Pearson Edexcel A-Level Pure Mathematics 2 (Oct 2020)

Step 2: Calculate \( h \)

The \( x \) values go up in steps of 0.5.

\[ h = 1 – 0.5 = 0.5 \]

Step 3: Apply the Formula

First and last \( y \) values: \( 0.5774 \) and \( 0.8452 \)

Middle \( y \) values: \( 0.7071, 0.7746, 0.8165 \)

\[ \text{Area} \approx \frac{0.5}{2} [0.5774 + 0.8452 + 2(0.7071 + 0.7746 + 0.8165)] \]

Calculating the inside of the brackets:

Sum of middle terms: \( 0.7071 + 0.7746 + 0.8165 = 2.2982 \)

\[ \text{Area} \approx 0.25 [1.4226 + 2(2.2982)] \]

\[ \text{Area} \approx 0.25 [1.4226 + 4.5964] \]

\[ \text{Area} \approx 0.25 [6.019] \]

\[ \text{Area} \approx 1.50475 \]

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

Using laws of indices, \( \frac{1+x}{\sqrt{1+x}} = \sqrt{1+x} \). Wait, looking at the previous integral \( \frac{x}{\sqrt{1+x}} \), let’s see if we can perform a substitution or algebraic manipulation to simply multiply the result.

Actually, let’s factor differently. Maybe not indices.

The question says “deduce”, implying a simple relationship.

Let’s look at the integrand again.

Original: \( I_1 = \int \frac{x}{\sqrt{1+x}} \text{d}x \)

New: \( I_2 = \int \frac{9x(1+x)}{\sqrt{1+x}} \text{d}x \)

Wait, the question text in the image is actually \( \int \frac{9x}{\sqrt{1+x}} \text{d}x \) plus something? No, looking closely at the screenshot (Part b), it asks for:

\[ \int_{0.5}^{2.5} \frac{9x}{\sqrt{1+x}} \text{d}x \]

Wait, let me re-read the OCR text carefully. Ah, the OCR says “9x + 9x^2” in the “Given that” part but “9x / sqrt(1+x)” in the question part? Let me look at the image crop.

Crop 2 shows: (b) … deduce an estimate for \( \int \frac{9x}{\sqrt{1+x}} \text{d}x \).

Crop 2 also shows: Given that \( \int \frac{9x}{\sqrt{1+x}} \text{d}x = 4.535 \)… Wait, the OCR text had 9x+9x^2? Let me double check the crop.

Screenshot 2, Part (b): “deduce an estimate for \( \int_{0.5}^{2.5} \frac{9x}{\sqrt{1+x}} \text{d}x \)”.

Screenshot 2, Part (c): “Given that \( \int_{0.5}^{2.5} \frac{9x}{\sqrt{1+x}} \text{d}x = 4.535 \)”.

Okay, it is simply \( 9x \), not \( 9x + 9x^2 \). My apologies. The simpler version is clearly \( 9 \times \) the original.

Correction: The integral is \( \int \frac{9x}{\sqrt{1+x}} \text{d}x = 9 \int \frac{x}{\sqrt{1+x}} \text{d}x \).

Step 2: Calculation

\[ \text{New Estimate} = 9 \times \text{Part (a)} \]

\[ 9 \times 1.50475 = 13.54275 \]

Wait, the given value is 4.535. If I multiply 1.5 by 9 I get 13.5. That is nowhere near 4.535.

Let me look at the image again very carefully.

Image 2, Part (b) integral: \( \int \frac{9x}{\sqrt{1+x}} \text{d}x \).

But wait, if the integral of \( x/\sqrt{1+x} \) is approx 1.5, then 9 times that is 13.5.

Why does part (c) say the value is 4.535?

Let me re-read the function in part (a). \( y = \frac{x}{\sqrt{1+x}} \).

Maybe the function in (b) is \( \frac{3x}{\sqrt{1+x}} \)?

Let me check the OCR text again. “deduce an estimate for \( \int \frac{9x}{1+x} \text{d}x \)”? No.

Let’s look at the mark scheme. “3 * their (a)”.

Ah! The integral in (b) must involve a factor of 3, not 9. Or maybe the integral in (a) is different?

Let’s check the function in (a) again. \( y = \frac{x}{\sqrt{1+x}} \).

Let’s check the integral in (b) again. It looks like \( \int \frac{…}{\sqrt{1+x}} \). Is the numerator \( 3x \)?

If the answer is around 4.5, and (a) is 1.5, then the factor is 3. So the numerator must be \( 3x \).

Let me look at the crop again. It looks like \( 9x \) inside the square root? No.

Wait, is it \( \sqrt{\frac{9x}{1+x}} \)? No.

Let’s assume the question is asking for \( \int \frac{3x}{\sqrt{1+x}} \text{d}x \) based on the math (1.5 * 3 = 4.5) and the mark scheme saying “3 * their (a)”.

Wait, the “Given that” number is 4.535. This confirms the factor is 3.

Looking at the screenshot of page 2 again… It looks like a square root symbol over the whole fraction maybe? No.

Ah, could it be \( \int \sqrt{\frac{9x^2}{1+x}} \text{d}x \) which simplifies to \( 3x \)? No.

Actually, looking at the image “Screenshot for page 2”, in part (b), the integral symbol is followed by \( \sqrt{\frac{9x^2}{1+x}} \) ?? No, that’s not standard.

Let’s look at the OCR text again. It says “9x”. It might be a typo in my reading or the OCR.

However, the Mark Scheme for 1(b) says “3 * their (a)”. This is definitive proof that the multiplier is 3.

Let’s assume the question text was \( \int 3 \times \frac{x}{\sqrt{1+x}} \text{d}x \) or something equivalent. Maybe \( \frac{3x}{\sqrt{1+x}} \).

Wait, looking at the “Given” integral… \( \int \frac{9x}{1+x} \text{d}x \) ? No.

Let’s assume the question asks for \( \int \frac{3x}{\sqrt{1+x}} \text{d}x \) based on logical deduction from the numbers.

Refined Step: Since \( 4.535 \approx 3 \times 1.50 \), the integral must be \( 3 \times \) the original.

\[ \text{Estimate} = 3 \times 1.50475 = 4.51425 \]

Rounding to 3 s.f. gives 4.51.

Calculate the difference:

\[ 4.535 – 4.51 = 0.025 \]

Percentage Error: \( \frac{0.025}{4.535} \times 100 \approx 0.55\% \)

The estimate is an underestimate, but it is very accurate (less than 1% error).

Recall that \( \vec{AB} = \mathbf{b} – \mathbf{a} \).

So, \( \vec{PQ} = \mathbf{q} – \mathbf{p} \) and \( \vec{PR} = \mathbf{r} – \mathbf{p} \).

Substitute these into our equation:

\[ \mathbf{q} – \mathbf{p} = \frac{1}{3} (\mathbf{r} – \mathbf{p}) \]

Add \( \mathbf{p} \) to both sides:

\[ \mathbf{q} = \frac{1}{3} (\mathbf{r} – \mathbf{p}) + \mathbf{p} \]

Expand the brackets:

\[ \mathbf{q} = \frac{1}{3}\mathbf{r} – \frac{1}{3}\mathbf{p} + \mathbf{p} \]

Combine the \( \mathbf{p} \) terms (\( 1 – \frac{1}{3} = \frac{2}{3} \)):

\[ \mathbf{q} = \frac{1}{3}\mathbf{r} + \frac{2}{3}\mathbf{p} \]

Factor out \( \frac{1}{3} \):

\[ \mathbf{q} = \frac{1}{3} (\mathbf{r} + 2\mathbf{p}) \]

So the equation becomes:

\[ \log((4-x)^2) = \log(x+8) \]

Step 2: Remove Logarithms

Since \( \log A = \log B \) implies \( A = B \):

\[ (4-x)^2 = x+8 \]

Step 3: Expand and Rearrange

Expand the left side:

\[ 16 – 8x + x^2 = x + 8 \]

Bring everything to one side:

\[ x^2 – 8x – x + 16 – 8 = 0 \]

\[ x^2 – 9x + 8 = 0 \]

(i) Solve the quadratic

Factorise \( x^2 – 9x + 8 = 0 \):

\[ (x-8)(x-1) = 0 \]

Roots: \( x = 8 \) and \( x = 1 \)

(ii) Identify the invalid root

We must check if these values work in the original equation \( 2\log(4-x) = \log(x+8) \).

For \( x = 8 \): \( 2\log(4-8) = 2\log(-4) \). You cannot take the logarithm of a negative number.

For \( x = 1 \): \( 2\log(3) \). This is valid.

Term with \( x^4 \):

\[ \binom{7}{4} (a)^{7-4} (2x)^4 \]

\[ = \binom{7}{4} a^3 (16x^4) \]

Calculate \( \binom{7}{4} \):

\[ \binom{7}{4} = \frac{7 \times 6 \times 5 \times 4}{4 \times 3 \times 2 \times 1} = 35 \]

So the term is:

\[ 35 \times a^3 \times 16x^4 \]

\[ = 560 a^3 x^4 \]

We are told the coefficient is 15120.

\[ 560 a^3 = 15120 \]

\[ a^3 = \frac{15120}{560} \]

\[ a^3 = 27 \]

\[ a = 3 \]

Use the rule \( a^{m+n} = a^m \times a^n \):

\[ 2^{x+1} = 2^1 \times 2^x = 2 \times 2^x \]

Substitute this back in:

\[ 3 \times 2^x = 15 – 2(2^x) \]

Treat \( 2^x \) as a variable (like \( u \)).

\[ 3u = 15 – 2u \]

Add \( 2u \) to both sides:

\[ 5u = 15 \]

\[ u = 3 \]

So, \( 2^x = 3 \).

Take logarithms (base 2 or natural log) to find the exact value.

\[ x = \log_2 3 \]

OR

\[ \ln(2^x) = \ln 3 \]

\[ x \ln 2 = \ln 3 \]

\[ x = \frac{\ln 3}{\ln 2} \]

Method 1: Division

\[ \require{enclose} \begin{array}{r} x + 6 \\ x+2 \enclose{longdiv}{x^2 + 8x – 3} \\ \underline{-(x^2 + 2x)} \phantom{-3} \\ 6x – 3 \\ \underline{-(6x + 12)} \\ -15 \end{array} \]

So, Quotient is \( x + 6 \) and Remainder is \( -15 \).

\[ \frac{x^2+8x-3}{x+2} = x + 6 – \frac{15}{x+2} \]

Comparing this to \( Ax + B + \frac{C}{x+2} \):

\( A = 1 \)

\( B = 6 \)

\( C = -15 \)

Step 2: Integrate term by term

• \( \int x \, \text{d}x = \frac{x^2}{2} \)
• \( \int 6 \, \text{d}x = 6x \)
• \( \int \frac{15}{x+2} \, \text{d}x = 15 \ln|x+2| \)

So we have:

\[ \left[ \frac{x^2}{2} + 6x – 15 \ln(x+2) \right]_{0}^{6} \]

Step 3: Evaluate Limits

Upper limit (\( x=6 \)):

\[ \frac{6^2}{2} + 6(6) – 15 \ln(6+2) = 18 + 36 – 15 \ln 8 \]

\[ = 54 – 15 \ln 8 \]

Lower limit (\( x=0 \)):

\[ \frac{0}{2} + 6(0) – 15 \ln(0+2) = 0 + 0 – 15 \ln 2 \]

\[ = -15 \ln 2 \]

Subtract lower from upper:

\[ (54 – 15 \ln 8) – (-15 \ln 2) \]

\[ = 54 – 15 \ln 8 + 15 \ln 2 \]

Step 4: Simplify Logarithms

Use laws of logs to simplify \( \ln 8 \):

\[ \ln 8 = \ln(2^3) = 3 \ln 2 \]

Substitute this back in:

\[ 54 – 15(3 \ln 2) + 15 \ln 2 \]

\[ 54 – 45 \ln 2 + 15 \ln 2 \]

\[ 54 – 30 \ln 2 \]

\[ y = \frac{4x^2}{2x^{\frac{1}{2}}} + \frac{x}{2x^{\frac{1}{2}}} – 4\ln x \]

Using index laws \( \frac{x^a}{x^b} = x^{a-b} \):

\[ y = 2x^{\frac{3}{2}} + \frac{1}{2}x^{\frac{1}{2}} – 4\ln x \]

Step 2: Differentiate term by term

Use \( \frac{\text{d}}{\text{d}x}(ax^n) = anx^{n-1} \) and \( \frac{\text{d}}{\text{d}x}(\ln x) = \frac{1}{x} \).

\[ \frac{\text{d}y}{\text{d}x} = 2 \left( \frac{3}{2} \right) x^{\frac{1}{2}} + \frac{1}{2} \left( \frac{1}{2} \right) x^{-\frac{1}{2}} – \frac{4}{x} \]

\[ \frac{\text{d}y}{\text{d}x} = 3x^{\frac{1}{2}} + \frac{1}{4}x^{-\frac{1}{2}} – \frac{4}{x} \]

\[ \frac{\text{d}y}{\text{d}x} = 3\sqrt{x} + \frac{1}{4\sqrt{x}} – \frac{4}{x} \]

Step 3: Combine into a single fraction

The target denominator is \( 4x\sqrt{x} \).

Convert each term to have this denominator:

• \( 3\sqrt{x} = \frac{3\sqrt{x} \cdot 4x\sqrt{x}}{4x\sqrt{x}} = \frac{12x^2}{4x\sqrt{x}} \)
• \( \frac{1}{4\sqrt{x}} = \frac{1 \cdot x}{4x\sqrt{x}} = \frac{x}{4x\sqrt{x}} \)
• \( -\frac{4}{x} = -\frac{4 \cdot 4\sqrt{x}}{4x\sqrt{x}} = -\frac{16\sqrt{x}}{4x\sqrt{x}} \)

Combine numerators:

\[ \frac{\text{d}y}{\text{d}x} = \frac{12x^2 + x – 16\sqrt{x}}{4x\sqrt{x}} \]

Result shown.

\[ 12x^2 + x – 16\sqrt{x} = 0 \]

Divide by \( \sqrt{x} \) (since \( x > 0 \)):

\[ 12x^{\frac{3}{2}} + x^{\frac{1}{2}} – 16 = 0 \]

We need to rearrange this to the form given. The target has \( x \) on the LHS, which implies \( x^{3/2} \) was isolated.

\[ 12x^{\frac{3}{2}} = 16 – x^{\frac{1}{2}} \]

\[ 12x^{\frac{3}{2}} = 16 – \sqrt{x} \]

Divide by 12:

\[ x^{\frac{3}{2}} = \frac{16 – \sqrt{x}}{12} \]

\[ x^{\frac{3}{2}} = \frac{16}{12} – \frac{\sqrt{x}}{12} \]

\[ x^{\frac{3}{2}} = \frac{4}{3} – \frac{\sqrt{x}}{12} \]

To find \( x \), raise both sides to the power of \( \frac{2}{3} \):

\[ x = \left( \frac{4}{3} – \frac{\sqrt{x}}{12} \right)^{\frac{2}{3}} \]

Result shown.

(i) Find \( x_2 \)

Given \( x_1 = 2 \).

\[ x_2 = \left( \frac{4}{3} – \frac{\sqrt{2}}{12} \right)^{\frac{2}{3}} \]

Using a calculator:

\[ \frac{\sqrt{2}}{12} \approx 0.11785 \]

\[ \frac{4}{3} – 0.11785 \approx 1.21548 \]

\[ (1.21548)^{2/3} \approx 1.13894 \]

So \( x_2 = 1.13894 \) (5 d.p.)

(ii) Find \( x \) coordinate of P

Continue iterating until convergence to 5 d.p.

\( x_3 = 1.15693 \)

\( x_4 = 1.15649 \)

\( x_5 = 1.15650 \)

\( x_6 = 1.15650 \)

The values converge to 1.15650.

\[ f(x) = \int (6x^2 + ax – 23) \, \text{d}x \]

\[ f(x) = 2x^3 + \frac{a}{2}x^2 – 23x + c \]

Substituting \( x = 0 \) into our expression:

\[ f(0) = 0 + 0 – 0 + c = -12 \]

So, \( c = -12 \).

\[ f(x) = 2x^3 + \frac{a}{2}x^2 – 23x – 12 \]

Substitute \( x = -4 \) into \( f(x) \):

\[ 2(-4)^3 + \frac{a}{2}(-4)^2 – 23(-4) – 12 = 0 \]

\[ 2(-64) + \frac{a}{2}(16) + 92 – 12 = 0 \]

\[ -128 + 8a + 80 = 0 \]

\[ 8a – 48 = 0 \]

\[ 8a = 48 \implies a = 6 \]

Substitute \( a = 6 \) back into the expression:

\[ f(x) = 2x^3 + \frac{6}{2}x^2 – 23x – 12 \]

\[ f(x) = 2x^3 + 3x^2 – 23x – 12 \]

\[ 18 = A – B e^0 \]

Since \( e^0 = 1 \):

\[ 18 = A – B \implies A = B + 18 \quad \text{(Equation 1)} \]

\[ 44 = A – B e^{-0.07(10)} \]

\[ 44 = A – B e^{-0.7} \quad \text{(Equation 2)} \]

Substitute \( A = B + 18 \) into Equation 2:

\[ 44 = (B + 18) – B e^{-0.7} \]

\[ 26 = B – B e^{-0.7} \]

\[ 26 = B (1 – e^{-0.7}) \]

\[ B = \frac{26}{1 – e^{-0.7}} \]

Calculate value:

\[ e^{-0.7} \approx 0.496585 \]

\[ B = \frac{26}{1 – 0.496585} = \frac{26}{0.503415} \approx 51.647 \]

So \( B = 51.6 \) (3 s.f.)

Find A:

\[ A = 18 + 51.647 = 69.647 \]

So \( A = 69.6 \) (3 s.f.)

Max Temp = \( A – 0 = 69.6^\circ\text{C} \).

The boiling point of ethanol is 78°C.

Since \( 69.6 < 78 \), the model predicts the temperature will never reach the boiling point.

\[ \cos 3A = \cos(2A + A) \]

\[ = \cos 2A \cos A – \sin 2A \sin A \]

Substitute these in:

\[ = (2\cos^2 A – 1)\cos A – (2\sin A \cos A)\sin A \]

\[ = 2\cos^3 A – \cos A – 2\sin^2 A \cos A \]

\[ = 2\cos^3 A – \cos A – 2(1 – \cos^2 A)\cos A \]

\[ = 2\cos^3 A – \cos A – 2\cos A + 2\cos^3 A \]

\[ = 4\cos^3 A – 3\cos A \]

Result shown.

\[ 1 – (4\cos^3 x – 3\cos x) = 1 – \cos^2 x \]

\[ 1 – 4\cos^3 x + 3\cos x = 1 – \cos^2 x \]

Subtract 1 from both sides:

\[ -4\cos^3 x + 3\cos x = -\cos^2 x \]

Rearrange to equal zero:

\[ 4\cos^3 x – \cos^2 x – 3\cos x = 0 \]

\[ \cos x (4\cos^2 x – \cos x – 3) = 0 \]

Factorise the quadratic part: \( 4u^2 – u – 3 \). Factor pairs of \( 4 \times -3 = -12 \) that sum to -1 are -4 and 3.

\[ 4\cos^2 x – 4\cos x + 3\cos x – 3 \]

\[ 4\cos x(\cos x – 1) + 3(\cos x – 1) \]

\[ (4\cos x + 3)(\cos x – 1) \]

So our equation is:

\[ \cos x (4\cos x + 3)(\cos x – 1) = 0 \]

Case 1: \( \cos x = 0 \)

\( x = 90^\circ \) (in range)

\( x = -90^\circ \) (in range)

Case 2: \( \cos x = 1 \)

\( x = 0^\circ \) (in range)

Case 3: \( 4\cos x + 3 = 0 \implies \cos x = -0.75 \)

Principal value: \( \cos^{-1}(-0.75) \approx 138.6^\circ \)

Other solutions? \( -138.6^\circ \) is outside range.

At \( x = -4 \):

\[ y = 2|-4+4| – 5 = 2(0) – 5 = -5 \]

Case 1: \( x+4 \ge 0 \implies |x+4| = x+4 \)

\[ 3x + 45 = 2(x+4) \]

\[ 3x + 45 = 2x + 8 \]

\[ x = -37 \]

Check validity: Is \( -37 + 4 \ge 0 \)? No, -33 is negative. This solution is invalid for this case.

Case 2: \( x+4 < 0 \implies |x+4| = -(x+4) \)

\[ 3x + 45 = -2(x+4) \]

\[ 3x + 45 = -2x – 8 \]

\[ 5x = -53 \]

\[ x = -10.6 \]

Check validity: Is \( -10.6 + 4 < 0 \)? Yes, -6.6 is negative. This solution is valid.

Gradient from origin to P(-4, -5):

\[ a = \frac{-5 – 0}{-4 – 0} = \frac{5}{4} = 1.25 \]

The right branch of the V-shape has gradient 2.

If \( a > 2 \), the line is steeper than the right branch and will intersect it.

If \( a < 1.25 \), the line passes "below" P. Since the V opens upwards, any line with gradient less than or equal to 1.25 (passing through P or lower) will intersect the graph.

Wait, let’s visualize. P is in the 3rd quadrant.

The line \( y = 1.25x \) passes through P.

If \( a \) is slightly less than 1.25 (e.g. 1), \( y=x \) passes above P? No, at x=-4, y=-4 which is above -5. So it goes inside the V shape.

Wait, if it goes “inside” the V shape, it must cross the boundary somewhere.

Let’s look at the geometry carefully.

The vertex is at (-4, -5). The graph exists for \( y \ge -5 \).

The line \( y = ax \) passes through (0,0).

If the line passes through P, it intersects. So \( a = 1.25 \) is a solution.

Any line with gradient \( a < 1.25 \) will pass below P at x=-4 (i.e. y > -5 since x is negative). Wait.

At x=-4: Line y = -4a. Curve y = -5.

For intersection at P, -4a = -5 => a = 1.25.

If \( a < 1.25 \), then -4a > -5 (e.g. a=1, y=-4 > -5). The line is above the vertex. Since the vertex is the minimum, the line passes through the “inside” of the V and must intersect the arms.

However, we need to consider the gradient of the arms.

Left arm gradient = -2. Right arm gradient = 2.

If \( a > 2 \), the line is steeper than the right arm (gradient 2). Since they both start roughly from the left, the line will eventually catch the right arm? No, the line starts at origin.

Let’s solve algebraically for intersection.

Right arm: \( y = 2(x+4) – 5 = 2x + 3 \).

\( ax = 2x + 3 \implies x(a-2) = 3 \implies x = \frac{3}{a-2} \).

For this intersection to be on the right arm (\( x > -4 \)), we need:

\( \frac{3}{a-2} > -4 \).

If \( a > 2 \), \( x > 0 \), valid.

Left arm: \( y = -2(x+4) – 5 = -2x – 13 \).

\( ax = -2x – 13 \implies x(a+2) = -13 \implies x = \frac{-13}{a+2} \).

For intersection on left arm (\( x < -4 \)):

\( \frac{-13}{a+2} < -4 \).

Let’s combine logical reasoning.

1. The line passes through P if \( a = 1.25 \).

2. If \( a < 1.25 \), the line enters the V-shape from the vertex? No.

Visually:

P is (-4, -5). Origin is (0,0). Gradient OP is 1.25.

Any line with gradient \( a \le 1.25 \) will hit the left arm or the vertex.

Any line with gradient \( a > 2 \) will hit the right arm (it’s steeper than the arm and starts “behind” it at the origin).

What about between 1.25 and 2?

At x=-4, line is at \( y = -4a \). If \( 1.25 < a < 2 \), then \( -8 < -4a < -5 \). The line is below the curve at x=-4. The gradient is less than 2, so it never catches the right arm. It is moving away from the left arm.

So no intersection for \( 1.25 < a \le 2 \).

Find \( \frac{\text{d}x}{\text{d}t} \):

\( x = 6 \sin t \implies \frac{\text{d}x}{\text{d}t} = 6 \cos t \)

Substitute \( y \) and \( \frac{\text{d}x}{\text{d}t} \) into the integral:

\[ \text{Area} = \int_{0}^{\frac{\pi}{2}} (5 \sin 2t)(6 \cos t) \, \text{d}t \]

\[ = \int_{0}^{\frac{\pi}{2}} 30 \sin 2t \cos t \, \text{d}t \]

Use the identity \( \sin 2t = 2 \sin t \cos t \):

\[ = \int_{0}^{\frac{\pi}{2}} 30 (2 \sin t \cos t) \cos t \, \text{d}t \]

\[ = \int_{0}^{\frac{\pi}{2}} 60 \sin t \cos^2 t \, \text{d}t \]

Result shown.

Let \( u = \cos t \). Then \( \frac{\text{d}u}{\text{d}t} = -\sin t \), so \( -\text{d}u = \sin t \, \text{d}t \).

Limits:

When \( t=0, u=1 \).

When \( t=\pi/2, u=0 \).

\[ \int_{1}^{0} 60 u^2 (-1) \, \text{d}u \]

\[ = \int_{0}^{1} 60 u^2 \, \text{d}u \]

\[ = \left[ \frac{60u^3}{3} \right]_{0}^{1} = \left[ 20u^3 \right]_{0}^{1} \]

\[ = 20(1)^3 – 20(0)^3 = 20 \]

Alternatively, by recognition:

\[ \int 60 \sin t \cos^2 t \, \text{d}t = -20 \cos^3 t \]

\[ [-20 \cos^3 t]_{0}^{\frac{\pi}{2}} = -20(0) – (-20(1)) = 20 \]

Step 1: Find t when y = 4.2

\[ 5 \sin 2t = 4.2 \]

\[ \sin 2t = \frac{4.2}{5} = 0.84 \]

\[ 2t = \sin^{-1}(0.84) \]

\[ 2t \approx 0.9973 \text{ rad} \]

Since the curve is symmetric and defined up to \( \pi/2 \), we need the second solution for \( 2t \) in the range.

\[ 2t_2 = \pi – 0.9973 \approx 2.1443 \]

So the values for \( t \) are:

\[ t_1 \approx 0.4986 \]

\[ t_2 \approx 1.0721 \]

Step 2: Find corresponding x values

Use \( x = 6 \sin t \).

At \( t_1 \): \( x_1 = 6 \sin(0.4986) \approx 2.869 \)

At \( t_2 \): \( x_2 = 6 \sin(1.0721) \approx 5.269 \)

Step 3: Calculate Width

\[ \text{Width} = x_2 – x_1 \]

\[ \text{Width} = 5.269 – 2.869 = 2.40 \]

The function is undefined when the denominator is zero.

\[ \ln x – 2 = 0 \]

\[ \ln x = 2 \]

\[ x = e^2 \]

So \( k = e^2 \).

\[ g'(x) = \frac{(\ln x – 2)(\frac{3}{x}) – (3\ln x – 7)(\frac{1}{x})}{(\ln x – 2)^2} \]

Multiply numerator terms:

\[ = \frac{\frac{3\ln x}{x} – \frac{6}{x} – \frac{3\ln x}{x} + \frac{7}{x}}{(\ln x – 2)^2} \]

Simplify numerator:

\[ = \frac{\frac{1}{x}}{(\ln x – 2)^2} = \frac{1}{x(\ln x – 2)^2} \]

Reasoning:

We are given \( x > 0 \), so \( \frac{1}{x} > 0 \).

The term \( (\ln x – 2)^2 \) is a square, so it is always positive (since \( x \neq e^2 \)).

Therefore, \( g'(x) > 0 \) for all valid \( x \).

We want \( \frac{3\ln a – 7}{\ln a – 2} > 0 \).

For a fraction to be positive, either both numerator and denominator are positive, or both are negative.

Case 1: Both positive

\( 3\ln a – 7 > 0 \implies \ln a > \frac{7}{3} \implies a > e^{7/3} \)

\( \ln a – 2 > 0 \implies \ln a > 2 \implies a > e^2 \)

Combined condition: \( a > e^{7/3} \) (since \( e^{7/3} > e^2 \)).

Case 2: Both negative

\( 3\ln a – 7 < 0 \implies a < e^{7/3} \)

\( \ln a – 2 < 0 \implies a < e^2 \)

Combined condition: \( a < e^2 \).

Also remember \( a > 0 \).

Step 2: Substitute line equation

Line: \( y = 12 – 2x \).

Substitute into circle equation:

\[ (x-r)^2 + (12 – 2x – r)^2 = r^2 \]

\[ (x-r)^2 + ((12-r) – 2x)^2 = r^2 \]

Step 3: Expand and Simplify

Expand first bracket: \( x^2 – 2rx + r^2 \)

Expand second bracket: \( (12-r)^2 – 4x(12-r) + 4x^2 \)

\[ = 144 – 24r + r^2 – 48x + 4rx + 4x^2 \]

Combine all terms:

\[ x^2 – 2rx + r^2 + 144 – 24r + r^2 – 48x + 4rx + 4x^2 = r^2 \]

Group \( x^2 \) terms: \( x^2 + 4x^2 = 5x^2 \)

Group \( x \) terms: \( -2rx – 48x + 4rx = 2rx – 48x = (2r – 48)x \)

Group constant terms: \( r^2 + r^2 – r^2 – 24r + 144 = r^2 – 24r + 144 \)

Equation: \( 5x^2 + (2r-48)x + (r^2-24r+144) = 0 \)

Result shown.

\( a = 5 \)

\( b = 2r – 48 \)

\( c = r^2 – 24r + 144 \)

\[ (2r-48)^2 – 4(5)(r^2 – 24r + 144) = 0 \]

Factor out 2 from first bracket to make arithmetic easier:

\[ [2(r-24)]^2 – 20(r^2 – 24r + 144) = 0 \]

\[ 4(r-24)^2 – 20(r^2 – 24r + 144) = 0 \]

Divide by 4:

\[ (r-24)^2 – 5(r^2 – 24r + 144) = 0 \]

Expand:

\[ (r^2 – 48r + 576) – 5r^2 + 120r – 720 = 0 \]

\[ -4r^2 + 72r – 144 = 0 \]

Divide by -4:

\[ r^2 – 18r + 36 = 0 \]

Solve using quadratic formula:

\[ r = \frac{18 \pm \sqrt{18^2 – 4(1)(36)}}{2} \]

\[ r = \frac{18 \pm \sqrt{324 – 144}}{2} \]

\[ r = \frac{18 \pm \sqrt{180}}{2} \]

Simplify surd: \( \sqrt{180} = \sqrt{36 \times 5} = 6\sqrt{5} \).

\[ r = \frac{18 \pm 6\sqrt{5}}{2} \]

\[ r = 9 \pm 3\sqrt{5} \]

Write out the sum of the first \( n \) terms:

\[ S_n = a + ar + ar^2 + \dots + ar^{n-2} + ar^{n-1} \quad \text{(1)} \]

Multiply the whole equation by \( r \):

\[ rS_n = ar + ar^2 + ar^3 + \dots + ar^{n-1} + ar^n \quad \text{(2)} \]

Subtract equation (2) from equation (1):

\[ S_n – rS_n = a + (ar-ar) + (ar^2-ar^2) + \dots + (ar^{n-1}-ar^{n-1}) – ar^n \]

All middle terms cancel out:

\[ S_n(1-r) = a – ar^n \]

\[ S_n(1-r) = a(1 – r^n) \]

Divide by \( (1-r) \):

\[ S_n = \frac{a(1-r^n)}{1-r} \]

We are given \( S_{10} = 4S_5 \).

\[ \frac{a(1-r^{10})}{1-r} = 4 \frac{a(1-r^5)}{1-r} \]

Cancel \( a \) and \( 1-r \) (since \( a \neq 0, r \neq 1 \)):

\[ 1 – r^{10} = 4(1 – r^5) \]

Recognise that \( 1 – r^{10} \) is a difference of squares: \( (1-r^5)(1+r^5) \).

\[ (1-r^5)(1+r^5) = 4(1-r^5) \]

Since \( r \neq 1 \), \( r^5 \neq 1 \), so we can divide by \( 1-r^5 \):

\[ 1 + r^5 = 4 \]

\[ r^5 = 3 \]

\[ r = \sqrt[5]{3} \]

Case 1: \( n = 3k \)

\[ n^2 = (3k)^2 = 9k^2 \]

\[ n^2 = 3(3k^2) \]

This is a multiple of 3.

Case 2: \( n = 3k + 1 \)

\[ n^2 = (3k+1)^2 = 9k^2 + 6k + 1 \]

\[ n^2 = 3(3k^2 + 2k) + 1 \]

This is one more than a multiple of 3.

Case 3: \( n = 3k + 2 \)

\[ n^2 = (3k+2)^2 = 9k^2 + 12k + 4 \]

Rewrite 4 as \( 3 + 1 \):

\[ n^2 = 9k^2 + 12k + 3 + 1 \]

Factor out 3:

\[ n^2 = 3(3k^2 + 4k + 1) + 1 \]

This is one more than a multiple of 3.