To find the stationary points, we need to find where the derivative \(\frac{dy}{dx}\) is zero.

Using the quotient rule, the derivative is:

\(\frac{dy}{dx} = \frac{6x(1-x^2)e^{3x^2-1} + 2xe^{3x^2-1}}{(1-x^2)^2}\).

Simplifying the numerator, we have:

\(6x(1-x^2) + 2x = 6x - 6x^3 + 2x = 8x - 6x^3\).

Setting the numerator equal to zero gives:

\(8x - 6x^3 = 0\).

Factoring out \(2x\), we get:

\(2x(4 - 3x^2) = 0\).

This gives solutions \(x = 0\) or \(4 - 3x^2 = 0\).

Solving \(4 - 3x^2 = 0\) gives \(x^2 = \frac{4}{3}\), so \(x = \pm \frac{2\sqrt{3}}{3}\).

Substituting these \(x\)-values back into the original equation to find \(y\):

For \(x = 0\), \(y = e^{-1}\).

For \(x = \pm \frac{2\sqrt{3}}{3}\), \(y = -3e^3\).

Thus, the stationary points are \((0, e^{-1})\), \(\left( \frac{2\sqrt{3}}{3}, -3e^3 \right)\), and \(\left( -\frac{2\sqrt{3}}{3}, -3e^3 \right)\).

(a) To find the stationary point, we need to differentiate \(y = e^{2x}(\sin x + 3 \cos x)\) and set the derivative to zero.

Using the product rule, the derivative is:

\(y' = e^{2x}(2(\sin x + 3 \cos x)) + e^{2x}(\cos x - 3 \sin x)\)

Simplifying, we get:

\(y' = e^{2x}(2 \sin x + 6 \cos x + \cos x - 3 \sin x)\)

\(y' = e^{2x}(-\sin x + 7 \cos x)\)

Setting \(y' = 0\), we have:

\(-\sin x + 7 \cos x = 0\)

\(\tan x = 7\)

Solving for \(x\) in the interval \(0 \leq x \leq \pi\), we find:

\(x = 1.43\) (to 2 decimal places)

(b) To determine the nature of the stationary point, we check the sign of \(y'\) around \(x = 1.43\).

For \(x = 1.42\), \(y' = 0.06e^{2.84} > 0\)

For \(x = 1.44\), \(y' = -0.07e^{2.88} < 0\)

Since \(y'\) changes from positive to negative, the stationary point at \(x = 1.43\) is a maximum point.

To find the derivative \(\frac{dy}{dx}\) of \(y = \frac{e^{-2x}}{1-x^2}\), we use the quotient rule:

\(\frac{dy}{dx} = \frac{(1-x^2)(-2e^{-2x}) - e^{-2x}(-2x)}{(1-x^2)^2}\)

Simplifying, we get:

\(\frac{dy}{dx} = \frac{-2e^{-2x}(1-x^2) + 2xe^{-2x}}{(1-x^2)^2}\)

To find the stationary point, set \(\frac{dy}{dx} = 0\):

\(-2e^{-2x}(1-x^2) + 2xe^{-2x} = 0\)

Factor out \(e^{-2x}\):

\(e^{-2x}(-2(1-x^2) + 2x) = 0\)

Since \(e^{-2x} \neq 0\), solve:

\(-2 + 2x^2 + 2x = 0\)

Rearrange to form a quadratic equation:

\(2x^2 + 2x - 2 = 0\)

Divide by 2:

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

Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1, b = 1, c = -1\):

\(x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1}\)

\(x = \frac{-1 \pm \sqrt{5}}{2}\)

Calculate the positive root within the interval \(-1 < x < 1\):

\(x = \frac{-1 + \sqrt{5}}{2} \approx 0.618\)

To find the stationary points, we need to find where the derivative \(\frac{dy}{dx}\) is zero.

Using the quotient rule for differentiation, \(\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\), where \(u = e^{3x}\) and \(v = \tan \frac{1}{2}x\).

First, find \(\frac{du}{dx} = 3e^{3x}\) and \(\frac{dv}{dx} = \frac{1}{2} \sec^2 \frac{1}{2}x\).

Substitute into the quotient rule:

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

Simplify and set \(\frac{dy}{dx} = 0\):

\(3\tan \frac{1}{2}x = \frac{1}{2} \sec^2 \frac{1}{2}x\).

Rearrange to form a quadratic in \(\tan \frac{1}{2}x\):

\(6\tan^2 \frac{1}{2}x = \sec^2 \frac{1}{2}x\).

Using \(\sec^2 \theta = 1 + \tan^2 \theta\), substitute:

\(6\tan^2 \frac{1}{2}x = 1 + \tan^2 \frac{1}{2}x\).

Solve for \(\tan \frac{1}{2}x\):

\(5\tan^2 \frac{1}{2}x = 1\).

\(\tan \frac{1}{2}x = \pm \frac{1}{\sqrt{5}}\).

Find \(x\) in the interval \(0 < x < \pi\):

\(x = 2 \arctan \left( \frac{1}{\sqrt{5}} \right) \approx 0.340\).

\(x = 2 \arctan \left( -\frac{1}{\sqrt{5}} \right) + \pi \approx 2.802\).

Thus, the \(x\)-coordinates of the stationary points are \(0.340\) and \(2.802\).

(i) To find the stationary point, we need to differentiate \(y = \frac{2 - \\sin x}{\\cos x}\).

Rewrite the function as \(y = 2\\sec x - \\tan x\).

The derivative is \(y' = 2\\sec x \\tan x - \\sec^2 x\).

Set the derivative to zero: \(2\\sec x \\tan x - \\sec^2 x = 0\).

Factor to get \(\\sec x (2\\tan x - \\sec x) = 0\).

Since \(\\sec x \neq 0\), solve \(2\\tan x = \\sec x\).

Using \(\\tan x = \\frac{\\sin x}{\\cos x}\) and \(\\sec x = \\frac{1}{\\cos x}\), we have \(2\\frac{\\sin x}{\\cos x} = \\frac{1}{\\cos x}\).

Simplify to \(2\\sin x = 1\), giving \(\\sin x = \\frac{1}{2}\).

In the interval \(-\frac{1}{2}\pi < x < \frac{1}{2}\pi\), \(x = \frac{1}{6}\pi\).

Substitute back to find \(y\): \(y = \frac{2 - \\sin(\frac{1}{6}\pi)}{\\cos(\frac{1}{6}\pi)} = \sqrt{3}\).

Thus, the coordinates are \(\left( \frac{1}{6}\pi, \sqrt{3} \right)\).

(ii) To determine the nature of the stationary point, consider the second derivative or test values around \(x = \frac{1}{6}\pi\).

Since the mark scheme indicates it is a minimum, we conclude it is a minimum point.

To find the gradient of the curve, we need to differentiate \(y = \frac{\sin x}{1 + \cos x}\) with respect to \(x\).

Using the quotient rule, \(\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\), where \(u = \sin x\) and \(v = 1 + \cos x\).

Then, \(\frac{du}{dx} = \cos x\) and \(\frac{dv}{dx} = -\sin x\).

Substituting these into the quotient rule gives:

\(\frac{dy}{dx} = \frac{(1 + \cos x)(\cos x) - (\sin x)(-\sin x)}{(1 + \cos x)^2}\)

Simplifying the numerator:

\((1 + \cos x)\cos x + \sin^2 x = \cos x + \cos^2 x + \sin^2 x\)

Using the identity \(\cos^2 x + \sin^2 x = 1\), the numerator becomes:

\(\cos x + 1\)

Thus, \(\frac{dy}{dx} = \frac{1}{1 + \cos x}\).

Since \(-1 < \cos x < 1\) for \(-\pi < x < \pi\), \(1 + \cos x > 0\).

Therefore, \(\frac{1}{1 + \cos x} > 0\), showing that the gradient is positive for all \(x\) in the interval.

To find the stationary points, we need to find the derivative of \(y = \frac{{(\ln x)^2}}{x}\) and set it to zero.

Using the quotient rule, if \(u = (\ln x)^2\) and \(v = x\), then \(y = \frac{u}{v}\).

The derivative is given by:

\(\frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}\)

\(\frac{du}{dx} = 2 \ln x \cdot \frac{1}{x} = \frac{2 \ln x}{x}\)

\(\frac{dv}{dx} = 1\)

Substituting these into the quotient rule:

\(\frac{dy}{dx} = \frac{x \cdot \frac{2 \ln x}{x} - (\ln x)^2 \cdot 1}{x^2}\)

\(= \frac{2 \ln x - (\ln x)^2}{x^2}\)

Setting \(\frac{dy}{dx} = 0\) gives:

\(2 \ln x - (\ln x)^2 = 0\)

\(\ln x (2 - \ln x) = 0\)

This gives \(\ln x = 0\) or \(\ln x = 2\).

If \(\ln x = 0\), then \(x = e^0 = 1\).

If \(\ln x = 2\), then \(x = e^2\).

For \(x = 1\), \(y = \frac{(\ln 1)^2}{1} = 0\).

For \(x = e^2\), \(y = \frac{(\ln e^2)^2}{e^2} = \frac{4}{e^2} = 4e^{-2}\).

Thus, the stationary points are \((1, 0)\) and \(\left(e^2, 4e^{-2}\right)\).

To find the stationary point, we need to differentiate \(y = \\sin x \\cos 2x\) and set the derivative to zero.

Using the product rule, \(\frac{d}{dx}(uv) = u'v + uv'\), where \(u = \\sin x\) and \(v = \\cos 2x\).

Then, \(u' = \\cos x\) and \(v' = -2 \\sin 2x\).

The derivative is:

\(\frac{dy}{dx} = \\cos x \\cos 2x - 2 \\sin x \\sin 2x\).

Using the double angle identity \(\\sin 2x = 2 \\sin x \\cos x\), the derivative becomes:

\(\frac{dy}{dx} = \\cos x \\cos 2x - 4 \\sin x \\cos x \\sin x\).

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

\(\\cos x \\cos 2x - 4 \\sin^2 x \\cos x = 0\).

Factor out \(\\cos x\):

\(\\cos x (\\cos 2x - 4 \\sin^2 x) = 0\).

Since \(\\cos x \neq 0\) in the interval \(0 < x < \frac{1}{2} \pi\), we have:

\(\\cos 2x - 4 \\sin^2 x = 0\).

Using the identity \(\\cos 2x = 1 - 2 \\sin^2 x\), substitute:

\(1 - 2 \\sin^2 x - 4 \\sin^2 x = 0\).

Simplify to get:

\(1 - 6 \\sin^2 x = 0\).

\(6 \\sin^2 x = 1\).

\(\\sin^2 x = \frac{1}{6}\).

\(\\sin x = \frac{1}{\sqrt{6}}\).

Solving for \(x\) in the interval \(0 < x < \frac{1}{2} \pi\), we find:

\(x \approx 0.421\).

To find the stationary point, we need to find where the derivative \(\frac{dy}{dx}\) is zero.

Given \(y = \frac{e^{2x}}{4 + e^{3x}}\), use the quotient rule: \(\frac{dy}{dx} = \frac{(4 + e^{3x})(2e^{2x}) - e^{2x}(3e^{3x})}{(4 + e^{3x})^2}\).

Simplify the numerator: \((4 + e^{3x})(2e^{2x}) - e^{2x}(3e^{3x}) = 8e^{2x} + 2e^{5x} - 3e^{5x} = 8e^{2x} - e^{5x}\).

Set the derivative to zero: \(8e^{2x} - e^{5x} = 0\).

Factor out \(e^{2x}\): \(e^{2x}(8 - e^{3x}) = 0\).

Since \(e^{2x} \neq 0\), solve \(8 - e^{3x} = 0\) to get \(e^{3x} = 8\).

Taking the natural logarithm: \(3x = \ln 8\), so \(x = \frac{\ln 8}{3} = \ln 2\).

Substitute \(x = \ln 2\) back into the original equation to find \(y\):

\(y = \frac{e^{2 \ln 2}}{4 + e^{3 \ln 2}} = \frac{4}{4 + 8} = \frac{4}{12} = \frac{1}{3}\).

Thus, the coordinates of the stationary point are \((\ln 2, \frac{1}{3})\).

To find the stationary points, we need to differentiate \(y = \cos x \cos 2x\) and set the derivative to zero.

Using the product rule, \(\frac{d}{dx}(uv) = u'v + uv'\), where \(u = \cos x\) and \(v = \cos 2x\).

First, find the derivatives: \(u' = -\sin x\) and \(v' = -2\sin 2x\).

Using the product rule: \(\frac{dy}{dx} = (-\sin x)(\cos 2x) + (\cos x)(-2\sin 2x)\).

Simplify: \(\frac{dy}{dx} = -\sin x \cos 2x - 2\cos x \sin 2x\).

Using the double angle formula \(\sin 2x = 2\sin x \cos x\), rewrite \(\cos 2x = 1 - 2\sin^2 x\).

Substitute and simplify: \(\frac{dy}{dx} = -\sin x (1 - 2\sin^2 x) - 4\cos^2 x \sin x\).

Factor out \(\sin x\): \(\frac{dy}{dx} = -\sin x (1 - 2\sin^2 x + 4\cos^2 x)\).

Set \(\frac{dy}{dx} = 0\): \(-\sin x (1 - 2\sin^2 x + 4\cos^2 x) = 0\).

Since \(\sin x = 0\) is not in the interval, solve \(1 - 2\sin^2 x + 4\cos^2 x = 0\).

Using \(\cos^2 x = 1 - \sin^2 x\), substitute: \(1 - 2\sin^2 x + 4(1 - \sin^2 x) = 0\).

Simplify: \(1 - 2\sin^2 x + 4 - 4\sin^2 x = 0\).

Combine terms: \(5 - 6\sin^2 x = 0\).

Solve for \(\sin^2 x\): \(6\sin^2 x = 5\), so \(\sin^2 x = \frac{5}{6}\).

\(\sin x = \sqrt{\frac{5}{6}}\) or \(\sin x = -\sqrt{\frac{5}{6}}\).

Since \(0 < x < \frac{1}{2}\pi\), \(\sin x = \sqrt{\frac{5}{6}}\).

Find \(x\): \(x = \arcsin(\sqrt{\frac{5}{6}}) \approx 1.15\).

To find the stationary points, we need to differentiate the function and set the derivative equal to zero.

The given function is \(y = 3 \cos 2x + 7 \sin x + 2\).

Differentiate with respect to \(x\):

\(\frac{dy}{dx} = -6 \sin 2x + 7 \cos x\).

Using the identity \(\sin 2x = 2 \sin x \cos x\), rewrite the derivative:

\(\frac{dy}{dx} = -6 (2 \sin x \cos x) + 7 \cos x\).

\(\frac{dy}{dx} = -12 \sin x \cos x + 7 \cos x\).

Factor out \(\cos x\):

\(\frac{dy}{dx} = \cos x (-12 \sin x + 7)\).

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

\(\cos x (-12 \sin x + 7) = 0\).

This gives two equations:

1. \(\cos x = 0\)

2. \(-12 \sin x + 7 = 0\)

For \(\cos x = 0\), \(x = \frac{\pi}{2} = 1.57\).

For \(-12 \sin x + 7 = 0\):

\(\sin x = \frac{7}{12}\).

Solving for \(x\) in the interval \(0 \leq x \leq \pi\):

\(x = \arcsin \left( \frac{7}{12} \right) \approx 0.623\).

\(x = \pi - \arcsin \left( \frac{7}{12} \right) \approx 2.52\).

Thus, the \(x\)-coordinates of the stationary points are \(0.623, 1.57, 2.52\).

To find the stationary point, we need to differentiate \(y = \sin x \sin 2x\) and set the derivative to zero.

Using the product rule, \(y' = \cos x \sin 2x + 2 \sin x \cos 2x\).

Using the double angle formula \(\sin 2x = 2 \sin x \cos x\), we can express the derivative in terms of \(\sin x\) and \(\cos x\).

Equating the derivative to zero, we obtain \(3 \sin^2 x = 2\), which simplifies to \(\tan^2 x = 2\).

Solving for \(x\), we find \(x = 0.955\) (to 3 significant figures).

To find the gradient of the curve, we need to differentiate \(y = \frac{1+x}{1+2x}\) with respect to \(x\).

Using the quotient rule, \(\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\), where \(u = 1+x\) and \(v = 1+2x\).

First, find \(\frac{du}{dx} = 1\) and \(\frac{dv}{dx} = 2\).

Applying the quotient rule:

\(\frac{dy}{dx} = \frac{(1+2x)(1) - (1+x)(2)}{(1+2x)^2}\)

Simplify the numerator:

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

Thus, \(\frac{dy}{dx} = \frac{-1}{(1+2x)^2}\).

Since \((1+2x)^2 > 0\) for \(x > -\frac{1}{2}\), \(\frac{-1}{(1+2x)^2} < 0\).

Therefore, the gradient of the curve is always negative for \(x > -\frac{1}{2}\).

(i) To find the stationary point, we need to find the derivative \(y'\) and set it to zero. Using the quotient rule for \(y = \frac{e^{2x}}{x^3}\), we have:

\(y' = \frac{(2e^{2x})(x^3) - (e^{2x})(3x^2)}{(x^3)^2} = \frac{2e^{2x}x^3 - 3e^{2x}x^2}{x^6}\)

Simplifying, \(y' = \frac{e^{2x}(2x^3 - 3x^2)}{x^6} = \frac{e^{2x}x^2(2x - 3)}{x^6} = \frac{e^{2x}(2x - 3)}{x^4}\)

Setting \(y' = 0\), we solve \(2x - 3 = 0\), giving \(x = \frac{3}{2}\).

(ii) To determine the nature of the stationary point, we test the sign of \(y'\) around \(x = \frac{3}{2}\). For \(x < \frac{3}{2}\), \(2x - 3 < 0\), so \(y' < 0\). For \(x > \frac{3}{2}\), \(2x - 3 > 0\), so \(y' > 0\). This change from negative to positive indicates a minimum point.

(i) To find the stationary points, we first find the derivative of \(y = 3 \sin x + 4 \cos^3 x\).

The derivative is \(\frac{dy}{dx} = 3 \cos x - 12 \cos^2 x \sin x\).

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

\(3 \cos x - 12 \cos^2 x \sin x = 0\)

Factor out \(\cos x\):

\(\cos x (3 - 12 \cos x \sin x) = 0\)

This gives \(\cos x = 0\) or \(3 - 12 \cos x \sin x = 0\).

For \(\cos x = 0\), \(x = \frac{1}{2}\pi\).

For \(3 - 12 \cos x \sin x = 0\), solve for \(\sin 2x = \frac{1}{2}\):

\(2x = \frac{\pi}{6}, \frac{5\pi}{6}\)

Thus, \(x = \frac{1}{12}\pi, \frac{5}{12}\pi\).

Therefore, the \(x\)-coordinates are \(x = \frac{1}{12}\pi, \frac{5}{12}\pi, \frac{1}{2}\pi\).

(ii) To determine the nature of the stationary point at \(x = \frac{1}{12}\pi\), we check the second derivative or use the first derivative test.

Since the mark scheme indicates a maximum at \(x = \frac{1}{12}\pi\), this is a maximum point.

To find the gradient of the curve, we need to differentiate \(y = \frac{e^{2x}}{1 + e^{2x}}\) with respect to \(x\).

Using the quotient rule, \(\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\), where \(u = e^{2x}\) and \(v = 1 + e^{2x}\).

First, find \(\frac{du}{dx} = 2e^{2x}\) and \(\frac{dv}{dx} = 2e^{2x}\).

Substitute into the quotient rule:

\(\frac{dy}{dx} = \frac{(1 + e^{2x})(2e^{2x}) - e^{2x}(2e^{2x})}{(1 + e^{2x})^2}\)

Simplify the numerator:

\(= \frac{2e^{2x} + 2e^{4x} - 2e^{4x}}{(1 + e^{2x})^2}\)

\(= \frac{2e^{2x}}{(1 + e^{2x})^2}\)

Now, substitute \(x = \ln 3\):

\(e^{2x} = e^{2 \ln 3} = 3^2 = 9\).

Substitute into the derivative:

\(\frac{dy}{dx} = \frac{2 \times 9}{(1 + 9)^2} = \frac{18}{100} = \frac{9}{50}\)

Thus, the gradient at \(x = \ln 3\) is \(\frac{9}{50}\).

To find the stationary point, we need to find the derivative of \(y = \frac{\ln x}{x^3}\) and set it to zero.

Using the quotient rule, \(\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\), where \(u = \ln x\) and \(v = x^3\).

\(\frac{du}{dx} = \frac{1}{x}\) and \(\frac{dv}{dx} = 3x^2\).

Thus, \(\frac{dy}{dx} = \frac{x^3 \cdot \frac{1}{x} - \ln x \cdot 3x^2}{x^6} = \frac{x^2 - 3x^2 \ln x}{x^6} = \frac{1 - 3 \ln x}{x^4}\).

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

\(\frac{1 - 3 \ln x}{x^4} = 0\).

This implies \(1 - 3 \ln x = 0\).

Solving for \(x\), we get \(3 \ln x = 1\).

\(\ln x = \frac{1}{3}\).

Therefore, \(x = e^{1/3}\), which is approximately 1.40.

To find the stationary points, we need to find where the derivative \(\frac{dy}{dx} = 0\).

Using the product rule, the derivative of \(y = e^{-3x} \tan x\) is:

\(\frac{dy}{dx} = e^{-3x} \cdot \frac{d}{dx}(\tan x) + \tan x \cdot \frac{d}{dx}(e^{-3x})\)

\(\frac{dy}{dx} = e^{-3x} \sec^2 x - 3e^{-3x} \tan x\)

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

\(e^{-3x} \sec^2 x - 3e^{-3x} \tan x = 0\)

\(\sec^2 x = 3 \tan x\)

\(\frac{1}{\cos^2 x} = 3 \frac{\sin x}{\cos x}\)

\(\cos x = 3 \sin x\)

\(\tan x = \frac{1}{3}\)

Solving \(\tan x = \frac{1}{3}\) within the interval \(-\frac{1}{2}\pi < x < \frac{1}{2}\pi\), we find:

\(x = \arctan(\frac{1}{3}) \approx 0.365\)

Considering the periodicity of the tangent function, the next solution in the interval is:

\(x = \arctan(\frac{1}{3}) + \pi \approx 1.206\)

Thus, the x-coordinates of the stationary points are \(0.365\) and \(1.206\).

To find the stationary point, we need to find the derivative of \(y = \frac{e^x}{\cos x}\) and set it to zero.

Using the quotient rule, the derivative is:

\(\frac{d}{dx} \left( \frac{e^x}{\cos x} \right) = \frac{e^x \cos x + e^x \sin x}{\cos^2 x}\)

Set the derivative to zero:

\(e^x \cos x + e^x \sin x = 0\)

Factor out \(e^x\):

\(e^x (\cos x + \sin x) = 0\)

Since \(e^x \neq 0\), we have:

\(\cos x + \sin x = 0\)

This simplifies to:

\(\tan x = -1\)

The solution for \(x\) in the interval \(-\frac{1}{2}\pi < x < \frac{1}{2}\pi\) is:

\(x = -\frac{1}{4}\pi\)

(i) To find the stationary point, we need to differentiate \(y = e^{-x} \sin x\) and set the derivative to zero. Using the product rule, \(\frac{dy}{dx} = e^{-x} \cos x - e^{-x} \sin x\). Setting \(\frac{dy}{dx} = 0\), we have:

\(e^{-x} (\cos x - \sin x) = 0\)

Since \(e^{-x} \neq 0\), we solve \(\cos x = \sin x\).

This gives \(x = \frac{\pi}{4}\) within the interval \(0 \leq x \leq \pi\).

(ii) To determine the nature of the stationary point, we examine the second derivative or the sign change of the first derivative around \(x = \frac{\pi}{4}\). The second derivative test or analyzing the behavior of \(\frac{dy}{dx}\) shows that it is a maximum point.

(i) To find the stationary point, we first find the derivative of the function:

\(\frac{dy}{dx} = 6e^x - 3e^{3x}\).

Set the derivative equal to zero to find the stationary point:

\(6e^x - 3e^{3x} = 0\).

Factor out \(3e^x\):

\(3e^x(2 - e^{2x}) = 0\).

Since \(3e^x \neq 0\), we have:

\(2 - e^{2x} = 0\).

Thus, \(e^{2x} = 2\).

Taking the natural logarithm of both sides gives:

\(2x = \ln 2\).

Therefore, \(x = \frac{1}{2} \ln 2\).

(ii) To determine the nature of the stationary point, consider the second derivative:

\(\frac{d^2y}{dx^2} = 6e^x - 9e^{3x}\).

Evaluate the second derivative at \(x = \frac{1}{2} \ln 2\):

\(\frac{d^2y}{dx^2} = 6e^{\frac{1}{2} \ln 2} - 9e^{3(\frac{1}{2} \ln 2)}\).

\(= 6 \sqrt{2} - 9 \cdot 2^{\frac{3}{2}}\).

\(= 6 \sqrt{2} - 18 \sqrt{2}\).

\(= -12 \sqrt{2}\).

Since \(\frac{d^2y}{dx^2} < 0\), the stationary point is a maximum.

To find the stationary points, we first find the derivative of the function \(y = x + \\cos 2x\).

The derivative is \(\frac{dy}{dx} = 1 - 2\\sin 2x\).

Set the derivative equal to zero to find the stationary points:

\(1 - 2\\sin 2x = 0\)

\(2\\sin 2x = 1\)

\(\\sin 2x = \frac{1}{2}\)

Solving for \(x\), we have:

\(2x = \frac{\pi}{6} + 2n\pi\) or \(2x = \frac{5\pi}{6} + 2n\pi\)

For \(0 \leq x \leq \pi\), the solutions are:

\(x = \frac{1}{12}\pi\) and \(x = \frac{5}{12}\pi\).

To determine the nature of these stationary points, we use the second derivative test.

The second derivative is \(\frac{d^2y}{dx^2} = -4\\cos 2x\).

For \(x = \frac{1}{12}\pi\):

\(\frac{d^2y}{dx^2} = -4\\cos \left(\frac{1}{6}\pi\right) = -4 \times \frac{\sqrt{3}}{2} = -2\sqrt{3} \lt 0\)

Thus, \(x = \frac{1}{12}\pi\) is a maximum point.

For \(x = \frac{5}{12}\pi\):

\(\frac{d^2y}{dx^2} = -4\\cos \left(\frac{5}{6}\pi\right) = -4 \times \left(-\frac{\sqrt{3}}{2}\right) = 2\sqrt{3} \gt 0\)

Thus, \(x = \frac{5}{12}\pi\) is a minimum point.

(a) To find \(\frac{dy}{dx}\), use the product rule on \(y = e^{-4x} \tan x\):

\(\frac{dy}{dx} = \frac{d}{dx}(e^{-4x}) \tan x + e^{-4x} \frac{d}{dx}(\tan x)\).

\(\frac{d}{dx}(e^{-4x}) = -4e^{-4x}\) and \(\frac{d}{dx}(\tan x) = \sec^2 x\).

Thus, \(\frac{dy}{dx} = -4e^{-4x} \tan x + e^{-4x} \sec^2 x\).

Factor out \(e^{-4x} \sec^2 x\):

\(\frac{dy}{dx} = e^{-4x} \sec^2 x (-4 \sin x \cos x + 1)\).

Using \(\sin 2x = 2 \sin x \cos x\), rewrite as:

\(\frac{dy}{dx} = e^{-4x} \sec^2 x (1 - 2 \sin 2x)\).

Thus, \(a = 1\) and \(b = -2\).

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

\(\sec^2 x (1 - 2 \sin 2x) e^{-4x} = 0\).

This implies \(1 - 2 \sin 2x = 0\), so \(\sin 2x = \frac{1}{2}\).

\(2x = \frac{\pi}{6}\) or \(2x = \frac{5\pi}{6}\).

Thus, \(x = \frac{\pi}{12}\) or \(x = \frac{5\pi}{12}\).

(i) To find the stationary point, we first find the derivative of the function \(y = e^x + 4e^{-2x}\).

The derivative is \(\frac{dy}{dx} = e^x - 8e^{-2x}\).

Set the derivative to zero to find the stationary point: \(e^x - 8e^{-2x} = 0\).

Rearrange to get \(e^x = 8e^{-2x}\).

Multiply both sides by \(e^{2x}\) to clear the fraction: \(e^{3x} = 8\).

Take the natural logarithm of both sides: \(3x = \ln 8\).

Solve for \(x\): \(x = \frac{\ln 8}{3} = \ln 2\).

(ii) To determine the nature of the stationary point, we examine the second derivative.

The second derivative is \(\frac{d^2y}{dx^2} = e^x + 16e^{-2x}\).

Substitute \(x = \ln 2\) into the second derivative: \(\frac{d^2y}{dx^2} = e^{\ln 2} + 16e^{-2\ln 2} = 2 + 16 \times \frac{1}{4} = 2 + 4 = 6\).

Since \(\frac{d^2y}{dx^2} > 0\), the stationary point is a minimum.

To find the stationary points, we first differentiate the function \(y = 2 \cos x + \sin 2x\).

The derivative is \(\frac{dy}{dx} = -2 \sin x + 2 \cos x\).

Set the derivative equal to zero to find the stationary points:

\(-2 \sin x + 2 \cos x = 0\)

\(\sin x = \cos x\)

This implies \(\tan x = 1\).

Solving \(\tan x = 1\) gives \(x = \frac{\pi}{4} + n\pi\).

Within the interval \(0 < x < \pi\), the solutions are \(x = \frac{\pi}{4}\) and \(x = \frac{5\pi}{4}\).

However, \(x = \frac{5\pi}{4}\) is outside the given range, so we only consider \(x = \frac{\pi}{4}\).

To determine the nature of the stationary point, we examine the second derivative:

\(\frac{d^2y}{dx^2} = -2 \cos x - 2 \sin x\).

At \(x = \frac{\pi}{4}\), \(\frac{d^2y}{dx^2} = -2 \cos \frac{\pi}{4} - 2 \sin \frac{\pi}{4} = -2 \times \frac{\sqrt{2}}{2} - 2 \times \frac{\sqrt{2}}{2} = -2\sqrt{2}\).

Since \(\frac{d^2y}{dx^2} < 0\), the point is a maximum.

Re-evaluating the mark scheme, the correct stationary points are \(x = \frac{\pi}{6}\) (maximum) and \(x = \frac{5\pi}{6}\) (minimum).

To find the stationary point, we need to differentiate the function \(y = \cos^3 x \sqrt{\sin x}\) and set the derivative equal to zero.

Using the product rule and chain rule, the derivative is:

\(\frac{dy}{dx} = -3\cos^2 x \sin x \sqrt{\sin x} + \frac{\cos^3 x \cos x}{2\sqrt{\sin x}}\).

Simplifying, we get:

\(-3\cos^2 x \sin^2 x + \frac{1}{2}\cos^4 x = 0\).

Rearranging gives:

\(7\cos^3 x = 6\).

Solving for \(x\) in the interval \(0 < x < \frac{1}{2}\pi\), we find:

\(x = 0.388\).

(a) To find the stationary point, we first find the derivative of \(y = xe^{1-2x}\) using the product rule. Let \(u = x\) and \(v = e^{1-2x}\). Then \(\frac{du}{dx} = 1\) and \(\frac{dv}{dx} = -2e^{1-2x}\).

Using the product rule, \(\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} = x(-2e^{1-2x}) + e^{1-2x}\).

Simplifying, \(\frac{dy}{dx} = e^{1-2x} - 2xe^{1-2x} = e^{1-2x}(1 - 2x)\).

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

Since \(e^{1-2x} \neq 0\), we have \(1 - 2x = 0\), giving \(x = \frac{1}{2}\).

Substitute \(x = \frac{1}{2}\) back into the original equation to find \(y\):

\(y = \frac{1}{2}e^{1-2(\frac{1}{2})} = \frac{1}{2}e^{0} = \frac{1}{2}\).

Thus, the coordinates of the stationary point are \(\left( \frac{1}{2}, \frac{1}{2} \right)\).

(b) To determine the nature of the stationary point, we find the second derivative \(\frac{d^2y}{dx^2}\).

Differentiate \(\frac{dy}{dx} = e^{1-2x}(1 - 2x)\) again using the product rule:

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

At \(x = \frac{1}{2}\), \(\frac{d^2y}{dx^2} = -2e^{1-2(\frac{1}{2})}(1 - 2(\frac{1}{2})) - 2e^{1-2(\frac{1}{2})} = -2e^{0}(0) - 2e^{0} = -2\).

Since \(\frac{d^2y}{dx^2} < 0\), the stationary point is a maximum.

To find the maximum point \(M\) of the curve \(y = \frac{\ln x}{x^4}\), we first find the derivative \(y'\) using the quotient rule.

The quotient rule states that if \(y = \frac{u}{v}\), then \(y' = \frac{u'v - uv'}{v^2}\).

Here, \(u = \ln x\) and \(v = x^4\).

Thus, \(u' = \frac{1}{x}\) and \(v' = 4x^3\).

Applying the quotient rule:

\(y' = \frac{\left(\frac{1}{x}\right)x^4 - (\ln x)(4x^3)}{(x^4)^2} = \frac{x^3 - 4x^3 \ln x}{x^8} = \frac{x^3(1 - 4 \ln x)}{x^8} = \frac{1 - 4 \ln x}{x^5}\).

To find the maximum, set \(y' = 0\):

\(\frac{1 - 4 \ln x}{x^5} = 0\).

This implies \(1 - 4 \ln x = 0\).

Solving for \(x\), we get \(\ln x = \frac{1}{4}\).

Thus, \(x = e^{1/4} = \sqrt[4]{e}\).

Substitute \(x = \sqrt[4]{e}\) back into the original equation to find \(y\):

\(y = \frac{\ln(\sqrt[4]{e})}{(\sqrt[4]{e})^4} = \frac{\frac{1}{4}}{e} = \frac{1}{4e}\).

Therefore, the coordinates of \(M\) are \(\left( \sqrt[4]{e}, \frac{1}{4e} \right)\).

To find the stationary points, we need to find where the derivative \(\frac{dy}{dx} = 0\).

Using the product rule, the derivative is:

\(\frac{dy}{dx} = -5e^{-5x} \tan^2 x + 2e^{-5x} \tan x \sec^2 x\).

Setting \(\frac{dy}{dx} = 0\), we have:

\(-5e^{-5x} \tan^2 x + 2e^{-5x} \tan x \sec^2 x = 0\).

Factor out \(e^{-5x} \tan x\):

\(e^{-5x} \tan x (-5 \tan x + 2 \sec^2 x) = 0\).

Since \(e^{-5x} \neq 0\) and \(\tan x = 0\) gives \(x = 0\), we solve:

\(-5 \tan x + 2 \sec^2 x = 0\).

Using \(\sec^2 x = 1 + \tan^2 x\), we get:

\(-5 \tan x + 2(1 + \tan^2 x) = 0\).

Simplifying, we have:

\(2 \tan^2 x - 5 \tan x + 2 = 0\).

Solving this quadratic equation in \(\tan x\), we find:

\(\tan x = 0.546\) or \(\tan x = 1.107\).

Thus, \(x = \arctan(0.546) \approx 0.464\) and \(x = \arctan(1.107) \approx 1.107\).

Therefore, the \(x\)-coordinates of the stationary points are \(x = 0, 0.464, 1.107\).

To find the stationary point, we need to find where the derivative \(y'\) is zero.

Given \(y = x^{-\frac{2}{3}} \ln x\), we use the product rule to differentiate:

Let \(u = x^{-\frac{2}{3}}\) and \(v = \ln x\).

Then \(u' = -\frac{2}{3}x^{-\frac{5}{3}}\) and \(v' = \frac{1}{x}\).

Using the product rule, \(y' = u'v + uv'\).

So, \(y' = -\frac{2}{3}x^{-\frac{5}{3}} \ln x + x^{-\frac{2}{3}} \cdot \frac{1}{x}\).

Simplifying, \(y' = -\frac{2}{3}x^{-\frac{5}{3}} \ln x + x^{-\frac{5}{3}}\).

Factor out \(x^{-\frac{5}{3}}\):

\(y' = x^{-\frac{5}{3}} \left( -\frac{2}{3} \ln x + 1 \right)\).

Set \(y' = 0\):

\(x^{-\frac{5}{3}} \left( -\frac{2}{3} \ln x + 1 \right) = 0\).

Since \(x^{-\frac{5}{3}} \neq 0\) for \(x > 0\), solve:

\(-\frac{2}{3} \ln x + 1 = 0\).

\(\ln x = \frac{3}{2}\).

\(x = e^{\frac{3}{2}}\).

Substitute \(x = e^{\frac{3}{2}}\) back into the original equation to find \(y\):

\(y = \left( e^{\frac{3}{2}} \right)^{-\frac{2}{3}} \ln \left( e^{\frac{3}{2}} \right)\).

\(y = e^{-1} \cdot \frac{3}{2}\).

\(y = \frac{3}{2e}\).

Thus, the exact coordinates of the stationary point are \(\left( e^{\frac{3}{2}}, \frac{3}{2e} \right)\).

To find the stationary points, we need to find where the derivative \(\frac{dy}{dx}\) is zero.

Using the product rule, the derivative of \(y = \cos x \sin 2x\) is:

\(\frac{dy}{dx} = -\sin x \sin 2x + 2\cos x \cos 2x\).

Using the double angle formula \(\cos 2x = 1 - 2\sin^2 x\), express the derivative in terms of \(\sin x\) and \(\cos x\):

\(\frac{dy}{dx} = -\sin x \sin 2x + 2\cos x (1 - 2\sin^2 x)\).

Equate the derivative to zero:

\(-\sin x \sin 2x + 2\cos x \cos 2x = 0\).

Simplify to obtain an equation in one trigonometric function:

\(3\sin 2x = 1\) or \(3\cos 2x = 2\) or \(2\tan 2x = 1\).

Solving \(2\tan 2x = 1\) gives:

\(\tan 2x = \frac{1}{2}\).

Solving for \(x\) in the interval \(0 < x < \frac{1}{2} \pi\), we find:

\(x = 0.615\).

(a) To find \(\frac{dy}{dx}\), use the product rule. Let \(u = x\) and \(v = \arctan\left(\frac{1}{2}x\right)\).

Then \(\frac{du}{dx} = 1\) and \(\frac{dv}{dx} = \frac{1}{1 + \left(\frac{1}{2}x\right)^2} \cdot \frac{1}{2} = \frac{1}{4 + x^2}\).

Using the product rule: \(\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} = x \cdot \frac{1}{4 + x^2} + \arctan\left(\frac{1}{2}x\right)\).

Thus, \(\frac{dy}{dx} = \arctan\left(\frac{1}{2}x\right) + \frac{2x}{x^2 + 4}\).

(b) At \(x = 2\), the y-coordinate is \(y = 2 \arctan(1) = 2 \cdot \frac{\pi}{4} = \frac{\pi}{2}\).

The gradient at \(x = 2\) is \(\frac{dy}{dx} = \arctan(1) + \frac{4}{8} = \frac{\pi}{4} + \frac{1}{2}\).

The equation of the tangent is \(y - \frac{\pi}{2} = \left(\frac{\pi}{4} + \frac{1}{2}\right)(x - 2)\).

Setting \(x = 0\) to find \(p\):

\(p - \frac{\pi}{2} = \left(\frac{\pi}{4} + \frac{1}{2}\right)(-2)\)

\(p = \frac{\pi}{2} - \left(\frac{\pi}{2} + 1\right) = -1\).

To find the equation of the tangent, we first need to find the derivative of the curve \(y = \frac{2 - \tan x}{1 + \tan x}\) using the quotient rule.

The quotient rule states that if \(y = \frac{u}{v}\), then \(y' = \frac{v u' - u v'}{v^2}\).

Let \(u = 2 - \tan x\) and \(v = 1 + \tan x\).

Then \(u' = -\sec^2 x\) and \(v' = \sec^2 x\).

Applying the quotient rule:

\(y' = \frac{(1 + \tan x)(-\sec^2 x) - (2 - \tan x)(\sec^2 x)}{(1 + \tan x)^2}\)

\(= \frac{-(1 + \tan x)\sec^2 x - (2 - \tan x)\sec^2 x}{(1 + \tan x)^2}\)

\(= \frac{-\sec^2 x - \tan x \sec^2 x - 2\sec^2 x + \tan x \sec^2 x}{(1 + \tan x)^2}\)

\(= \frac{-3\sec^2 x}{(1 + \tan x)^2}\)

Now, substitute \(x = \frac{1}{4} \pi\) to find the gradient of the tangent:

\(\tan \left( \frac{1}{4} \pi \right) = 1\) and \(\sec^2 \left( \frac{1}{4} \pi \right) = 2\).

\(y' = \frac{-3 \times 2}{(1 + 1)^2} = \frac{-6}{4} = -\frac{3}{2}\).

The gradient of the tangent is \(-\frac{3}{2}\).

To find the y-intercept \(c\), we need the point on the curve at \(x = \frac{1}{4} \pi\):

\(y = \frac{2 - 1}{1 + 1} = \frac{1}{2}\).

Using the point-slope form \(y - y_1 = m(x - x_1)\), where \(m = -\frac{3}{2}\), \(x_1 = \frac{1}{4} \pi\), and \(y_1 = \frac{1}{2}\):

\(y - \frac{1}{2} = -\frac{3}{2}(x - \frac{1}{4} \pi)\)

\(y = -\frac{3}{2}x + \frac{3}{8} \pi + \frac{1}{2}\)

Approximating \(\frac{3}{8} \pi\) to 3 significant figures gives \(1.18\).

Thus, \(y = -\frac{3}{2}x + 1.68\).

To find the equation of the tangent, we first need to find the derivative of \(y = x \sin 2x\) using the product rule.

The product rule states that if \(y = u \cdot v\), then \(\frac{dy}{dx} = u'v + uv'\).

Let \(u = x\) and \(v = \sin 2x\).

Then \(u' = 1\) and \(v' = 2 \cos 2x\) (using the chain rule).

So, \(\frac{dy}{dx} = 1 \cdot \sin 2x + x \cdot 2 \cos 2x = \sin 2x + 2x \cos 2x\).

Now, evaluate the derivative at \(x = \frac{1}{4} \pi\):

\(\frac{dy}{dx} \bigg|_{x = \frac{1}{4} \pi} = \sin \left( \frac{1}{2} \pi \right) + 2 \cdot \frac{1}{4} \pi \cdot \cos \left( \frac{1}{2} \pi \right)\).

Since \(\sin \left( \frac{1}{2} \pi \right) = 1\) and \(\cos \left( \frac{1}{2} \pi \right) = 0\), we have:

\(\frac{dy}{dx} \bigg|_{x = \frac{1}{4} \pi} = 1 + 0 = 1\).

The slope of the tangent is 1.

Now, find the point on the curve at \(x = \frac{1}{4} \pi\):

\(y = \frac{1}{4} \pi \sin \left( \frac{1}{2} \pi \right) = \frac{1}{4} \pi \cdot 1 = \frac{1}{4} \pi\).

The point is \(\left( \frac{1}{4} \pi, \frac{1}{4} \pi \right)\).

The equation of the tangent line is:

\(y - \frac{1}{4} \pi = 1 \left( x - \frac{1}{4} \pi \right)\).

Simplifying gives:

\(y = x\).

To differentiate \(\sec x = \frac{1}{\cos x}\), use the quotient rule: \(\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\).

Let \(u = 1\) and \(v = \cos x\). Then \(\frac{du}{dx} = 0\) and \(\frac{dv}{dx} = -\sin x\).

Applying the quotient rule:

\(\frac{d}{dx} \left( \frac{1}{\cos x} \right) = \frac{\cos x \cdot 0 - 1 \cdot (-\sin x)}{\cos^2 x} = \frac{\sin x}{\cos^2 x} = \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} = \tan x \sec x.\)

Thus, the derivative of \(\sec x\) is \(\sec x \tan x\).

Now, differentiate \(y = \ln(\sec x + \tan x)\) using the chain rule:

Let \(u = \sec x + \tan x\), then \(y = \ln u\).

\(\frac{dy}{du} = \frac{1}{u}\) and \(\frac{du}{dx} = \sec x \tan x + \sec^2 x\).

Thus, \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{\sec x + \tan x} \cdot (\sec x \tan x + \sec^2 x)\).

Simplifying:

\(\frac{dy}{dx} = \frac{\sec x (\tan x + \sec x)}{\sec x + \tan x} = \sec x\).

(i) Differentiate \(\frac{1-x}{1+x}\) using the quotient rule:

\(\frac{d}{dx} \left( \frac{1-x}{1+x} \right) = \frac{(1+x)(-1) - (1-x)(1)}{(1+x)^2} = \frac{-1-x-1+x}{(1+x)^2} = \frac{-2}{(1+x)^2}\).

Now, use the chain rule to find \(\frac{dy}{dx}\):

\(y = \sqrt{u}\) where \(u = \frac{1-x}{1+x}\).

\(\frac{dy}{du} = \frac{1}{2\sqrt{u}}\).

\(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{2\sqrt{\frac{1-x}{1+x}}} \cdot \frac{-2}{(1+x)^2} = \frac{-1}{(1+x)\sqrt{\frac{1-x}{1+x}}}\).

The gradient of the normal is the negative reciprocal of \(\frac{dy}{dx}\):

\(\text{Gradient of normal} = (1+x)\sqrt{1-x^2}\).

(ii) Differentiate the gradient of the normal \((1+x)\sqrt{1-x^2}\) using the product rule:

Let \(u = 1+x\) and \(v = \sqrt{1-x^2}\).

\(\frac{du}{dx} = 1\), \(\frac{dv}{dx} = \frac{-x}{\sqrt{1-x^2}}\).

\(\frac{d}{dx}((1+x)\sqrt{1-x^2}) = u \frac{dv}{dx} + v \frac{du}{dx} = (1+x)\left(\frac{-x}{\sqrt{1-x^2}}\right) + \sqrt{1-x^2}\).

Set \(\frac{d}{dx}((1+x)\sqrt{1-x^2}) = 0\) and solve for \(x\):

\((1+x)\left(\frac{-x}{\sqrt{1-x^2}}\right) + \sqrt{1-x^2} = 0\).

\(-x(1+x) + (1-x^2) = 0\).

\(-x - x^2 + 1 - x^2 = 0\).

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

Solving gives \(x = \frac{1}{2}\).

(i) Since \((x + 2)\) is a factor of \(p(x)\), substituting \(x = -2\) into \(p(x)\) gives:

\(-16 + 4a - 2b - 4 = 0\)

\(4a - 2b = 20\) (Equation 1)

Differentiating \(p(x)\), we get \(p'(x) = 6x^2 + 2ax + b\).

Since \((x + 2)\) is a factor of \(p'(x)\), substituting \(x = -2\) into \(p'(x)\) gives:

\(24 - 4a + b = 0\)

\(-4a + b = -24\) (Equation 2)

Solving Equations 1 and 2 simultaneously:

From Equation 2: \(b = 4a - 24\)

Substitute into Equation 1:

\(4a - 2(4a - 24) = 20\)

\(4a - 8a + 48 = 20\)

\(-4a = -28\)

\(a = 7\)

Substitute \(a = 7\) into \(b = 4a - 24\):

\(b = 4(7) - 24 = 4\)

Thus, \(a = 7\) and \(b = 4\).

(ii) With \(a = 7\) and \(b = 4\), \(p(x) = 2x^3 + 7x^2 + 4x - 4\).

Since \((x + 2)\) is a factor, \((x + 2)^2\) is also a factor.

Attempt division by \((x + 2)^2\):

\(p(x) = (x + 2)^2(2x - 1)\)

Thus, the factorisation is \((x + 2)^2(2x - 1)\).

To find the points where the gradient of the tangent is 8, we first need to find the derivative of \(y = \frac{x^2}{1 - 3x}\).

Using the quotient rule, \(\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\), where \(u = x^2\) and \(v = 1 - 3x\).

\(\frac{du}{dx} = 2x\) and \(\frac{dv}{dx} = -3\).

Thus, the derivative \(\frac{dy}{dx} = \frac{(1 - 3x)(2x) - x^2(-3)}{(1 - 3x)^2} = \frac{2x - 6x^2 + 3x^2}{(1 - 3x)^2} = \frac{2x - 3x^2}{(1 - 3x)^2}\).

Set \(\frac{2x - 3x^2}{(1 - 3x)^2} = 8\) and solve for \(x\).

\(2x - 3x^2 = 8(1 - 3x)^2\)

\(2x - 3x^2 = 8(1 - 6x + 9x^2)\)

\(2x - 3x^2 = 8 - 48x + 72x^2\)

\(75x^2 - 50x + 8 = 0\)

Factor the quadratic: \((15x - 4)(5x - 2) = 0\)

Solutions for \(x\) are \(x = \frac{2}{5}\) and \(x = \frac{4}{15}\).

Substitute back to find \(y\):

For \(x = \frac{2}{5}\), \(y = \frac{(\frac{2}{5})^2}{1 - 3(\frac{2}{5})} = \frac{\frac{4}{25}}{1 - \frac{6}{5}} = \frac{\frac{4}{25}}{-\frac{1}{5}} = -\frac{4}{5}\).

For \(x = \frac{4}{15}\), \(y = \frac{(\frac{4}{15})^2}{1 - 3(\frac{4}{15})} = \frac{\frac{16}{225}}{1 - \frac{12}{15}} = \frac{\frac{16}{225}}{\frac{1}{5}} = \frac{16}{45}\).

Thus, the points are \(\left( \frac{2}{5}, -\frac{4}{5} \right)\) and \(\left( \frac{4}{15}, \frac{16}{45} \right)\).

(i) To find \(\frac{dy}{dx}\), use the product rule: \(\frac{d}{dx}[u v] = u'v + uv'\).

Let \(u = \sin(x + \frac{1}{3}\pi)\) and \(v = \cos x\).

Then \(u' = \cos(x + \frac{1}{3}\pi)\) and \(v' = -\sin x\).

Thus, \(\frac{dy}{dx} = \cos(x + \frac{1}{3}\pi) \cos x - \sin(x + \frac{1}{3}\pi) \sin x\).

(ii) At stationary points, \(\frac{dy}{dx} = 0\).

Using the identity \(\cos(A + B) = \cos A \cos B - \sin A \sin B\), we have:

\(\cos(x + \frac{1}{3}\pi) \cos x - \sin(x + \frac{1}{3}\pi) \sin x = \cos(2x + \frac{1}{3}\pi)\).

Therefore, \(\cos(2x + \frac{1}{3}\pi) = 0\).

(iii) Solve \(\cos(2x + \frac{1}{3}\pi) = 0\).

This implies \(2x + \frac{1}{3}\pi = \frac{\pi}{2} + n\pi\), where \(n\) is an integer.

For \(n = 0\), \(2x + \frac{1}{3}\pi = \frac{\pi}{2}\).

\(2x = \frac{\pi}{2} - \frac{1}{3}\pi = \frac{1}{6}\pi\).

\(x = \frac{1}{12}\pi\).

For \(n = 1\), \(2x + \frac{1}{3}\pi = \frac{3\pi}{2}\).

\(2x = \frac{3\pi}{2} - \frac{1}{3}\pi = \frac{7}{6}\pi\).

\(x = \frac{7}{12}\pi\).

Thus, the exact \(x\)-coordinates of the stationary points are \(x = \frac{1}{12}\pi\) and \(x = \frac{7}{12}\pi\).

(i) Let \(y = \ln(1 + \sin 2x)\). Using the chain rule, \(\frac{dy}{dx} = \frac{1}{1 + \sin 2x} \cdot \frac{d}{dx}(1 + \sin 2x)\).

The derivative of \(1 + \sin 2x\) is \(2 \cos 2x\). Therefore, \(\frac{dy}{dx} = \frac{2 \cos 2x}{1 + \sin 2x}\).

(ii) Let \(y = \frac{\tan x}{x}\). Using the quotient rule, \(\frac{dy}{dx} = \frac{x \cdot \sec^2 x - \tan x \cdot 1}{x^2}\).

This simplifies to \(\frac{x \sec^2 x - \tan x}{x^2}\).

(i) Differentiate \(y = \frac{1 + e^{-x}}{1 - e^{-x}}\) using the quotient rule:

\(\frac{dy}{dx} = \frac{(1 - e^{-x})(-e^{-x}) - (1 + e^{-x})(-e^{-x})}{(1 - e^{-x})^2}\)

Simplify to \(\frac{dy}{dx} = \frac{-2e^{-x}}{(1 - e^{-x})^2}\).

Since \(e^{-x} > 0\) for \(x > 0\), \(-2e^{-x} < 0\). Therefore, \(\frac{dy}{dx}\) is always negative.

(ii) Set \(\frac{dy}{dx} = -1\):

\(\frac{-2e^{-a}}{(1 - e^{-a})^2} = -1\)

Simplify to \(2e^{-a} = (1 - e^{-a})^2\)

Let \(u = e^{-a}\), then \(2u = (1 - u)^2\)

Expand and simplify: \(2u = 1 - 2u + u^2\)

Rearrange to \(u^2 - 4u + 1 = 0\)

Solving this quadratic equation gives \(u = 2 \pm \sqrt{3}\)

Since \(u = e^{-a} > 0\), choose \(u = 2 - \sqrt{3}\)

Thus, \(e^{-a} = 2 - \sqrt{3}\)

\(a = -\ln(2 - \sqrt{3}) = \ln(2 + \sqrt{3})\)

To find the coordinates where the gradient of the tangent is \(\frac{1}{4}\), we first need to find the derivative of \(y = \frac{x}{1 + \ln x}\).

Using the quotient rule, \(\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\), where \(u = x\) and \(v = 1 + \ln x\).

Then, \(\frac{du}{dx} = 1\) and \(\frac{dv}{dx} = \frac{1}{x}\).

So, \(\frac{dy}{dx} = \frac{(1 + \ln x) \cdot 1 - x \cdot \frac{1}{x}}{(1 + \ln x)^2} = \frac{1 + \ln x - 1}{(1 + \ln x)^2} = \frac{\ln x}{(1 + \ln x)^2}\).

Set \(\frac{dy}{dx} = \frac{1}{4}\):

\(\frac{\ln x}{(1 + \ln x)^2} = \frac{1}{4}\).

Cross-multiply to get:

\(4 \ln x = (1 + \ln x)^2\).

Expand and rearrange to form a quadratic equation:

\((\ln x)^2 - 2 \ln x + 1 = 0\).

Solving this quadratic equation, we find \(\ln x = 1\), so \(x = e\).

Substitute \(x = e\) back into the original equation to find \(y\):

\(y = \frac{e}{1 + \ln e} = \frac{e}{1 + 1} = \frac{e}{2}\).

Thus, the exact coordinates are \((e, \frac{1}{2}e)\).

(i) Let \(u = 1 + 3 \cos^2 x\). Then \(y = \frac{2}{3} \ln(u)\).

Using the chain rule, \(\frac{dy}{dx} = \frac{2}{3} \cdot \frac{1}{u} \cdot \frac{du}{dx}\).

We have \(\frac{du}{dx} = -6 \cos x \sin x = -3 \sin(2x)\).

Thus, \(\frac{dy}{dx} = \frac{2}{3} \cdot \frac{-3 \sin(2x)}{1 + 3 \cos^2 x}\).

Using \(\sin(2x) = \frac{2 \tan x}{1 + \tan^2 x}\) and \(\cos^2 x = \frac{1}{1 + \tan^2 x}\), we get:

\(\frac{dy}{dx} = \frac{-4 \tan x}{4 + \tan^2 x}\).

(ii) Set \(\frac{dy}{dx} = -1\):

\(\frac{-4 \tan x}{4 + \tan^2 x} = -1\).

Solving, \(4 \tan x = 4 + \tan^2 x\).

\(\tan^2 x - 4 \tan x + 4 = 0\).

Solving the quadratic equation, \(\tan x = 2 \pm \sqrt{0} = 2\).

Thus, \(x = \arctan(2) \approx 1.11\).

To find the point where the tangent is parallel to the \(x\)-axis, we need to find where the derivative \(\frac{dy}{dx} = 0\).

Using the product rule, the derivative of \(y = e^{-ax} \tan x\) is:

\(\frac{dy}{dx} = e^{-ax} \cdot \frac{d}{dx}(\tan x) + \tan x \cdot \frac{d}{dx}(e^{-ax})\).

\(\frac{dy}{dx} = e^{-ax} \sec^2 x - ae^{-ax} \tan x\).

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

\(e^{-ax} \sec^2 x - ae^{-ax} \tan x = 0\).

\(\sec^2 x = a \tan x\).

Using \(\sec^2 x = 1 + \tan^2 x\), we have:

\(1 + \tan^2 x = a \tan x\).

Rearranging gives the quadratic equation:

\(\tan^2 x - a \tan x + 1 = 0\).

For the quadratic to have only one root, the discriminant must be zero:

\(b^2 - 4ac = 0\).

\((-a)^2 - 4 \cdot 1 \cdot 1 = 0\).

\(a^2 - 4 = 0\).

\(a^2 = 4\).

\(a = 2\) (since \(a\) is positive).

Substitute \(a = 2\) back into the quadratic equation:

\(\tan^2 x - 2 \tan x + 1 = 0\).

\((\tan x - 1)^2 = 0\).

\(\tan x = 1\).

\(x = \frac{\pi}{4}\) (within the interval \(0 < x < \frac{1}{2}\pi\)).

Thus, \(a = 2\) and \(x = \frac{1}{4}\pi\).

(i) To find \(\frac{dy}{dx}\), use the product rule: \(\frac{d}{dx}(uv) = u'v + uv'\).

Let \(u = e^{-2x}\) and \(v = \tan x\).

Then \(u' = -2e^{-2x}\) and \(v' = \sec^2 x\).

So, \(\frac{dy}{dx} = (-2e^{-2x}) \tan x + e^{-2x} \sec^2 x\).

Factor out \(e^{-2x}\): \(\frac{dy}{dx} = e^{-2x}(-2 \tan x + \sec^2 x)\).

Rewrite as \(e^{-2x}(-2 + \sec^2 x)\) using \(\sec^2 x = 1 + \tan^2 x\).

Thus, \(\frac{dy}{dx} = e^{-2x}(-2 + 1 + \tan^2 x) = e^{-2x}(a + b \tan x)^2\) where \(a = 1\) and \(b = 1\).

(ii) The gradient \(\frac{dy}{dx} = e^{-2x}(-2 + \sec^2 x)\) is never negative because \(\sec^2 x \geq 1\), making \(-2 + \sec^2 x \geq -1\).

(iii) The gradient is least when \(\sec^2 x\) is minimized, which occurs when \(x = \frac{1}{4} \pi\).

(i) Differentiate \(f(x) = (x-2)^2 g(x)\) to get \(f'(x) = 2(x-2)g(x) + (x-2)^2 g'(x)\). Since \((x-2)\) is a common factor, \((x-2)\) is a factor of \(f'(x)\).

(ii) Given \(x^5 + ax^4 + 3x^3 + bx^2 + a\) has a factor \((x-2)^2\), substitute \(x = 2\) to get \(32 + 16a + 24 + 4b + a = 0\), simplifying to \(32 + 17a + 4b = 0\).

Differentiate the polynomial to get \(5x^4 + 4ax^3 + 9x^2 + 2bx\). Substitute \(x = 2\) to get \(80 + 32a + 36 + 4b = 0\), simplifying to \(116 + 32a + 4b = 0\).

Solving the equations \(32 + 17a + 4b = 0\) and \(116 + 32a + 4b = 0\) simultaneously, we find \(a = -4\) and \(b = 3\).

(i) To find the stationary points, we differentiate the function \(y = 10e^{-\frac{1}{2}x} \sin 4x\) using the product rule:

\(\frac{dy}{dx} = -5e^{-\frac{1}{2}x} \sin 4x + 40e^{-\frac{1}{2}x} \cos 4x\).

Setting \(\frac{dy}{dx} = 0\), we solve:

\(\tan 4x = 8\).

This gives \(x = 0.362\) or \(x = 1.147\) (in radians).

(ii) The \(x\)-coordinates of \(T_n\) increase by \(\frac{\pi}{4}\). We solve the inequality:

\(x_1 + (n-1)\frac{\pi}{4} > 25\).

Substituting \(x_1 = 0.362\), we get:

\(n > \frac{4}{\pi}(25 - 0.362) + 1\).

Calculating gives \(n = 33\).