):

\[ y – 5 = \frac{13}{11}(x – (-2)) \] \[ 11(y – 5) = 13(x + 2) \] \[ 11y – 55 = 13x + 26 \] \[ 13x – 11y + 81 = 0 \]

✓ (3 marks)

✏ Working:

\[ R\cos\alpha = 2 \] \[ R\sin\alpha = 8 \]

Calculate \( R \):

\[ R = \sqrt{2^2 + 8^2} = \sqrt{4+64} = \sqrt{68} = 2\sqrt{17} \]

Calculate \( \alpha \):

\[ \tan\alpha = \frac{8}{2} = 4 \] \[ \alpha = \tan^{-1}(4) \approx 1.326 \text{ rad} \]

Expression: \( 2\sqrt{17}\cos(\theta – 1.326) \)

✓ (3 marks)

✏ Working:

For \( n=9 \):

\[ S_9 = \frac{9}{2}(2\cos x + 8\sin x) \] \[ S_9 = 4.5(2\cos x + 8\sin x) \]

Notice the term in the bracket is exactly the expression from part (a).

\[ S_9 = 4.5(2\sqrt{17}\cos(x – 1.326)) \] \[ S_9 = 9\sqrt{17}\cos(x – 1.326) \]

(i) Maximum value:

The maximum of cosine is 1. So max \( S_9 = 9\sqrt{17} \).

(ii) Smallest positive \( x \):

Max occurs when \( \cos(x – 1.326) = 1 \), so \( x – 1.326 = 0 \).

\[ x = 1.33 \text{ rad} \]

✓ (3 marks)

✏ Working:

\[ x = t^2 + 6t – 16 = (t+3)^2 – 25 \] \[ x + 25 = (t+3)^2 \] \[ \sqrt{x+25} = t+3 \]

Now look at \( y = 6\ln(t+3) \). We can substitute \( t+3 \).

\[ y = 6\ln(\sqrt{x+25}) \] \[ y = 6\ln((x+25)^{1/2}) \] \[ y = 6 \times \frac{1}{2} \ln(x+25) \] \[ y = 3\ln(x+25) \]

So \( A = 3, B = 25 \).

✓ (3 marks)

✏ Working:

At y-axis, \( x = 0 \):

\[ t^2 + 6t – 16 = 0 \] \[ (t+8)(t-2) = 0 \]

Since \( t > -3 \), we must have \( t = 2 \).

Coordinates of P when \( t=2 \):

\[ y = 6\ln(2+3) = 6\ln 5 \]

Find gradient:

\[ \frac{dx}{dt} = 2t + 6 \] \[ \frac{dy}{dt} = \frac{6}{t+3} \] \[ \frac{dy}{dx} = \frac{6/(t+3)}{2t+6} = \frac{6}{2(t+3)^2} = \frac{3}{(t+3)^2} \]

At \( t=2 \):

\[ m = \frac{3}{(2+3)^2} = \frac{3}{25} \]

Equation of tangent:

\[ y – 6\ln 5 = \frac{3}{25}(x – 0) \] \[ 25y – 150\ln 5 = 3x \] \[ -3x + 25y = 150\ln 5 \] \[ 25y – 3x = 150\ln 5 \]

✓ (4 marks)

✏ Working:

\[ \frac{3kx – 18}{(x+4)(x-2)} = \frac{A}{x+4} + \frac{B}{x-2} \] \[ 3kx – 18 = A(x-2) + B(x+4) \]

Let \( x = 2 \):

\[ 6k – 18 = 6B \implies B = k – 3 \]

Let \( x = -4 \):

\[ -12k – 18 = -6A \implies A = 2k + 3 \]

So \( f(x) = \frac{2k+3}{x+4} + \frac{k-3}{x-2} \)

✓ (3 marks)

✏ Working:

\[ \int_{-3}^{1} \left( \frac{2k+3}{x+4} + \frac{k-3}{x-2} \right) dx \] \[ = [(2k+3)\ln|x+4| + (k-3)\ln|x-2|]_{-3}^{1} \]

Substitute limits:

Upper (1): \( (2k+3)\ln 5 + (k-3)\ln 1 = (2k+3)\ln 5 \)

Lower (-3): \( (2k+3)\ln 1 + (k-3)\ln 5 = (k-3)\ln 5 \)

Subtract:

\[ (2k+3)\ln 5 – (k-3)\ln 5 = 21 \] \[ \ln 5 [ (2k+3) – (k-3) ] = 21 \] \[ \ln 5 [ k + 6 ] = 21 \] \[ k + 6 = \frac{21}{\ln 5} \] \[ k = \frac{21}{\ln 5} – 6 \]

✓ (4 marks)

✏ Working:

Volume of water \( V = 20 \times 10 \times h = 200h \).

Differentiating w.r.t \( h \): \( \frac{dV}{dh} = 200 \).

Model says rate of change of \( V \) is inversely proportional to \( \sqrt{h} \):

\[ \frac{dV}{dt} = \frac{k}{\sqrt{h}} \]

Using chain rule: \( \frac{dh}{dt} = \frac{dh}{dV} \times \frac{dV}{dt} \)

\[ \frac{dh}{dt} = \frac{1}{200} \times \frac{k}{\sqrt{h}} = \frac{k/200}{\sqrt{h}} \]

Let \( \lambda = k/200 \), then \( \frac{dh}{dt} = \frac{\lambda}{\sqrt{h}} \).

✓ (3 marks)

✏ Working:

\[ \sqrt{h} \frac{dh}{dt} = \lambda \] \[ \int h^{1/2} dh = \int \lambda dt \] \[ \frac{2}{3} h^{3/2} = \lambda t + c \] \[ h^{3/2} = \frac{3\lambda}{2} t + \frac{3c}{2} \]

Let \( A = \frac{3\lambda}{2} \) and \( B = \frac{3c}{2} \). So \( h^{3/2} = At + B \).

Use conditions:

1. \( t=0, h=1.44 \):

\[ (1.44)^{3/2} = A(0) + B \implies 1.2^3 = 1.728 = B \]

2. \( t=8, h=3.24 \):

\[ (3.24)^{3/2} = A(8) + 1.728 \] \[ 1.8^3 = 8A + 1.728 \] \[ 5.832 = 8A + 1.728 \] \[ 8A = 4.104 \] \[ A = 0.513 \]

Equation: \( h^{3/2} = 0.513t + 1.728 \)

✓ (5 marks)

✏ Working:

Tank is full when \( h = 5 \).

\[ 5^{3/2} = 0.513t + 1.728 \] \[ 11.18… = 0.513t + 1.728 \] \[ 0.513t = 9.452… \] \[ t \approx 18.4 \text{ mins} \]

✓ (2 marks)

✏ Working:

At \( t=0 \):

\[ N_A = |0-3| + 4 = 3 + 4 = 7 \] \[ N_B = 8 – |0-6| = 8 – 6 = 2 \]

Difference = \( 7 – 2 = 5 \) (thousands).

✓ (2 marks)

✏ Working:

The graph for A stops decreasing and starts increasing at its minimum point.

Vertex of \( N_A = |t-3|+4 \) is at \( t=3 \).

So \( T=3 \).

Reason: This is the minimum point of the curve where the trend reverses.

✓ (2 marks)

✏ Working:

Solve \( N_A = N_B \):

\[ |t-3| + 4 = 8 – |2t-6| \] \[ |t-3| + |2(t-3)| = 4 \] \[ |t-3| + 2|t-3| = 4 \] \[ 3|t-3| = 4 \] \[ |t-3| = \frac{4}{3} \]

So \( t-3 = \frac{4}{3} \) or \( t-3 = -\frac{4}{3} \).

\( t = 3 + 1.33 = 4.33 \) or \( t = 3 – 1.33 = 1.67 \).

Exact values: \( \frac{13}{3} \) and \( \frac{5}{3} \).

Looking at the graph, \( N_A \) is above \( N_B \) (the blue V is higher than the red inverted V) OUTSIDE these intersection points.

Range: \( \{t : 0 \leq t < \frac{5}{3}\} \cup \{t : t > \frac{13}{3}\} \)

✓ (5 marks)

✏ Working:

\[ (3+x)^{-2} = 3^{-2}(1 + \frac{x}{3})^{-2} = \frac{1}{9}(1 + \frac{x}{3})^{-2} \]

Expand \( (1+y)^{-2} = 1 – 2y + \frac{(-2)(-3)}{2}y^2 + \dots \)

\[ = 1 – 2(\frac{x}{3}) + 3(\frac{x}{3})^2 \] \[ = 1 – \frac{2x}{3} + \frac{3x^2}{9} = 1 – \frac{2x}{3} + \frac{x^2}{3} \]

Multiply by \( \frac{1}{9} \):

\[ \frac{1}{9} – \frac{2x}{27} + \frac{x^2}{27} \]

✓ (4 marks)

✏ Working:

Integrate \( 6x \times (\text{expansion}) \):

\[ \int_{0.2}^{0.4} 6x(\frac{1}{9} – \frac{2x}{27} + \frac{x^2}{27}) dx \] \[ = \int_{0.2}^{0.4} (\frac{2x}{3} – \frac{4x^2}{9} + \frac{2x^3}{9}) dx \] \[ = [\frac{x^2}{3} – \frac{4x^3}{27} + \frac{x^4}{18}]_{0.2}^{0.4} \]

Evaluate limits:

Upper (0.4): \( \frac{0.16}{3} – \frac{0.256}{27} + \frac{0.0256}{18} \approx 0.04527 \)

Lower (0.2): \( \frac{0.04}{3} – \frac{0.032}{27} + \frac{0.0016}{18} \approx 0.012237 \)

Difference \( \approx 0.03304 \).

✓ (4 marks)

✏ Working:

\[ \int \frac{6(u-3)}{u^2} du = \int (\frac{6}{u} – \frac{18}{u^2}) du \] \[ = 6\ln|u| + 18u^{-1} \] \[ = [ 6\ln|3+x| + \frac{18}{3+x} ]_{0.2}^{0.4} \]

Upper (0.4): \( 6\ln(3.4) + \frac{18}{3.4} = 6\ln(3.4) + \frac{90}{17} \)

Lower (0.2): \( 6\ln(3.2) + \frac{18}{3.2} = 6\ln(3.2) + \frac{90}{16} \)

Subtract:

\[ 6(\ln 3.4 – \ln 3.2) + (\frac{90}{17} – \frac{45}{8}) \] \[ = 6\ln(\frac{3.4}{3.2}) + (\frac{720 – 765}{136}) \] \[ = 6\ln(\frac{17}{16}) – \frac{45}{136} \]

✓ (5 marks)

✏ Working:

RHS: \( 8\sin 2\theta(1+\tan^2\theta) = 8(2\sin\theta\cos\theta)(\sec^2\theta) \)

\[ = 16\sin\theta\cos\theta \frac{1}{\cos^2\theta} = 16\frac{\sin\theta}{\cos\theta} = 16\tan\theta \]

So equation becomes:

\[ 2\tan\theta(8\cos\theta + 23\sin^2\theta) = 16\tan\theta \]

Rearrange:

\[ 2\tan\theta(8\cos\theta + 23\sin^2\theta – 8) = 0 \]

Using \( \tan\theta = \frac{\sin\theta}{\cos\theta} \):

\[ 2\frac{\sin\theta}{\cos\theta}(8\cos\theta + 23(1-\cos^2\theta) – 8) = 0 \]

Multiply by \( \cos^2\theta \) (to form \( \sin 2\theta \)):

\[ \sin 2\theta (23\cos^2\theta – 8\cos\theta – 15) = 0 \]

✓ (3 marks)

✏ Working:

1. \( \sin 2x = 0 \Rightarrow 2x = 720, 900, 1080 \Rightarrow x = 360, 450, 540 \). Note: \( x \neq 450 \) (undefined tan).

2. \( 23\cos^2 x – 8\cos x – 15 = 0 \)

\[ (23\cos x + 15)(\cos x – 1) = 0 \]

\( \cos x = 1 \Rightarrow x = 360 \).

\( \cos x = -15/23 \). Inverse cos gives \( 130.7^\circ \). In range \( 360 \leq x \leq 540 \), \( x = 360 + 130.7 = 490.7^\circ \).

✓ (4 marks)

✏ Working:

Assume \( \sin x – \cos x < 1 \).

Squaring both sides (careful with inequalities, but assume positive first or analyze terms):

\[ (\sin x – \cos x)^2 < 1 \] \[ \sin^2 x - 2\sin x\cos x + \cos^2 x < 1 \] \[ (\sin^2 x + \cos^2 x) - \sin 2x < 1 \] \[ 1 - \sin 2x < 1 \] \[ -\sin 2x < 0 \] \[ \sin 2x > 0 \]

However, \( x \) is obtuse (\( 90^\circ < x < 180^\circ \)).

So \( 2x \) is in the range \( 180^\circ < 2x < 360^\circ \) (3rd/4th quadrant).

In this range, \( \sin 2x \) MUST be negative.

This contradicts \( \sin 2x > 0 \).

Therefore, the assumption is false, and \( \sin x – \cos x \geq 1 \).

✓ (3 marks)