Pearson Edexcel GCE A-Level Pure Mathematics 2 (October 2021)

Step 1: Understanding the Formulas

We are dealing with an arithmetic series. We recall the formula for the \( n \)-th term:

\( u_n = a + (n-1)d \)

where \( a \) is the first term and \( d \) is the common difference.

Step 1: Choosing the Formula

We need the sum of the first 500 terms, \( S_{500} \). The formula for the sum of an arithmetic series is:

\( S_n = \frac{n}{2}[2a + (n-1)d] \)

We have \( n = 500 \), \( a = 16 \), and \( d = 0.4 \).

Step 1: Analyze the quadratic function

\( f(x) = 7 – 2x^2 \). Since \( x^2 \ge 0 \) for all real \( x \), it follows that \( -2x^2 \le 0 \).

Therefore, the maximum value occurs when \( x = 0 \), giving \( f(0) = 7 \). All other values will be less than 7.

Step 1: Calculate inner function \( f(1.8) \)

We work from the inside out. First find \( f(1.8) \).

Step 1: Set \( y = g(x) \) and rearrange

Let \( y = \frac{3x}{5x – 1} \). We need to solve for \( x \) in terms of \( y \).

We use the subtraction law of logarithms: \( \log_a A – \log_a B = \log_a \left(\frac{A}{B}\right) \).

Recall that \( \log_a x = n \) implies \( x = a^n \).

When \( \theta \) is small (in radians), we can assume:

• \( \sin x \approx x \)
• \( \cos x \approx 1 – \frac{x^2}{2} \)

Apply these to the terms in the expression.

For \( 4 \sin\left(\frac{\theta}{2}\right) \): Let \( x = \frac{\theta}{2} \). Then \( \sin\left(\frac{\theta}{2}\right) \approx \frac{\theta}{2} \).

For \( 3 \cos^2\theta \): We know \( \cos\theta \approx 1 – \frac{\theta^2}{2} \), so \( \cos^2\theta \approx \left(1 – \frac{\theta^2}{2}\right)^2 \).

Alternatively, use the identity \( \cos^2\theta = 1 – \sin^2\theta \approx 1 – \theta^2 \), or just expand the squared approximation and ignore higher powers (like \( \theta^4 \)).

A stationary point occurs when \( \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \).

Substitute \( x = 1 \) into the first derivative.

Method 1: Check second derivative

Substitute \( x = 1 \) into \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} \).

Step 1: Use geometry of a straight line

Angles on a straight line sum to \( \pi \) radians (180°).

The line \( AOF \) passes through the centre. The angles around \( O \) on this side are \( \angle AOB \), \( \angle BOC \), \( \angle COD \), \( \angle DOE \), and \( \angle EOF \).

However, from the diagram and the construction of sectors, \( O, B, C \) lie on a line and \( O, E, D \) lie on a line (radii of the sectors align to form the continuous shape boundaries). Wait, let’s verify. The shape is O-A-B-C… this implies the perimeter goes from B to C. Since C is at distance \( 2r \) and B is at distance \( r \), and they are connected, the simplest logical construction for a “fan” logo is that they share the same radial angle.

Thus, the total angle \( \pi \) is split into three parts: the left sector angle, the middle sector angle \( \theta \), and the right sector angle.

Since sector \( OFE \) is congruent to \( OAB \), let \( \angle AOB = \alpha \). Then \( \angle EOF = \alpha \).

Therefore: \( \alpha + \theta + \alpha = \pi \).

Step 1: Identify components

The total area consists of:

• Area of sector \( OAB \) (radius \( r \))
• Area of sector \( OFE \) (radius \( r \))
• Area of sector \( ODC \) (radius \( 2r \))? Wait.

Let’s look at the shape. The logo is the shaded region or the bounded shape. It includes the two small wings and the large top part. Actually, looking at the boundaries, the region is composed of the two small sectors plus the large sector? No.

Looking at the connectivity: O-A-B-C-D-E-F-O. The area enclosed by this perimeter.

It consists of:

1. Sector \( OAB \) 2. Sector \( OFE \) 3. The region bounded by \( BC \), \( DE \), arc \( CD \) and the lines \( OB, OE \).

Actually, it’s simpler. It’s the large sector \( ODC \) (which goes down to the origin) PLUS the two side sectors? No, that would overlap.

Let’s assume the standard composition: The shape is the union of Sector OAB, Sector OFE, and Sector ODC. But do they overlap?

Based on the angles: Sector OAB is from angle \( \pi \) to \( \pi – \frac{\pi-\theta}{2} \). Sector ODC is from angle … to ….

Actually, the “Sector ODC” has radius \( 2r \). It covers the middle angle \( \theta \).

The “Sector OAB” covers the side angle. They do not overlap in angle.

So Area = Area(Sector OAB) + Area(Sector OFE) + Area(Sector ODC).

Step 1: Identify boundary lengths

Perimeter = \( OA + \text{arc } AB + BC + \text{arc } CD + DE + \text{arc } EF + FO \)

• \( OA = r \)
• \( FO = r \)
• \( BC = 2r – r = r \) (Since \( OC=2r, OB=r \))
• \( DE = 2r – r = r \)
• Arc \( AB = r \times \text{angle} = r \left( \frac{\pi – \theta}{2} \right) \)
• Arc \( EF = r \left( \frac{\pi – \theta}{2} \right) \)
• Arc \( CD = (2r) \times \theta = 2r\theta \)

Step 1: Find the gradient function \( \frac{\mathrm{d}y}{\mathrm{d}x} \)

Differentiate the equation of the curve.

Step 1: Find y-intercepts

The line \( l \) meets the y-axis when \( x = 0 \).

For line \( y = 2x – 23 \): at \( x=0, y = -23 \).

For curve \( C \): at \( x=0, y = -23 \).

Step 1: Setup the Integral

The region \( R \) is bounded by the curve and the line from \( x = 0 \) to \( x = 5 \).

Looking at the graph (or checking values), in this interval, the curve is above the line? Let’s check \( x=1 \).

Curve: \( 1 – 10 + 27 – 23 = -5 \).

Line: \( 2(1) – 23 = -21 \).

\(-5 > -21\), so Curve is on top.

Area \( = \int_{0}^{5} (\text{Curve} – \text{Line}) \, \mathrm{d}x \).

Step 1: Differentiate term by term w.r.t \( x \)

Remember that \( y \) is a function of \( x \), so we use the chain rule for \( y \) terms and product rule for \( xy \).

• \( px^3 \rightarrow 3px^2 \)
• \( qxy \rightarrow q(1 \cdot y + x \frac{\mathrm{d}y}{\mathrm{d}x}) \) (Product Rule)
• \( 3y^2 \rightarrow 6y \frac{\mathrm{d}y}{\mathrm{d}x} \) (Chain Rule)
• \( 26 \rightarrow 0 \)

Step 1: Use Point \( P(-1, -4) \) in Curve Equation

Since P is on the curve, its coordinates verify the equation.

Step 2: Use Gradient of Normal

The normal equation is \( 19x + 26y + 123 = 0 \).

Rearrange to find gradient of normal \( m_N \):

\[ 26y = -19x – 123 \] \[ y = -\frac{19}{26}x – \dots \]

So \( m_N = -\frac{19}{26} \).

The tangent gradient \( m_T \) is perpendicular to normal, so \( m_T = \frac{26}{19} \).

✓ (B1)

Step 3: Equate derivative to tangent gradient

Substitute \( x = -1, y = -4 \) into derivative from part (a).

Step 4: Solve Simultaneous Equations

(1) \( p = 4q + 22 \)

Substitute into (2):

Let’s write out the first few terms to identify the pattern.

Note that \( \cos(180n)^\circ \) alternates between 1 and -1.

• \( n=1 \): \( \cos(180) = -1 \) (Note: summation starts at \( n=2 \))
• \( n=2 \): \( \cos(360) = 1 \)
• \( n=3 \): \( \cos(540) = -1 \)
• \( n=4 \): \( \cos(720) = 1 \)

The term is \( \left(\frac{3}{4}\right)^n \times (-1)^n \) or \( \left(\frac{3}{4}\right)^n \times 1 \) depending on \( n \).

Wait, \( \cos(180n) = (-1)^n \).

Geometric Series Sum to Infinity: \( S_\infty = \frac{a}{1-r} \).

First term (at \( n=2 \)):

\[ a = \left(-\frac{3}{4}\right)^2 = \frac{9}{16} \]

Common ratio:

\[ r = -\frac{3}{4} \]

Check convergence: \( |r| < 1 \), so it converges.

Step 1: Take logs of both sides

We start with \( T = al^b \).

Step 1: Relate to Line Equation

The equation is in the form \( Y = mX + c \), where:

• \( Y = \log_{10} T \)
• \( X = \log_{10} l \)
• Gradient \( m = b \)
• Y-intercept \( c = \log_{10} a \)

Step 2: Calculate Gradient \( b \)

Using points \( (-0.7, 0) \) and \( (0.21, 0.45) \):

Step 3: Calculate Intercept \( \log_{10} a \)

Use equation \( Y = bX + c \) and point \( (-0.7, 0) \):

In the model \( T = al^b \), if \( l = 1 \), then \( T = a(1)^b = a \).

Step 1: Transformations

Start with \( y = |2x – 3k| \).

• \( y = -|2x – 3k| \): Reflection in the \( x \)-axis (inverted V shape).
• \( y = k – |2x – 3k| \): Translation up by \( k \).

The original vertex was at \( 2x – 3k = 0 \Rightarrow x = \frac{3k}{2} \), \( y = 0 \).

The new vertex (maximum) is at \( x = \frac{3k}{2} \), \( y = k \).

Step 2: Intercepts

y-intercept (set \( x = 0 \)):

\[ y = k – |2(0) – 3k| = k – |-3k| = k – 3k = -2k \]

Coordinate: \( (0, -2k) \).

x-intercepts (set \( y = 0 \)):

\[ 0 = k – |2x – 3k| \Rightarrow |2x – 3k| = k \] \[ 2x – 3k = k \quad \text{or} \quad 2x – 3k = -k \] \[ 2x = 4k \Rightarrow x = 2k \] \[ 2x = 2k \Rightarrow x = k \]

Coordinates: \( (k, 0) \) and \( (2k, 0) \).

Solve \( k – |2x – 3k| > x – k \).

Rearrange: \( 2k – x > |2x – 3k| \).

This splits into two linear inequalities or can be solved by squaring (careful with signs) or considering cases.

Case 1: \( 2x – 3k \ge 0 \) (i.e., \( x \ge 1.5k \))

\[ 2k – x > 2x – 3k \] \[ 5k > 3x \Rightarrow x < \frac{5k}{3} \]

Combined with \( x \ge 1.5k \): \( 1.5k \le x < \frac{5k}{3} \).

Case 2: \( 2x – 3k < 0 \) (i.e., \( x < 1.5k \))

\[ 2k – x > -(2x – 3k) \] \[ 2k – x > -2x + 3k \] \[ x > k \]

Combined with \( x < 1.5k \): \( k < x < 1.5k \).

Equation: \( y = 3 – 5f\left(\frac{1}{2}x\right) \).

Let’s track the maximum point of \( f(x) \), which is \( (1.5k, k) \).

1. Inside transformation \( f(\frac{1}{2}x) \): Horizontal stretch scale factor 2.

\( x \)-coordinate becomes \( 2 \times 1.5k = 3k \).

2. Outside transformation \( \times -5 \): Vertical stretch s.f. 5 and reflection in x-axis.

\( y \)-coordinate becomes \( -5 \times k = -5k \). (Max becomes Min).

3. Outside transformation \( +3 \): Shift up 3.

\( y \)-coordinate becomes \( -5k + 3 \).

So the maximum point of \( f \) becomes the minimum point of the new graph.

Step 1: Differentiate Substitution

\( u = 1 + x^{1/2} \)

\[ \frac{\mathrm{d}u}{\mathrm{d}x} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}} \] \[ \mathrm{d}x = 2\sqrt{x} \, \mathrm{d}u \]

Since \( \sqrt{x} = u – 1 \), we have \( \mathrm{d}x = 2(u-1) \, \mathrm{d}u \).

Step 2: Change Limits

When \( x = 0 \), \( u = 1 + \sqrt{0} = 1 \). So \( p = 1 \).

When \( x = 16 \), \( u = 1 + \sqrt{16} = 1 + 4 = 5 \). So \( q = 5 \).

Step 3: Substitute into Integral

Also \( x = (u-1)^2 \).

\[ \int \frac{x}{u} (2(u-1) \, \mathrm{d}u) \] \[ \int \frac{(u-1)^2}{u} \cdot 2(u-1) \, \mathrm{d}u \] \[ \int \frac{2(u-1)^3}{u} \, \mathrm{d}u \]

✓ (M1) (A1)

Step 1: Expand and Simplify

\[ (u-1)^3 = u^3 – 3u^2 + 3u – 1 \] \[ \frac{2(u^3 – 3u^2 + 3u – 1)}{u} = 2\left( u^2 – 3u + 3 – \frac{1}{u} \right) \] \[ = 2u^2 – 6u + 6 – \frac{2}{u} \]

Step 1: Differentiate \( x \) and \( y \) w.r.t \( \theta \)

\( x = \sin 2\theta \)

\[ \frac{\mathrm{d}x}{\mathrm{d}\theta} = 2\cos 2\theta \]

\( y = \text{cosec}^3 \theta = (\sin \theta)^{-3} \)

Chain rule: \( -3(\sin \theta)^{-4} (\cos \theta) \)

\[ \frac{\mathrm{d}y}{\mathrm{d}\theta} = -3 \text{cosec}^4 \theta \cos \theta = -3 \text{cosec}^3 \theta \cot \theta \]

✓ (B1) (M1)

Step 2: Apply Formula

\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y/\mathrm{d}\theta}{\mathrm{d}x/\mathrm{d}\theta} \] \[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{-3 \text{cosec}^3 \theta \cot \theta}{2 \cos 2\theta} \]

✓ (A1)

Step 1: Find \( \theta \) when \( y = 8 \)

\[ \text{cosec}^3 \theta = 8 \] \[ \text{cosec} \theta = \sqrt[3]{8} = 2 \] \[ \sin \theta = \frac{1}{2} \]

Since \( 0 < \theta < \frac{\pi}{2} \), \( \theta = \frac{\pi}{6} \) (30 degrees).

✓ (M1)

Step 2: Substitute \( \theta \) into gradient expression

At \( \theta = 30^\circ \):

• \( \text{cosec} 30^\circ = 2 \)
• \( \cot 30^\circ = \sqrt{3} \)
• \( \cos 60^\circ = 0.5 \)

\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{-3(2)^3 (\sqrt{3})}{2(0.5)} \] \[ = \frac{-3(8)(\sqrt{3})}{1} \] \[ = -24\sqrt{3} \]

✓ (M1) (A1)

Step 1: Rate of Volume Change

\( \frac{\mathrm{d}V}{\mathrm{d}t} = \text{Rate In} – \text{Rate Out} \)

\[ \frac{\mathrm{d}V}{\mathrm{d}t} = 0.48 – 0.1h \]

Step 2: Relate Volume to Height

Volume of cuboid = Area of Base × Height.

Base Area = \( 8 \times 3 = 24 \) m².

\( V = 24h \).

Differentiating w.r.t \( h \): \( \frac{\mathrm{d}V}{\mathrm{d}h} = 24 \).

Step 3: Chain Rule

\[ \frac{\mathrm{d}h}{\mathrm{d}t} = \frac{\mathrm{d}h}{\mathrm{d}V} \times \frac{\mathrm{d}V}{\mathrm{d}t} \] \[ \frac{\mathrm{d}h}{\mathrm{d}t} = \frac{1}{24} (0.48 – 0.1h) \]

✓ (M1)

Step 4: Rearrange to match target form

\[ 24 \frac{\mathrm{d}h}{\mathrm{d}t} = 0.48 – 0.1h \]

Multiply by 50 to clear decimals:

\[ 24(50) \frac{\mathrm{d}h}{\mathrm{d}t} = 0.48(50) – 0.1h(50) \] \[ 1200 \frac{\mathrm{d}h}{\mathrm{d}t} = 24 – 5h \]

✓ (A1)

Step 1: Separate Variables

\[ \frac{1200}{24 – 5h} \mathrm{d}h = \mathrm{d}t \]

Step 2: Integrate

\[ \int \frac{1200}{24 – 5h} \mathrm{d}h = \int 1 \mathrm{d}t \] \[ 1200 \left( \frac{\ln|24 – 5h|}{-5} \right) = t + C \] \[ -240 \ln|24 – 5h| = t + C \]

✓ (M1) (A1)

Step 3: Use Initial Conditions

At \( t = 0 \), \( h = 2 \).

\[ -240 \ln|24 – 5(2)| = 0 + C \] \[ C = -240 \ln(14) \]

Step 4: Rearrange for \( h \)

\[ -240 \ln(24 – 5h) = t – 240 \ln 14 \]

Divide by -240:

\[ \ln(24 – 5h) = -\frac{t}{240} + \ln 14 \]

Exponentiate:

\[ 24 – 5h = e^{-\frac{t}{240} + \ln 14} \] \[ 24 – 5h = e^{\ln 14} \cdot e^{-\frac{t}{240}} \] \[ 24 – 5h = 14 e^{-\frac{t}{240}} \] \[ 5h = 24 – 14 e^{-\frac{t}{240}} \] \[ h = 4.8 – 2.8 e^{-\frac{t}{240}} \]

✓ (dM1) (A1)

Comparing to \( h = A + Be^{-kt} \):

\( A = 4.8 \), \( B = -2.8 \), \( k = \frac{1}{240} \).

The maximum height of the tank is 5m.

Look at the equation for \( h \): \( h = 4.8 – 2.8 e^{-kt} \).

As \( t \to \infty \), \( e^{-kt} \to 0 \).

So \( h \to 4.8 \).

Since the limiting height is 4.8m, which is less than 5m, the tank will never become full.

Equation: \( H = 3 + 2(2\cos(0.5t) – \sin(0.5t)) \).

Using part (a), \( 2\cos(0.5t) – \sin(0.5t) = \sqrt{5} \cos(0.5t + 0.464) \).

\[ H = 3 + 2(\sqrt{5} \cos(0.5t + 0.464)) \] \[ H = 3 + 2\sqrt{5} \cos(0.5t + 0.464) \]

(i) Max Height:

Max value of cosine is 1.

\[ H_{\text{max}} = 3 + 2\sqrt{5} \approx 7.47 \text{ m} \]

(ii) Time of Max:

Cosine is 1 when angle is 0, \( 2\pi \), etc.

\[ 0.5t + 0.464 = 0 \] (Negative time, reject) \[ 0.5t + 0.464 = 2\pi \] \[ 0.5t = 2\pi – 0.464 \] \[ 0.5t = 5.819 \] \[ t = 11.6 \text{ s} \]

✓ (M1) (A1)

Below water means \( H < 0 \).

\[ 3 + 2\sqrt{5} \cos(0.5t + 0.464) < 0 \] \[ \cos(0.5t + 0.464) < -\frac{3}{2\sqrt{5}} \] \[ \cos(0.5t + 0.464) < -0.6708 \]

Find Critical Angles:

\[ \arccos(-0.6708) = 2.306 \text{ rads} \]

Cos is negative in quadrants 2 and 3.

Angles: \( 2.306 \) and \( 2\pi – 2.306 = 3.977 \).

So \( 2.306 < 0.5t + 0.464 < 3.977 \).

Solve for \( t \):

\[ 1.842 < 0.5t < 3.513 \] \[ 3.684 < t < 7.026 \]

Duration \( T = 7.026 – 3.684 = 3.34 \text{ s} \).

✓ (M1) (A1)