2025 Cambridge IGCSE Mathematics - Additional (0606) Practice Paper with Answers

Let \(y = (2x-3)(4x+1)^{\frac{1}{2}}\). Using the product rule, \(\frac{\text{d}y}{\text{dx}} = u \frac{\text{d}v}{\text{dx}} + v \frac{\text{d}u}{\text{dx}}\), where: \(u = 2x-3 \implies \frac{\text{d}u}{\text{dx}} = 2\) and \(v = (4x+1)^{\frac{1}{2}} \implies \frac{\text{d}v}{\text{dx}} = 2(4x+1)^{-\frac{1}{2}}\). Therefore, \(\frac{\text{d}y}{\text{dx}} = 2(2x-3)(4x+1)^{-\frac{1}{2}} + 2(4x+1)^{\frac{1}{2}}\). Substituting \(x = 2\) gives \(\frac{\text{d}y}{\text{dx}} = 1 \cdot \frac{2}{3} + 3 \cdot 2 = \frac{2}{3} + 6 = \frac{20}{3}\).

M1: For applying product rule correctly. A1: For correct differentiation of \((4x+1)^{1/2}\) obtaining \(2(4x+1)^{-1/2}\). A1: For correct full derivative expression. M1: For substituting \(x=2\) into their derivative. A1: For the correct final exact value of \(\frac{20}{3}\).

The equation can be rewritten as \(3 \cdot (3^x)^2 - 10(3^x) + 3 = 0\). Let \(y = 3^x\), which gives \(3y^2 - 10y + 3 = 0\). Factorising gives \((3y - 1)(y - 3) = 0\), so \(y = \frac{1}{3}\) or \(y = 3\). Since \(y = 3^x\), we have \(3^x = 3^{-1} \implies x = -1\), or \(3^x = 3^1 \implies x = 1\).

M1: For writing the equation as a quadratic in \(3^x\). M1: For attempting to factorise or solve the quadratic. A1: For obtaining \(y = \frac{1}{3}\) and \(y = 3\). M1: For solving exponential equations \(3^x = \text{constant}\). A1: For both \(x = -1\) and \(x = 1\) correct.

Using the identity \(\cos^2 \theta = 1 - \sin^2 \theta\), we get \(2(1 - \sin^2 \theta) + 3\sin \theta = 3\). This simplifies to \(2\sin^2 \theta - 3\sin \theta + 1 = 0\). Factorising the quadratic yields \((2\sin \theta - 1)(\sin \theta - 1) = 0\). Thus, \(\sin \theta = \frac{1}{2}\) or \(\sin \theta = 1\). In the range \(0^\circ \le \theta \le 360^\circ\), \(\sin \theta = \frac{1}{2}\) gives \(\theta = 30^\circ, 150^\circ\), and \(\sin \theta = 1\) gives \(\theta = 90^\circ\). The solutions are \(\theta = 30^\circ, 90^\circ, 150^\circ\).

M1: For using the identity \(\cos^2 \theta = 1 - \sin^2 \theta\). M1: For rearranging into standard quadratic form. M1: For solving to find \(\sin \theta = \frac{1}{2}\) and \(\sin \theta = 1\). A1: For \(\theta = 90^\circ\). A1: For \(\theta = 30^\circ\) and \(\theta = 150^\circ\).

Since the terms form a geometric progression, \(\frac{k-3}{k-1} = \frac{k-4.2}{k-3}\). Cross-multiplying: \((k-3)^2 = (k-1)(k-4.2)\), which expands to \(k^2 - 6k + 9 = k^2 - 5.2k + 4.2\). Solving for \(k\) gives \(0.8k = 4.8 \implies k = 6\). The first term is \(a = 6 - 1 = 5\) and the common ratio is \(r = \frac{6-3}{6-1} = 0.6\). The sum to infinity is \(S_{\infty} = \frac{a}{1-r} = \frac{5}{1-0.6} = 12.5\).

M1: For setting up the ratio equation. M1: For expanding and solving for \(k\). A1: For \(k = 6\). M1: For finding \(a = 5\) and \(r = 0.6\). A1: For applying the sum to infinity formula. A1: For correct sum to infinity of \(12.5\).

Completing the square on the circle's equation gives \((x-2)^2 + (y+3)^2 = 25\), so the center is \(C(2, -3)\). The gradient of the radius \(CP\) is \(m_{CP} = \frac{1 - (-3)}{5 - 2} = \frac{4}{3}\). The tangent is perpendicular to the radius, so its gradient is \(m_t = -\frac{3}{4}\). The equation of the tangent is \(y - 1 = -\frac{3}{4}(x - 5)\), which simplifies to \(3x + 4y - 19 = 0\).

M1: For completing the square to find the center. A1: For center \(C(2, -3)\). M1: For finding the gradient of the radius. M1: For finding the gradient of the tangent. M1: For forming the equation of the line. A1: For \(3x + 4y - 19 = 0\).

Equating the line and curve equations gives \(mx + 5 = x^2 + 4x + 9\), which simplifies to \(x^2 + (4-m)x + 4 = 0\). For no intersection, the discriminant must be negative: \(b^2 - 4ac < 0 \implies (4-m)^2 - 4(1)(4) < 0\). This gives \((4-m)^2 < 16\), so \(-4 < 4-m < 4\). Solving this inequality yields \(0 < m < 8\).

M1: For equating line and curve and rearranging into standard quadratic form. A1: For correct quadratic. M1: For using \(b^2 - 4ac < 0\). A1: For getting \((4-m)^2 < 16\). M1: For solving the inequality. A1: For correct final range \(0 < m < 8\).

Let \(Y = \lg y\) and \(X = x^2\). The gradient of the line is \(m = \frac{13 - 5}{6 - 2} = 2\). Using \((2, 5)\), the line equation is \(Y - 5 = 2(X - 2) \implies Y = 2X + 1\). Substituting back: \(\lg y = 2x^2 + 1\). In exponential form, \(y = 10^{2x^2 + 1} = 10^1 \cdot (10^2)^{x^2} = 10 \cdot 100^{x^2}\). Thus, the expression is \(y = 10 \cdot 100^{x^2}\).

M1: For calculating the gradient of the line as 2. M1: For finding the line equation \(\lg y = 2x^2 + c\). A1: For \(\lg y = 2x^2 + 1\). M1: For converting to exponential base 10 form. M1: For using index laws to separate terms. A1: For the final correct expression \(y = 10 \cdot 100^{x^2}\).

(a) Let \(y = \frac{2x + 5}{x - 3}\). Then \(y(x - 3) = 2x + 5 \implies xy - 3y = 2x + 5 \implies xy - 2x = 3y + 5 \implies x(y - 2) = 3y + 5 \implies x = \frac{3y + 5}{y - 2}\). Thus, \(f^{-1}(x) = \frac{3x + 5}{x - 2}\). (b) The domain of \(f^{-1}\) is the range of \(f\). Since \(f(x) = 2 + \frac{11}{x - 3}\) and \(x > 3\), we have \(f(x) > 2\). Therefore, the domain of \(f^{-1}\) is \(x > 2\).

M1: For setting \(y = f(x)\) and multiplying. M1: For collecting terms in \(x\) and factorising. A1: For isolating \(x\) correctly. A1: For correct inverse expression. M1: For realizing the domain is the range of \(f\). A1: For correct domain \(x > 2\).

(a) To find the coordinates of \(P\), set \(y = 0\):
\(4e^{2x} - 5 = 0\)
\(e^{2x} = \frac{5}{4}\)
\(2x = \ln\left(\frac{5}{4}\right)\)
\(x = \frac{1}{2}\ln\left(\frac{5}{4}\right)\)
So, \(P\) has coordinates \(\left(\frac{1}{2}\ln\left(\frac{5}{4}\right), 0\right)\).

(b) To find the coordinates of \(Q\), set \(x = 0\):
\(y = 4e^{0} - 5 = 4 - 5 = -1\)
So, \(Q\) has coordinates \((0, -1)\).

Differentiating \(y = 4e^{2x} - 5\) with respect to \(x\):
\(\frac{dy}{dx} = 8e^{2x}\)
At \(Q(0, -1)\), the gradient of the tangent is:
\(m_t = 8e^{0} = 8\)
Therefore, the gradient of the normal is:
\(m_n = -\frac{1}{m_t} = -\frac{1}{8}\)

Using the point-slope form with \(Q(0, -1)\):
\(y - (-1) = -\frac{1}{8}(x - 0)\)
\(y + 1 = -\frac{1}{8}x\)
\(y = -\frac{1}{8}x - 1\) (or \(x + 8y + 8 = 0\))

M1: Sets \(y = 0\) and attempts to solve for \(x\) using logarithms.
A1: Correct coordinates for \(P\): \(\left(\frac{1}{2}\ln\left(\frac{5}{4}\right), 0\right)\) or equivalent.
B1: Correct coordinates for \(Q(0, -1)\).
M1: Correctly differentiates to find \(\frac{dy}{dx} = 8e^{2x}\).
M1: Evaluates gradient at \(x = 0\) and uses \(m_1 m_2 = -1\) to find the gradient of the normal.
A1: Correct equation of the normal (e.g., \(y = -\frac{1}{8}x - 1\) or \(x + 8y + 8 = 0\)).

From the first equation, use the laws of logarithms:
\(\log_2\left(\frac{x}{y}\right) = 2\)
\(\frac{x}{y} = 2^2 = 4\)
\(x = 4y\) [Equation 1]

From the second equation, express all terms with base 3:
\(3^{2x} \times (3^2)^y = (3^3)^{x-1}\)
\(3^{2x} \times 3^{2y} = 3^{3x - 3}\)
\(3^{2x + 2y} = 3^{3x - 3}\)

Equating exponents:
\(2x + 2y = 3x - 3\)
\(x - 2y = 3\) [Equation 2]

Substitute Equation 1 into Equation 2:
\(4y - 2y = 3\)
\(2y = 3\)
\(y = 1.5\) (or \(\frac{3}{2}\))

Find \(x\):
\(x = 4(1.5) = 6\)

Since both \(x\) and \(y\) are positive, the solutions are valid for the logarithmic terms.

M1: Applies log division law and removes log to obtain \(x = 4y\).
M1: Converts exponential equation into base 3 terms: \(3^{2x} \times 3^{2y} = 3^{3x-3}\).
A1: Obtains the correct linear equation \(2x + 2y = 3x - 3\) (or equivalent).
M1: Substitutes \(x = 4y\) into their linear equation to solve for one variable.
A1: Obtains \(y = 1.5\) (or \(\frac{3}{2}\)).
A1: Obtains \(x = 6\).

Use the double-angle identity: \(\cos(2\theta) = 1 - 2\sin^2(\theta)\).

Substitute into the equation:
\((1 - 2\sin^2(\theta)) + 5\sin(\theta) - 3 = 0\)
\(-2\sin^2(\theta) + 5\sin(\theta) - 2 = 0\)

Multiply by \(-1\) to make the leading coefficient positive:
\(2\sin^2(\theta) - 5\sin(\theta) + 2 = 0\)

Factorise the quadratic expression:
\((2\sin(\theta) - 1)(\sin(\theta) - 2) = 0\)

This yields two possible cases:
1) \(2\sin(\theta) - 1 = 0 \Rightarrow \sin(\theta) = \frac{1}{2}\)
2) \(\sin(\theta) - 2 = 0 \Rightarrow \sin(\theta) = 2\)

For \(\sin(\theta) = 2\), there are no solutions since the range of the sine function is \([-1, 1]\).

For \(\sin(\theta) = \frac{1}{2}\) in the interval \(0 \le \theta \le 2\pi\):
The reference angle is \(\theta = \frac{\pi}{6}\).
Since sine is positive in the first and second quadrants:
\(\theta = \frac{\pi}{6}\)
\(\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}\)

Thus, the solutions are \(\theta = \frac{\pi}{6}\) and \(\theta = \frac{5\pi}{6}\).

M1: Uses the identity \(\cos(2\theta) = 1 - 2\sin^2(\theta)\) to obtain a quadratic in \(\sin(\theta)\).
A1: Obtains the correct quadratic equation: \(2\sin^2(\theta) - 5\sin(\theta) + 2 = 0\).
M1: Factorises the quadratic correctly into two linear factors.
B1: Explains or indicates that \(\sin(\theta) = 2\) has no solutions.
A1: Identifies the first solution \(\theta = \frac{\pi}{6}\).
A1: Identifies the second solution \(\theta = \frac{5\pi}{6}\).

(a) Complete the square for the \(x\) and \(y\) terms in the circle's equation:
\((x^2 - 6x) + (y^2 + 8y) = 11\)
\((x - 3)^2 - 9 + (y + 4)^2 - 16 = 11\)
\((x - 3)^2 + (y + 4)^2 = 36\)

Comparing with the standard equation of a circle, \((x - h)^2 + (y - k)^2 = r^2\):
Center \(C = (3, -4)\)
Radius \(r = \sqrt{36} = 6\).

(b) Substitute \(y = mx + 2\) into the circle equation:
\(x^2 + (mx + 2)^2 - 6x + 8(mx + 2) - 11 = 0\)
\(x^2 + m^2x^2 + 4mx + 4 - 6x + 8mx + 16 - 11 = 0\)
\((1 + m^2)x^2 + (12m - 6)x + 9 = 0\)

For the line to be a tangent to the circle, the quadratic equation must have exactly one real root, so its discriminant must be zero (\(b^2 - 4ac = 0\)):
\((12m - 6)^2 - 4(1 + m^2)(9) = 0\)
\((12m - 6)^2 - 36(1 + m^2) = 0\)

Notice that \((12m - 6)^2 = [6(2m - 1)]^2 = 36(2m - 1)^2\). Substitute this back:
\(36(2m - 1)^2 - 36(1 + m^2) = 0\)

Divide the entire equation by 36:
\((2m - 1)^2 - (1 + m^2) = 0\)
\(4m^2 - 4m + 1 - 1 - m^2 = 0\)
\(3m^2 - 4m = 0\)
\(m(3m - 4) = 0\)

Therefore, the possible values of the constant \(m\) are \(m = 0\) or \(m = \frac{4}{3}\).

M1: Completes the square for \(x\) and \(y\) to find the center or radius.
A1: Correct center \(C(3, -4)\) and radius \(r = 6\).
M1: Substitutes \(y = mx + 2\) into the circle equation to form a quadratic in \(x\).
A1: Obtains the correct quadratic equation: \((1+m^2)x^2 + (12m-6)x + 9 = 0\).
M1: Sets the discriminant to zero and attempts to solve the resulting equation for \(m\).
A1: Obtains the correct values: \(m = 0\) and \(m = \frac{4}{3}\).

To find the stationary points, we first find the first derivative of the function \(y\) with respect to \(x\):

\(\frac{dy}{dx} = \frac{1}{3} \cdot 3(2x - 1)^2 \cdot 2 - 8 = 2(2x - 1)^2 - 8\).

Set \(\frac{dy}{dx} = 0\):

\(2(2x - 1)^2 - 8 = 0\)

\((2x - 1)^2 = 4\)

Taking the square root of both sides:

\(2x - 1 = 2\) or \(2x - 1 = -2\)

From \(2x - 1 = 2\), we get \(2x = 3 \implies x = 1.5\).

From \(2x - 1 = -2\), we get \(2x = -1 \implies x = -0.5\).

Now, calculate the corresponding \(y\)-coordinates:

For \(x = 1.5\):

\(y = \frac{1}{3}(2(1.5) - 1)^3 - 8(1.5) = \frac{1}{3}(2)^3 - 12 = \frac{8}{3} - 12 = -\frac{28}{3} \approx -9.33\).

For \(x = -0.5\):

\(y = \frac{1}{3}(2(-0.5) - 1)^3 - 8(-0.5) = \frac{1}{3}(-2)^3 + 4 = -\frac{8}{3} + 4 = \frac{4}{3} \approx 1.33\).

Next, find the second derivative to determine the nature of these points:

\(\frac{d^2y}{dx^2} = 4(2x - 1) \cdot 2 = 8(2x - 1)\).

Evaluate at \(x = 1.5\):

\(\frac{d^2y}{dx^2} = 8(2(1.5) - 1) = 16 > 0\).

Since the second derivative is positive, \((1.5, -\frac{28}{3})\) (or \((1.5, -9.33)\)) is a minimum point.

Evaluate at \(x = -0.5\):

\(\frac{d^2y}{dx^2} = 8(2(-0.5) - 1) = -16 < 0\).

Since the second derivative is negative, \((-0.5, \frac{4}{3})\) (or \((-0.5, 1.33)\)) is a maximum point.

M1: Attempt to differentiate the function using the chain rule.
A1: Correct derivative \(\frac{dy}{dx} = 2(2x - 1)^2 - 8\).
M1: Setting \(\frac{dy}{dx} = 0\) and attempting to solve for \(x\).
A1: Correct \(x\)-values: \(x = 1.5\) and \(x = -0.5\).
A1: Correct corresponding \(y\)-values: \(-\frac{28}{3}\) (or \(-9.33\)) and \(\frac{4}{3}\) (or \(1.33\)).
M1: Correctly finding the second derivative \(\frac{d^2y}{dx^2} = 8(2x - 1)\) (or using alternative sign-table method) to determine the nature of the stationary points.
A1: Correctly concluding that \((1.5, -9.33)\) is a minimum point and \((-0.5, 1.33)\) is a maximum point.

First, change the base of \(\log_{25}(x+2)\) to base 5:

\(\log_{25}(x+2) = \frac{\log_5(x+2)}{\log_5 25} = \frac{\log_5(x+2)}{2} = \frac{1}{2}\log_5(x+2)\).

Substitute this back into the original equation:

\(\log_5(x + 6) - \frac{1}{2}\log_5(x + 2) = 1\).

Multiply the entire equation by 2 to clear the fraction:

\(2\log_5(x + 6) - \log_5(x + 2) = 2\).

Apply the power law of logarithms to the first term:

\(\log_5((x + 6)^2) - \log_5(x + 2) = 2\).

Apply the quotient law to combine the logarithms:

\(\log_5\left(\frac{(x+6)^2}{x+2}\right) = 2\).

Convert the logarithmic equation into its exponential form:

\(\frac{(x+6)^2}{x+2} = 5^2\)

\(\frac{x^2 + 12x + 36}{x + 2} = 25\)

Multiply both sides by \(x+2\):

\(x^2 + 12x + 36 = 25(x + 2)\)

\(x^2 + 12x + 36 = 25x + 50\)

Rearrange into a quadratic equation set to zero:

\(x^2 - 13x - 14 = 0\).

Factorize the quadratic equation:

\((x - 14)(x + 1) = 0\).

This gives \(x = 14\) or \(x = -1\).

Now we must verify if these values are valid in the original logarithmic expression:
- For \(x = 14\): \(x+6 = 20 > 0\) and \(x+2 = 16 > 0\) (valid).
- For \(x = -1\): \(x+6 = 5 > 0\) and \(x+2 = 1 > 0\) (valid).

Thus, both solutions are valid.

M1: For applying the change of base formula correctly to convert \(\log_{25}(x+2)\) into base 5.
A1: Obtaining \(\frac{1}{2}\log_5(x+2)\).
M1: Using the power law of logarithms to rewrite \(2\log_5(x+6)\) as \(\log_5(x+6)^2\).
M1: Using the quotient law of logarithms to combine the terms on the left side.
A1: Correctly writing the equation in exponential form: \(\frac{(x+6)^2}{x+2} = 25\).
M1: For simplifying and obtaining the correct quadratic equation \(x^2 - 13x - 14 = 0\) and solving it.
A1: Finding the two values \(x = 14\) and \(x = -1\) (with verification of both).

To solve \(3\cos(2\theta) + 7\sin\theta - 5 = 0\), we first use the double-angle identity for cosine: \(\cos(2\theta) = 1 - 2\sin^2\theta\).

Substitute this identity into the equation:

\(3(1 - 2\sin^2\theta) + 7\sin\theta - 5 = 0\)

Expand the expression:

\(3 - 6\sin^2\theta + 7\sin\theta - 5 = 0\)

Simplify and rearrange into a quadratic equation in terms of \(\sin\theta\):

\(-6\sin^2\theta + 7\sin\theta - 2 = 0\)

Multiply by \(-1\):

\(6\sin^2\theta - 7\sin\theta + 2 = 0\)

Factorize the quadratic equation:

\((2\sin\theta - 1)(3\sin\theta - 2) = 0\)

This gives two cases:

Case 1: \(2\sin\theta - 1 = 0 \implies \sin\theta = \frac{1}{2}\)
Within the range \(0^\circ \le \theta \le 360^\circ\):
\(\theta = 30^\circ\) or \(\theta = 180^\circ - 30^\circ = 150^\circ\).

Case 2: \(3\sin\theta - 2 = 0 \implies \sin\theta = \frac{2}{3}\)
Find the principal angle:
\(\theta = \sin^{-1}\left(\frac{2}{3}\right) \approx 41.8^\circ\) (to 1 decimal place).

The second solution in the second quadrant is:
\(\theta = 180^\circ - 41.81^\circ \approx 138.2^\circ\) (to 1 decimal place).

Combining all solutions, we have:
\(\theta = 30^\circ, 41.8^\circ, 138.2^\circ, 150^\circ\).

M1: For using the identity \(\cos(2\theta) = 1 - 2\sin^2\theta\) to rewrite the equation in terms of \(\sin\theta\) only.
A1: Correct quadratic equation \(6\sin^2\theta - 7\sin\theta + 2 = 0\).
M1: Factorizing or solving the quadratic equation to find values for \(\sin\theta\).
A1: Finding the two values \(\sin\theta = \frac{1}{2}\) and \(\sin\theta = \frac{2}{3}\).
A1: Finding \(\theta = 30^\circ\) and \(\theta = 150^\circ\).
A1: Finding the principal angle \(\theta = 41.8^\circ\) (accept \(41.8\)).
A1: Finding the second quadrant angle \(\theta = 138.2^\circ\) (accept \(138\)).

The general term in the binomial expansion of \((a + b)^n\) is given by:

\(T_{r+1} = \binom{n}{r} a^{n-r} b^r\).

Here, \(a = x^2\), \(b = -\frac{k}{x} = -k x^{-1}\), and \(n = 8\). Substituting these in:

\(T_{r+1} = \binom{8}{r} (x^2)^{8-r} (-k x^{-1})^r\)

\(T_{r+1} = \binom{8}{r} x^{16-2r} (-k)^r x^{-r}\)

\(T_{r+1} = \binom{8}{r} (-k)^r x^{16-3r}\).

We are looking for the term containing \(x^7\), so we set the exponent of \(x\) equal to 7:

\(16 - 3r = 7\)

\(3r = 9 \implies r = 3\).

Now substitute \(r = 3\) back into the term to find the coefficient:

\(T_4 = \binom{8}{3} (-k)^3 x^7 = 56 \cdot (-k^3) x^7 = -56k^3 x^7\).

The coefficient is given as \(-1512\), so:

\(-56k^3 = -1512\)

\(k^3 = \frac{-1512}{-56}\)

\(k^3 = 27\)

Taking the cube root of both sides, we get:

\(k = 3\).

M1: Writing down the general term of the binomial expansion with general index \(r\).
A1: Simplifying the powers of \(x\) to obtain the index of \(x\) as \(16 - 3r\).
M1: Setting \(16 - 3r = 7\) and solving to find \(r = 3\).
A1: Finding the binomial coefficient \(\binom{8}{3} = 56\).
M1: Equating the coefficient containing \(k\) to \(-1512\), taking into account the negative sign from the power of \(-k\).
A1: Correct equation \(-56k^3 = -1512\) leading to \(k^3 = 27\).
A1: Correctly solving to get \(k = 3\).

(a) To find the center and the radius, we rewrite the equation by completing the square for both \(x\) and \(y\):

\(x^2 - 4x + y^2 + 6y = 12\)

\((x - 2)^2 - 4 + (y + 3)^2 - 9 = 12\)

\((x - 2)^2 + (y + 3)^2 = 12 + 4 + 9\)

\((x - 2)^2 + (y + 3)^2 = 25\).

Comparing this to the standard circle equation \((x-h)^2 + (y-k)^2 = r^2\):

- The coordinates of the center \(C\) are \((2, -3)\).
- The radius \(r\) is \(\sqrt{25} = 5\).

(b) The point \(P(5, 1)\) lies on the circle. The line from the center \(C(2, -3)\) to \(P(5, 1)\) is the radius. The gradient of this radius \(CP\) is:

\(m_{\text{radius}} = \frac{1 - (-3)}{5 - 2} = \frac{4}{3}\).

Since the tangent is perpendicular to the radius at the point of contact, the gradient of the tangent \(m_{\text{tangent}}\) is:

\(m_{\text{tangent}} = -\frac{1}{m_{\text{radius}}} = -\frac{3}{4}\).

Using the equation of a line with gradient \(-\frac{3}{4}\) passing through \(P(5, 1)\):

\(y - 1 = -\frac{3}{4}(x - 5)\)

Multiply by 4 to clear the fraction:

\(4(y - 1) = -3(x - 5)\)

\(4y - 4 = -3x + 15\)

\(3x + 4y - 19 = 0\).

M1: For attempting to complete the square for both \(x\) and \(y\).
A1: Correct coordinates of the center \((2, -3)\) and radius \(r = 5\).
M1: Attempting to calculate the gradient of the radius \(CP\) using the center and \(P(5, 1)\).
A1: Correct gradient of radius as \(\frac{4}{3}\).
M1: Using \(m_1 m_2 = -1\) to find the gradient of the tangent as \(-\frac{3}{4}\).
M1: Attempting to write down the equation of a straight line passing through \((5, 1)\) with their tangent gradient.
A1: Correct equation of the tangent in any equivalent form, e.g. \(3x + 4y - 19 = 0\) or \(y = -\frac{3}{4}x + \frac{19}{4}\).

To find where the line and curve do not intersect, we set their equations equal to each other:

\(2kx - 3 = x^2 + 2x + k^2\).

Rearrange this into a standard quadratic equation of the form \(ax^2 + bx + c = 0\):

\(x^2 + 2x - 2kx + k^2 + 3 = 0\)

\(x^2 + (2 - 2k)x + (k^2 + 3) = 0\).

Here, the coefficients are:
- \(a = 1\)
- \(b = 2 - 2k\)
- \(c = k^2 + 3\).

For the line and curve not to intersect, the quadratic equation must have no real roots. This means the discriminant must be less than zero (\(b^2 - 4ac < 0\)):

\((2 - 2k)^2 - 4(1)(k^2 + 3) < 0\)

Expand both parts of the inequality:

\((4 - 8k + 4k^2) - 4(k^2 + 3) < 0\)

\(4 - 8k + 4k^2 - 4k^2 - 12 < 0\).

Simplify by combining like terms:

\(-8k - 8 < 0\).

Solve the linear inequality for \(k\):

\(-8k < 8\)

Divide by \(-8\) and reverse the inequality sign:

\(k > -1\).

M1: For equating the line and the curve and attempting to rearrange into a quadratic equation.
A1: Correct quadratic equation in standard form, e.g. \(x^2 + (2-2k)x + (k^2+3) = 0\).
M1: Setting up the discriminant inequality \(b^2 - 4ac < 0\) for no intersection.
A1: Correct substitution of coefficients into the discriminant: \((2-2k)^2 - 4(k^2+3) < 0\).
M1: Correctly expanding and simplifying the discriminant to a linear expression.
A1: Obtaining \(-8k - 8 < 0\).
A1: Correct final range of values \(k > -1\).

(a) Let \(Y = \lg y\) and \(X = x^2\). The relationship between \(Y\) and \(X\) is linear, represented by \(Y = mX + c\).

The gradient \(m\) of the line passing through \((2, 5)\) and \((6, 13)\) is:

\(m = \frac{13 - 5}{6 - 2} = \frac{8}{4} = 2\).

Using the point-slope form with the point \((2, 5)\):

\(Y - 5 = 2(X - 2)\)

\(Y - 5 = 2X - 4\)

\(Y = 2X + 1\).

Substituting back \(Y = \lg y\) and \(X = x^2\):

\(\lg y = 2x^2 + 1\).

(b) To express \(y\) in terms of \(x\), rewrite the logarithmic equation in exponential form (using base 10):

\(y = 10^{2x^2 + 1}\).

Using the laws of indices:

\(y = 10^1 \cdot 10^{2x^2}\)

\(y = 10 \cdot (10^2)^{x^2}\)

\(y = 10 \cdot 100^{x^2}\).

This is in the form \(y = A \cdot B^{x^2}\), where \(A = 10\) and \(B = 100\).

M1: Finding the gradient of the linear equation using \(m = \frac{13 - 5}{6 - 2}\).
A1: Correct gradient of \(m = 2\).
M1: Substituting gradient and one point to find the vertical intercept.
A1: Correct expression \(\lg y = 2x^2 + 1\).
M1: Converting the logarithmic equation into exponential form base 10.
A1: Correct power manipulation showing \(10^{2x^2 + 1} = 10^1 \cdot 10^{2x^2}\).
A1: Correctly determining \(A = 10\) and \(B = 100\) to give \(y = 10 \cdot 100^{x^2}\).

(a) To find \(\mathrm{f}^{-1}(x)\), let \(y = \mathrm{f}(x)\):

\(y = \frac{2x + 5}{x - 3}\).

Rearrange the equation to make \(x\) the subject:

\(y(x - 3) = 2x + 5\)

\(xy - 3y = 2x + 5\)

Collect terms involving \(x\) on one side:

\(xy - 2x = 3y + 5\)

Factor out \(x\):

\(x(y - 2) = 3y + 5\)

\(x = \frac{3y + 5}{y - 2}\).

Therefore, replacing \(y\) with \(x\):

\(\mathrm{f}^{-1}(x) = \frac{3x + 5}{x - 2}\) for \(x \neq 2\).

(b) To solve \(\mathrm{f}^2(x) = 5\), we write it as \(\mathrm{f}(\mathrm{f}(x)) = 5\).

Applying the inverse function to both sides:

\(\mathrm{f}(x) = \mathrm{f}^{-1}(5)\).

First, compute the value of \(\mathrm{f}^{-1}(5)\) using the expression from part (a):

\(\mathrm{f}^{-1}(5) = \frac{3(5) + 5}{5 - 2} = \frac{15 + 5}{3} = \frac{20}{3}\).

Now, solve \(\mathrm{f}(x) = \frac{20}{3}\):

\[ \frac{2x + 5}{x - 3} = \frac{20}{3} \]

Cross-multiply to solve for \(x\):

\(3(2x + 5) = 20(x - 3)\)

\(6x + 15 = 20x - 60\)

Subtract \(6x\) from both sides and add \(60\) to both sides:

\(15 + 60 = 20x - 6x\)

\(75 = 14x\)

\(x = \frac{75}{14}\) (or approximately \(5.36\)).

M1: Setting \(y = \frac{2x+5}{x-3}\) and attempting to isolate the \(x\)-terms.
A1: Factorizing to get \(x(y-2) = 3y+5\) or equivalent.
A1: Correct expression \(\mathrm{f}^{-1}(x) = \frac{3x+5}{x-2}\) (stating the domain \(x \neq 2\) is optional but preferred).
M1: Realizing that \(\mathrm{f}^2(x) = 5\) is equivalent to \(\mathrm{f}(x) = \mathrm{f}^{-1}(5)\) (or attempting to find composite function \(\mathrm{f}(\mathrm{f}(x))\) algebraically).
A1: Finding \(\mathrm{f}^{-1}(5) = \frac{20}{3}\).
M1: Setting up the linear equation \(\frac{2x+5}{x-3} = \frac{20}{3}\) and cross-multiplying.
A1: Correct value \(x = \frac{75}{14}\) (accept equivalent fractions or 3 s.f. decimal \(5.36\)).

First, integrate \(\frac{\text{d}y}{\text{d}x}\) to find the general equation of the curve: \
\(y = \int \left( 4(3x+1)^{-\frac{1}{2}} + e^{2x} \right) \text{d}x\) \
\(y = 4 \cdot \frac{(3x+1)^{\frac{1}{2}}}{\frac{1}{2} \cdot 3} + \frac{1}{2}e^{2x} + C\) \
\(y = \frac{8}{3}\sqrt{3x+1} + \frac{1}{2}e^{2x} + C\) \
\
Now, use the given point \((0, 5)\) to find the constant of integration, \(C\): \
\(5 = \frac{8}{3}\sqrt{3(0)+1} + \frac{1}{2}e^{2(0)} + C\) \
\(5 = \frac{8}{3}(1) + \frac{1}{2}(1) + C\) \
\(5 = \frac{16+3}{6} + C\) \
\(5 = \frac{19}{6} + C \implies C = \frac{11}{6}\) \
\
This gives the equation of the curve as: \
\(y = \frac{8}{3}\sqrt{3x+1} + \frac{1}{2}e^{2x} + \frac{11}{6}\) \
\
Substitute \(x = 1\) into the equation to find \(y\): \
\(y = \frac{8}{3}\sqrt{3(1)+1} + \frac{1}{2}e^{2(1)} + \frac{11}{6}\) \
\(y = \frac{8}{3}(2) + \frac{1}{2}e^2 + \frac{11}{6}\) \
\(y = \frac{16}{3} + \frac{11}{6} + \frac{1}{2}e^2\) \
\(y = \frac{32 + 11}{6} + \frac{1}{2}e^2\) \
\(y = \frac{43}{6} + \frac{1}{2}e^2\)

M1: Attempt to integrate the gradient function with at least one term correct. \
A1: Correct integration of the first term to get \(\frac{8}{3}(3x+1)^{\frac{1}{2}}\). \
A1: Correct integration of the second term to get \(\frac{1}{2}e^{2x}\). \
M1: Substitute \(x = 0\) and \(y = 5\) into their integrated expression containing \(C\). \
A1: Correct value of \(C = \frac{11}{6}\). \
M1: Substitute \(x = 1\) into their curve equation. \
A1: Correct exact value of \(y = \frac{43}{6} + \frac{1}{2}e^2\) (or any equivalent exact fraction/decimal form, approx 10.9).

Let the center of the circle be \((h, k)\). Since the center lies on the line \(y = 2x - 3\), we have: \
\(k = 2h - 3\) \
\
Because \(P(1, 2)\) and \(Q(5, 4)\) lie on the circle, the distance from the center to \(P\) is equal to the distance from the center to \(Q\): \
\((h-1)^2 + (k-2)^2 = (h-5)^2 + (k-4)^2\) \
\
Substitute \(k = 2h - 3\) into this equation: \
\((h-1)^2 + (2h-3-2)^2 = (h-5)^2 + (2h-3-4)^2\) \
\((h-1)^2 + (2h-5)^2 = (h-5)^2 + (2h-7)^2\) \
\
Expand both sides: \
\((h^2 - 2h + 1) + (4h^2 - 20h + 25) = (h^2 - 10h + 25) + (4h^2 - 28h + 49)\) \
\(5h^2 - 22h + 26 = 5h^2 - 38h + 74\) \
\
Simplify by subtracting \(5h^2\) from both sides: \
\(-22h + 26 = -38h + 74\) \
\(16h = 48\) \
\(h = 3\) \
\
Now, find \(k\): \
\(k = 2(3) - 3 = 3\) \
So, the center of the circle is \((3, 3)\). \
\
Next, find the radius squared, \(r^2\), using point \(P(1, 2)\): \
\(r^2 = (3-1)^2 + (3-2)^2 = 2^2 + 1^2 = 5\) \
\
Write the standard equation of the circle: \
\((x-3)^2 + (y-3)^2 = 5\) \
\
Expand to find the general form: \
\(x^2 - 6x + 9 + y^2 - 6y + 9 = 5\) \
\(x^2 + y^2 - 6x - 6y + 13 = 0\)

M1: Set up distance equation between the center \((h, k)\) and the points \(P\) and \(Q\). \
M1: Substitute \(k = 2h - 3\) (or equivalent relationship) into the distance equation. \
A1: Obtain a correct simplified linear equation in one variable, e.g., \(16h = 48\). \
A1: Correctly find the center as \((3, 3)\). \
M1: Use their center and either point \(P\) or \(Q\) to find the radius squared, \(r^2 = 5\). \
M1: Expand \((x-h)^2 + (y-k)^2 = r^2\) into the general polynomial form. \
A1: Correct final equation: \(x^2 + y^2 - 6x - 6y + 13 = 0\).

Divide both sides of the equation by \(2\cos\left(2\theta - \frac{\pi}{6}\right)\) (assuming it is non-zero) to express the equation in terms of tangent: \
\(\frac{3\sin\left(2\theta - \frac{\pi}{6}\right)}{\cos\left(2\theta - \frac{\pi}{6}\right)} = 2\) \
\(3\tan\left(2\theta - \frac{\pi}{6}\right) = 2\) \
\(\tan\left(2\theta - \frac{\pi}{6}\right) = \frac{2}{3}\) \
\
Let \(\phi = 2\theta - \frac{\pi}{6}\). Since \(0 \le \theta \le \pi\), we must determine the range for \(\phi\): \
\(0 \le 2\theta \le 2\pi\) \
\(-\frac{\pi}{6} \le 2\theta - \frac{\pi}{6} \le 2\pi - \frac{\pi}{6}\) \
\(-\frac{\pi}{6} \le \phi \le \frac{11\pi}{6}\) \
In decimal form, this interval is approximately \(-0.524 \le \phi \le 5.760\). \
\
Now, solve \(\tan\phi = \frac{2}{3}\): \
Find the basic angle \(\alpha = \tan^{-1}\left(\frac{2}{3}\right) \approx 0.5880\) radians. \
\
Since the tangent is positive, \(\phi\) lies in Quadrant 1 and Quadrant 3. \
Quadrant 1: \(\phi = 0.5880\) \
Quadrant 3: \(\phi = \pi + 0.5880 \approx 3.7296\) \
Both values lie within the interval \([-0.524, 5.760]\). \
\
Now solve for \(\theta\) using \(2\theta = \phi + \frac{\pi}{6}\): \
\
For \(\phi = 0.5880\): \
\(2\theta = 0.5880 + \frac{\pi}{6}\) \
\(2\theta \approx 0.5880 + 0.5236 = 1.1116\) \
\(\theta \approx 0.5558 \approx 0.556\) radians. \
\
For \(\phi = 3.7296\): \
\(2\theta = 3.7296 + \frac{\pi}{6}\) \
\(2\theta \approx 3.7296 + 0.5236 = 4.2532\) \
\(\theta \approx 2.1266 \approx 2.13\) radians.

M1: Use the identity \(\tan x = \frac{\sin x}{\cos x}\) to obtain \(\tan\left(2\theta - \frac{\pi}{6}\right) = \frac{2}{3}\). \
M1: State the correct interval for the substituted angle, e.g., \(-\frac{\pi}{6} \le 2\theta - \frac{\pi}{6} \le \frac{11\pi}{6}\) (or equivalent decimal interval). \
M1: Find the basic angle \(\alpha \approx 0.588\) radians. \
A1: Obtain the first correct angle for \(2\theta - \frac{\pi}{6} = 0.588\) and second correct angle \(2\theta - \frac{\pi}{6} = 3.73\). \
M1: Correct process to solve for \(\theta\) from their values of \(\phi\). \
A1: One correct value of \(\theta\) (either \(0.556\) or \(2.13\)) rounded to 3 significant figures. \
A1: Both correct values of \(\theta\) (\(0.556\) and \(2.13\)) with no extra values in the range.