2025 Cambridge IAL Mathematics (9709) 模擬試題連答案詳解

Let \( a \) be the first term and \( r \) be the common ratio.
We are given:
\( u_3 = a r^2 = 24 \)
\( u_6 = a r^5 = 3 \)

Dividing these equations:
\( \frac{a r^5}{a r^2} = \frac{3}{24} \)
\( r^3 = \frac{1}{8} \implies r = \frac{1}{2} \)

Substitute \( r = \frac{1}{2} \) back to find \( a \):
\( a \left(\frac{1}{2}\right)^2 = 24 \implies \frac{a}{4} = 24 \implies a = 96 \)

The sum to infinity is given by:
\( S_{\infty} = \frac{a}{1 - r} = \frac{96}{1 - \frac{1}{2}} = \frac{96}{\frac{1}{2}} = 192 \).

M1: Set up equations for \( u_3 \) and \( u_6 \) and solve to find \( r \).
A1: Correctly find \( a = 96 \) and \( r = \frac{1}{2} \).
A1: Calculate the correct sum to infinity of \( 192 \).

Using the identity \( \sin^2 x = 1 - \cos^2 x \), we substitute this into the equation:
\( 3(1 - \cos^2 x) - 5\cos x - 1 = 0 \)
\( 3 - 3\cos^2 x - 5\cos x - 1 = 0 \)
\( 3\cos^2 x + 5\cos x - 2 = 0 \)

We can factorise this quadratic in terms of \( \cos x \):
\( (3\cos x - 1)(\cos x + 2) = 0 \)

This gives two possible equations:
1) \( \cos x = \frac{1}{3} \)
2) \( \cos x = -2 \) (which has no real solutions since \( -1 \le \cos x \le 1 \))

For \( \cos x = \frac{1}{3} \) in the interval \( 0^\circ \le x \le 180^\circ \):
\( x = \cos^{-1}\left(\frac{1}{3}\right) \approx 70.5^\circ \) (to 1 decimal place).

M1: Use \( \sin^2 x = 1 - \cos^2 x \) to obtain a quadratic equation in terms of \( \cos x \).
M1: Factorise or solve the quadratic equation to get \( \cos x = \frac{1}{3} \).
A1: Obtain the correct angle of \( 70.5^\circ \) (or \( 70.5 \)).

To find the inverse function, we set \( y = \text{f}(x) \) and rearrange to express \( x \) in terms of \( y \):
\( y = \frac{4}{x-2} + 3 \)

Subtract \( 3 \) from both sides:
\( y - 3 = \frac{4}{x-2} \)

Take the reciprocal of both sides (or multiply by \( x-2 \) and divide by \( y-3 \)):
\( x - 2 = \frac{4}{y-3} \)

Add \( 2 \) to both sides:
\( x = \frac{4}{y-3} + 2 \)

Replacing \( y \) with \( x \) gives the inverse function:
\( \text{f}^{-1}(x) = \frac{4}{x-3} + 2 \) (or equivalently, \( \text{f}^{-1}(x) = \frac{2x - 2}{x-3} \)).

M1: Set \( y = \frac{4}{x-2} + 3 \) and subtract \( 3 \) from both sides.
M1: Rearrange the equation to isolate \( x \) on one side.
A1: State the correct inverse function \( \text{f}^{-1}(x) = \frac{4}{x-3} + 2 \) (or equivalent form).

To find the stationary point, we first differentiate \( y = 4x + 9x^{-1} \) with respect to \( x \):
\( \frac{\text{d}y}{\text{d}x} = 4 - 9x^{-2} = 4 - \frac{9}{x^2} \)

At a stationary point, \( \frac{\text{d}y}{\text{d}x} = 0 \):
\( 4 - \frac{9}{x^2} = 0 \implies \frac{9}{x^2} = 4 \implies x^2 = \frac{9}{4} \)

Since \( x > 0 \), we take the positive square root:
\( x = \frac{3}{2} = 1.5 \)

Now, substitute \( x = 1.5 \) back into the equation of the curve to find the corresponding \( y \)-coordinate:
\( y = 4(1.5) + \frac{9}{1.5} = 6 + 6 = 12 \)

Therefore, the coordinates of the stationary point are \( (1.5, 12) \).

M1: Differentiate the expression to obtain \( 4 - \frac{9}{x^2} \).
M1: Set the derivative equal to zero and solve for \( x \) (accepting only the positive value \( x = 1.5 \)).
A1: Find the correct coordinates \( (1.5, 12) \).

The area of the region is given by the definite integral:
\( A = \int_{1}^{5} (3x + 1)^{\frac{1}{2}} \, \text{d}x \)

To integrate, we use the reverse chain rule:
\( \int (3x + 1)^{\frac{1}{2}} \, \text{d}x = \frac{(3x + 1)^{\frac{3}{2}}}{\frac{3}{2} \times 3} = \frac{2}{9} (3x + 1)^{\frac{3}{2}} \)

Now, apply the limits from \( x = 1 \) to \( x = 5 \):
\( A = \left[ \frac{2}{9} (3x + 1)^{\frac{3}{2}} \right]_{1}^{5} \)
\( A = \frac{2}{9} (3(5) + 1)^{\frac{3}{2}} - \frac{2}{9} (3(1) + 1)^{\frac{3}{2}} \)
\( A = \frac{2}{9} (16)^{\frac{3}{2}} - \frac{2}{9} (4)^{\frac;{3}{2}} \)

Since \( 16^{\frac;{3}{2}} = 64 \) and \( 4^{\frac;{3}{2}} = 8 \):
\( A = \frac{2}{9} (64) - \frac{2}{9} (8) = \frac{2}{9} (64 - 8) = \frac{2}{9} (56) = \frac{112}{9} \).

M1: Integrate \( (3x + 1)^{\frac{1}{2}} \) to get \( k(3x+1)^{\frac{3}{2}} \) where \( k \) is a constant.
A1: Correct integration to obtain \( \frac{2}{9} (3x + 1)^{\frac{3}{2}} \).
A1: Correctly substitute limits to obtain the exact value of \( \frac{112}{9} \) (or \( 12\frac{4}{9} \)).

For the line to be a tangent to the curve, the equation formed by equating them must have exactly one real solution (i.e., its discriminant must be zero).

Equating the equations of the line and the curve:
\( mx - 5 = x^2 - 4x + 4 \)

Rearranging into a standard quadratic form \( ax^2 + bx + c = 0 \):
\( x^2 - (4 + m)x + 9 = 0 \)

Identify the coefficients:
\( a = 1 \), \( b = -(4 + m) \), and \( c = 9 \).

For tangency, we set the discriminant \( b^2 - 4ac = 0 \):
\( (-(4 + m))^2 - 4(1)(9) = 0 \)
\( (4 + m)^2 - 36 = 0 \)
\( (4 + m)^2 = 36 \)

Taking the square root of both sides:
\( 4 + m = 6 \implies m = 2 \)
\( 4 + m = -6 \implies m = -10 \)

Thus, the two possible values of \( m \) are \( 2 \) and \( -10 \).

M1: Equate the line and curve equations and rearrange to form a quadratic equation in \( x \).
M1: Set the discriminant \( b^2 - 4ac = 0 \) to obtain a quadratic equation in \( m \).
A1: Correctly solve to find the two values \( m = 2 \) and \( m = -10 \).

We use the formulas for the perimeter and area of a sector of a circle:
1) Perimeter: \( P = 2r + r\theta = 20 \implies r\theta = 20 - 2r \)
2) Area: \( A = \frac{1}{2} r^2 \theta = 24 \implies r^2\theta = 48 \)

We can rewrite the area equation as:
\( r(r\theta) = 48 \)

Substitute \( r\theta = 20 - 2r \) into this equation:
\( r(20 - 2r) = 48 \)
\( 20r - 2r^2 = 48 \)
\( 2r^2 - 20r + 48 = 0 \)

Divide the entire quadratic equation by \( 2 \):
\( r^2 - 10r + 24 = 0 \)

Factorise the quadratic equation:
\( (r - 4)(r - 6) = 0 \)

This gives two possible values for the radius:
\( r = 4 \) or \( r = 6 \).

M1: Set up two equations for the perimeter and area in terms of \( r \) and \( \theta \).
M1: Substitute one equation into the other to eliminate \( \theta \) and form a quadratic equation in terms of \( r \).
A1: Solve the quadratic equation correctly to find \( r = 4 \) and \( r = 6 \).

For the quadratic equation to have no real roots, its discriminant must be strictly less than zero:
\( b^2 - 4ac < 0 \)

Identify the coefficients from the given equation:
\( a = k+1 \), \( b = -4 \), and \( c = k-2 \).

Substitute these into the discriminant inequality:
\( (-4)^2 - 4(k+1)(k-2) < 0 \)
\( 16 - 4(k^2 - k - 2) < 0 \)

Divide the entire inequality by \( 4 \):
\( 4 - (k^2 - k - 2) < 0 \)
\( 4 - k^2 + k + 2 < 0 \)
\( -k^2 + k + 6 < 0 \)

Multiply by \( -1 \) and reverse the inequality sign:
\( k^2 - k - 6 > 0 \)

Factorise the quadratic expression:
\( (k - 3)(k + 2) > 0 \)

Thus, the solution to the inequality is:
\( k < -2 \) or \( k > 3 \).

M1: Set up the discriminant inequality \( b^2 - 4ac < 0 \) with correct coefficients.
M1: Simplify to a three-term quadratic inequality in \( k \) (e.g. \( k^2 - k - 6 > 0 \)) and find its critical values.
A1: State the correct final set of values: \( k < -2 \) or \( k > 3 \) (or equivalent notation).

To find the area of the region, we evaluate the definite integral of the curve between the given limits: \(\text{Area} = \int_{1}^{5} (2x - 1)^{\frac{3}{2}} \, dx\). Using the integration rule \(\int (ax+b)^n \, dx = \frac{(ax+b)^{n+1}}{a(n+1)}\), we obtain the integrated term: \(\left[ \frac{(2x - 1)^{\frac{5}{2}}}{2 \times \frac{5}{2}} \right]_{1}^{5} = \left[ \frac{1}{5}(2x - 1)^{\frac{5}{2}} \right]_{1}^{5}\). Substituting the upper limit \(x = 5\) and the lower limit \(x = 1\), we get: \(\frac{1}{5}(2(5) - 1)^{\frac{5}{2}} - \frac{1}{5}(2(1) - 1)^{\frac{5}{2}} = \frac{1}{5}(9)^{\frac{5}{2}} - \frac{1}{5}(1)^{\frac{5}{2}}\). Since \(9^{\frac{5}{2}} = 243\) and \(1^{\frac{5}{2}} = 1\), this simplifies to: \(\frac{243}{5} - \frac{1}{5} = \frac{242}{5} = 48.4\).

M1: For integrating to obtain a term of the form \(k(2x-1)^{5/2}\), where \(k\) is a constant. A1: For obtaining the correct integrated expression \(\frac{1}{5}(2x-1)^{5/2}\). A1: For substituting the limits correctly and obtaining the final exact area of \(48.4\) or \(\frac{242}{5}\).

Let the first term be \( a \) and the common ratio be \( r \). The sum to infinity is given by:
\( S_\infty = \frac{a}{1-r} = 12 \implies 1-r = \frac{a}{12} \implies r = 1 - \frac{a}{12} \)

The second term is given by:
\( u_2 = ar = 2.25 = \frac{9}{4} \)

Substitute the expression for \( r \) into the second equation:
\( a \left( 1 - \frac{a}{12} \right) = \frac{9}{4} \)
\( a - \frac{a^2}{12} = \frac{9}{4} \)

Multiply the entire equation by 12 to clear the denominators:
\( 12a - a^2 = 27 \implies a^2 - 12a + 27 = 0 \)

Factorise the quadratic equation:
\( (a-3)(a-9) = 0 \)

Thus, the two possible values for the first term are \( a = 3 \) or \( a = 9 \).

Both values yield valid common ratios \( r = \frac{3}{4} \) and \( r = \frac{1}{4} \) respectively, which satisfy the convergence condition \( |r| < 1 \).

M1: Writes down equations for the sum to infinity and the second term in terms of \( a \) and \( r \).
M1: Obtains a quadratic equation in one variable by substitution.
A1: Correctly simplifies to the quadratic equation \( a^2 - 12a + 27 = 0 \) (or equivalent).
A1: Correctly solves to find the two possible values of the first term: 3 and 9.

Divide the entire equation by \( \cos^2\theta \) (under the assumption \( \cos\theta \ne 0 \)):
\( 3 - 5\tan\theta - 2\tan^2\theta = 0 \)

Rearranging into standard quadratic form:
\( 2\tan^2\theta + 5\tan\theta - 3 = 0 \)

Factorise the quadratic equation:
\( (2\tan\theta - 1)(\tan\theta + 3) = 0 \)

This gives two possible values for \( \tan\theta \):
\( \tan\theta = 0.5 \) or \( \tan\theta = -3 \)

For the interval \( 0^\circ \le \theta \le 180^\circ \):
- From \( \tan\theta = 0.5 \):
\( \theta = \tan^{-1}(0.5) \approx 26.6^\circ \)
- From \( \tan\theta = -3 \):
\( \theta = 180^\circ - \tan^{-1}(3) \approx 180^\circ - 71.6^\circ = 108.4^\circ \)

So, the solutions are \( 26.6^\circ \) and \( 108.4^\circ \).

M1: Divides by \( \cos^2\theta \) to obtain a quadratic equation in \( \tan\theta \).
A1: Solves or factorises the quadratic to find \( \tan\theta = 0.5 \) and \( \tan\theta = -3 \).
A1: Obtains \( \theta = 26.6^\circ \) (accept \( 26.56^\circ \) or better, rounded to 1 d.p.).
A1: Obtains \( \theta = 108.4^\circ \) (accept \( 108.43^\circ \) or better, rounded to 1 d.p.). Deduct 1 mark overall if any extra solutions are provided within the range.

First, we find the composite function \( fg(x) \) by substituting \( g(x) \) into \( f(x) \):
\( fg(x) = f(g(x)) = \frac{4}{2(x^2 + 1) - 3} \)

Simplify the denominator:
\( fg(x) = \frac{4}{2x^2 + 2 - 3} = \frac{4}{2x^2 - 1} \)

Now, we set this expression equal to \( \frac{1}{2} \):
\( \frac{4}{2x^2 - 1} = \frac{1}{2} \)

Cross-multiply to solve for \( x \):
\( 2x^2 - 1 = 8 \implies 2x^2 = 9 \implies x^2 = 4.5 = \frac{9}{2} \)

Since the domain of \( g \) is restricted to \( x \ge 0 \), we take the positive square root:
\( x = \sqrt{\frac{9}{2}} = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2} \)

M1: Substitutes the expression for \( g(x) \) into \( f(x) \) to form the composite function.
A1: Correctly simplifies the composite function to \( \frac{4}{2x^2 - 1} \).
M1: Sets the composite function equal to \( 0.5 \) and solves for \( x^2 \).
A1: Identifies the correct positive exact root \( x = \frac{3\sqrt{2}}{2} \) (or equivalent, e.g. \( \sqrt{4.5} \)), explicitly or implicitly rejecting the negative root.

Rewrite the curve equation by dividing each term by \( \sqrt{x} \):
\( y = \frac{2x}{\sqrt{x}} + \frac{3}{\sqrt{x}} = 2x^{1/2} + 3x^{-1/2} \)

Differentiate with respect to \( x \):
\( \frac{dy}{dx} = 2 \left( \frac{1}{2}x^{-1/2} \right) + 3 \left( -\frac{1}{2}x^{-3/2} \right) = x^{-1/2} - \frac{3}{2}x^{-3/2} \)

Set \( \frac{dy}{dx} = 0 \) to find the stationary point:
\( x^{-1/2} - \frac{3}{2}x^{-3/2} = 0 \implies x^{-1/2} = \frac{3}{2}x^{-3/2} \)

Multiply both sides by \( x^{3/2} \):
\( x = 1.5 \)

Substitute \( x = 1.5 \) back into the original curve equation to find the y-coordinate:
\( y = \frac{2(1.5) + 3}{\sqrt{1.5}} = \frac{6}{\sqrt{1.5}} = 2\sqrt{6} \approx 4.90 \)

Now, find the second derivative to determine the nature of the stationary point:
\( \frac{d^2y}{dx^2} = -\frac{1}{2}x^{-3/2} + \frac{9}{4}x^{-5/2} \)

Evaluate \( \frac{d^2y}{dx^2} \) at \( x = 1.5 \):
\( \frac{d^2y}{dx^2}\Big|_{x=1.5} = -\frac{1}{2}(1.5)^{-3/2} + \frac{9}{4}(1.5)^{-5/2} \approx -0.272 + 0.816 = 0.544 > 0 \)

Since the second derivative is greater than zero, the stationary point is a minimum. Therefore, the stationary point is \( (1.5, 2\sqrt{6}) \) and it is a minimum.

M1: Expresses \( y \) as separate fractional powers of \( x \) and attempts differentiation.
A1: Sets \( \frac{dy}{dx} = 0 \) and correctly solves to find \( x = 1.5 \).
A1: Obtains the correct y-coordinate \( 2\sqrt{6} \) (or \( \sqrt{24} \), or \( 4.90 \) to 3 s.f.).
A1: Correctly evaluates the second derivative (or uses another valid method) at \( x = 1.5 \) to show that the point is a minimum.

Find the x-coordinates of the points of intersection of the curve and the line:
\( 4x - x^2 = x \implies x^2 - 3x = 0 \implies x(x-3) = 0 \)

The intersection points are at \( x = 0 \) and \( x = 3 \).

The area \( A \) of the enclosed region is given by the integral of the upper function minus the lower function:
\( A = \int_{0}^{3} \left( (4x - x^2) - x \right) dx = \int_{0}^{3} (3x - x^2) dx \)

Integrate with respect to \( x \):
\( A = \left[ \frac{3}{2}x^2 - \frac{1}{3}x^3 \right]_{0}^{3} \)

Evaluate the definite integral by substituting the limits of integration:
\( A = \left( \frac{3}{2}(3)^2 - \frac{1}{3}(3)^3 \right) - (0) = \frac{27}{2} - 9 = 13.5 - 9 = 4.5 \)

Thus, the area of the enclosed region is \( 4.5 \).

M1: Equates the line and curve equations to find the correct limits of integration: \( x = 0 \) and \( x = 3 \).
M1: Sets up the correct definite integral of the difference of the two functions: \( \int_{0}^{3} (3x - x^2) dx \).
A1: Integrates correctly to obtain \( \frac{3}{2}x^2 - \frac{1}{3}x^3 \).
A1: Evaluates using the limits correctly to obtain the final answer \( 4.5 \) (or equivalent fraction \( \frac{9}{2} \)).

We can solve this problem using either a geometric or an algebraic approach.

**Method 1: Geometric Approach**
Complete the square for the circle equation to find its center and radius:
\( (x-3)^2 - 9 + (y+4)^2 - 16 = 0 \implies (x-3)^2 + (y+4)^2 = 25 \)

So, the circle has center \( (3, -4) \) and radius \( R = 5 \).

The equation of the line is \( 2x - y + c = 0 \).
For the line to be tangent to the circle, the perpendicular distance from the center \( (3, -4) \) to the line must be equal to the radius \( 5 \):
\( \frac{|2(3) - (-4) + c|}{\sqrt{2^2 + (-1)^2}} = 5 \)
\( \frac{|10 + c|}{\sqrt{5}} = 5 \implies |10 + c| = 5\sqrt{5} \)

This gives two solutions:
\( 10 + c = 5\sqrt{5} \implies c = -10 + 5\sqrt{5} \)
\( 10 + c = -5\sqrt{5} \implies c = -10 - 5\sqrt{5} \)

**Method 2: Algebraic Approach**
Substitute \( y = 2x + c \) into the circle's equation:
\( x^2 + (2x+c)^2 - 6x + 8(2x+c) = 0 \)
\( x^2 + 4x^2 + 4cx + c^2 - 6x + 16x + 8c = 0 \)
\( 5x^2 + (4c+10)x + (c^2+8c) = 0 \)

For tangency, the discriminant \( B^2 - 4AC = 0 \):
\( (4c+10)^2 - 4(5)(c^2+8c) = 0 \)
\( 16c^2 + 80c + 100 - 20c^2 - 160c = 0 \)
\( -4c^2 - 80c + 100 = 0 \implies c^2 + 20c - 25 = 0 \)

Solving this quadratic equation in \( c \) using the quadratic formula:
\( c = \frac{-20 \pm \sqrt{400 - 4(1)(-25)}}{2} = \frac{-20 \pm \sqrt{500}}{2} = -10 \pm 5\sqrt{5} \)

M1: Completes the square to find center and radius of the circle, OR substitutes \( y = 2x+c \) to form a quadratic in \( x \).
M1: Uses the perpendicular distance formula set equal to radius, OR uses the discriminant condition \( B^2 - 4AC = 0 \) for tangency.
A1: Correctly forms the equation \( |10 + c| = 5\sqrt{5} \) or \( c^2 + 20c - 25 = 0 \).
A1: Obtains both correct exact values of \( c \): \( -10 + 5\sqrt{5} \) and \( -10 - 5\sqrt{5} \).

Let the angle of the sector be \( \theta \) radians.

The perimeter of the sector, which consists of two radii and the arc length, is given by:
\( P = 2r + r\theta = 20 \implies r\theta = 20 - 2r \implies \theta = \frac{20-2r}{r} \)

The area of the sector is given by:
\( A = \frac{1}{2}r^2\theta \)

Substitute the expression for \( \theta \) into the area formula:
\( A = \frac{1}{2}r^2 \left( \frac{20-2r}{r} \right) = \frac{1}{2}r(20-2r) = 10r - r^2 \)

To find the maximum possible area, differentiate \( A \) with respect to \( r \) and set the derivative to 0:
\( \frac{dA}{dr} = 10 - 2r = 0 \implies r = 5 \)

Verify that this value of \( r \) gives a maximum:
\( \frac{d^2A}{dr^2} = -2 < 0 \), which confirms a maximum.

Now, calculate the maximum area:
\( A_{\text{max}} = 10(5) - (5)^2 = 50 - 25 = 25 \text{ cm}^2 \).

M1: Expresses the perimeter equation \( 2r + r\theta = 20 \) and rearranges to express \( A \) in terms of \( r \) only.
A1: Obtains the correct quadratic function for the area: \( A = 10r - r^2 \).
M1: Differentiates and sets \( \frac{dA}{dr} = 0 \) to find the stationary value of \( r \).
A1: Correctly calculates the maximum area as \( 25 \) (or \( 25 \text{ cm}^2 \)).

To find where the line and curve intersect, set their equations equal to each other:
\( (k+1)x = x^2 + 2kx + k + 2 \)

Rearrange the terms into standard quadratic form \( Ax^2 + Bx + C = 0 \):
\( x^2 + 2kx - (k+1)x + k + 2 = 0 \)
\( x^2 + (k-1)x + (k+2) = 0 \)

For the line and curve not to intersect, the quadratic equation must have no real roots. Therefore, its discriminant must be negative:
\( B^2 - 4AC < 0 \)

Substitute the coefficients \( A = 1 \), \( B = k-1 \), and \( C = k+2 \):
\( (k-1)^2 - 4(1)(k+2) < 0 \)
\( k^2 - 2k + 1 - 4k - 8 < 0 \)
\( k^2 - 6k - 7 < 0 \)

Factorise the quadratic inequality:
\( (k-7)(k+1) < 0 \)

The critical values are \( k = 7 \) and \( k = -1 \).

Since the inequality is less than zero, the solution range is:
\( -1 < k < 7 \).

M1: Sets up the intersection equation and rearranges it into standard quadratic form.
A1: Identifies the correct coefficients and sets up the discriminant inequality: \( (k-1)^2 - 4(1)(k+2) < 0 \).
M1: Factorises the resulting quadratic expression to find the critical values \( k = -1 \) and \( k = 7 \).
A1: Expresses the final answer as the correct inequality range: \( -1 < k < 7 \) (or interval notation \( (-1, 7) \)).

We use integration by parts: \(\int u \frac{dv}{dx} \, dx = uv - \int v \frac{du}{dx} \, dx\). Let \(u = \ln(2x+1)\) and \(\frac{dv}{dx} = (2x+1)^{-2}\). Then \(\frac{du}{dx} = \frac{2}{2x+1}\) and \(v = -\frac{1}{2}(2x+1)^{-1} = -\frac{1}{2(2x+1)}\). Applying the integration by parts formula: \(\int_{0}^{2} \frac{\ln(2x+1)}{(2x+1)^2} \, dx = \left[ -\frac{\ln(2x+1)}{2(2x+1)} \right]_0^2 - \int_{0}^{2} \left( -\frac{1}{2(2x+1)} \right) \left( \frac{2}{2x+1} \right) \, dx\) \(= \left[ -\frac{\ln(2x+1)}{2(2x+1)} \right]_0^2 + \int_{0}^{2} (2x+1)^{-2} \, dx\) \(= \left[ -\frac{\ln(2x+1)}{2(2x+1)} - \frac{1}{2(2x+1)} \right]_0^2\). Evaluating this expression at the limits: At the upper limit \(x = 2\): \(-\frac{\ln 5}{10} - \frac{1}{10}\). At the lower limit \(x = 0\): \(-\frac{\ln 1}{2} - \frac{1}{2} = -\frac{1}{2}\). Subtracting the lower limit value from the upper limit value: \(\left( -\frac{\ln 5}{10} - \frac{1}{10} \right) - \left( -\frac{1}{2} \right) = \frac{4}{10} - \frac{1}{10}\ln 5 = \frac{2}{5} - \frac{1}{10}\ln 5\). Thus, \(a = \frac{2}{5}\) and \(b = -\frac{1}{10}\).

M1: For choosing integration by parts with \(u = \ln(2x+1)\) and \(v' = (2x+1)^{-2}\).
B1: For finding the correct derivative \(u' = \frac{2}{2x+1}\).
A1: For finding the correct integral \(v = -\frac{1}{2(2x+1)}\).
M1: For applying the integration by parts formula correctly.
A1: For completing the integration to obtain \(-\frac{\ln(2x+1)}{2(2x+1)} - \frac{1}{2(2x+1)}\).
M1: For substituting limits \(2\) and \(0\) correctly into their integrated expression.
A1: For obtaining either limit value correctly: \(-\frac{1}{10}\ln 5 - \frac{1}{10}\) or \(-\frac{1}{2}\).
A1: For the final exact answer \(\frac{2}{5} - \frac{1}{10}\ln 5\) (or equivalent with \(a = 2/5\) and \(b = -1/10\)).

(a) The inequality \(|z - (2+2i)| \le 2\) represents the region inside and on a circle with center \(C(2, 2)\) and radius \(2\). The inequality \(\frac{1}{4}\pi \le \arg(z-1) \le \frac{1}{2}\pi\) represents the region between two half-lines starting from the point \(A(1, 0)\). One half-line goes in the direction of \(\pi/4\) (gradient 1) and the other vertically upwards. The required region is the intersection of these two areas.

(b) The value of \(|z|\) represents the distance from the origin \(O(0,0)\) to \(z\). The point furthest from the origin on the circle lies on the line of symmetry passing through \(O\) and the center \(C(2,2)\), which has equation \(y = x\). The maximum distance is \(OC + R\) where \(R\) is the radius. Distance \(OC = \sqrt{2^2 + 2^2} = 2\sqrt{2}\). Radius is \(2\). So the maximum distance is \(2\sqrt{2} + 2\). The point corresponding to this maximum distance is \(z_{max} = (2 + \sqrt{2}) + (2 + \sqrt{2})i\). For this point, \(z_{max} - 1 = (1 + \sqrt{2}) + (2 + \sqrt{2})i\). Since the imaginary part \(2+\sqrt{2}\) is greater than the real part \(1+\sqrt{2}\), the argument of \(z_{max}-1\) is greater than \(\pi/4\), and since both parts are positive, the argument is less than \(\pi/2\). Thus, this point lies within the shaded region. The maximum value is \(2\sqrt{2} + 2\).

(a)
B1: For sketching a circle with center \(2+2i\) and radius \(2\) (must pass through \((2,0)\) and \((0,2)\)).
B1: For drawing the two half-lines starting from \(1\) on the real axis, one vertical and one at \(\pi/4\).
B1: For shading the correct region inside the circle and between the two half-lines.
B1: For showing correct boundary conventions (solid boundaries since the inequalities are non-strict).

(b)
M1: For identifying that the maximum value of \(|z|\) is on the line through the origin and the center of the circle.
M1: For calculating the distance from the origin to the center of the circle as \(2\sqrt{2}\).
M1: For adding the radius \(2\) to this distance.
A1: For justifying that this point lies within the region and stating the exact maximum value \(2\sqrt{2} + 2\).