Past Exam Questions

First, find \(\frac{dy}{d\theta}\):

\(\frac{dy}{d\theta} = 1 - 2\sin \theta\).

Next, find \(\frac{dx}{d\theta}\) using the quotient rule:

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

Let \(u = \cos \theta\) and \(v = 2 - \sin \theta\).

Then \(\frac{du}{d\theta} = -\sin \theta\) and \(\frac{dv}{d\theta} = -\cos \theta\).

Using the quotient rule, \(\frac{dx}{d\theta} = \frac{v \frac{du}{d\theta} - u \frac{dv}{d\theta}}{v^2}\).

\(\frac{dx}{d\theta} = \frac{(2 - \sin \theta)(-\sin \theta) + \cos \theta \cdot \cos \theta}{(2 - \sin \theta)^2}\).

\(\frac{dx}{d\theta} = \frac{-2\sin \theta + \sin^2 \theta + \cos^2 \theta}{(2 - \sin \theta)^2}\).

Since \(\cos^2 \theta + \sin^2 \theta = 1\),

\(\frac{dx}{d\theta} = \frac{-2\sin \theta + 1}{(2 - \sin \theta)^2}\).

Now, use \(\frac{dy}{dx} = \frac{dy}{d\theta} \div \frac{dx}{d\theta}\):

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

\(\frac{dy}{dx} = (2 - \sin \theta)^2\).

(i) Differentiate \(x = a \cos^3 t\) and \(y = a \sin^3 t\) with respect to \(t\):

\(\frac{dx}{dt} = -3a \cos^2 t \sin t\)

\(\frac{dy}{dt} = 3a \sin^2 t \cos t\)

Then, \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3a \sin^2 t \cos t}{-3a \cos^2 t \sin t} = -\frac{\tan t}{\cos t} = -\tan t \sec t\).

(ii) The equation of the tangent is given by:

\(\frac{y - a \sin^3 t}{x - a \cos^3 t} = \frac{dy}{dx} = -\tan t \sec t\)

Rearranging gives:

\(x \sin t + y \cos t = a \sin t \cos t\).

(iii) The tangent meets the \(x\)-axis when \(y = 0\):

\(x \sin t = a \sin t \cos t \Rightarrow x = a \cos t\).

The tangent meets the \(y\)-axis when \(x = 0\):

\(y \cos t = a \sin t \cos t \Rightarrow y = a \sin t\).

The length \(XY\) is:

\(\sqrt{(a \cos t)^2 + (a \sin t)^2} = \sqrt{a^2 (\cos^2 t + \sin^2 t)} = a\).

First, find \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\):

\(\frac{dx}{d\theta} = a(2 - 2\cos 2\theta) = 2a(1 - \cos 2\theta)\)

\(\frac{dy}{d\theta} = a(2\sin 2\theta)\)

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

\(\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{2a\sin 2\theta}{2a(1 - \cos 2\theta)}\)

Simplify the expression:

\(\frac{dy}{dx} = \frac{\sin 2\theta}{1 - \cos 2\theta}\)

Using the identity \(\sin 2\theta = 2\sin \theta \cos \theta\) and \(1 - \cos 2\theta = 2\sin^2 \theta\), we have:

\(\frac{dy}{dx} = \frac{2\sin \theta \cos \theta}{2\sin^2 \theta} = \frac{\cos \theta}{\sin \theta} = \cot \theta\)

First, find \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\):

\(\frac{dx}{d\theta} = 2 + 2\cos 2\theta\)

\(\frac{dy}{d\theta} = 2\sin 2\theta\)

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

\(\frac{dy}{dx} = \frac{dy}{d\theta} \div \frac{dx}{d\theta} = \frac{2\sin 2\theta}{2 + 2\cos 2\theta}\)

Simplify the expression:

\(\frac{dy}{dx} = \frac{2\sin 2\theta}{2(1 + \cos 2\theta)} = \frac{\sin 2\theta}{1 + \cos 2\theta}\)

Using the identity \(\sin 2\theta = 2\sin \theta \cos \theta\) and \(1 + \cos 2\theta = 2\cos^2 \theta\), we have:

\(\frac{dy}{dx} = \frac{2\sin \theta \cos \theta}{2\cos^2 \theta} = \frac{\sin \theta}{\cos \theta} = \tan \theta\)

(a) Differentiate \(x = te^{2t}\) with respect to \(t\):

\(\frac{dx}{dt} = e^{2t} + 2te^{2t}\).

Differentiate \(y = t^2 + t + 3\) with respect to \(t\):

\(\frac{dy}{dt} = 2t + 1\).

Use the chain rule to find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2t + 1}{e^{2t}(1 + 2t)}\).

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

(b) At \(t = -1\), find \(x\) and \(y\):

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

The gradient of the tangent is \(e^{-2(-1)} = e^2\), so the gradient of the normal is \(-\frac{1}{e^2}\).

The equation of the normal is \(y - 3 = -\frac{1}{e^2}(x + \frac{1}{e^2})\).

Substitute \(x = 0\) to find \(y\):

\(y = 3 - \frac{1}{e^4}\).

Thus, the normal passes through \(\left( 0, 3 - \frac{1}{e^4} \right)\).

First, find \(\\frac{dx}{dt}\):

\(\\frac{dx}{dt} = \\frac{d}{dt}(2t - \\tan t) = 2 - \\sec^2 t\).

Next, find \(\\frac{dy}{dt}\):

\(\\frac{dy}{dt} = \\frac{d}{dt}(\\ln(\\sin 2t)) = \\frac{1}{\\sin 2t} \\cdot \\frac{d}{dt}(\\sin 2t) = \\frac{1}{\\sin 2t} \\cdot 2 \\cos 2t = \\frac{2 \\cos 2t}{\\sin 2t}\).

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

\(\\frac{dy}{dx} = \\frac{dy}{dt} \\div \\frac{dx}{dt} = \\frac{2 \\cos 2t}{\\sin 2t} \\div (2 - \\sec^2 t)\).

Simplify using trigonometric identities:

\(\\frac{dy}{dx} = \\frac{2 \\cos 2t}{\\sin 2t} \\cdot \\frac{1}{2 - \\sec^2 t}\).

Using the double angle formula, \(\\sin 2t = 2 \\sin t \\cos t\), and \(\\cos 2t = \\cos^2 t - \\sin^2 t\), we simplify:

\(\\frac{dy}{dx} = \\frac{\\cos t}{\\sin t} = \\cot t\).

(a) To find \(\frac{dy}{dx}\), we use the chain rule: \(\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}\).

First, find \(\frac{dx}{dt}\):

\(x = \frac{1}{\cos t} \Rightarrow \frac{dx}{dt} = \sec t \tan t\).

Next, find \(\frac{dy}{dt}\):

\(y = \ln \tan t \Rightarrow \frac{dy}{dt} = \frac{1}{\tan t} \cdot \sec^2 t = \frac{\sec^2 t}{\tan t}\).

Now, \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac{\sec^2 t}{\tan t}}{\sec t \tan t} = \frac{\sec t}{\tan^2 t}\).

Since \(\tan t = \frac{\sin t}{\cos t}\), we have \(\tan^2 t = \frac{\sin^2 t}{\cos^2 t}\).

Thus, \(\frac{dy}{dx} = \frac{\sec t \cos^2 t}{\sin^2 t} = \frac{\cos t}{\sin^2 t}\).

(b) When \(y = 0\), \(\ln \tan t = 0 \Rightarrow \tan t = 1 \Rightarrow t = \frac{\pi}{4}\).

At \(t = \frac{\pi}{4}\), \(x = \frac{1}{\cos \frac{\pi}{4}} = \sqrt{2}\).

The gradient \(\frac{dy}{dx} = \frac{\cos t}{\sin^2 t} = \frac{\cos \frac{\pi}{4}}{\sin^2 \frac{\pi}{4}} = \frac{\frac{1}{\sqrt{2}}}{\left(\frac{1}{\sqrt{2}}\right)^2} = \sqrt{2}\).

The equation of the tangent is \(y - 0 = \sqrt{2}(x - \sqrt{2})\).

Simplifying gives \(y = \sqrt{2}x - 2\).

First, find \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\):

\(\frac{dx}{d\theta} = \\sin \theta\)

\(\frac{dy}{d\theta} = -\\sin \theta + \frac{1}{2} \\sin 2\theta\)

Use the chain rule to find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{dy}{d\theta} \div \frac{dx}{d\theta} = \frac{-\\sin \theta + \frac{1}{2} \\sin 2\theta}{\\sin \theta}\)

Simplify using the double angle identity \(\sin 2\theta = 2\\sin \theta \\cos \theta\):

\(\frac{dy}{dx} = \frac{-\\sin \theta + \\sin \theta \\cos \theta}{\\sin \theta} = -1 + \\cos \theta\)

Use the identity \(\\cos \theta = 1 - 2\\sin^2 \left( \frac{1}{2} \theta \right)\):

\(\frac{dy}{dx} = -1 + 1 - 2\\sin^2 \left( \frac{1}{2} \theta \right) = -2\\sin^2 \left( \frac{1}{2} \theta \right)\)

(a) To find \(\frac{dy}{dx}\), we first find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).

\(\frac{dx}{dt} = 1 + \frac{1}{t+2}\).

Using the product rule, \(\frac{dy}{dt} = e^{-2t} - 2(t-1)e^{-2t} = e^{-2t} - 2te^{-2t} + 2e^{-2t} = (3 - 2t)e^{-2t}\).

Now, \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{(3 - 2t)e^{-2t}}{1 + \frac{1}{t+2}} = \frac{(3 - 2t)(t + 2)}{t + 3} e^{-2t}\).

(b) A stationary point occurs when \(\frac{dy}{dx} = 0\).

Setting \((3 - 2t)(t + 2) = 0\), we solve for \(t\):

\(3 - 2t = 0 \Rightarrow t = \frac{3}{2}\).

Substitute \(t = \frac{3}{2}\) into \(y = (t - 1)e^{-2t}\):

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

(a) Differentiate \(x\) and \(y\) with respect to \(t\):

\(\frac{dx}{dt} = \frac{3}{2+3t}\)

\(\frac{dy}{dt} = \frac{2}{(2+3t)^2}\)

The gradient \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2}{3(2+3t)}\).

Since \(2+3t > 0\), the gradient is always positive.

(b) At the y-axis, \(x = 0\), so \(\ln(2+3t) = 0\) implies \(2+3t = 1\), hence \(t = -\frac{1}{3}\).

Substitute \(t = -\frac{1}{3}\) into \(y = \frac{t}{2+3t}\):

\(y = \frac{-\frac{1}{3}}{1} = -\frac{1}{3}\).

The point is \((0, -\frac{1}{3})\).

The gradient at this point is \(\frac{2}{3}\).

The equation of the tangent is \(y + \frac{1}{3} = \frac{2}{3}x\).

Rearrange to get \(3y = 2x - 1\).

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 \(2.802\).

Answer: The stationary point is \((\frac{\pi}{6},\sqrt{3})\), and it is a minimum point.

Rewrite the function as

\(y=\frac{2-\sin x}{\cos x}=2\sec x-\tan x.\)

Differentiate:

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

For a stationary point, set \(\dfrac{dy}{dx}=0\):

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

Factorising,

\(\sec x(2\tan x-\sec x)=0.\)

Since \(\sec x\ne0\),

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

Using \(\tan x=\dfrac{\sin x}{\cos x}\) and \(\sec x=\dfrac{1}{\cos x}\),

\(2\sin x=1.\)

Hence

\(\sin x=\frac12.\)

In the interval \(-\dfrac{\pi}{2}<x<\dfrac{\pi}{2}\), this gives

\(x=\frac{\pi}{6}.\)

Substitute into the curve:

\(y=\frac{2-\frac12}{\frac{\sqrt3}{2}}=\sqrt3.\)

So the stationary point is

\(\left(\frac{\pi}{6},\sqrt3\right).\)

To determine its nature, check the sign of \(\dfrac{dy}{dx}\). For example, at \(x=0\),

\(2\sec x\tan x-\sec^2x=-1<0,\)

while for a value just greater than \(\dfrac{\pi}{6}\), the derivative is positive. Therefore the gradient changes from negative to positive, so the stationary point is a minimum.

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