(a) Since \((x^2+x+1)\) is a factor of \(p(x)\), divide \(p(x)\) by \(x^2+x+1\).

\[ p(x)=2x^4+ax^3+4x^2+bx-3. \]

Using polynomial division, the remainder is

\[ (b-2)x+(a-7). \]

For \(x^2+x+1\) to be a factor, the remainder must be zero.

Therefore

\[ b-2=0 \]

and

\[ a-7=0. \]

So

\[ a=7,\quad b=2. \]

Answer: \(a=7\), \(b=2\).

(b) With \(a=7\) and \(b=2\),

\[ p(x)=2x^4+7x^3+4x^2+2x-3. \]

Substitute \(x=-3\):

\[ p(-3)=2(-3)^4+7(-3)^3+4(-3)^2+2(-3)-3. \]

\[ =162-189+36-6-3=0. \]

Since \(p(-3)=0\), by the factor theorem, \((x+3)\) is a factor of \(p(x)\).

(a) Sketch the graphs

\[ y=\ln x \]

and

\[ y=\operatorname{cosec}\frac12x \]

for \(0<x<\pi\).

On this interval, \(y=\ln x\) is increasing.

Also, \(\sin\frac12x\) is increasing on \(0<x<\pi\), so \(y=\operatorname{cosec}\frac12x\) is decreasing.

Therefore the two graphs can meet at most once.

Since the sketch shows one intersection, the equation has exactly one root in \(0<x<\pi\).

(b) Let

\[ f(x)=\ln x-\operatorname{cosec}\frac12x. \]

At \(x=2.6\),

\[ f(2.6)=\ln(2.6)-\operatorname{cosec}(1.3)=-0.0823\ldots \]

At \(x=2.9\),

\[ f(2.9)=\ln(2.9)-\operatorname{cosec}(1.45)=0.0574\ldots \]

Since there is a change of sign, the root lies between \(2.6\) and \(2.9\).

(c) Using

\[ x_{n+1}=\exp\left(\operatorname{cosec}\frac12x_n\right), \]

and starting with \(x_0=2.6\):

\[ x_1=2.82306 \]

\[ x_2=2.75335 \]

\[ x_3=2.77082 \]

\[ x_4=2.76609 \]

\[ x_5=2.76734 \]

\[ x_6=2.76701 \]

\[ x_7=2.76710 \]

Therefore the root correct to \(3\) decimal places is

\[ x=2.767. \]

Answer: \(2.767\).

Start with

\[ ye^{3x}\frac{dy}{dx}=x(y+5). \]

Separate the variables:

\[ \frac{y}{y+5}\frac{dy}{dx}=xe^{-3x}. \]

So

\[ \int \frac{y}{y+5}\,dy=\int xe^{-3x}\,dx. \]

For the left-hand side, write

\[ \frac{y}{y+5}=1-\frac{5}{y+5}. \]

Hence

\[ \int \frac{y}{y+5}\,dy=y-5\ln(y+5). \]

For the right-hand side, integrate by parts:

\[ \int xe^{-3x}\,dx=-\frac13xe^{-3x}-\frac19e^{-3x}. \]

Therefore

\[ y-5\ln(y+5)=-\frac13xe^{-3x}-\frac19e^{-3x}+C. \]

Use \(y=0\) when \(x=0\):

\[ 0-5\ln5=-\frac19+C. \]

So

\[ C=\frac19-5\ln5. \]

Therefore

\[ y-5\ln(y+5)=-\frac13xe^{-3x}-\frac19e^{-3x}+\frac19-5\ln5. \]

Equivalently,

\[ y-5\ln\left(\frac{y+5}{5}\right)=\frac19-\left(\frac{x}{3}+\frac19\right)e^{-3x}. \]

(a) Use

\[ x=\sqrt3\tan u. \]

Then

\[ dx=\sqrt3\sec^2u\,du. \]

Also,

\[ 3+x^2=3+3\tan^2u=3\sec^2u. \]

and

\[ x^3=(\sqrt3\tan u)^3=3\sqrt3\tan^3u. \]

Therefore

\[ \frac{x^3}{3+x^2}\,dx = \frac{3\sqrt3\tan^3u}{3\sec^2u}\cdot\sqrt3\sec^2u\,du. \]

\[ =3\tan^3u\,du. \]

When \(x=1\),

\[ 1=\sqrt3\tan u \]

so

\[ \tan u=\frac1{\sqrt3} \]

and

\[ u=\frac{\pi}{6}. \]

When \(x=3\),

\[ 3=\sqrt3\tan u \]

so

\[ \tan u=\sqrt3 \]

and

\[ u=\frac{\pi}{3}. \]

Hence

\[ I=\int_{\frac16\pi}^{\frac13\pi}3\tan^3u\,du. \]

(b) It is easier to divide first:

\[ \frac{x^3}{3+x^2}=x-\frac{3x}{x^2+3}. \]

So

\[ I=\int_1^3\left(x-\frac{3x}{x^2+3}\right)\,dx. \]

\[ I=\left[\frac{x^2}{2}-\frac32\ln(x^2+3)\right]_1^3. \]

\[ I=\left(\frac92-\frac32\ln12\right)-\left(\frac12-\frac32\ln4\right). \]

\[ I=4-\frac32\ln3. \]

Answer: \(4-\frac32\ln3\).

(a) Start with

\[ y^2=k\frac{x-2}{x+2}. \]

Differentiate implicitly:

\[ 2y\frac{dy}{dx} = k\frac{(x+2)-(x-2)}{(x+2)^2}. \]

So

\[ 2y\frac{dy}{dx}=\frac{4k}{(x+2)^2}. \]

From the original equation,

\[ k=\frac{y^2(x+2)}{x-2}. \]

Substitute this into the derivative equation:

\[ 2y\frac{dy}{dx} = \frac{4}{(x+2)^2}\cdot\frac{y^2(x+2)}{x-2}. \]

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

\[ 2y\frac{dy}{dx}=\frac{4y^2}{x^2-4}. \]

Therefore

\[ \frac{dy}{dx}=\frac{2y}{x^2-4}. \]

(b) When \(k=5\) and \(x=3\),

\[ y^2=5\frac{3-2}{3+2}=1. \]

So

\[ y=1\quad\text{or}\quad y=-1. \]

Using

\[ \frac{dy}{dx}=\frac{2y}{x^2-4}, \]

at \(x=3\):

\[ \frac{dy}{dx}=\frac{2y}{9-4}=\frac{2y}{5}. \]

So the two gradients are

\[ m_1=\frac25,\quad m_2=-\frac25. \]

The tangents make angles

\[ \tan^{-1}\left(\frac25\right) \]

and

\[ -\tan^{-1}\left(\frac25\right) \]

with the positive \(x\)-axis.

Therefore the angle between the tangents is

\[ 2\tan^{-1}\left(\frac25\right). \]

Answer: \(2\tan^{-1}\left(\frac25\right)\).